WO2018213998A1 - Signal transmission method and device - Google Patents

Signal transmission method and device Download PDF

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Publication number
WO2018213998A1
WO2018213998A1 PCT/CN2017/085384 CN2017085384W WO2018213998A1 WO 2018213998 A1 WO2018213998 A1 WO 2018213998A1 CN 2017085384 W CN2017085384 W CN 2017085384W WO 2018213998 A1 WO2018213998 A1 WO 2018213998A1
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WO
WIPO (PCT)
Prior art keywords
data
bits
parameter
duty ratios
receiving end
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PCT/CN2017/085384
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French (fr)
Chinese (zh)
Inventor
罗鹏飞
姜彤
董晨
李强
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华为技术有限公司
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Application filed by 华为技术有限公司 filed Critical 华为技术有限公司
Priority to CN201780091111.XA priority Critical patent/CN110651436B/en
Priority to PCT/CN2017/085384 priority patent/WO2018213998A1/en
Publication of WO2018213998A1 publication Critical patent/WO2018213998A1/en

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    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04BTRANSMISSION
    • H04B10/00Transmission systems employing electromagnetic waves other than radio-waves, e.g. infrared, visible or ultraviolet light, or employing corpuscular radiation, e.g. quantum communication
    • H04B10/11Arrangements specific to free-space transmission, i.e. transmission through air or vacuum

Definitions

  • the present application relates to the field of communications technologies, and in particular, to a signal transmission method and apparatus.
  • VLC Visible Light Communication
  • the VLC technology is a method of transmitting a signal to be transmitted by a light-emitting diode (LED) lamp in a blinking manner, and the camera extracts information from the captured video frame during the process of capturing the LED light. To get the signal sent by the LED light.
  • LED light-emitting diode
  • VLC technology has greater bandwidth potential, higher security, and is a green communication technology.
  • the LED lights need to be less than half of the camera frame rate ( ⁇ 10fps to 30fps) flashes light and dark to send a signal.
  • the critical flicker frequency (CFF) seen by the naked eye is generally 100 fps. Therefore, if the LED light flashes at a frequency of 10 fps to 30 fps, the human eye will see the blinking state of the LED light, which is detrimental to the human eyesight.
  • USOOK Undersampled frequency shift ON-OFF keying
  • the final LED can emit a light signal that does not flicker, and allows the camera to acquire 1 bit from the received two consecutive frames of pictures, thereby achieving data transmission efficiency of 0.5 bits/frame.
  • the UFSOOK technology can realize the flicker-free camera communication, only 0.5 bits of information in the original signal can be obtained in the image obtained by the camera once exposed, so the data transmission efficiency is low.
  • the present application provides a signal transmission method and apparatus to solve the problem of low data transmission efficiency.
  • the present application provides a signal transmission method, including: acquiring, by a transmitting end, X first duty ratios, Y second duty ratios, and Z third duty ratios, the Y second duty ratios Each of the second duty ratios is obtained by mapping N bits in the first data to be transmitted, and each of the Z third duty ratios is to be M-bit mapping in the transmitted second data, X ⁇ 1, X is an integer, Y ⁇ 0, Y is an integer, Z ⁇ 0, Z is an integer, N ⁇ 1, N is an integer, M ⁇ 1,M An integer; the transmitting end generates X under-sampled pulse width modulation (UPWM) symbols corresponding to the X first duty ratios, and Y UPWMs corresponding to the Y second duty ratios a symbol, and Z UPWM symbols corresponding to the Z third duty cycles; the transmitting end sequentially transmits the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
  • UPWM pulse width modulation
  • the signal transmission method by mapping the first data and the second data to be transmitted to corresponding duty ratios, one or more bit information can be transmitted for each duty ratio.
  • the first data and the second data are transmitted by transmitting a UPWM signal corresponding to the duty ratio.
  • the transmitting end and the receiving end perform camera communication, the receiving end can collect one duty piece information through one frame image, thereby extracting one or more bit information corresponding to the duty ratio, thereby improving data. Transmission efficiency.
  • the transmitting end can map the first data to the second duty ratio, and enable each of the second duty ratios to transmit N pieces of bit information in the first data, thereby improving data transmission. effectiveness.
  • the transmitting end can map the second data to the third duty ratio, and enable each third duty ratio to transmit M pieces of bit information in the second data, thereby improving data transmission. effectiveness.
  • the present application provides a signal transmission method, including: receiving, by a receiving end, consecutive X first signals to obtain a first parameter, X ⁇ 1, where X is an integer; and the receiving end detects continuous according to the second parameter.
  • Y second signals, Y ⁇ 0, Y is an integer; the receiving end performs a first processing operation on the Y second signals according to the first parameter to obtain Y second percentages; the receiving end uses the The second parameter demodulates the Y second percentages to obtain first data; the receiving end detects consecutive Z third signals according to the third parameter, Z ⁇ 0, Z is an integer; the receiving end uses The first parameter performs the first processing operation on the Z third signals to obtain Z third percentages; the receiving end demodulates the Z third percentages by using the third parameter to obtain the first Two data.
  • the receiving end receives the first signal and obtains the detection according to the first signal.
  • the first processing operation can be performed on the second signal and the third signal by using the first parameter to obtain a percentage corresponding to each signal.
  • the second percentage corresponding to each second signal and the third percentage corresponding to the third signal may be restored to one or more bit information, that is, the receiving end can receive from the received
  • Each of the second signal and the third signal is restored to one or more bit information, thereby improving data transmission efficiency.
  • the method further includes: the receiving end performing the first processing operation on the X first signals, to obtain X first percentage.
  • the second parameter includes a first mapping rule, where the first mapping rule includes a one-to-one correspondence between different n groups of bits and different n percentages, where the n groups of bits are Each of the set of bits includes N bits; the receiving end demodulates the Y second percentages by using the second parameter to obtain the first data, including: the receiving end according to the first mapping rule Each of the second percentages of the Y second percentages is mapped to N bits in the first data; the receiving end is determined by the second second signal according to the receiving order of the Y second signals The first data is obtained by arranging N bits for mapping.
  • the third parameter includes a second mapping rule, where the second mapping rule includes a one-to-one correspondence between different m groups of bits and different m percentages, where the m group of bits is Each group of bits includes M bits; the receiving end demodulates the Z third percentages by using the third parameter to obtain second data, including: the receiving end according to the second mapping rule Each third percentage of the Z third percentages is mapped to M bits in the second data; the receiving end is determined by the third third percentage according to the receiving order of the Z third signals
  • the second data is obtained by arranging M bits for mapping.
  • the receiving end performs the first processing operation on the X first signals by using the first parameter, and after obtaining the first first percentages, the method further includes: the receiving end is configured according to the preset third And mapping a rule and a first percentage of the t percentages of the X first percentages to obtain a third data; wherein the third mapping rule includes the X first percentages a one-to-one correspondence between the different first ordering ratios of the t percentages and the y group bits, each of the y group bits includes R bits, the third data For one of the y group bits, t ⁇ X, t is an integer, R ⁇ 1, and R is an integer.
  • the application provides a transmitting end, including: a processing unit, configured to acquire X first duty ratios, Y second duty ratios, and Z third duty ratios, where the second second portions Each second duty ratio in the air ratio is obtained by mapping N bits in the first data to be transmitted, and each of the Z third duty ratios is from the second data to be transmitted
  • a processing unit configured to acquire X first duty ratios, Y second duty ratios, and Z third duty ratios, where the second second portions
  • Each second duty ratio in the air ratio is obtained by mapping N bits in the first data to be transmitted, and each of the Z third duty ratios is from the second data to be transmitted
  • M bit maps X ⁇ 1, X is an integer, Y ⁇ 0, Y is an integer, Z ⁇ 0, Z is an integer, N ⁇ 1, N is an integer, M ⁇ 1, M is an integer
  • the unit is further configured to generate X undersampled pulse width modulated UPWM symbols corresponding to the X first duty ratios, Y UPWM symbols corresponding to the
  • the processing unit acquires the Y second duty ratios, specifically: according to the order of the bits in the first data, grouping the N bits into the first data.
  • the processing unit acquires the Z third duty ratios, specifically, according to the order of the bits in the second data, grouping the M bits into the second data.
  • the application provides a receiving end, including: a processing unit, configured to detect consecutive X first signals to obtain a first parameter, X ⁇ 1, where X is an integer; the processing unit is further configured to a second parameter, detecting consecutive Y second signals, Y ⁇ 0, Y being an integer; the processing unit is further configured to perform a first processing operation on the Y second signals according to the first parameter, to obtain Y first a second percentage; the processing unit is further configured to demodulate the Y second percentages by using the second parameter to obtain first data; the processing unit is further configured to detect continuous according to the third parameter Z third signals, Z ⁇ 0, Z is an integer; the processing unit is further configured to perform the first processing operation on the Z third signals by using the first parameter, to obtain Z third percentages; The processing unit is further configured to demodulate the Z third percentages by using the third parameter to obtain second data.
  • the processing unit is further configured to perform the first processing operation on the X first signals to obtain X first percentages.
  • the second parameter includes a first mapping rule, where the first mapping rule includes a one-to-one correspondence between different n groups of bits and different n percentages, where the n groups of bits are Each of the set of bits includes N bits; the processing unit demodulates the Y second percentages by using the second parameter to obtain the first data, and specifically includes: Ys according to the first mapping rule Each second percentage of the second percentage is mapped to N bits in the first data; according to the receiving order of the Y second signals, N of each second percentage will be mapped The bits are arranged to obtain the first data.
  • the third parameter includes a second mapping rule, where the second mapping rule includes a one-to-one correspondence between different m groups of bits and different m percentages, where the m group of bits is Each of the set of bits includes M bits; the processing unit demodulates the Z third percentages by using the third parameter to obtain the second data, specifically, including: the Z according to the second mapping rule Each of the third percentages is mapped to M bits in the second data; according to the order of receiving the Z third signals, M of each of the third percentages will be mapped The bits are arranged to obtain the second data.
  • the processing unit is further configured to perform the first on the X first signals by using the first parameter. Processing operation, after obtaining the first percentages of X, acquiring the third according to the preset third mapping rule and the first percentage of the mutually different first percentages of the X first percentages data;
  • the third mapping rule includes a one-to-one correspondence between the y different order of the first percentage of the X first percentages and the y group bits.
  • Each group of bits in the group bit includes R bits
  • the third data is a group of the y group bits, t ⁇ X, t is an integer, R ⁇ 1, and R is an integer.
  • the X UPWM symbols, the Y UPWM symbols, and each of the Z UPWM symbols comprise a k-segment first waveform and a k-segment second a waveform
  • the first waveform is a PWM waveform with an average duty ratio of D
  • the second waveform is a PWM waveform with an average duty ratio of 1-D
  • each of the first waveforms of the first waveform of the k-segment is followed by A second waveform of the second waveform of the k segment is adjacent, k ⁇ 1, k is an integer, and 0 ⁇ D ⁇ 100%.
  • the first waveform includes consecutive J1 first sub-waveforms; the duty ratio of the J1 first sub-waveforms is a uniform D; or the first waveform is included in any first preset duration Ti
  • the average duty ratio of the J2 first sub-waveforms is D1, and the absolute value of the difference between D1 and D is less than or equal to the first preset value, J2 ⁇ J1.
  • each of the first sub-waveforms of the J1 first sub-waveforms is a pulse waveform.
  • the first preset value is 0.
  • the second waveform includes consecutive J3 second sub-waveforms; the duty ratio of the J3 second sub-waveforms is 1-D; or the second waveform is within any first preset duration Ti
  • the average duty ratio of the included J4 second sub-waveforms is D2, and the absolute value of the difference between D2 and 1-D is less than or equal to the second preset value, J4 ⁇ J3.
  • each of the J3 second sub-waveforms is a pulse waveform.
  • the second preset value is 0.
  • each UPWM symbol satisfies at least one of the following four conditions: 1.
  • the total duration of the k first waveforms is T/2; 3.
  • the durations of the first waveform and the second waveform are both less than or equal to the second preset duration; 4.
  • the duration of the first waveform of each segment and the first The absolute value of the difference between the durations of the second waveforms adjacent to one waveform is less than or equal to the third predetermined value.
  • the transmitting end when transmitting each UPWM symbol, the transmitting end sends the first waveform of the k segment and the second waveform of the k segment alternately, so the transmitting end transmits each UPWM. In the case of symbols, the flicker problem is avoided while maintaining the average power.
  • the X first duty ratios include p duty ratios, L1 minimum duty ratios, and L2 maximums a duty ratio, the minimum duty ratio and the maximum duty ratio are preset, the minimum duty ratio being less than any one of the p duty ratios, the maximum duty ratio being greater than the p Any one of the duty ratios, L1 ⁇ 0, L1 is an integer, L2 ⁇ 0, L2 is an integer, p ⁇ 1, and p is an integer.
  • the X first duty ratios indicate a second parameter, a third parameter, and/or third data; wherein the second parameter is used to assist the receiving end to restore the Y second duty ratios to the The first data is used by the auxiliary receiving end to restore the Z third duty ratios to the second data.
  • the second parameter, the third parameter, and/or the third data are transmitted through the X first duty ratios, and the auxiliary receiving end restores the first data and the second data, thereby improving the resolution of the receiving end.
  • the correct rate of one data and the second data is transmitted through the X first duty ratios, and the auxiliary receiving end restores the first data and the second data, thereby improving the resolution of the receiving end.
  • the order of the first duty ratios of the t different ones of the X first duty ratios is according to the third Data and a preset third mapping rule, where the third mapping rule includes a one-to-one correspondence between y different ordering sequences of t different first duty ratios and y group bits, the y group Each set of bits in the bit includes R bits, the third data is a set of the y set of bits, t ⁇ X, t is an integer, R ⁇ 1, and R is an integer.
  • the first data includes a third parameter, where the third parameter is used to assist the receiving end to restore the Z third duty cycles to the second data.
  • the second parameter is preset; or the second parameter is obtained by the receiving end according to the X first percentages.
  • the third parameter is preset; or the third parameter is obtained by the receiving end according to the X first percentages.
  • the second data and the third parameter are used by the auxiliary receiving end to restore the first data and the second data, thereby improving the correctness rate of the first data and the second data by the receiving end.
  • the X first signals include L1 minimum values, L2 maximum values, and p signals except the minimum value and the maximum value, L1 ⁇ 0, L1 is an integer, L2 ⁇ 0, and L2 is an integer. p ⁇ 1, and p is an integer.
  • the first parameter includes: a maximum value, a minimum value, a non-linear value sequence, and/or phase error indication information in the X first signals.
  • the first processing operation includes: nonlinear compensation, normalization, and/or phase compensation operation.
  • the first signal, the second signal, and the third signal are luminance value signals, or the first signal, the second signal, and the third signal are amplitude signals.
  • the present application further provides a transmitting apparatus, including: a processor, a memory, and a transceiver; the processor may execute a program or an instruction stored in the memory, thereby implementing various implementations in the first aspect.
  • the signal transmission method including: a processor, a memory, and a transceiver; the processor may execute a program or an instruction stored in the memory, thereby implementing various implementations in the first aspect.
  • the present application further provides a receiving apparatus, including: a processor, a memory, and a transceiver; the processor may execute a program or an instruction stored in the memory, thereby implementing various implementations in the second aspect.
  • the signal transmission method including: a processor, a memory, and a transceiver; the processor may execute a program or an instruction stored in the memory, thereby implementing various implementations in the second aspect.
  • the present application further provides a storage medium, where the computer storage medium can store a program, and the program can be implemented to implement some or all of the steps in the embodiments of the signal transmission method provided by the present application.
  • the present application further provides a communication system, including the transmitting apparatus according to any one of the third aspect or the third aspect, and the implementation of any one of the fourth aspect or the fourth aspect
  • FIG. 1 is a block diagram of a communication system provided by the present application.
  • FIG. 2 is a schematic structural diagram 1 of a transmitting end of the present application
  • FIG. 3 is a schematic structural diagram 1 of a receiving end of the present application.
  • FIG. 4 is a flow chart of an embodiment of a signal transmission method according to the present application.
  • FIG. 5A is a schematic diagram 1 of a UPWM symbol provided by the present application.
  • FIG. 5B is a schematic diagram 2 of a UPWM symbol provided by the present application.
  • 5C is a schematic diagram 3 of a UPWM symbol provided by the present application.
  • FIG. 5D is a schematic diagram 4 of a UPWM symbol provided by the present application.
  • 6A is a schematic structural diagram 2 of a transmitting end of the present application.
  • 6B is a schematic structural diagram 3 of a transmitting end of the present application.
  • 6C is a schematic structural diagram 4 of a transmitting end of the present application.
  • FIG. 7A is a schematic structural view 2 of a receiving end of the present application.
  • 7B is a schematic structural diagram 3 of a receiving end of the present application.
  • FIG. 7C is a schematic structural diagram 4 of a receiving end of the present application.
  • the words “exemplary” or “such as” are used to mean an example, illustration, or illustration. Any embodiment or design described as “exemplary” or “for example” in the embodiments of the present application should not be construed as preferred or advantageous over other embodiments or designs. Rather, the use of the words “exemplary” or “such as” is intended to present the concepts in a particular manner.
  • the signal transmission method provided by the present application provided by the present application can be applied to various communication systems.
  • it may be an OCC communication system or a radio frequency communication system.
  • a communication system provided by the present application includes at least one transmitting end and at least one receiving end.
  • the transmitting end may be an OC-enabled lighting fixture, a front and rear headlight of a car, a traffic signal, etc.
  • the receiving end may be a built-in camera with an OCC function, a smartphone, a tablet computer, a surveillance camera, and an in-vehicle driving record. Instrument and so on.
  • the sender may also be a terminal having a radio frequency function.
  • the transmitting end includes a bus, a processor, a memory, and a communication interface.
  • the processor is a control center of the transmitting end, and connects various parts of the entire transmitting end by using various interfaces and lines, by running or executing an application and/or an operating system stored in the memory, and calling data stored in the memory, Perform various functions of the sender and process data to monitor the sender as a whole.
  • the processor may include digital signal processor devices, microprocessor devices, analog to digital converters, digital to analog converters, etc., which are capable of distributing the control and signal processing functions of the transmitting end in accordance with their respective capabilities.
  • the communication interface can include a radio frequency (RF) circuit that can be used to send and receive information and process the received information to the processor.
  • the RF circuit includes, but is not limited to, an antenna, at least one amplifier, a transceiver, a coupler, an LNA (low noise amplifier), a duplexer, etc., and communicates with other devices through the wireless communication.
  • the wireless communication may use any communication standard or protocol, including but not limited to a global system of mobile communication (GSM), a general packet radio service (GPRS), and code division multiple access ( Code division multiple access (CDMA), wideband code division multiple access (WCDMA), LTE (long term evolution, long term evolution), Wi-Fi or low power Wi-Fi, and WLAN technology.
  • GSM global system of mobile communication
  • GPRS general packet radio service
  • CDMA Code division multiple access
  • WCDMA wideband code division multiple access
  • LTE long term evolution, long term evolution
  • Wi-Fi wireless fidelity
  • WLAN technology wireless technology
  • the sender also Input/out
  • the receiving end may include a communication interface, a processor, a memory, and a bus
  • the processor includes an image processor, a digital signal processor device, a microprocessor device, an analog to digital converter, a digital to analog converter, and the like.
  • the bus is used to connect the processor, the memory, and the communication interface, and implement data transfer between the processor, the memory, and the communication interface.
  • the processor receives the command from the communication interface via the bus, decrypts the received command, performs calculation or data processing according to the decrypted command, and transmits the processed data from the communication interface to the other device through the bus.
  • the memory includes program modules, data modules, and the like.
  • the program modules may be comprised of software, firmware, hardware, or at least two of them for storing applications and operating systems.
  • the communication interface can be connected to other external network element nodes by wirelessly connecting to the network to complete data transmission and reception.
  • the receiving end also includes other input/output devices such as a camera or the like.
  • FIG. 4 is a flowchart of an embodiment of a signal transmission method provided by the present application, where the method includes the following steps:
  • Step 401 The transmitting end acquires X first duty ratios, Y second duty ratios, and Z third duty ratios, and each of the second second duty ratios is to be sent
  • the N bit maps in the first data are obtained, and each of the Z third duty ratios is obtained by mapping M bits in the second data to be transmitted, X ⁇ 1, X is Integer, Y ⁇ 0, Y is an integer, Z ⁇ 0, Z is an integer, N ⁇ 1, N is an integer, M ⁇ 1, and M is an integer.
  • the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may respectively correspond to the preamble sequence, the frame header, and the frame payload in one frame transmitted by the transmitting end.
  • X UPWM symbols corresponding to the X first duty ratios may be used as a preamble sequence of the frame, the first data is used as the frame header of the frame, and the second data is used as the payload of the frame.
  • the preamble sequence may be used to indicate that the receiving end is about to have data transmission, so that the receiving end can advance It is ready to receive data, and at the same time, realize frame synchronization between the receiving end and the transmitting end.
  • the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may respectively correspond to a preamble sequence and a frame in one frame sent by the transmitting end.
  • Y ⁇ 0 and Z ⁇ 0 it means that there is a preamble sequence, a frame header and a payload in a frame.
  • one frame sent by the transmitting end may include other sequences in addition to the preamble sequence, the frame header and the payload, for example, a sequence for channel estimation, phase error measurement, and the like.
  • the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may respectively correspond to a portion of one frame transmitted by the transmitting end.
  • the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may also indicate mutually independent information.
  • the X first duty ratios may indicate third data, and the first data, the second data, and the third data are independent of each other.
  • the manner in which the sending end acquires the Y second duty ratios may be:
  • the transmitting end divides the bits in the first data into Y groups according to the order of the bits in the first data, in groups of N bits. And then mapping the Y group bits to the Y second duty ratios according to a preset first mapping rule, where the first mapping rule includes a different one of n groups of bits and different n duty ratios Correspondence relationship.
  • n duty ratios in the first mapping rule may be equally spaced, or may be non-equally distributed.
  • four different duty ratios may be ⁇ 20%, 40. %, 60%, 80% ⁇ , or ⁇ 10%, 40%, 60%, 70% ⁇ .
  • N log 2 n.
  • n is the UPWM order used when modulating the first data.
  • mapping rules can be set corresponding to one n value.
  • the selected first mapping rule may be as shown in Table 1 or Table 2:
  • the binary "10010111” can be divided into “10”, “01”, “01” and “11”, a total of 4 groups, if according to Table 1
  • the first mapping rule shown, then the four sets of bits are mapped to "60%, 40%, 40%, and 80%, respectively.”
  • the manner in which the transmitting end acquires the Z third duty ratios may be:
  • the transmitting end divides the bits in the second data into Z groups according to the order of the bits in the second data, in groups of M bits. And then mapping the Z group of bits to the Z third duty ratios according to a preset second mapping rule, where the second mapping rule includes different ones of m groups of bits and different m duty ratios Correspondence relationship.
  • different m duty ratios in the second mapping rule may be equally spaced or may be distributed at different intervals.
  • Different duty cycles can be ⁇ 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80% ⁇ , or ⁇ 0%, 20%, 30%, 40%, 50%, 60%, 70%, 100% ⁇ .
  • M log 2 m
  • m is the UPWM order used to modulate the second data
  • Each of the different m sets of bits in the second mapping rule includes M bits.
  • mapping rules can be set corresponding to one m value.
  • the selected second mapping rule can be as shown in Table 3, Table 4, and Table 5:
  • Bit combination Duty cycle 000 10% 001 20% 010 30% 011 40% 100 50% 101 60% 110 70% 111 80%
  • Bit combination Duty cycle 000 80% 001 70% 010 60% 011 50% 100 40% 101 30% 110 20% 111 10%
  • Bit combination Duty cycle 000 10% 001 80% 010 20% 011 70% 100 30% 101 60% 110 40% 111 50%
  • the binary sequence "001011010" of the second data, the divided "001", “011”, and “010", respectively, can be mapped. It is 20%, 40% and 30%.
  • mapping rules when multiple mapping rules are set corresponding to one m value, the transmitting end can randomly select A mapping rule modulates the second data, and the mapping rule can be changed at any time during the communication process.
  • the modulation orders m and n may be fixed values, or may be determined by the transmitting end according to measurement parameters sent by the receiving end.
  • the measurement parameters may include a received signal to noise ratio, a bit error rate, a detection signal, and the like at the receiving end.
  • the first data when the first data is related to the second data, for example, the first data and the second data are respectively used as a frame header and a payload of the same frame transmitted by the transmitting end, the first data may be used to include a third parameter, where The third parameter is used to assist the receiving end to restore the Z third duty ratios to the second data.
  • the third parameter may include the value of Z, the value of m, and/or the label of the second mapping rule, and the like.
  • the first mapping rule adopts a mapping rule as shown in Table 1
  • the second mapping rule adopts a mapping rule as shown in Table 3.
  • the duty ratio used by the first mapping rule is ⁇ 20%, 40%, 60%, 80% ⁇ is a subset of the ⁇ 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80% ⁇ duty cycle used by the second mapping rule.
  • the X first duty cycles may include p duty cycles, L1 minimum duty cycles, and L2 maximum duty cycles, the minimum duty cycles and the maximum duty cycles being preset
  • the minimum duty ratio is less than any one of the p duty ratios
  • the maximum duty ratio is greater than any one of the p duty ratios, L1 ⁇ 0, L1 is an integer, L2 ⁇ 0, L2 is an integer, p ⁇ 1, and p is an integer.
  • p may be a preset constant
  • p duty ratios, L1 minimum duty ratios, and L2 maximum duty ratios among the X first duty ratios may be arranged in a preset different arrangement order and send. Different indication information is indicated by different arrangement order.
  • the X first duty ratios may be used to allow the receiving end to perform frame synchronization, obtain phase error information, parameters of nonlinear curve information, and may also be used to indicate second parameters, third parameters, and/or Three data.
  • the second parameter is used to assist the receiving end to restore the Y second duty ratios to the first data, and the second parameter includes the value of Y, the value of n, and/or the first mapping rule, and the like.
  • the correspondence between the order of the X first duty ratios and the indication information may be as shown in Table 6:
  • the transmitting end may determine X first duty ratios according to Table 6 above. ⁇ 0%, 80%, 60%, 40%, 20%, 100% ⁇ .
  • the order of the first duty ratios of the t different ones of the X first duty ratios is according to the third data and the preset Obtaining a third mapping rule, where the third mapping rule includes a one-to-one correspondence between y different arrangement orders of t different first duty ratios and y group bits, each of the y group bits A group of bits includes R bits, the third data is a group of the y group bits, t ⁇ X, t is an integer, R ⁇ 1, and R is an integer.
  • the transmitting end may determine that the X first duty ratios are ⁇ 0%, 60%, 40%, 20%, 80%. , 100% ⁇ .
  • Scene 1 L1+L2>0, Y>0, Z>0.
  • the X first duty ratios correspond to the preamble sequence, the Y second duty ratios serve as frame headers, and the Z third duty ratios serve as loads.
  • the duty cycle sequence of the transmitting end before UPWM modulation (including 6 first duty ratios, 5 second duty ratios, and 10 third duty ratios) is: ⁇ 0%, 20%, 40%, 80%, 60%, 100%, 20%, 80%, 20%, 80%, 20%, 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20% , 40% ⁇ .
  • the transmitting end needs to transmit data "011110" through X first duty ratios.
  • the transmitting end determines that "011110” can be obtained by combining two third data “011” and "110".
  • the transmitting end can obtain the duty cycle sequence corresponding to the data "011110” (including two sets of six first duty ratios) of ⁇ 0%, 40%, 20%, 80%, 60%, 100. %, 0%, 60%, 20%, 40%, 80%, 100% ⁇ , so the transmitter can perform UPWM modulation on the 12 duty cycle sequences, and then sequentially transmit them in order.
  • the X first duty cycles correspond to a preamble sequence.
  • the second duty cycle is ⁇ 20%, 80%, 20%, 80%, 20% ⁇ .
  • the p duty ratios are ⁇ 20%, 32%, 44%, 56%, 68%, 80% ⁇ .
  • the duty cycle of the transmitter before UPWM modulation (including 6 first duty cycles, 5 second duty cycles, and 10 third duty cycles) is: ⁇ 20%, 32%, 68% , 56%, 44%, 80%, 20%, 80%, 20%, 80%, 20%, 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20 %, 40% ⁇ .
  • the transmitting end needs to transmit data "011110" through X first duty ratios.
  • the transmitting end determines that "011110” can be obtained by combining two third data “011” and "110".
  • the transmitting end can obtain the duty cycle sequence corresponding to the data "011110” (including two sets of six first duty ratios) of ⁇ 20%, 44%, 32%, 68%, 56%, 80. %, 20%, 44%, 68%, 56%, 32%, 80% ⁇ , so the transmitter can UPWM modulate the 12 duty cycle sequences, and then send them sequentially.
  • Step 402 The transmitting end generates X under-sampled pulse width modulated UPWM symbols corresponding to the X first duty ratios, and Y UPWM symbols corresponding to the Y second duty ratios, and the Z Z UPWM symbols corresponding to the third duty cycle.
  • each of the U UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols includes a k-segment first waveform and a k-segment second waveform, k ⁇ 1, k being an integer.
  • Each of the first waveforms of the first waveform of the k-segment is adjacent to a second waveform of the second waveform of the k-segment, that is, the first waveform of the k-segment and the second waveform of the k-segment, according to "1st The first waveform, the first second waveform, the second first waveform, the second second waveform, ..., the kth first waveform, and the kth second waveform are sequentially arranged.
  • the first waveform is a PWM waveform with an average duty ratio of D
  • the second waveform is a PWM waveform with an average duty ratio of 1-D, 0 ⁇ D ⁇ 100%.
  • the first waveform may include consecutive J1 first sub-waveforms, and each of the first sub-waveforms of the J1 first sub-waveforms is a pulse waveform, that is, a first sub-waveform is a complete PWM The waveform within the period.
  • the duty ratio of the J1 first sub-waveforms is D; or the average duty ratio of the J2 first sub-waveforms included in the first waveform in any first preset duration Ti is D1, D1 and D
  • the absolute value of the difference is less than or equal to the first preset value, J2 ⁇ J1.
  • the PWM period of each of the first sub-waveforms may be the same or different.
  • the second waveform may include consecutive J3 second sub-waveforms, each of the J3 second sub-waveforms being a pulse waveform, ie, a second sub-waveform is a complete PWM The waveform within the period.
  • the duty ratio of the J3 second sub-waveforms is 1-D; or the average duty ratio of the J4 second sub-waveforms included in the second waveform in any first preset duration Ti is D2, D2 and The absolute value of the difference of 1-D is less than or equal to the second preset value, J4 ⁇ J3.
  • the PWM period of each of the second sub-waveforms may be the same or different.
  • the UPWM symbol has a duration of T, and the waveform of the UPWM symbol can be as shown in FIGS. 5A-5D.
  • each of the first sub-waveforms in the first waveform has the same PWM period, both of which are T1, and each of the first sub-waveforms has a duty ratio of 20%, and each second sub-waveform
  • the PWM period of the waveform is the same, both are T3, and the duty ratio of each second sub-waveform is 80%.
  • T1 may be equal to T3 or may not be equal.
  • the PWM periods of the first sub-waveforms in the first waveform are different, as shown in FIG. 5B, the PWM period of the first first sub-waveform is T1, and the PWM period of the second first sub-waveform is T2, T1 ⁇ T2.
  • the duty cycle of each first sub-waveform is 20%.
  • PWM period of each second sub-waveform in the second waveform Different, as shown in FIG. 5B, the PWM period of the first second sub-waveform is T3, and the PWM period of the second second sub-waveform is T4, T3 ⁇ T4.
  • the duty cycle of each first sub-waveform is 80%.
  • the first preset value is 3%
  • the second preset value is 0.
  • Each of the first sub-waveforms in the first waveform has the same PWM period, both of which are T1.
  • the duty ratios of the first sub-waveforms may be the same or different.
  • Each of the second sub-waveforms has the same PWM period, both of which are T3.
  • the duty ratios of the second sub-waveforms may be the same or different. Among them, T1 may be equal to T3 or may not be equal.
  • the first preset value is 0, and the second preset value is 4%.
  • the PWM period of each first sub-waveform in the first waveform is different.
  • the PWM period of the first first sub-waveform is T1
  • the PWM period of the second first sub-waveform is T2, T1 ⁇ T2.
  • the duty ratios of the first sub-waveforms may be the same or different.
  • the PWM period of each second sub-waveform in the second waveform is different. As shown in FIG.
  • the PWM period of the first second sub-waveform is T3, and the PWM period of the second second sub-waveform is T4, T3 ⁇ T4.
  • the duty ratios of the second sub-waveforms may be the same or different.
  • the X UPWM symbols, the Y UPWM symbols, and each of the Z UPWM symbols meet at least one of the following four conditions:
  • the total duration of the k first waveforms is T/2.
  • the duration of the first waveform and the second waveform are both less than or equal to the second preset duration.
  • the absolute value of the difference between the duration of the first waveform of each segment and the duration of the second waveform adjacent to the first waveform is less than or equal to a third predetermined value.
  • a UPWM symbol corresponding to a second duty ratio of 30% is taken as an example.
  • Fc 50 Hz
  • T 0.02 s.
  • the total duration of the 10 first waveforms and the total duration of the 10 second waveforms are both 0.01 s. If the second preset duration is 0.0012 s, the duration of the first waveform and the second waveform generated by the transmitting end are both less than 0.0012 s. If the third preset value is 0.0002 s, the absolute value of the duration difference between any adjacent first waveform and the second waveform is small. It is equal to 0.0002s.
  • the durations of any of the first waveform and the second waveform are equal.
  • the UPWM symbol designed in the present application can enable the transmitting end to avoid the problem of flicker while maintaining the average power in the process of transmitting the UPWM symbol.
  • Step 403 The transmitting end sequentially transmits the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
  • the transmitting end first transmits X UPWM symbols, then sends the Y UPWM symbols immediately, and finally sends the Z UPWM symbols.
  • the transmitting end may transmit the UPWM symbol by using an LED lamp.
  • the transmitting end may send the UPWM symbol through the radio frequency module.
  • Step 404 The receiving end detects consecutive X first signals to obtain a first parameter.
  • the X first signals include L1 minimum values, L2 maximum values, and p signals except the minimum value and the maximum value, L1 ⁇ 0, L1 is an integer, L2 ⁇ 0, L2 is an integer, p ⁇ 1, p is an integer.
  • the first parameter may include, but is not limited to, a maximum value, a minimum value, a non-linear value sequence, and/or phase error indication information among the X first signals.
  • Example 1 based on the UPWM signal sent by the transmitting end in the above scenario 1, the receiving end receives the luminance signal or the amplitude signal respectively corresponding to each UPWM signal.
  • the luminance signal Let's take the luminance signal as an example:
  • the receiving end detects the LED luminance signal (for example, RGB value) from the captured video frame, there is no phase error, then it is assumed that the received luminance signal is ⁇ 100, 220, 265, 326, 298, 350. , 220, 326, 220, 326, 220, 220, 265, 298, 326, 220, 265, 298, 326, 220, 265 ⁇ .
  • the receiving end determines by detection that 350 is the maximum value of the received signal and 100 is the minimum value of the received signal.
  • the receiving end normalizes ⁇ 100, 220, 265, 326, 298, 350 ⁇ using the above maximum and minimum values to obtain ⁇ 0%, 48%, 66%, 90%, 79%, 100% ⁇ . Then, by arranging ⁇ 0%, 48%, 66%, 90%, 79%, 100% ⁇ from low to high, a sequence of nonlinear values is obtained ⁇ 0%, 48%, 66%, 79%, 90% , 100% ⁇ .
  • the receiver uses the resulting non-linear value sequence ⁇ 0%, 48%, 66%, 79%, 90%, 100% ⁇ for ⁇ 0%, 48%, 66%, 90%, 79%, 100% ⁇ performed nonlinear compensation, resulting in ⁇ 0%, 20%, 40%, 80%, 60%, 100% ⁇ .
  • ⁇ 0%, 20%, 40%, 80%, 60%, 100% ⁇ By comparing ⁇ 0%, 20%, 40%, 80%, 60%, 100% ⁇ with Table 9 above, it is determined that the presence in Table 9 is ⁇ 0%, 20%, 40%, 80%, 60%, 100 6 percent of the %7 sequences in the same order. Therefore, the receiving end can determine that there is no phase error.
  • the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be ⁇ 350, 326, 298, 220, 265, 100, 326, 220, 326. , 220, 326, 326, 298, 265, 220, 326, 298, 265, 220, 326, 298 ⁇ .
  • the receiving end determines by detection that 350 is the maximum value of the received signal and 100 is the minimum value of the received signal.
  • the receiving end normalizes ⁇ 350, 326, 298, 220, 265, 100 ⁇ using the above maximum and minimum values to obtain ⁇ 100%, 90%, 79%, 48%, 66%, 0% ⁇ . Then, by arranging ⁇ 100%, 90%, 79%, 48%, 66%, 0% ⁇ from low to high, a sequence of nonlinear values is obtained ⁇ 0%, 48%, 66%, 79%, 90% , 100% ⁇ .
  • ⁇ 100%, 80%, 60%, 20%, 40%, 0% ⁇ By comparing ⁇ 100%, 80%, 60%, 20%, 40%, 0% ⁇ with Table 9 above, it is determined that there is no ⁇ 100%, 80%, 60%, 20%, 40%, 6 percent of the sequences in 0% ⁇ are in the same order.
  • the obtained ⁇ 0%, 20%, 40%, 80%, 60% , 100% ⁇ exists in Table 9, so that the receiving end can determine that there is a phase error.
  • Step 405 The receiving end detects consecutive Y second signals according to the second parameter.
  • the receiving end acquires a second parameter for detecting consecutive Y second signals.
  • the second parameter may be a preset parameter, or the receiving end acquires according to the X first percentages.
  • the first percentage of X is obtained after the receiving end performs the first processing operation on the X signals.
  • the first processing operation may include non-linear compensation, normalization, and/or phase compensation operations.
  • nonlinear compensation, normalization, and phase compensation operations are performed on the six first signals, and six first percentages are obtained as ⁇ 0. %, 20%, 40%, 80%, 60%, 100% ⁇ .
  • the receiving end determines, according to Y in the second parameter, that the five consecutive signals after the six first signals received by the receiving end are the second signals.
  • the five second signals are ⁇ 220, 326, 220, 326, 220 ⁇ .
  • the six first signals are ⁇ 350, 326, 298, 220, 265, 100 ⁇
  • the five second signals are ⁇ 326, 220, 326, 220, 326 ⁇ .
  • Step 406 The receiving end performs a first processing operation on the Y second signals according to the first parameter, to obtain Y second percentages.
  • the receiving end performing the first processing operation on the five second signals according to the first parameter may include: receiving, according to the 350 and 100 pairs ⁇ 220, 326, 220, 326, 220 ⁇ normalized to obtain ⁇ 8%, 90%, 48%, 90%, 48% ⁇ .
  • the receiving end performing the first processing operation on the five second signals according to the first parameter may include: receiving, according to the 350 and 100 pairs, ⁇ 326, 220 , 326, 220, 326 ⁇ normalized to obtain ⁇ 90%, 48%, 90%, 48%, 90% ⁇ .
  • Non-linear processing of ⁇ 90%, 48%, 90%, 48%, 90% ⁇ according to the nonlinear numerical sequence ⁇ 0%, 48%, 66%, 79%, 90%, 100% ⁇ , yielding ⁇ 80% , 20%, 80%, 20%, 80% ⁇ .
  • the second phase error indication parameter it is determined that there is a phase error, so the phase compensation operation is performed on ⁇ 80%, 20%, 80%, 20%, 80% ⁇ , and 5 second percentages are obtained ⁇ 20%, 80%, 20%, 80%, 20% ⁇ .
  • Step 407 The receiving end demodulates the Y second percentages by using the second parameter to obtain first data.
  • the receiving end may map each second percentage of the Y second percentages to the first data according to the first mapping rule. N bits. Then, according to the receiving order of the Y second signals, N bits are mapped by each second percentage to obtain the first data.
  • Step 408 The receiving end detects consecutive Z third signals according to the third parameter.
  • the receiving end can acquire a third parameter for detecting consecutive Z third signals.
  • the third parameter may be preset, or may be obtained by the receiving end according to the X first percentages, or may be obtained by the receiving end from the first data.
  • the six first percentages obtained by the receiving end are ⁇ 0%, 20%, 40%, 80%, 60%, 100% ⁇ .
  • the receiving end determines, according to Z in the third parameter, that the consecutive 10 signals are the third signal after the 5 second signals received by the receiving end.
  • the ten third signals are ⁇ 220, 265, 298, 326, 220, 265, 298, 326, 220. , 265 ⁇ .
  • the five second signals are ⁇ 326, 220, 326, 220, 326 ⁇
  • the ten third signals are ⁇ 326, 298, 265, 220, 326, 298, 265, 220, 326, 298 ⁇ .
  • Step 409 The receiving end performs the first processing operation on the Z third signals by using the first parameter, to obtain Z third percentages.
  • the receiving end performing the first processing operation on the ten third signals according to the first parameter may include: receiving, according to the 350 and 100 pairs, ⁇ 220, 265, 298, 326, 220 , 265, 298, 326, 220, 265 ⁇ normalized to give ⁇ 48%, 66%, 79%, 90%, 48%, 66%, 79%, 90%, 48%, 66% ⁇ .
  • nonlinear numerical sequence ⁇ 0%, 48%, 66%, 79%, 90%, 100% ⁇ for ⁇ 48%, 66%, 79%, 90%, 48%, 66%, 79%, 90%, 48%, 66% ⁇ nonlinear processing, resulting in 10 third percentages ⁇ 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40% ⁇ .
  • the first phase error indication The number determines that there is no phase error, so the receiver does not need to perform phase compensation.
  • the receiving end performs the first processing operation on the ten third signals according to the first parameter, which may include: receiving end Normalized according to 350 and 100 for ⁇ 326,298,265,220,326,298,265,220,326,298 ⁇ , yielding ⁇ 90%, 79%, 66%, 48%, 90%, 79 %, 66%, 48%, 90%, 79% ⁇ .
  • non-linear numerical sequence ⁇ 0%, 48%, 66%, 79%, 90%, 100% ⁇ for ⁇ 90%, 79%, 66%, 48%, 90%, 79%, 66%, 48%, 90%, 79% ⁇ were subjected to nonlinear processing to obtain ⁇ 80%, 60%, 40%, 20%, 80%, 60%, 40%, 20%, 80%, 60% ⁇ .
  • the second phase error indication parameter it is determined that there is a phase error, so phase compensation operation is performed on ⁇ 80%, 60%, 40%, 20%, 80%, 60%, 40%, 20%, 80%, 60% ⁇ , Get 10 third percentages ⁇ 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40% ⁇ .
  • Step 410 The receiving end demodulates the Z third percentages by using the third parameter to obtain second data.
  • the receiving end may map each third percentage of the Z third percentages to the second data according to the second mapping rule. M bits. Then, according to the order of receiving the Z third signals, M bits are mapped out by each third percentage to obtain the second data.
  • the receiving end can map 10 third percentages ⁇ 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40% ⁇ according to the mapping rule 41, respectively. It is ⁇ 00,01,10,11,00,01,10,11,00,01 ⁇ .
  • ⁇ 00, 01, 10, 11, 00, 01, 10, 11, 00, 01 ⁇ are arranged to obtain the first data "00011011000110110001".
  • X first percentages may also be used to indicate the third data.
  • the receiving end may further acquire the third data according to the preset third mapping rule and the first percentage order of the t first different percentages among the X first percentages.
  • the receiving end receives the luminance signal corresponding to each UPWM signal.
  • the receiving end detects the LED luminance signal from the captured video frame, there is no phase error, then it is assumed that the received luminance signal is ⁇ 100, 265, 220, 326, 298, 350, 100, 298, 220 , 265, 326, 350 ⁇ .
  • the receiving end normalizes the two sets of first signals according to the maximum value and the minimum value, and obtains ⁇ 0%, 66%, 48%, 90%, 79%, 100%, 0% ⁇ , ⁇ 79%, 48%, 66%, 90%, 100% ⁇ .
  • a sequence obtained by normalizing any one of the two sets of the first signals is selected to determine a sequence of non-linear values. For example, ⁇ 0%, 66%, 48%, 90%, 79%, 100% ⁇ is arranged from low to high, resulting in a sequence of nonlinear values ⁇ 0%, 48%, 66%, 79%, 90% , 100% ⁇ .
  • the receiving end can determine that ⁇ 0%, 40%, 20%, 80%, 60%, 100% ⁇ is a set of first percentages, and according to Table 10, the six first percentage maps can be determined.
  • the third data is "011".
  • ⁇ 0%, 60%, 20%, 40%, 80%, 100% ⁇ is a set of first percentages, and according to Table 10, it can be determined that the third data of the six first percentage maps is "110" ".
  • the receiving end arranges "011" and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
  • the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be ⁇ 350, 298, 326, 220, 265, 100, 350, 265, 326.
  • the receiving end determines by detection that 350 is the maximum value of the received signal, and 100 is the minimum value of the received signal. According to the above positions of the maximum value and the minimum value, it can be determined that the continuously received ⁇ 350, 298, 326, 220, 265, 100 ⁇ is a set of 6 first signals, ⁇ 350, 265, 326, 298, 220, 100 ⁇ is another set of 6 first signals.
  • L1 1 minimum 100
  • L2 1 maximum 350
  • p 4 other signals.
  • the receiving end normalizes the two sets of first signals according to the maximum value and the minimum value, and obtains ⁇ 100%, 79%, 90%, 48%, 66%, 0% ⁇ , ⁇ 79%, 48%, 66%, 90%, 100% ⁇ .
  • a sequence obtained by normalizing any one of the two sets of the first signals is selected to determine a sequence of non-linear values. For example, ⁇ 100%, 79%, 90%, 48%, 66%, 0% ⁇ is arranged from low to high, resulting in a non-linear numerical sequence ⁇ 0%, 48%, 66%, 79%, 90% , 100% ⁇ .
  • Phase compensation operation for each percentage in ⁇ 100%, 60%, 80%, 20%, 40%, 0% ⁇ , ⁇ 100%, 40%, 80%, 60%, 20%, 0% ⁇ After that, the obtained ⁇ 0%, 20%, 40%, 80%, 60%, 100% ⁇ , ⁇ 100%, 40%, 80%, 60%, 20%, 0% ⁇ are all present in Table 10. . Therefore, the receiving end can determine that there is a phase error. Then the receiving end can determine that ⁇ 0%, 40%, 20%, 80%, 60%, 100% ⁇ is a set of first percentages, and according to Table 10, the six first percentage maps can be determined.
  • the third data is "011".
  • ⁇ 0%, 60%, 20%, 40%, 80%, 100% ⁇ is a set of first percentages, and according to Table 10, it can be determined that the third data of the six first percentage maps is "110" ".
  • the receiving end arranges "011” and "110” according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
  • the first processing operations are performed on the X first signals, the Y second signals, and the Z third signals, the first processing may be performed separately or in series. No restrictions.
  • the serial execution may be performed by the receiving end performing the first processing operation on all the signals acquired by the receiving end in the process of acquiring the first parameter, and performing the first processing operation separately when performing the first processing operation.
  • the normalized, nonlinearly compensated, and/or phase compensated operations obtain a percentage of all signals and then determine which are the second percentage corresponding to the second signal and which are the third percentages corresponding to the third signal.
  • Example 3 based on the UPWM signal sent by the transmitting end in the above scenario 3, the receiving end receives the luminance signal or the amplitude signal corresponding to each UPWM signal respectively.
  • the luminance signal as an example:
  • the receiving end detects the LED luminance signal from the captured video frame without phase error, it is assumed that the received luminance signal is ⁇ 220, 249, 310, 292, 272, 326, 220, 326, 220. , 326, 220, 220, 265, 298, 326, 220, 265, 298, 326, 220, 265 ⁇ .
  • the receiving end determines by detection 326 that it is the maximum value of the received signal, and 220 is the minimum value of the received signal. According to the above positions of the maximum value and the minimum value, it can be determined that the continuously received ⁇ 220, 249, 310, 292, 272, 326 ⁇ is 6 first signals.
  • the receiving end is ⁇ 220, 249, 310, 292, 272, 326, 220, 326, 220, 326, 220, 220, 265, 298, 326, 220, 265, 298, 326, 220 according to the maximum value and the minimum value.
  • the receiving end uses the obtained nonlinear numerical sequence ⁇ 0%, 27%, 50%, 68%, 85%, 100% ⁇ for ⁇ 0%, 27%, 85%, 68%, 50%, 100%, 0% , 100%, 0%, 100%, 0%, 0%, 42%, 74%, 100%, 0%, 42%, 74%, 100%, 0%, 42% ⁇ nonlinear compensation, get: ⁇ 0%, 20%, 80%, 60%, 40%, 100%, 0%, 100%, 0%, 100%, 0%, 100%, 0%, 0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33% ⁇ .
  • the receiving end can determine that ⁇ 0%, 20%, 80%, 60%, 40%, 100% ⁇ is 6 first percentages.
  • the receiving end can determine that the five consecutive signals after the six first signals are ⁇ 220, 326, 220, 326, 220 ⁇ as the second signal, corresponding to ⁇ 0%, 100%, 0%, 100%, 0. % ⁇ is 5 second percentages.
  • mapping rule 21 (shown in Table 14 below), ⁇ 0%, 100%, 0%, 100%, 0% ⁇ is mapped, and the first data is obtained as: 01010.
  • the second data is: 00011011000110110001.
  • the third mapping rule (ie, Table 13) stored in the receiving end may be a normalized result of the third mapping rule (ie, Table 11) stored in the sending end, that is, the X numbers in Table 13.
  • a percentage is obtained by normalizing the X first duty cycles in Table 11.
  • the mapping rule 21 (ie, Table 14) stored in the receiving end may be a normalized result of the mapping rule 21 (ie, Table 8) stored in the transmitting end, that is, the percentage in Table 14 is the duty ratio in Table 8. Obtained after normalization.
  • the mapping rule 41 (ie, Table 15) stored in the receiving end may be a normalized result of the mapping rule 41 (ie, Table 1) stored in the transmitting end, that is, the percentage in Table 15 is the duty ratio in Table 1. Obtained after normalization.
  • the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be ⁇ 326, 310, 249, 272, 292, 220, 326, 220, 326. , 220, 326, 326, 298, 265, 220, 326, 298, 265, 220, 326, 298 ⁇ .
  • the receiving end determines by detection 326 that it is the maximum value of the received signal, and 220 is the minimum value of the received signal. According to the positions of the above maximum and minimum values, it can be determined that the continuously received ⁇ 326, 310, 249, 272, 292, 220 ⁇ is 6 first signals.
  • the receiving end is based on the maximum value and the minimum value, and is ⁇ 326, 310, 249, 272, 292, 220, 326, 220, 326, 220, 326, 326, 298, 265, 220, 326, 298, 265, 220, 326,298 ⁇ normalized to get ⁇ 100%, 85%, 27%, 50%, 68%, 0%, 100%, 0%, 100%, 0%, 100%, 100%, 74%, 42%, 0%, 100%, 74%, 42%, 0%, 100%, 74%, 42%, 0%, 100%, 74% ⁇ .
  • the normalized ⁇ 100%, 85%, 27%, 50%, 68%, 0% ⁇ are arranged from low to high, resulting in a non-linear numerical sequence ⁇ 0%, 27%, 50%, 68%, 85%, 100% ⁇ .
  • nonlinear numerical sequence ⁇ 0%, 27%, 50%, 68%, 85%, 100% ⁇ for ⁇ 100%, 85%, 27%, 50%, 68%, 0%, 100%, 0 %, 100%, 0%, 100%, 100%, 74%, 42%, 0%, 100%, 74%, 42%, 0%, 100%, 74% ⁇ nonlinear compensation, get ⁇ 100% , 80%, 20%, 40%, 60%, 0%, 100%, 0%, 100%, 0%, 100%, 100%, 67%, 33%, 0%, 100%, 67%, 33 %, 0%, 100%, 67% ⁇ .
  • the sequence ⁇ 100%, 80%, 20%, 40%, 60%, 0% ⁇ obtained by normalizing and non-linearly compensating the six first signals is compared with Table 13 below, and it is determined that there is no ⁇ 100%, 80%, 20%, 40%, 60%, 0% ⁇
  • the sequence is the same in order, then the receiving end is ⁇ 100%, 80%, 20%, 40%, 60%, 0%, 100 %, 0%, 100%, 0%, 100%, 100%, 67%, 33%, 0%, 100%, 67%, 33%, 0%, 100%, 67% ⁇ for phase compensation operation, ⁇ 0%, 20%, 80%, 60%, 40%, 100%, 0%, 100%, 0%, 100%, 0%, 0%, 33%, 67%, 100%, 0%, 33 %, 67%, 100%, 0%, 33% ⁇ .
  • mapping rule 21 ⁇ 0%, 100%, 0%, 100%, 0% ⁇ is mapped, and the first data is obtained as: 01010.
  • the receiving end can determine 10 consecutive signals ⁇ 326, 298, 265, 220, 326, 298, 265 after 5 second signals ⁇ 326, 220, 326, 220, 326 ⁇ , 220, 326, 298 ⁇ is the third signal, and the corresponding phase-compensated ⁇ 0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33% ⁇ is 10 third percentages.
  • Example 4 Based on the UPWM signal sent by the transmitting end in the above scenario 4, the receiving end receives the luminance signal or the amplitude signal corresponding to each UPWM signal respectively. Let's take the luminance signal as an example:
  • the receiving end detects the LED luminance signal from the captured video frame without phase error, it is assumed that the received luminance signal is ⁇ 220, 272, 249, 310, 292, 326, 220, 272, 310.
  • the determination 326 is the maximum of the received signals and 220 is the minimum of the received signals. According to the positions of the maximum and minimum values, ⁇ 220, 272, 249, 310, 292, 326 ⁇ , ⁇ 220, 272, 310, 292, 249, 326 ⁇ can be determined as two sets of first signals.
  • a sequence of non-linear values can be determined from a sequence obtained by normalizing a set of first signals in two sets. For example, a low-to-high alignment of ⁇ 0%, 50%, 27%, 85%, 68%, 100% ⁇ yields a sequence of nonlinear values ⁇ 0%, 27%, 50%, 68%, 85% , 100% ⁇ .
  • the receiving end can determine that ⁇ 0%, 40%, 20%, 80%, 60%, 100% ⁇ is a set of first percentages, and according to Table 16, the six first percentage maps can be determined.
  • the third data is "011".
  • ⁇ 0%, 60%, 20%, 40%, 80%, 100% ⁇ is a set of first percentages, and according to Table 16, it can be determined that the third data of the six first percentage maps is "110" ".
  • the receiving end arranges "011” and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
  • the third mapping rule (ie, Table 16) stored in the receiving end may be a normalized result of the third mapping rule (ie, Table 2) stored in the transmitting end, that is, the X numbers in Table 16. A percentage is obtained by normalizing the X first duty cycles in Table 12.
  • the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be ⁇ 326, 292, 310, 249, 272, 220, 326, 292, 249 , 272, 310, 220 ⁇ .
  • the receiving end determines by detection 326 that it is the maximum value of the received signal, and 220 is the minimum value of the received signal. According to the positions of the above maximum and minimum values, it can be determined that the continuously received ⁇ 326, 292, 310, 249, 272, 220 ⁇ is a set of six first signals, ⁇ 326, 292, 249, 272, 310, 220 ⁇ is another set of 6 first signals.
  • the receiving end is based on the maximum and minimum values, for ⁇ 326, 292, 310, 249, 272, 220, 326, 292, 249, 272, 310, 220 ⁇ normalized to get ⁇ 100%, 68%, 85%, 27%, 50%, 0%, 100%, 68%, 27%, 50%, 85%, 0% ⁇ .
  • a sequence obtained by normalizing any one of the two sets of the first signals is selected to determine a sequence of non-linear values. For example, ⁇ 100%, 68%, 85%, 27%, 50%, 0% ⁇ is arranged from low to high, resulting in a sequence of nonlinear values ⁇ 0%, 27%, 50%, 68%, 85% , 100% ⁇ .
  • the receiving end can determine that ⁇ 0%, 40%, 20%, 80%, 60%, 100% ⁇ is a set of first percentages, and according to Table 16, the six first percentage maps can be determined.
  • the third data is "011".
  • ⁇ 0%, 60%, 20%, 40%, 80%, 100% ⁇ is a set of first percentages, and according to Table 16, it can be determined that the third data of the six first percentage maps is "110" ".
  • the receiving end arranges "011” and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
  • each duty ratio can be transmitted by one or one.
  • the first data and the second data are transmitted by transmitting a UPWM signal corresponding to the duty ratio.
  • the receiving end can collect one duty piece information through one frame image, thereby extracting one or more bit information corresponding to the duty ratio, thereby improving data. Transmission efficiency.
  • each network element such as a transmitting end, a receiving end, etc.
  • each network element includes hardware structures and/or software modules corresponding to performing respective functions.
  • the present application can be implemented in a combination of hardware or hardware and computer software in combination with the elements and algorithm steps of the various examples described in the embodiments disclosed herein. Whether a function is implemented in hardware or computer software to drive hardware depends on the specific application and design constraints of the solution. A person skilled in the art can use different methods to implement the described functions for each particular application, but such implementation should not be considered to be beyond the scope of the present application.
  • the application may divide the function module by the sending end, the receiving end, and the like according to the foregoing method example.
  • each function module may be divided according to each function, or two or more functions may be integrated into one place.
  • the above integrated modules can be implemented in the form of hardware or in the form of software functional modules. It should be noted that the division of modules in the present application is schematic, and is only a logical function division, and may be further divided in actual implementation.
  • FIG. 6A shows a possible structural diagram of a transmitting end involved in the foregoing embodiment, where the transmitting end includes: a processing unit and a sending unit.
  • the processing unit is configured to support the transmitting end to perform steps 401-402 in FIG. 4; the sending unit is configured to support the transmitting end to perform step 403 in FIG. 4. All the related content of the steps involved in the foregoing method embodiments may be referred to the functional descriptions of the corresponding functional modules, and details are not described herein again.
  • FIG. 6B shows a possible structural diagram of the transmitting end involved in the above embodiment.
  • the transmitting end includes a processing module 602 and a communication module 603.
  • the processing module 602 is configured to control and manage the actions of the sender.
  • the processing module 602 is configured to support the sender to perform steps 401-403 in FIG. 4, and/or other processes for the techniques described herein.
  • the communication module 603 is configured to support communication between the sender and other network entities, such as communication with the functional modules or network entities shown in FIG.
  • the sending end may further include a storage module 601 for storing program code and data of the transmitting end.
  • the processing module 602 can be a processor or a controller, for example, a central processing unit (CPU), a general-purpose processor, a digital signal processor (DSP), and an application-specific integrated circuit (Application-Specific). Integrated Circuit (ASIC), Field Programmable Gate Array (FPGA) or other programmable logic device, transistor logic device, hardware component, or any combination thereof. It is possible to implement or carry out the various illustrative logical blocks, modules and circuits described in connection with the present disclosure.
  • the processor may also be a combination of computing functions, for example, including one or more microprocessor combinations, a combination of a DSP and a microprocessor, and the like.
  • the communication module 603 can be a transceiver, a transceiver circuit, a communication interface, or the like.
  • the storage module 601 can be a memory.
  • the transmitting end of the present application may be the transmitting end shown in FIG. 6C.
  • the transmitting end includes a processor 612, a transceiver 613, a memory 611, and a bus 614.
  • the transceiver 613, the processor 612, and the memory 611 are connected to each other through a bus 614.
  • the bus 614 may be a Peripheral Component Interconnect (PCI) bus or an Extended Industry Standard Architecture (EISA) bus.
  • PCI Peripheral Component Interconnect
  • EISA Extended Industry Standard Architecture
  • the bus can be divided into an address bus, a data bus, a control bus, and the like. For ease of representation, only one thick line is shown in Figure 6C, but it does not mean that there is only one bus or one type of bus.
  • FIG. 7A shows a possible structural diagram of the receiving end involved in the foregoing embodiment, where the receiving end includes: a processing unit.
  • the processing unit is configured to support the receiving end to perform steps 403-409 in FIG. Wherein all the steps involved in the above method embodiments are related
  • the content can be referred to the functional description of the corresponding function module, and will not be described here.
  • FIG. 7B shows a possible structural diagram of the receiving end involved in the above embodiment.
  • the receiving end includes a processing module 702 and a communication module 703.
  • the processing module 702 is configured to control and manage the actions of the receiving end.
  • the processing module 702 is configured to support the receiving end to perform steps 403-409 in FIG. 4, and/or other processes for the techniques described herein.
  • the communication module 703 is configured to support communication between the receiving end and other network entities, such as communication with the functional modules or network entities shown in FIG.
  • the receiving end may further include a storage module 701 for storing program codes and data of the receiving end.
  • the processing module 702 can be a processor or a controller, such as a CPU, a general purpose processor, a DSP, an ASIC, an FPGA or other programmable logic device, a transistor logic device, a hardware component, or any combination thereof. It is possible to implement or carry out the various illustrative logical blocks, modules and circuits described in connection with the present disclosure.
  • the processor may also be a combination of computing functions, for example, including one or more microprocessor combinations, a combination of a DSP and a microprocessor, and the like.
  • the communication module 703 can be a transceiver, a transceiver circuit, a communication interface, or the like.
  • the storage module 701 can be a memory.
  • the processing module 702 is a processor
  • the communication module 703 is a transceiver
  • the storage module 701 is a memory
  • the receiving end of the present application may be the receiving end shown in FIG. 7C.
  • the receiving end includes a processor 712, a transceiver 713, a memory 711, and a bus 714.
  • the transceiver 713, the processor 712, and the memory 711 are connected to each other through a bus 714; the bus 714 may be a PCI bus or an EISA bus or the like.
  • the bus can be divided into an address bus, a data bus, a control bus, and the like. For ease of representation, only one thick line is shown in Figure 7C, but it does not mean that there is only one bus or one type of bus.
  • the steps of a method or algorithm described in connection with the present disclosure may be implemented in a hardware or may be implemented by a processor executing software instructions.
  • the software instructions can be composed of corresponding software modules, which can be stored in random access memory (RAM), flash memory, read only memory (ROM), and erasable programmable read only memory (Erasable).
  • PROM Programmable ROM
  • EEPROM electrically erasable programmable read only memory
  • registers hard disk, removable hard disk, compact disk read only (CD-ROM) or any other form of storage medium known in the art.
  • An exemplary storage medium is coupled to the processor to enable the processor to read information from, and write information to, the storage medium.
  • the storage medium can also be an integral part of the processor.
  • the processor and the storage medium can be located in an ASIC. Additionally, the ASIC can be located in a core network interface device.
  • the processor and the storage medium may also exist as discrete components in the core network interface device.
  • the present invention further provides a computer storage medium, wherein the computer storage medium may store a program, where the program may include part or all of the embodiments of the resource scheduling method provided by the present invention.
  • the storage medium may be a magnetic disk, an optical disk, a read-only memory (English: read-only memory, abbreviated as: ROM) or a random access memory (English: random access memory, abbreviation: RAM).
  • the present application also provides a communication system including a transmitting end as shown in FIG. 6A, 6B or 6C, and a receiving end as shown in FIG. 7A, 7B or 7C.

Abstract

The present application provides a signal transmission method and device, which relates to the technical field of communications, and can solve the problem of low data transmission efficiency. The method comprises: a transmitting end acquiring X first duty ratios, Y second duty ratios, and Z third duty ratios, wherein each of the second duty ratios is obtained by mapping N bits in first data to be transmitted, each of the third duty ratios is obtained by mapping M bits in second data to be transmitted, X is an integer greater than or equal to 1, Y is an integer greater than or equal to 0, Z is an integer greater than or equal to 0, N is an integer greater than or equal to 1, and M is an integer greater than or equal to 1; the transmitting end generating X UPWM symbols corresponding to the X first duty ratios, Y UPWM symbols corresponding to the Y second duty ratios, and Z UPWM symbols corresponding to the Z third duty ratios; and the transmitting end sequentially transmitting the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.

Description

一种信号传输方法及装置Signal transmission method and device 技术领域Technical field
本申请涉及通信技术领域,尤其涉及一种信号传输方法及装置。The present application relates to the field of communications technologies, and in particular, to a signal transmission method and apparatus.
背景技术Background technique
在通信技术领域中,通过可见光通信(Visible Light Communication,VLC)技术实现相机通信(Optical Camera Communications,OCC)。VLC技术是一种通过发光半导体(light-emitting diode,LED)灯将待发送的信号以明暗闪烁的方式发送出去,相机在拍摄该LED灯的过程中,通过对拍摄到的视频帧进行信息提取,以获取LED灯发送的信号。VLC技术与传统的基于射频的通信技术相比较,具有更大的带宽潜力、更高的安全性,以及是一种绿色环保的通信技术。In the field of communication technology, camera communication (Octical Camera Communications, OCC) is realized by Visible Light Communication (VLC) technology. The VLC technology is a method of transmitting a signal to be transmitted by a light-emitting diode (LED) lamp in a blinking manner, and the camera extracts information from the captured video frame during the process of capturing the LED light. To get the signal sent by the LED light. Compared with traditional RF-based communication technologies, VLC technology has greater bandwidth potential, higher security, and is a green communication technology.
一般情况下,相机和LED灯进行相机通信时,由于相机的帧率较低(<60fps),为了保证相机能够正确接收LED灯发送的信息,LED灯需要以相机帧率的一半以下的频率(<10fps~30fps)进行明暗闪烁,来发送信号。而肉眼可见的闪烁截止频率(critical flicker frequency,CFF)一般为100fps。因此,若LED灯按照10fps~30fps的频率进行明暗闪烁,人的肉眼则会看到LED灯的闪烁状态,有损于人的视力。In general, when the camera and LED lights are used for camera communication, because the camera's frame rate is low (<60fps), in order to ensure that the camera can correctly receive the information sent by the LED lights, the LED lights need to be less than half of the camera frame rate ( <10fps to 30fps) flashes light and dark to send a signal. The critical flicker frequency (CFF) seen by the naked eye is generally 100 fps. Therefore, if the LED light flashes at a frequency of 10 fps to 30 fps, the human eye will see the blinking state of the LED light, which is detrimental to the human eyesight.
基于欠采样的频率频移开关键控(Undersampled frequency shift ON-OFF keying,UFSOOK)通过对基带数据进行具有频移键控副载波调制,使用两个不同载波频率来表示比特“1”和比特“0”。其中两个频率均高于CFF,并且每个已调信号的持续时间均为2/Fc,其中Fc是接收端的帧率。最终LED可发射不闪烁的光信号,并且可让相机可从接收到的连续两帧图片中获取1个比特,从而实现了0.5比特/帧的数据传输效率。Undersampled frequency shift ON-OFF keying (UFSOOK) uses frequency shift keying subcarrier modulation for baseband data, using two different carrier frequencies to represent bits "1" and bits" 0". Two of the frequencies are higher than CFF, and the duration of each modulated signal is 2/Fc, where Fc is the frame rate of the receiving end. The final LED can emit a light signal that does not flicker, and allows the camera to acquire 1 bit from the received two consecutive frames of pictures, thereby achieving data transmission efficiency of 0.5 bits/frame.
然而,采用UFSOOK技术虽然能够实现无闪烁相机通信,但相机每进行曝光一次得到的图像中,仅能得到原始信号中的0.5个比特的信息,因此数据的传输效率较低。However, although the UFSOOK technology can realize the flicker-free camera communication, only 0.5 bits of information in the original signal can be obtained in the image obtained by the camera once exposed, so the data transmission efficiency is low.
发明内容Summary of the invention
本申请提供了一种信号传输方法及装置,以解决数据的传输效率较低的问题。The present application provides a signal transmission method and apparatus to solve the problem of low data transmission efficiency.
第一方面,本申请提供一种信号传输方法,包括:发送端获取X个第一占空比、Y个第二占空比和Z个第三占空比,该Y个第二占空比中的每个第二占空比由待发送的第一数据中的N个比特映射所得,该Z个第三占空比中的每个第三占空比由待 发送的第二数据中的M个比特映射所得,X≥1,X为整数,Y≥0,Y为整数,Z≥0,Z为整数,N≥1,N为整数,M≥1,M为整数;该发送端生成与该X个第一占空比对应的X个基于欠采样脉冲宽度调制(Undersampled Pulse width modulation,UPWM)符号、与该Y个第二占空比对应的Y个UPWM符号,和与该Z个第三占空比对应的Z个UPWM符号;该发送端依次发送该X个UPWM符号、该Y个UPWM符号以及该Z个UPWM符号。In a first aspect, the present application provides a signal transmission method, including: acquiring, by a transmitting end, X first duty ratios, Y second duty ratios, and Z third duty ratios, the Y second duty ratios Each of the second duty ratios is obtained by mapping N bits in the first data to be transmitted, and each of the Z third duty ratios is to be M-bit mapping in the transmitted second data, X≥1, X is an integer, Y≥0, Y is an integer, Z≥0, Z is an integer, N≥1, N is an integer, M≥1,M An integer; the transmitting end generates X under-sampled pulse width modulation (UPWM) symbols corresponding to the X first duty ratios, and Y UPWMs corresponding to the Y second duty ratios a symbol, and Z UPWM symbols corresponding to the Z third duty cycles; the transmitting end sequentially transmits the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
采用本申请提供的信号传输方法,通过将待发送的第一数据和第二数据映射为对应的占空比,使得每个占空比能够传递1个或1个以上比特信息。通过发送与占空比对应的UPWM信号来传输第一数据和第二数据。那么,当发送端和接收端进行相机通信时,接收端能够通过一帧图像采集到一个占空比信息,从而提取出该占空比对应的1个或1个以上比特信息,从而提高了数据传输效率。With the signal transmission method provided by the present application, by mapping the first data and the second data to be transmitted to corresponding duty ratios, one or more bit information can be transmitted for each duty ratio. The first data and the second data are transmitted by transmitting a UPWM signal corresponding to the duty ratio. Then, when the transmitting end and the receiving end perform camera communication, the receiving end can collect one duty piece information through one frame image, thereby extracting one or more bit information corresponding to the duty ratio, thereby improving data. Transmission efficiency.
可选的,当Y>0时,该发送端获取Y个第二占空比,包括:该发送端按照该第一数据中比特的排列顺序,以N个比特为一组,将该第一数据中的比特划分为Y组,N=log2n;该发送端根据预设的第一映射规则,将该Y组比特映射为该Y个第二占空比,该第一映射规则包括不同的n组比特与不同的n个占空比之间的一一对应关系。Optionally, when Y>0, the sending end acquires the Y second duty ratios, including: the sending end groups the N bits according to the order of the bits in the first data, and the first The bits in the data are divided into Y groups, N=log 2 n; the transmitting end maps the Y group bits to the Y second duty ratios according to a preset first mapping rule, and the first mapping rule includes different A one-to-one correspondence between n sets of bits and different n duty ratios.
在这种可选的方式中,发送端可以将第一数据映射为第二占空比,并且使得每个第二占空比能够传递第一数据中的N个比特信息,从而提高了数据传输效率。In this optional manner, the transmitting end can map the first data to the second duty ratio, and enable each of the second duty ratios to transmit N pieces of bit information in the first data, thereby improving data transmission. effectiveness.
可选的,当Z>0时,该发送端获取Z个第三占空比,包括:该发送端按照该第二数据中比特的排列顺序,以M个比特为一组,将该第二数据中的比特划分为Z组,M=log2m;该发送端根据预设的第二映射规则,将该Z组比特映射为该Z个第三占空比,该第二映射规则包括不同的m组比特与不同的m个占空比之间的一一对应关系。Optionally, when Z>0, the sending end acquires Z third duty ratios, including: the sending end groups the M bits according to the order of the bits in the second data, and the second The bits in the data are divided into Z groups, M=log 2 m; the transmitting end maps the Z group bits to the Z third duty ratios according to a preset second mapping rule, and the second mapping rule includes different A one-to-one correspondence between m sets of bits and different m duty cycles.
在这种可选的方式中,发送端可以将第二数据映射为第三占空比,并且使得每个第三占空比能够传递第二数据中的M个比特信息,从而提高了数据传输效率。In this optional manner, the transmitting end can map the second data to the third duty ratio, and enable each third duty ratio to transmit M pieces of bit information in the second data, thereby improving data transmission. effectiveness.
第二方面,本申请提供一种信号传输方法,包括:接收端检测连续的X个第一信号,以获取第一参数,X≥1,X为整数;该接收端根据第二参数,检测连续的Y个第二信号,Y≥0,Y为整数;该接收端根据该第一参数对该Y个第二信号进行第一处理操作,得到Y个第二百分比;该接收端使用该第二参数对该Y个第二百分比进行解调,得到第一数据;该接收端根据第三参数,检测连续的Z个第三信号,Z≥0,Z为整数;该接收端使用该第一参数对该Z个第三信号进行该第一处理操作,得到Z个第三百分比;该接收端使用该第三参数对该Z个第三百分比进行解调,得到第二数据。In a second aspect, the present application provides a signal transmission method, including: receiving, by a receiving end, consecutive X first signals to obtain a first parameter, X≥1, where X is an integer; and the receiving end detects continuous according to the second parameter. Y second signals, Y≥0, Y is an integer; the receiving end performs a first processing operation on the Y second signals according to the first parameter to obtain Y second percentages; the receiving end uses the The second parameter demodulates the Y second percentages to obtain first data; the receiving end detects consecutive Z third signals according to the third parameter, Z≥0, Z is an integer; the receiving end uses The first parameter performs the first processing operation on the Z third signals to obtain Z third percentages; the receiving end demodulates the Z third percentages by using the third parameter to obtain the first Two data.
采用本申请提供的方法,接收端在接收到第一信号,并根据对第一信号的检测获 得第一参数后,能够通过使用第一参数对第二信号以及第三信号进行第一处理操作,得到每个信号对应的百分比。其中,每个第二信号对应的第二百分比和第三信号对应的第三百分比都可被还原为1个或1个以上比特信息,也就是说,接收端能够从接收到的每个第二信号以及第三信号中还原为1个或1个以上比特信息,从而提高了数据传输效率。With the method provided by the application, the receiving end receives the first signal and obtains the detection according to the first signal. After the first parameter is obtained, the first processing operation can be performed on the second signal and the third signal by using the first parameter to obtain a percentage corresponding to each signal. The second percentage corresponding to each second signal and the third percentage corresponding to the third signal may be restored to one or more bit information, that is, the receiving end can receive from the received Each of the second signal and the third signal is restored to one or more bit information, thereby improving data transmission efficiency.
可选的,该接收端检测连续的X个第一信号,以获取第一参数之后,该方法还包括:该接收端对该X个第一信号进行该第一处理操作,得到X个第一百分比。Optionally, after the receiving end detects the consecutive X first signals to obtain the first parameter, the method further includes: the receiving end performing the first processing operation on the X first signals, to obtain X first percentage.
可选的,当Y>0时,该第二参数包括第一映射规则,该第一映射规则包括不同的n组比特与不同的n个百分比之间的一一对应关系,该n组比特中的每一组比特中包括N个比特;该接收端使用该第二参数对该Y个第二百分比进行解调,得到第一数据,包括:该接收端根据该第一映射规则将该Y个第二百分比中的每个第二百分比映射为该第一数据中的N个比特;该接收端根据该Y个第二信号的接收顺序,将由该每个第二百分比映射出N个比特进行排列,得到该第一数据。Optionally, when Y>0, the second parameter includes a first mapping rule, where the first mapping rule includes a one-to-one correspondence between different n groups of bits and different n percentages, where the n groups of bits are Each of the set of bits includes N bits; the receiving end demodulates the Y second percentages by using the second parameter to obtain the first data, including: the receiving end according to the first mapping rule Each of the second percentages of the Y second percentages is mapped to N bits in the first data; the receiving end is determined by the second second signal according to the receiving order of the Y second signals The first data is obtained by arranging N bits for mapping.
可选的,当Z>0时,该第三参数包括第二映射规则,该第二映射规则包括不同的m组比特与不同的m个百分比之间的一一对应关系,该m组比特中的每一组比特中包括M个比特;该接收端使用该第三参数对该Z个第三百分比进行解调,得到第二数据,包括:该接收端根据该第二映射规则将该Z个第三百分比中的每个第三百分比映射为该第二数据中的M个比特;该接收端根据该Z个第三信号的接收顺序,将由该每个第三百分比映射出M个比特进行排列,得到该第二数据。Optionally, when Z>0, the third parameter includes a second mapping rule, where the second mapping rule includes a one-to-one correspondence between different m groups of bits and different m percentages, where the m group of bits is Each group of bits includes M bits; the receiving end demodulates the Z third percentages by using the third parameter to obtain second data, including: the receiving end according to the second mapping rule Each third percentage of the Z third percentages is mapped to M bits in the second data; the receiving end is determined by the third third percentage according to the receiving order of the Z third signals The second data is obtained by arranging M bits for mapping.
可选的,该接收端使用该第一参数对该X个第一信号进行该第一处理操作,得到X个第一百分比之后,该方法还包括:该接收端根据预设的第三映射规则和该X个第一百分比中的t个互不相同的第一百分比的排列顺序,获取第三数据;其中,该第三映射规则包括该X个第一百分比中的t个互不相同的第一百分比的y种不同的排列顺序与y组比特之间的一一对应关系,该y组比特中的每一组比特包括R个比特,该第三数据为该y组比特中的一组,t≤X,t为整数,R≥1,R为整数。Optionally, the receiving end performs the first processing operation on the X first signals by using the first parameter, and after obtaining the first first percentages, the method further includes: the receiving end is configured according to the preset third And mapping a rule and a first percentage of the t percentages of the X first percentages to obtain a third data; wherein the third mapping rule includes the X first percentages a one-to-one correspondence between the different first ordering ratios of the t percentages and the y group bits, each of the y group bits includes R bits, the third data For one of the y group bits, t ≤ X, t is an integer, R ≥ 1, and R is an integer.
第三方面,本申请提供一种发送端,包括:处理单元,用于获取X个第一占空比、Y个第二占空比和Z个第三占空比,该Y个第二占空比中的每个第二占空比由待发送的第一数据中的N个比特映射所得,该Z个第三占空比中的每个第三占空比由待发送的第二数据中的M个比特映射所得,X≥1,X为整数,Y≥0,Y为整数,Z≥0,Z为整数,N≥1,N为整数,M≥1,M为整数;该处理单元,还用于生成与该X个第一占空比对应的X个基于欠采样的脉冲宽度调制UPWM符号、与该Y个第二占空比对应的Y个UPWM符号,和与该Z个第三占空比对应的Z个UPWM符号;发送单元,用于依次发送该X个UPWM符号、该Y个UPWM符号以及该Z个UPWM符号。 In a third aspect, the application provides a transmitting end, including: a processing unit, configured to acquire X first duty ratios, Y second duty ratios, and Z third duty ratios, where the second second portions Each second duty ratio in the air ratio is obtained by mapping N bits in the first data to be transmitted, and each of the Z third duty ratios is from the second data to be transmitted In the M bit maps, X≥1, X is an integer, Y≥0, Y is an integer, Z≥0, Z is an integer, N≥1, N is an integer, M≥1, M is an integer; The unit is further configured to generate X undersampled pulse width modulated UPWM symbols corresponding to the X first duty ratios, Y UPWM symbols corresponding to the Y second duty ratios, and the Z The Z UPWM symbols corresponding to the third duty ratio; the transmitting unit is configured to sequentially send the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
可选的,当Y>0时,该处理单元获取Y个第二占空比,具体包括:按照该第一数据中比特的排列顺序,以N个比特为一组,将该第一数据中的比特划分为Y组,N=log2n;根据预设的第一映射规则,将该Y组比特映射为该Y个第二占空比,该第一映射规则包括不同的n组比特与不同的n个占空比之间的一一对应关系。Optionally, when Y>0, the processing unit acquires the Y second duty ratios, specifically: according to the order of the bits in the first data, grouping the N bits into the first data. The bits are divided into Y groups, N=log 2 n; according to a preset first mapping rule, the Y group bits are mapped to the Y second duty ratios, and the first mapping rule includes different n groups of bits and A one-to-one correspondence between different n duty cycles.
可选的,当Z>0时,该处理单元获取Z个第三占空比,具体包括:按照该第二数据中比特的排列顺序,以M个比特为一组,将该第二数据中的比特划分为Z组,M=log2m;根据预设的第二映射规则,将该Z组比特映射为该Z个第三占空比,该第二映射规则包括不同的m组比特与不同的m个占空比之间的一一对应关系。Optionally, when Z>0, the processing unit acquires the Z third duty ratios, specifically, according to the order of the bits in the second data, grouping the M bits into the second data. The bits are divided into Z groups, M=log 2 m; according to a preset second mapping rule, the Z group bits are mapped to the Z third duty ratios, and the second mapping rule includes different m groups of bits and A one-to-one correspondence between different m duty cycles.
本申请提供的发送端的技术效果可以参见上述第一方面或第一方面的各个实现方式的技术效果,此处不再赘述。For the technical effects of the sending end provided by the present application, refer to the technical effects of the foregoing first aspect or the implementation manner of the first aspect, and details are not described herein again.
第四方面,本申请提供一种接收端,包括:处理单元,用于检测连续的X个第一信号,以获取第一参数,X≥1,X为整数;该处理单元,还用于根据第二参数,检测连续的Y个第二信号,Y≥0,Y为整数;该处理单元,还用于根据该第一参数对该Y个第二信号进行第一处理操作,得到Y个第二百分比;该处理单元,还用于使用该第二参数对该Y个第二百分比进行解调,得到第一数据;该处理单元,还用于根据第三参数,检测连续的Z个第三信号,Z≥0,Z为整数;该处理单元,还用于使用该第一参数对该Z个第三信号进行该第一处理操作,得到Z个第三百分比;该处理单元,还用于使用该第三参数对该Z个第三百分比进行解调,得到第二数据。In a fourth aspect, the application provides a receiving end, including: a processing unit, configured to detect consecutive X first signals to obtain a first parameter, X≥1, where X is an integer; the processing unit is further configured to a second parameter, detecting consecutive Y second signals, Y≥0, Y being an integer; the processing unit is further configured to perform a first processing operation on the Y second signals according to the first parameter, to obtain Y first a second percentage; the processing unit is further configured to demodulate the Y second percentages by using the second parameter to obtain first data; the processing unit is further configured to detect continuous according to the third parameter Z third signals, Z≥0, Z is an integer; the processing unit is further configured to perform the first processing operation on the Z third signals by using the first parameter, to obtain Z third percentages; The processing unit is further configured to demodulate the Z third percentages by using the third parameter to obtain second data.
可选的,该处理单元,还用于对该X个第一信号进行该第一处理操作,得到X个第一百分比。Optionally, the processing unit is further configured to perform the first processing operation on the X first signals to obtain X first percentages.
可选的,当Y>0时,该第二参数包括第一映射规则,该第一映射规则包括不同的n组比特与不同的n个百分比之间的一一对应关系,该n组比特中的每一组比特中包括N个比特;该处理单元使用该第二参数对该Y个第二百分比进行解调,得到第一数据,具体包括:根据该第一映射规则将该Y个第二百分比中的每个第二百分比映射为该第一数据中的N个比特;根据该Y个第二信号的接收顺序,将由该每个第二百分比映射出N个比特进行排列,得到该第一数据。Optionally, when Y>0, the second parameter includes a first mapping rule, where the first mapping rule includes a one-to-one correspondence between different n groups of bits and different n percentages, where the n groups of bits are Each of the set of bits includes N bits; the processing unit demodulates the Y second percentages by using the second parameter to obtain the first data, and specifically includes: Ys according to the first mapping rule Each second percentage of the second percentage is mapped to N bits in the first data; according to the receiving order of the Y second signals, N of each second percentage will be mapped The bits are arranged to obtain the first data.
可选的,当Z>0时,该第三参数包括第二映射规则,该第二映射规则包括不同的m组比特与不同的m个百分比之间的一一对应关系,该m组比特中的每一组比特中包括M个比特;该处理单元使用该第三参数对该Z个第三百分比进行解调,得到第二数据,具体包括:根据该第二映射规则将该Z个第三百分比中的每个第三百分比映射为该第二数据中的M个比特;根据该Z个第三信号的接收顺序,将由该每个第三百分比映射出M个比特进行排列,得到该第二数据。Optionally, when Z>0, the third parameter includes a second mapping rule, where the second mapping rule includes a one-to-one correspondence between different m groups of bits and different m percentages, where the m group of bits is Each of the set of bits includes M bits; the processing unit demodulates the Z third percentages by using the third parameter to obtain the second data, specifically, including: the Z according to the second mapping rule Each of the third percentages is mapped to M bits in the second data; according to the order of receiving the Z third signals, M of each of the third percentages will be mapped The bits are arranged to obtain the second data.
可选的,该处理单元,还用于在使用该第一参数对该X个第一信号进行该第一 处理操作,得到X个第一百分比之后,根据预设的第三映射规则和该X个第一百分比中的t个互不相同的第一百分比的排列顺序,获取第三数据;Optionally, the processing unit is further configured to perform the first on the X first signals by using the first parameter. Processing operation, after obtaining the first percentages of X, acquiring the third according to the preset third mapping rule and the first percentage of the mutually different first percentages of the X first percentages data;
其中,该第三映射规则包括该X个第一百分比中的t个互不相同的第一百分比的y种不同的排列顺序与y组比特之间的一一对应关系,该y组比特中的每一组比特包括R个比特,该第三数据为该y组比特中的一组,t≤X,t为整数,R≥1,R为整数。The third mapping rule includes a one-to-one correspondence between the y different order of the first percentage of the X first percentages and the y group bits. Each group of bits in the group bit includes R bits, the third data is a group of the y group bits, t≤X, t is an integer, R≥1, and R is an integer.
本申请提供的接收端的技术效果可以参见上述第二方面或第二方面的各个实现方式的技术效果,此处不再赘述。For technical effects of the receiving end provided by the present application, refer to the technical effects of the foregoing second aspect or the implementation manner of the second aspect, and details are not described herein again.
结合上述第一方面或第三方面,可选的,该X个UPWM符号、该Y个UPWM符号和该Z个UPWM符号中的每个UPWM符号,均包括k段第一波形和k段第二波形,该第一波形为平均占空比为D的PWM波形,该第二波形为平均占空比为1-D的PWM波形,该k段第一波形中的每段第一波形之后均与该k段第二波形中的一段第二波形相邻,k≥1,k为整数,0≤D≤100%。In combination with the above first aspect or the third aspect, optionally, the X UPWM symbols, the Y UPWM symbols, and each of the Z UPWM symbols comprise a k-segment first waveform and a k-segment second a waveform, the first waveform is a PWM waveform with an average duty ratio of D, and the second waveform is a PWM waveform with an average duty ratio of 1-D, and each of the first waveforms of the first waveform of the k-segment is followed by A second waveform of the second waveform of the k segment is adjacent, k≥1, k is an integer, and 0≤D≤100%.
可选的,该第一波形包含连续的J1个第一子波形;该J1个第一子波形的占空比为均D;或者,该第一波形在任意第一预设时长Ti内包含的J2个第一子波形的平均占空比为D1,D1与D的差值的绝对值小于或者等于第一预设值,J2<J1。Optionally, the first waveform includes consecutive J1 first sub-waveforms; the duty ratio of the J1 first sub-waveforms is a uniform D; or the first waveform is included in any first preset duration Ti The average duty ratio of the J2 first sub-waveforms is D1, and the absolute value of the difference between D1 and D is less than or equal to the first preset value, J2 < J1.
可选的,该J1个第一子波形中的每个第一子波形为一个脉冲波形。Optionally, each of the first sub-waveforms of the J1 first sub-waveforms is a pulse waveform.
可选的,第一预设值为0。Optionally, the first preset value is 0.
可选的,该第二波形包含连续的J3个第二子波形;该J3个第二子波形的占空比为均1-D;或者,该第二波形在任意第一预设时长Ti内包含的J4个第二子波形的平均占空比为D2,D2与1-D的差值的绝对值小于或者等于第二预设值,J4<J3。Optionally, the second waveform includes consecutive J3 second sub-waveforms; the duty ratio of the J3 second sub-waveforms is 1-D; or the second waveform is within any first preset duration Ti The average duty ratio of the included J4 second sub-waveforms is D2, and the absolute value of the difference between D2 and 1-D is less than or equal to the second preset value, J4<J3.
可选的,该J3个第二子波形中的每个第二子波形为一个脉冲波形。Optionally, each of the J3 second sub-waveforms is a pulse waveform.
可选的,第二预设值为0。Optionally, the second preset value is 0.
可选的,该每个UPWM符号均满足以下四个条件中的至少一个条件:一、该每个UPWM符号的时长均为T,T=1/Fc,Fc表示接收端的帧率;二、该k个第一波形的总时长为T/2;三、该第一波形和该第二波形的时长均小于或者等于第二预设时长;四、该每段第一波形的时长和与该第一波形相邻的第二波形的时长之间的差值的绝对值小于等于第三预设值。Optionally, each UPWM symbol satisfies at least one of the following four conditions: 1. The duration of each UPWM symbol is T, T=1/Fc, and Fc represents a frame rate of the receiving end; The total duration of the k first waveforms is T/2; 3. The durations of the first waveform and the second waveform are both less than or equal to the second preset duration; 4. The duration of the first waveform of each segment and the first The absolute value of the difference between the durations of the second waveforms adjacent to one waveform is less than or equal to the third predetermined value.
在上述四种可选的方式中,发送端在发送每个UPWM符号时,都是通过k段第一波形和k段第二波形交替发送的方式进行发送,因此,发送端在发送每个UPWM符号时,在保持平均功率不变的情况下避免出现闪烁问题。In the above four alternative manners, when transmitting each UPWM symbol, the transmitting end sends the first waveform of the k segment and the second waveform of the k segment alternately, so the transmitting end transmits each UPWM. In the case of symbols, the flicker problem is avoided while maintaining the average power.
可选的,该X个第一占空比包括p个占空比、L1个最小占空比以及L2个最大 占空比,该最小占空比和该最大占空比均为预设的,该最小占空比小于该p个占空比中的任意一个占空比,该最大占空比大于该p个占空比中的任意一个占空比,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。Optionally, the X first duty ratios include p duty ratios, L1 minimum duty ratios, and L2 maximums a duty ratio, the minimum duty ratio and the maximum duty ratio are preset, the minimum duty ratio being less than any one of the p duty ratios, the maximum duty ratio being greater than the p Any one of the duty ratios, L1 ≥ 0, L1 is an integer, L2 ≥ 0, L2 is an integer, p ≥ 1, and p is an integer.
可选的,该X个第一占空比指示第二参数、第三参数和/或第三数据;其中,该第二参数用于辅助接收端将该Y个第二占空比还原为该第一数据;该第三参数用于辅助接收端将该Z个第三占空比还原为该第二数据。Optionally, the X first duty ratios indicate a second parameter, a third parameter, and/or third data; wherein the second parameter is used to assist the receiving end to restore the Y second duty ratios to the The first data is used by the auxiliary receiving end to restore the Z third duty ratios to the second data.
在这种可选的方式中,通过X个第一占空比传递第二参数、第三参数和/或第三数据,辅助接收端还原第一数据以及第二数据,提高了接收端解析第一数据以及第二数据的正确率。In this optional manner, the second parameter, the third parameter, and/or the third data are transmitted through the X first duty ratios, and the auxiliary receiving end restores the first data and the second data, thereby improving the resolution of the receiving end. The correct rate of one data and the second data.
可选的,当该X个第一占空比用于指示该第三数据时,该X个第一占空比中的t个互不相同的第一占空比的排列顺序根据该第三数据和预设的第三映射规则得到,该第三映射规则包括t个互不相同的第一占空比的y种不同的排列顺序与y组比特之间的一一对应关系,该y组比特中的每一组比特包括R个比特,该第三数据为该y组比特中的一组,t≤X,t为整数,R≥1,R为整数。Optionally, when the X first duty ratios are used to indicate the third data, the order of the first duty ratios of the t different ones of the X first duty ratios is according to the third Data and a preset third mapping rule, where the third mapping rule includes a one-to-one correspondence between y different ordering sequences of t different first duty ratios and y group bits, the y group Each set of bits in the bit includes R bits, the third data is a set of the y set of bits, t ≤ X, t is an integer, R ≥ 1, and R is an integer.
可选的,该第一数据包括第三参数,该第三参数用于辅助接收端将该Z个第三占空比还原为该第二数据。Optionally, the first data includes a third parameter, where the third parameter is used to assist the receiving end to restore the Z third duty cycles to the second data.
结合上述第二方面或第四方面,可选的,该第二参数为预设的;或者,该第二参数为该接收端根据该X个第一百分比获取的。In combination with the foregoing second aspect or the fourth aspect, optionally, the second parameter is preset; or the second parameter is obtained by the receiving end according to the X first percentages.
可选的,该第三参数为预设的;或者,该第三参数为该接收端根据该X个第一百分比获取的。Optionally, the third parameter is preset; or the third parameter is obtained by the receiving end according to the X first percentages.
通过这两种可选的方式,通过第二参数、第三参数,辅助接收端还原第一数据以及第二数据,提高了接收端解析第一数据以及第二数据的正确率。In the two optional manners, the second data and the third parameter are used by the auxiliary receiving end to restore the first data and the second data, thereby improving the correctness rate of the first data and the second data by the receiving end.
可选的,该X个第一信号包括L1个最小值、L2个最大值以及除该最小值和最大值以外的p个信号,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。Optionally, the X first signals include L1 minimum values, L2 maximum values, and p signals except the minimum value and the maximum value, L1≥0, L1 is an integer, L2≥0, and L2 is an integer. p ≥ 1, and p is an integer.
可选的,该第一参数包括:该X个第一信号中的最大值、最小值、非线性数值序列和/或相位错误指示信息。Optionally, the first parameter includes: a maximum value, a minimum value, a non-linear value sequence, and/or phase error indication information in the X first signals.
可选的,该第一处理操作包括:非线性补偿、归一化和/或相位补偿操作。Optionally, the first processing operation includes: nonlinear compensation, normalization, and/or phase compensation operation.
可选的,该第一信号、该第二信号以及该第三信号为亮度值信号,或者,该第一信号、该第二信号以及该第三信号为幅值信号。Optionally, the first signal, the second signal, and the third signal are luminance value signals, or the first signal, the second signal, and the third signal are amplitude signals.
第五方面,本申请还提供了一种发送装置,包括:处理器、存储器及收发器;所述处理器可以执行所述存储器中所存储的程序或指令,从而实现以第一方面各种实现 方式所述信号传输方法。In a fifth aspect, the present application further provides a transmitting apparatus, including: a processor, a memory, and a transceiver; the processor may execute a program or an instruction stored in the memory, thereby implementing various implementations in the first aspect. The signal transmission method.
本申请提供的发送装置的技术效果可以参见上述第一方面或第一方面的各个实现方式的技术效果,此处不再赘述。For the technical effects of the sending device provided by the present application, refer to the technical effects of the foregoing first aspect or the implementation manner of the first aspect, and details are not described herein again.
第六方面,本申请还提供了一种接收装置,包括:处理器、存储器及收发器;所述处理器可以执行所述存储器中所存储的程序或指令,从而实现以第二方面各种实现方式所述信号传输方法。In a sixth aspect, the present application further provides a receiving apparatus, including: a processor, a memory, and a transceiver; the processor may execute a program or an instruction stored in the memory, thereby implementing various implementations in the second aspect. The signal transmission method.
本申请提供的接收装置的技术效果可以参见上述第二方面或第二方面的各个实现方式的技术效果,此处不再赘述。For technical effects of the receiving device provided by the present application, reference may be made to the technical effects of the foregoing second aspect or the implementation manner of the second aspect, and details are not described herein again.
第七方面,本申请还提供了一种存储介质,该计算机存储介质可存储有程序,该程序执行时可实现包括本申请提供的信号传输方法各实施例中的部分或全部步骤。In a seventh aspect, the present application further provides a storage medium, where the computer storage medium can store a program, and the program can be implemented to implement some or all of the steps in the embodiments of the signal transmission method provided by the present application.
第八方面,本申请还提供了一种通信系统,包括如第三方面或第三方面的任一种实现方式所述的发送装置,和如第四方面或第四方面的任一种实现方式所述的接收装置;或者包括如第五方面或第五方面的任一种实现方式所述的发送装置,和如第六方面或第六方面的任一种实现方式所述的接收装置。In an eighth aspect, the present application further provides a communication system, including the transmitting apparatus according to any one of the third aspect or the third aspect, and the implementation of any one of the fourth aspect or the fourth aspect The receiving device, or the transmitting device according to any one of the fifth aspect or the fifth aspect, and the receiving device according to any one of the sixth aspect or the sixth aspect.
附图说明DRAWINGS
图1为本申请提供的一种通信系统的框图;1 is a block diagram of a communication system provided by the present application;
图2为本申请一种发送端的结构示意图一;2 is a schematic structural diagram 1 of a transmitting end of the present application;
图3为本申请一种接收端的结构示意图一;3 is a schematic structural diagram 1 of a receiving end of the present application;
图4为本申请一种信号传输方法的一个实施例的流程图;4 is a flow chart of an embodiment of a signal transmission method according to the present application;
图5A为本申请提供的一种UPWM符号的示意图一;FIG. 5A is a schematic diagram 1 of a UPWM symbol provided by the present application; FIG.
图5B为本申请提供的一种UPWM符号的示意图二;FIG. 5B is a schematic diagram 2 of a UPWM symbol provided by the present application; FIG.
图5C为本申请提供的一种UPWM符号的示意图三;5C is a schematic diagram 3 of a UPWM symbol provided by the present application;
图5D为本申请提供的一种UPWM符号的示意图四;FIG. 5D is a schematic diagram 4 of a UPWM symbol provided by the present application; FIG.
图6A为本申请一种发送端的结构示意图二;6A is a schematic structural diagram 2 of a transmitting end of the present application;
图6B为本申请一种发送端的结构示意图三;6B is a schematic structural diagram 3 of a transmitting end of the present application;
图6C为本申请一种发送端的结构示意图四;6C is a schematic structural diagram 4 of a transmitting end of the present application;
图7A为本申请一种接收端的结构示意图二;7A is a schematic structural view 2 of a receiving end of the present application;
图7B为本申请一种接收端的结构示意图三; 7B is a schematic structural diagram 3 of a receiving end of the present application;
图7C为本申请一种接收端的结构示意图四。FIG. 7C is a schematic structural diagram 4 of a receiving end of the present application.
具体实施方式detailed description
本申请的说明书和权利要求书及上述附图中的术语“第一”、“第二”和“第三”等是用于区别不同对象,而不是用于限定特定顺序。The terms "first", "second", and "third" and the like in the specification and claims of the present application and the above drawings are used to distinguish different objects, and are not intended to limit the specific order.
在本申请实施例中,“示例性的”或者“例如”等词用于表示作例子、例证或说明。本申请实施例中被描述为“示例性的”或者“例如”的任何实施例或设计方案不应被解释为比其它实施例或设计方案更优选或更具优势。确切而言,使用“示例性的”或者“例如”等词旨在以具体方式呈现相关概念。In the embodiments of the present application, the words "exemplary" or "such as" are used to mean an example, illustration, or illustration. Any embodiment or design described as "exemplary" or "for example" in the embodiments of the present application should not be construed as preferred or advantageous over other embodiments or designs. Rather, the use of the words "exemplary" or "such as" is intended to present the concepts in a particular manner.
本文中字符“/”,一般表示前后关联对象是一种“或”的关系。本文中术语“和/或”,仅仅是一种描述关联对象的关联关系,表示可以存在三种关系,例如,A和/或B,可以表示:单独存在A,同时存在A和B,单独存在B这三种情况。The character "/" in this article generally indicates that the contextual object is an "or" relationship. The term "and/or" in this context is merely an association describing the associated object, indicating that there may be three relationships, for example, A and / or B, which may indicate that A exists separately, and both A and B exist, respectively. B these three situations.
本申请提供的本申请提供的一种信号传输方法可以应用于多种通信系统。例如,可以是OCC通信系统,也可以是射频通信系统。The signal transmission method provided by the present application provided by the present application can be applied to various communication systems. For example, it may be an OCC communication system or a radio frequency communication system.
示例性的,参见图1,为本申请提供的一种通信系统,包括至少一个发送端和至少一个接收端。其中,发送端可以是具有OCC功能的照明灯具、汽车的前后车灯、交通信号灯等,相应的,接收端可以是具有OCC功能的内置相机的智能手机、平板电脑、监控摄像头以及车内行车记录仪等。或者,发送端也可以是具有射频功能的终端。Illustratively, referring to FIG. 1, a communication system provided by the present application includes at least one transmitting end and at least one receiving end. The transmitting end may be an OC-enabled lighting fixture, a front and rear headlight of a car, a traffic signal, etc., and correspondingly, the receiving end may be a built-in camera with an OCC function, a smartphone, a tablet computer, a surveillance camera, and an in-vehicle driving record. Instrument and so on. Alternatively, the sender may also be a terminal having a radio frequency function.
示例性的,参见图2,发送端包括总线、处理器、存储器以及通信接口。其中,处理器是该发送端的控制中心,利用各种接口和线路连接整个发送端的各个部分,通过运行或执行存储在存储器内的应用程序和/或操作系统,以及调用存储在存储器内的数据,执行发送端的各种功能和处理数据,从而对发送端进行整体监控。处理器可以包括数字信号处理器设备、微处理器设备、模数转换器、数模转换器等等,这些设备能够根据各自的能力而分配发送端的控制和信号处理功能。通信接口可以包括射频(radio frequency,RF)电路,通信接口可用于收发信息,并将接收到的信息给处理器处理。其中,通常,RF电路包括但不限于天线、至少一个放大器、收发信机、耦合器、LNA(low noise amplifier,低噪声放大器)、双工器等,通过无线通信与网络与其他设备通信。其中,该无线通信可以使用任一通信标准或协议,包括但不限于全球移动通讯系统(global system of mobile communication,GSM)、通用分组无线服务(general packet radio service,GPRS)、码分多址(code division multiple access,CDMA)、宽带码分多址(wideband code division multiple access,WCDMA)、LTE(long term evolution,长期演进)、Wi-Fi或者低功耗Wi-Fi,以及WLAN技术等。此外,发送端还 可以包括输入/输出设备,例如LED灯或者其他闪烁灯等。Illustratively, referring to FIG. 2, the transmitting end includes a bus, a processor, a memory, and a communication interface. Wherein, the processor is a control center of the transmitting end, and connects various parts of the entire transmitting end by using various interfaces and lines, by running or executing an application and/or an operating system stored in the memory, and calling data stored in the memory, Perform various functions of the sender and process data to monitor the sender as a whole. The processor may include digital signal processor devices, microprocessor devices, analog to digital converters, digital to analog converters, etc., which are capable of distributing the control and signal processing functions of the transmitting end in accordance with their respective capabilities. The communication interface can include a radio frequency (RF) circuit that can be used to send and receive information and process the received information to the processor. Generally, the RF circuit includes, but is not limited to, an antenna, at least one amplifier, a transceiver, a coupler, an LNA (low noise amplifier), a duplexer, etc., and communicates with other devices through the wireless communication. The wireless communication may use any communication standard or protocol, including but not limited to a global system of mobile communication (GSM), a general packet radio service (GPRS), and code division multiple access ( Code division multiple access (CDMA), wideband code division multiple access (WCDMA), LTE (long term evolution, long term evolution), Wi-Fi or low power Wi-Fi, and WLAN technology. In addition, the sender also Input/output devices such as LED lights or other flashing lights can be included.
参见图3,接收端可以包括通信接口、处理器、存储器以及总线,其中处理器包括图像处理器、数字信号处理器设备、微处理器设备、模数转换器、数模转换器等。其中,总线用于连接处理器、存储器和通信接口,并且在处理器、存储器、和通信接口之间实现数据传输。处理器通过总线从通信接口接收到命令,解密接收到的命令,根据解密的命令执行计算或数据处理,以及将处理后的数据通过总线从通信接口发送至其他设备。存储器包括程序模块以及数据模块等。程序模块可以由软件、固件、硬件或其中的至少两种组成,用于存储应用程序以及操作系统。通信接口可以通过无线连接到网络以连接到外部其它的网元节点,完成数据收发。接收端还包括其他输入/输出装置,例如摄像头等。Referring to FIG. 3, the receiving end may include a communication interface, a processor, a memory, and a bus, wherein the processor includes an image processor, a digital signal processor device, a microprocessor device, an analog to digital converter, a digital to analog converter, and the like. The bus is used to connect the processor, the memory, and the communication interface, and implement data transfer between the processor, the memory, and the communication interface. The processor receives the command from the communication interface via the bus, decrypts the received command, performs calculation or data processing according to the decrypted command, and transmits the processed data from the communication interface to the other device through the bus. The memory includes program modules, data modules, and the like. The program modules may be comprised of software, firmware, hardware, or at least two of them for storing applications and operating systems. The communication interface can be connected to other external network element nodes by wirelessly connecting to the network to complete data transmission and reception. The receiving end also includes other input/output devices such as a camera or the like.
参见图4,为本申请提供的一种信号传输方法的一个实施例的流程图,该方法包括如下步骤:4 is a flowchart of an embodiment of a signal transmission method provided by the present application, where the method includes the following steps:
步骤401,发送端获取X个第一占空比、Y个第二占空比和Z个第三占空比,该Y个第二占空比中的每个第二占空比由待发送的第一数据中的N个比特映射所得,该Z个第三占空比中的每个第三占空比由待发送的第二数据中的M个比特映射所得,X≥1,X为整数,Y≥0,Y为整数,Z≥0,Z为整数,N≥1,N为整数,M≥1,M为整数。Step 401: The transmitting end acquires X first duty ratios, Y second duty ratios, and Z third duty ratios, and each of the second second duty ratios is to be sent The N bit maps in the first data are obtained, and each of the Z third duty ratios is obtained by mapping M bits in the second data to be transmitted, X≥1, X is Integer, Y≥0, Y is an integer, Z≥0, Z is an integer, N≥1, N is an integer, M≥1, and M is an integer.
在本申请中,X个第一占空比、Y个第二占空比和Z个第三占空比可以分别对应发送端发送的一个帧中的前导序列、帧头和帧的载荷。例如,在发送端发送的一个帧中,X个第一占空比对应的X个UPWM符号可以作为帧的前导序列,第一数据作为帧的帧头,第二数据作为帧的载荷。当X个第一占空比用于生成在发送第一数据和第二数据之前的发送的一串前导序列时,可以通过所述前导序列指示接收端即将有数据发送,以使得接收端能够提前准备进行数据接收,同时,实现接收端与发送端的帧同步。需要说明的是,Y和Z可以为0。当Y=0时,表示发送端没有第一数据,当Z=0时,表示发送端没有第二数据。示例性的,从帧结构角度来说,X个第一占空比、Y个第二占空比和Z个第三占空比可以分别对应于发送端发送的一个帧中的前导序列、帧头和帧的载荷,当Y=0且Z=0时,表示一个帧里面没有帧头和载荷,只有前导序列;当Y=0且Z≠0时,表示一个帧里面没有帧头,只有前导序列和载荷;当Y≠0且Z=0时,表示一个帧里面没有载荷只有前导序列和帧头,当Y≠0且Z≠0时,表示一个帧里面有前导序列、帧头和载荷。In the present application, the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may respectively correspond to the preamble sequence, the frame header, and the frame payload in one frame transmitted by the transmitting end. For example, in a frame sent by the transmitting end, X UPWM symbols corresponding to the X first duty ratios may be used as a preamble sequence of the frame, the first data is used as the frame header of the frame, and the second data is used as the payload of the frame. When the X first duty ratios are used to generate a series of preamble sequences transmitted before the first data and the second data are sent, the preamble sequence may be used to indicate that the receiving end is about to have data transmission, so that the receiving end can advance It is ready to receive data, and at the same time, realize frame synchronization between the receiving end and the transmitting end. It should be noted that Y and Z may be 0. When Y=0, it means that the sender has no first data, and when Z=0, it means that the sender has no second data. Exemplarily, from the perspective of a frame structure, the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may respectively correspond to a preamble sequence and a frame in one frame sent by the transmitting end. The header and frame payloads, when Y=0 and Z=0, indicate that there is no frame header and payload in a frame, only the preamble sequence; when Y=0 and Z≠0, it means that there is no frame header in one frame, only the preamble Sequence and load; when Y≠0 and Z=0, it means that there is no load in the frame only the preamble sequence and the frame header. When Y≠0 and Z≠0, it means that there is a preamble sequence, a frame header and a payload in a frame.
可选的,发送端发送的一个帧除了包括前导序列、帧头和载荷之外还可以包括其他序列,例如用于信道估计、相位错误测量等的序列。X个第一占空比、Y个第二占空比和Z个第三占空比可以分别对应发送端发送的一个帧中的一部分。例如,X个第一占空比、Y个第二占空比可以分别作为发送端发送的一个帧的前导序列和帧头,此 时Z=0;X个第一占空比、Y个第二占空比和Z个第三占空比也可以分别作为发送端发送的一个帧的前导序列、帧头和部分载荷;X个第一占空比和Z个第三占空比也可以分别作为发送端发送的一个帧中的其他序列和载荷,此时Y=0。Optionally, one frame sent by the transmitting end may include other sequences in addition to the preamble sequence, the frame header and the payload, for example, a sequence for channel estimation, phase error measurement, and the like. The X first duty ratios, the Y second duty ratios, and the Z third duty ratios may respectively correspond to a portion of one frame transmitted by the transmitting end. For example, the X first duty ratios and the Y second duty ratios may respectively serve as a preamble sequence and a frame header of a frame transmitted by the transmitting end, Z = 0; X first duty ratio, Y second duty ratio and Z third duty ratio may also be used as a preamble sequence, a frame header and a partial load of a frame transmitted by the transmitting end; The first duty cycle and the Z third duty cycles may also be used as other sequences and loads in one frame transmitted by the transmitting end, respectively, where Y=0.
可选的,X个第一占空比、Y个第二占空比和Z个第三占空比也可以分别指示相互独立的信息。例如,X个第一占空比可以指示第三数据,第一数据、第二数据和第三数据之间相互独立。Optionally, the X first duty ratios, the Y second duty ratios, and the Z third duty ratios may also indicate mutually independent information. For example, the X first duty ratios may indicate third data, and the first data, the second data, and the third data are independent of each other.
可以理解的是,上述对X个第一占空比、Y个第二占空比和Z个第三占空比所表示的信息仅为示例性的列举,X个第一占空比、Y个第二占空比和Z个第三占空比还可以用于指示具有其他含义的信息,对此,本申请不做限制。It can be understood that the information represented by the above X first duty ratio, Y second duty ratio and Z third duty ratio is only an exemplary enumeration, X first duty ratios, Y The second duty cycle and the Z third duty cycle may also be used to indicate information having other meanings, and the application is not limited thereto.
在一个示例中,当Y>0时,即表示发送端存在第一数据需要发送时,发送端获取Y个第二占空比的方式可以为:In an example, when Y>0, that is, when the first data needs to be sent on the transmitting end, the manner in which the sending end acquires the Y second duty ratios may be:
发送端按照第一数据中比特的排列顺序,以N个比特为一组,将该第一数据中的比特划分为Y组。然后根据预设的第一映射规则,将该Y组比特映射为该Y个第二占空比,该第一映射规则包括不同的n组比特与不同的n个占空比之间的一一对应关系。The transmitting end divides the bits in the first data into Y groups according to the order of the bits in the first data, in groups of N bits. And then mapping the Y group bits to the Y second duty ratios according to a preset first mapping rule, where the first mapping rule includes a different one of n groups of bits and different n duty ratios Correspondence relationship.
需要说明的是,第一映射规则中不同的n个占空比可以等间隔分布,也可以非等间隔分布,例如,当n=4时,4个不同占空比可为{20%、40%、60%、80%},或者为{10%、40%、60%、70%}。It should be noted that different n duty ratios in the first mapping rule may be equally spaced, or may be non-equally distributed. For example, when n=4, four different duty ratios may be {20%, 40. %, 60%, 80%}, or {10%, 40%, 60%, 70%}.
其中,N=log2n。n为调制第一数据时使用的UPWM阶数。Y的大小可以根据第一数据的比特数和N值确定。例如,假设第一数据为二进制“10010111”,当采用4阶UPWM调制(n=4)时,N=2。那么第一数据需要分为4组,每一组则包括2个比特。发送端可以选择与n=4对应的映射规则作为第一映射规则。Where N = log 2 n. n is the UPWM order used when modulating the first data. The size of Y can be determined based on the number of bits of the first data and the value of N. For example, assume that the first data is binary "10010111" and when using 4th order UPWM modulation (n=4), N=2. Then the first data needs to be divided into 4 groups, and each group includes 2 bits. The transmitting end may select a mapping rule corresponding to n=4 as the first mapping rule.
需要说明的是,对应于一个n值可以设置多个映射规则。例如,当发送端确定n=4时,选择的第一映射规则可以如表1或表2所示:It should be noted that multiple mapping rules can be set corresponding to one n value. For example, when the transmitting end determines that n=4, the selected first mapping rule may be as shown in Table 1 or Table 2:
表1Table 1
比特组合Bit combination 占空比Duty cycle
0000 20%20%
0101 40%40%
1010 60%60%
1111 80%80%
表2Table 2
比特组合Bit combination 占空比Duty cycle
0000 40%40%
0101 60%60%
1010 80%80%
1111 20%20%
可以理解的是,按照第一数据中“0”和“1”的排列顺序,二进制“10010111”可划分为“10”“01”“01”“11”,共4组,如果按照如表1所示的第一映射规则,那么这4组比特分别映射为“60%、40%、40%和80%”。It can be understood that, according to the order of "0" and "1" in the first data, the binary "10010111" can be divided into "10", "01", "01" and "11", a total of 4 groups, if according to Table 1 The first mapping rule shown, then the four sets of bits are mapped to "60%, 40%, 40%, and 80%, respectively."
在一个示例中,假设第一数据为二进制“1100100”,当采用4阶UPWM调制(n=4)时,N=2。由于第一数据长度为7,因此需要在第一数据中补充至少1个预设比特,使得Y为整数。例如,在“1100100”前补充1个预设比特“0”,并按照其中“0”和“1”的排列顺序将“01100100”划分为4组,每一组包括2个比特,依次为“01”“10”“01”“00”。进而发送端可以选择与n=4对应的映射规则作为第一映射规则将该4组比特映射为4个第二占空比。例如,按照如表2所示n=4的第一映射规则,“01”“10”“01”“00”即可分别被映射为“60%、80%、60%和40%”。In one example, assume that the first data is binary "1100100" and when using 4th order UPWM modulation (n=4), N=2. Since the first data length is 7, it is necessary to supplement at least one preset bit in the first data such that Y is an integer. For example, a preset bit "0" is added before "1100100", and "01100100" is divided into 4 groups according to the order of "0" and "1", and each group includes 2 bits, followed by " 01” “10” “01” “00”. Furthermore, the transmitting end may select the mapping rule corresponding to n=4 as the first mapping rule to map the four sets of bits into four second duty ratios. For example, according to the first mapping rule of n=4 as shown in Table 2, "01" "10" "01" "00" can be mapped to "60%, 80%, 60%, and 40%, respectively."
在一个示例中,当Z>0时,即表示发送端存在第二数据需要发送时,发送端获取Z个第三占空比的方式可以为:In an example, when Z>0, that is, when the second data needs to be sent on the transmitting end, the manner in which the transmitting end acquires the Z third duty ratios may be:
发送端按照第二数据中比特的排列顺序,以M个比特为一组,将该第二数据中的比特划分为Z组。然后根据预设的第二映射规则,将该Z组比特映射为该Z个第三占空比,该第二映射规则包括不同的m组比特与不同的m个占空比之间的一一对应关系。The transmitting end divides the bits in the second data into Z groups according to the order of the bits in the second data, in groups of M bits. And then mapping the Z group of bits to the Z third duty ratios according to a preset second mapping rule, where the second mapping rule includes different ones of m groups of bits and different m duty ratios Correspondence relationship.
在本申请中,第二映射规则中不同的m个占空比可以等间隔分布也可非等间隔分布。例如,m=4时4个不同占空比可以为{20%、40%、60%、80%},或者为{10%、40%、60%、70%},m=8时8个不同占空比可以为{10%、20%、30%、40%、50%、60%、70%、80%},或者为{0%、20%、30%、40%、50%、60%、70%、100%}。In the present application, different m duty ratios in the second mapping rule may be equally spaced or may be distributed at different intervals. For example, when m=4, 4 different duty cycles can be {20%, 40%, 60%, 80%}, or {10%, 40%, 60%, 70%}, 8 when m=8. Different duty cycles can be {10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%}, or {0%, 20%, 30%, 40%, 50%, 60%, 70%, 100%}.
其中,M=log2m,m为调制第二数据使用的UPWM阶数,Z的大小可以根据第二数据的比特数和M的值确定。例如,假设待发送的第二数据为二进制“001011010”,m=8,M=log2 8=3,那么,可以将该第二数据以每3个比特为一组,按照该第二数据中“0”和“1”的排列顺序,划分为“001”“011”“010”,共3组。 Where M=log 2 m, m is the UPWM order used to modulate the second data, and the size of Z can be determined according to the number of bits of the second data and the value of M. For example, if the second data to be transmitted is binary "001011010", m=8, and M=log 2 8=3, then the second data may be grouped into every 3 bits, according to the second data. The order of "0" and "1" is divided into "001", "011" and "010", and there are 3 groups.
第二映射规则中不同的m组比特中的每一组均包括M个比特。该m组比特为以M个比特进行随机组合所有可能的组合情况。例如,当M=2时,2个“0”和“1”能够组成m=4种组合,分别为“00”、“01”、“10”以及“11”。当M=3时,3个“0”和“1”能够组成m=8种组合,分别为“000”、“001”、“010”、“011”、“100”、“101”、“110”以及“111”。因此,该M组比特中的每一组比特都能够从该第二映射规则中查找到匹配的比特组合,进而查找到对应的占空比。Each of the different m sets of bits in the second mapping rule includes M bits. The m sets of bits are all possible combinations of random combinations of M bits. For example, when M=2, two "0"s and "1"s can constitute m=4 combinations, which are "00", "01", "10", and "11", respectively. When M=3, 3 “0” and “1” can form m=8 combinations, respectively “000”, “001”, “010”, “011”, “100”, “101”, “ 110" and "111". Therefore, each of the M sets of bits can find a matched bit combination from the second mapping rule, thereby finding a corresponding duty ratio.
需要说明的是,对应于一个m值可以设置多个映射规则。例如,当发送端确定m=8时,选择的第二映射规则可以如表3、表4以及表5所示:It should be noted that multiple mapping rules can be set corresponding to one m value. For example, when the sender determines that m=8, the selected second mapping rule can be as shown in Table 3, Table 4, and Table 5:
表3table 3
比特组合Bit combination 占空比Duty cycle
000000 10%10%
001001 20%20%
010010 30%30%
011011 40%40%
100100 50%50%
101101 60%60%
110110 70%70%
111111 80%80%
表4Table 4
比特组合Bit combination 占空比Duty cycle
000000 80%80%
001001 70%70%
010010 60%60%
011011 50%50%
100100 40%40%
101101 30%30%
110110 20%20%
111111 10%10%
表5table 5
比特组合Bit combination 占空比Duty cycle
000000 10%10%
001001 80%80%
010010 20%20%
011011 70%70%
100100 30%30%
101101 60%60%
110110 40%40%
111111 50%50%
例如,假设选择如表3所示的第二映射规则,那么,第二数据的二进制序列“001011010”,所划分得到的“001”、“011”、“010”,这三组比特分别可以映射为20%、40%和30%。For example, assuming that the second mapping rule as shown in Table 3 is selected, the binary sequence "001011010" of the second data, the divided "001", "011", and "010", respectively, can be mapped. It is 20%, 40% and 30%.
在本申请中,当对应于一个m值设置了多个映射规则时,发送端可以随机选择 一个映射规则对第二数据进行调制,也可以在通信过程中可随时更换映射规则。In the present application, when multiple mapping rules are set corresponding to one m value, the transmitting end can randomly select A mapping rule modulates the second data, and the mapping rule can be changed at any time during the communication process.
需要说明的是,在本申请中,调制阶数m和n可以是固定值,也可以是,也可以是发送端根据接收端发送的测量参数确定。该测量参数可以包括接收端的接收信噪比、误码率、探测信号等。It should be noted that, in the present application, the modulation orders m and n may be fixed values, or may be determined by the transmitting end according to measurement parameters sent by the receiving end. The measurement parameters may include a received signal to noise ratio, a bit error rate, a detection signal, and the like at the receiving end.
示例性的,发送端在获取Y个第二百分比、Z个第三百分比之前,接收由接收端发送的探测信号。然后发送端可以根据该探测信号计算发送端与该接收端之间的距离。当发送端与该接收端之间的距离小于或者等于3米时,发送端确定m=8,n=4;当发送端与该接收端之间的距离大于3米时,小于10米时,发送端确定m=4,n=2;当发送端确定发送端与该接收端之间的距离大于10米时,发送端确定m=2,n=2。Exemplarily, the transmitting end receives the sounding signal sent by the receiving end before acquiring the Y second percentages and the Z third percentages. Then, the transmitting end can calculate the distance between the transmitting end and the receiving end according to the detection signal. When the distance between the transmitting end and the receiving end is less than or equal to 3 meters, the transmitting end determines that m=8, n=4; when the distance between the transmitting end and the receiving end is greater than 3 meters, less than 10 meters, The transmitting end determines that m=4, n=2; when the transmitting end determines that the distance between the transmitting end and the receiving end is greater than 10 meters, the transmitting end determines that m=2, n=2.
可选的,当第一数据与第二数据相关,例如,第一数据和第二数据分别作为发送端传输的同一帧的帧头和载荷时,第一数据可以用于包括第三参数,该第三参数用于辅助接收端将Z个第三占空比还原为第二数据。例如,第三参数可以包Z的值、m的值和/或第二映射规则的标号等。Optionally, when the first data is related to the second data, for example, the first data and the second data are respectively used as a frame header and a payload of the same frame transmitted by the transmitting end, the first data may be used to include a third parameter, where The third parameter is used to assist the receiving end to restore the Z third duty ratios to the second data. For example, the third parameter may include the value of Z, the value of m, and/or the label of the second mapping rule, and the like.
在一个示例中,当第一数据和第二数据分别作为发送端传输的同一帧的帧头和载荷时,m≥n。当m>n时,第一映射规则所包含的n个占空比可以与第二映射规则所包含的m个占空比可以完全不相同,或者部分相同,或者该n个占空比是该m个占空比的子集。例如,第一映射规则采用如表1所示的映射规则,第二映射规则采用如表3所示的映射规则,第一映射规则使用的占空比为{20%、40%、60%、80%}是第二映射规则使用的占空比为{10%、20%、30%、40%、50%、60%、70%、80%}的子集。In one example, when the first data and the second data respectively serve as the frame header and payload of the same frame transmitted by the transmitting end, m ≥ n. When m>n, the n duty ratios included in the first mapping rule may be completely different from, or partially identical to, the m duty ratios included in the second mapping rule, or the n duty ratios are A subset of m duty cycles. For example, the first mapping rule adopts a mapping rule as shown in Table 1, and the second mapping rule adopts a mapping rule as shown in Table 3. The duty ratio used by the first mapping rule is {20%, 40%, 60%, 80%} is a subset of the {10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%} duty cycle used by the second mapping rule.
在一个示例中,X个第一占空比可以包括p个占空比、L1个最小占空比以及L2个最大占空比,该最小占空比和该最大占空比均为预设的,该最小占空比小于该p个占空比中的任意一个占空比,该最大占空比大于该p个占空比中的任意一个占空比,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。In one example, the X first duty cycles may include p duty cycles, L1 minimum duty cycles, and L2 maximum duty cycles, the minimum duty cycles and the maximum duty cycles being preset The minimum duty ratio is less than any one of the p duty ratios, and the maximum duty ratio is greater than any one of the p duty ratios, L1≥0, L1 is an integer, L2 ≥ 0, L2 is an integer, p ≥ 1, and p is an integer.
可选地,p可以为预设常数,X个第一占空比中的p个占空比、L1个最小占空比和L2个最大占空比可以按照预设的不同的排列顺序排列和发送。通过不同的排列顺序来指示不同的指示信息。Optionally, p may be a preset constant, and p duty ratios, L1 minimum duty ratios, and L2 maximum duty ratios among the X first duty ratios may be arranged in a preset different arrangement order and send. Different indication information is indicated by different arrangement order.
在本申请中,X个第一占空比可以用于让接收端进行帧同步,获取相位错误信息、非线性曲线信息的参数,也可以用于指示第二参数、第三参数和/或第三数据。其中,第二参数用于辅助接收端将Y个第二占空比还原为第一数据,第二参数包括Y的值、n的值和/或第一映射规则等。In the present application, the X first duty ratios may be used to allow the receiving end to perform frame synchronization, obtain phase error information, parameters of nonlinear curve information, and may also be used to indicate second parameters, third parameters, and/or Three data. The second parameter is used to assist the receiving end to restore the Y second duty ratios to the first data, and the second parameter includes the value of Y, the value of n, and/or the first mapping rule, and the like.
示例性的,X个第一占空比的排列顺序与指示信息之间的对应关系,可以如表6所示: Exemplarily, the correspondence between the order of the X first duty ratios and the indication information may be as shown in Table 6:
表6Table 6
Figure PCTCN2017085384-appb-000001
Figure PCTCN2017085384-appb-000001
例如,假设发送端确定第二参数为Y=5,n=2,采用第一映射规则2,第三参数为Z=0,那么发送端可以根据上述表6确定X个第一占空比为{0%,80%,60%,40%,20%,100%}。For example, suppose the transmitting end determines that the second parameter is Y=5, n=2, adopts the first mapping rule 2, and the third parameter is Z=0, then the transmitting end may determine X first duty ratios according to Table 6 above. {0%, 80%, 60%, 40%, 20%, 100%}.
当该X个第一占空比用于指示该第三数据时,该X个第一占空比中的t个互不相同的第一占空比的排列顺序根据该第三数据和预设的第三映射规则得到,该第三映射规则包括t个互不相同的第一占空比的y种不同的排列顺序与y组比特之间的一一对应关系,该y组比特中的每一组比特包括R个比特,该第三数据为该y组比特中的一组,t≤X,t为整数,R≥1,R为整数。When the X first duty ratios are used to indicate the third data, the order of the first duty ratios of the t different ones of the X first duty ratios is according to the third data and the preset Obtaining a third mapping rule, where the third mapping rule includes a one-to-one correspondence between y different arrangement orders of t different first duty ratios and y group bits, each of the y group bits A group of bits includes R bits, the third data is a group of the y group bits, t ≤ X, t is an integer, R ≥ 1, and R is an integer.
示例性的,第三映射规则可以如表7所示,其中X=6,t=4,R=3,t个互不相同的第一占空比分别是:20%、40%、60%、80%。 Exemplarily, the third mapping rule can be as shown in Table 7, where X=6, t=4, and R=3, and the first first duty ratios that are different from each other are: 20%, 40%, 60%. 80%.
表7Table 7
X个第一占空比X first duty cycles 第三数据Third data
0%40%20%60%80%100%0%40%20%60%80%100% 000000
0%40%60%20%80%100%0%40%60%20%80%100% 001001
0%40%60%80%20%100%0%40%60%80%20%100% 010010
0%40%20%80%60%100%0%40%20%80%60%100% 011011
0%40%80%20%60%100%0%40%80%20%60%100% 100100
0%40%80%60%20%100%0%40%80%60%20%100% 101101
0%60%20%40%80%100%0%60%20%40%80%100% 110110
0%60%40%20%80%100%0%60%40%20%80%100% 111111
例如,当第三数据为111时,那么根据上述表7所示的第三映射规则,发送端可以确定X个第一占空比为{0%,60%,40%,20%,80%,100%}。For example, when the third data is 111, then according to the third mapping rule shown in Table 7 above, the transmitting end may determine that the X first duty ratios are {0%, 60%, 40%, 20%, 80%. , 100%}.
下面将结合如下四种可能的场景,对X个第一占空比,Y个第二占空比以及Z个第三占空比的获取进行说明。The acquisition of X first duty ratios, Y second duty ratios, and Z third duty ratios will be described below in conjunction with the following four possible scenarios.
场景一:L1+L2>0,Y>0,Z>0。X个第一占空比对应前导序列,Y个第二占空比作为帧头,Z个第三占空比作为载荷。Scene 1: L1+L2>0, Y>0, Z>0. The X first duty ratios correspond to the preamble sequence, the Y second duty ratios serve as frame headers, and the Z third duty ratios serve as loads.
假设,待发送的第二数据为“00011011000110110001”,发送端确定使用4阶UPWM调制(即m=4),且采用编号为41的映射规则(如表1所示)。因此,发送端Z=10个第三占空比序列:{20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}。Assume that the second data to be transmitted is "00011011000110110001", the transmitting end determines to use 4th order UPWM modulation (ie, m=4), and uses the mapping rule numbered 41 (as shown in Table 1). Therefore, the transmitting terminal Z=10 third duty cycle sequences: {20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40%}.
假设,发送端确定使用2阶UPWM调制(n=2),且采用标号为21的映射规则(如表8所示),Y为预设值5。由于n=2,即每个第二占空比由1个比特映射所得,而Y=5,即第二数据应为一个5比特的数据,才能映射5个第二占空比。由于Z=10,10的二进制序列为“1010”,少一个比特位,因此可以在“1010”前补1个比特“0”,得到第二数据“01010”。即“01010”发送端发送的第一数据。进而,发送端根据“01010”,n=2,以及映射规则21,得到5个第二占空比{20%,80%,20%,80%,20%}。Assume that the transmitting end determines to use 2nd order UPWM modulation (n=2), and uses the mapping rule labeled 21 (as shown in Table 8), and Y is the preset value of 5. Since n=2, that is, each second duty ratio is obtained by mapping 1 bit, and Y=5, that is, the second data should be a 5-bit data, five second duty ratios can be mapped. Since Z=10, the binary sequence of 10 is "1010", one bit less, so one bit "0" can be added before "1010" to obtain the second data "01010". That is, the first data sent by the sender of "01010". Further, the transmitting end obtains five second duty ratios {20%, 80%, 20%, 80%, 20%} according to "01010", n=2, and mapping rule 21.
假设p=4,p个占空比为{20%、40%、60%、80%},最小占空比为0%,最大占空比为100%,L1=1,L2=1。根据预设的表9,可以确定当第二参数包括:Y=5,n=2, 映射规则21(即第一数据采用的第一映射规则),第三参数包括:m=4,映射规则41(即第二数据采用的第二映射规则)时,X个第一占空比为{0%,20%,40%,80%,60%,100%},X=6。Assuming p=4, p duty cycles are {20%, 40%, 60%, 80%}, the minimum duty cycle is 0%, the maximum duty cycle is 100%, L1=1, L2=1. According to the preset table 9, it can be determined that when the second parameter includes: Y=5, n=2, The mapping rule 21 (ie, the first mapping rule adopted by the first data), the third parameter includes: m=4, and when the mapping rule 41 (ie, the second mapping rule adopted by the second data), the X first duty ratios are {0%, 20%, 40%, 80%, 60%, 100%}, X=6.
因此发送端在UPWM调制前的占空比序列(包括6个第一占空比、5个第二占空比以及10个第三占空比)为:{0%,20%,40%,80%,60%,100%,20%,80%,20%,80%,20%,20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}。Therefore, the duty cycle sequence of the transmitting end before UPWM modulation (including 6 first duty ratios, 5 second duty ratios, and 10 third duty ratios) is: {0%, 20%, 40%, 80%, 60%, 100%, 20%, 80%, 20%, 80%, 20%, 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20% , 40%}.
表8Table 8
比特组合Bit combination 占空比Duty cycle
00 20%20%
11 80%80%
表9Table 9
Figure PCTCN2017085384-appb-000002
Figure PCTCN2017085384-appb-000002
场景二,L1+L2>0,Y=0,Z=0。X个第一占空比用于指示第三数据。Scene 2, L1+L2>0, Y=0, Z=0. The X first duty cycles are used to indicate the third data.
假设,发送端需要通过X个第一占空比来发送数据“011110”,根据表10,发送端确定可以通过两个第三数据“011”和“110”组合得到“011110”。发送端即可使用两组X(X=6)个第一占空比来映射数据“011110”。那么,根据表10,发送端可以得到数据“011110”对应的占空比序列(包括两组6个第一占空比)为{0%,40%,20%,80%,60%,100%,0%,60%,20%,40%,80%,100%},因此发送端可以对这12个占空比序列进行UPWM调制,再按照顺序依次发送。 It is assumed that the transmitting end needs to transmit data "011110" through X first duty ratios. According to Table 10, the transmitting end determines that "011110" can be obtained by combining two third data "011" and "110". The transmitting end can map the data "011110" using two sets of X (X = 6) first duty ratios. Then, according to Table 10, the transmitting end can obtain the duty cycle sequence corresponding to the data "011110" (including two sets of six first duty ratios) of {0%, 40%, 20%, 80%, 60%, 100. %, 0%, 60%, 20%, 40%, 80%, 100%}, so the transmitter can perform UPWM modulation on the 12 duty cycle sequences, and then sequentially transmit them in order.
需要注意的是,表10中X=6,t=4,t个互不相同的第一占空比分别为:20%、40%、60%、80%。It should be noted that in Table 10, X=6, t=4, and the first different duty ratios of t are different: 20%, 40%, 60%, and 80%, respectively.
表10Table 10
Figure PCTCN2017085384-appb-000003
Figure PCTCN2017085384-appb-000003
场景三:L1+L2=0,Y>0,Z>0。X个第一占空比对应于前导序列。Scene 3: L1+L2=0, Y>0, Z>0. The X first duty cycles correspond to a preamble sequence.
假设第二数据为“00011011000110110001”,发送端确定m=4,第二数据采用的第二映射规则为映射规则41,那么发送端通过对第二数据的映射,可以得到Z=10个第三占空比序列:{20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}。假设,发送端确定Y=5,n=2,且第一数据采用的第一映射规则为映射规则21,那么用于表示Z=10的第一数据即为:“01010”,从而得到5个第二占空比{20%,80%,20%,80%,20%}。Assuming that the second data is "00011011000110110001", the transmitting end determines m=4, and the second mapping rule adopted by the second data is the mapping rule 41, then the transmitting end can obtain Z=10 third occupying by mapping the second data. Empty ratio sequence: {20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40%}. It is assumed that the transmitting end determines that Y=5, n=2, and the first mapping rule adopted by the first data is the mapping rule 21, then the first data used to represent Z=10 is: “01010”, thereby obtaining 5 The second duty cycle is {20%, 80%, 20%, 80%, 20%}.
当L1+L2=0时,假设p=6,p个占空比为{20%,32%,44%,56%,68%,80%}。 根据预设的表11,可以确定当第二参数包括:Y=5,n=2,映射规则21(即第一数据采用的第一映射规则),第三参数包括:m=4,映射规则41(即第二数据采用的第二映射规则)时,X个第一占空比为{20%,32%,68%,56%,44%,80%},X=6。When L1+L2=0, assuming p=6, the p duty ratios are {20%, 32%, 44%, 56%, 68%, 80%}. According to the preset table 11, it can be determined that when the second parameter includes: Y=5, n=2, mapping rule 21 (ie, the first mapping rule adopted by the first data), and the third parameter includes: m=4, mapping rule 41 (ie, the second mapping rule adopted by the second data), the first duty ratios of X are {20%, 32%, 68%, 56%, 44%, 80%}, and X=6.
因此,发送端在UPWM调制前的占空比序列(包括6个第一占空比、5个第二占空比以及10个第三占空比)为:{20%,32%,68%,56%,44%,80%,20%,80%,20%,80%,20%,20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}。Therefore, the duty cycle of the transmitter before UPWM modulation (including 6 first duty cycles, 5 second duty cycles, and 10 third duty cycles) is: {20%, 32%, 68% , 56%, 44%, 80%, 20%, 80%, 20%, 80%, 20%, 20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20 %, 40%}.
表11Table 11
Figure PCTCN2017085384-appb-000004
Figure PCTCN2017085384-appb-000004
场景四:L1+L2=0,Y=0,Z=0。X个第一占空比用于指示第三数据。Scene 4: L1+L2=0, Y=0, Z=0. The X first duty cycles are used to indicate the third data.
假设,发送端需要通过X个第一占空比来发送数据“011110”,根据表12,发送端确定可以通过两个第三数据“011”和“110”组合得到“011110”。发送端即可使用两组X(X=6)个第一占空比来映射数据“011110”。那么,根据表12,发送端可以得到数据“011110”对应的占空比序列(包括两组6个第一占空比)为{20%,44%,32%,68%,56%,80%,20%,44%,68%,56%,32%,80%},因此发送端可以对这12个占空比序列进行UPWM调制,再按照顺序依次发送。It is assumed that the transmitting end needs to transmit data "011110" through X first duty ratios. According to Table 12, the transmitting end determines that "011110" can be obtained by combining two third data "011" and "110". The transmitting end can map the data "011110" using two sets of X (X = 6) first duty ratios. Then, according to Table 12, the transmitting end can obtain the duty cycle sequence corresponding to the data "011110" (including two sets of six first duty ratios) of {20%, 44%, 32%, 68%, 56%, 80. %, 20%, 44%, 68%, 56%, 32%, 80%}, so the transmitter can UPWM modulate the 12 duty cycle sequences, and then send them sequentially.
需要注意的,是表12中X=6,t=4,t个互不相同的第一占空比分别为:32%、44%、56%、68%。 It should be noted that in Table 12, X=6, t=4, and the first duty ratios of t different from each other are: 32%, 44%, 56%, and 68%, respectively.
表12Table 12
Figure PCTCN2017085384-appb-000005
Figure PCTCN2017085384-appb-000005
步骤402,发送端生成与该X个第一占空比对应的X个基于欠采样的脉冲宽度调制UPWM符号、与该Y个第二占空比对应的Y个UPWM符号,和与该Z个第三占空比对应的Z个UPWM符号。Step 402: The transmitting end generates X under-sampled pulse width modulated UPWM symbols corresponding to the X first duty ratios, and Y UPWM symbols corresponding to the Y second duty ratios, and the Z Z UPWM symbols corresponding to the third duty cycle.
其中,X个UPWM符号、Y个UPWM符号和Z个UPWM符号中的每个UPWM符号,均包括k段第一波形和k段第二波形,k≥1,k为整数。该k段第一波形中的每段第一波形之后均与该k段第二波形中的一段第二波形相邻,即该k段第一波形和该k段第二波形,按照“第1个第一波形,第1个第二波形,第2个第一波形,第2个第二波形,……,第k个第一波形,第k个第二波形”的顺序排列。Wherein, each of the U UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols includes a k-segment first waveform and a k-segment second waveform, k≥1, k being an integer. Each of the first waveforms of the first waveform of the k-segment is adjacent to a second waveform of the second waveform of the k-segment, that is, the first waveform of the k-segment and the second waveform of the k-segment, according to "1st The first waveform, the first second waveform, the second first waveform, the second second waveform, ..., the kth first waveform, and the kth second waveform are sequentially arranged.
需要说明的是,对于不同的UPWM符号,k的值可以相同,也可以不同。例如,该X个UPWM符号中的每个UPWM符号包括10(k=10)个第一波形和第二波形, 该Y个UPWM符号中的每个UPWM符号包括8(k=8)个第一波形和第二波形,该Z个UPWM符号中的每个UPWM符号包括14(k=14)个第一波形和第二波形,或者X个UPWM符号中的第一个UPWM符号包括10(k=10)个第一波形和第二波形,第二个UPWM符号包括12(k=12)个第一波形和第二波形;Y个UPWM符号中的第一个UPWM符号包括8(k=8)个第一波形和第二波形,第二个UPWM符号包括14(k=14)个第一波形和第二波形;Z个UPWM符号中的第一个UPWM符号包括9(k=9)个第一波形和第二波形,第二个UPWM符号包括11(k=11)个第一波形和第二波形。It should be noted that the values of k may be the same or different for different UPWM symbols. For example, each UPWM symbol of the X UPWM symbols includes 10 (k=10) first waveforms and second waveforms, Each of the Y UPWM symbols includes 8 (k=8) first waveforms and a second waveform, each UPWM symbol of the Z UPWM symbols comprising 14 (k=14) first waveforms and The second waveform, or the first UPWM symbol of the X UPWM symbols, includes 10 (k=10) first waveforms and second waveforms, and the second UPWM symbol includes 12 (k=12) first waveforms and Two waveforms; the first UPWM symbol of the Y UPWM symbols includes 8 (k=8) first waveforms and second waveforms, and the second UPWM symbol includes 14 (k=14) first waveforms and second waveforms The first UPWM symbol of the Z UPWM symbols includes 9 (k=9) first waveforms and second waveforms, and the second UPWM symbol includes 11 (k=11) first waveforms and second waveforms.
在本申请中,第一波形为平均占空比为D的PWM波形,第二波形为平均占空比为1-D的PWM波形,0≤D≤100%。例如,该X个第一占空比中的一个第一占空比为20%,那么与20%对应的UPWM符号中的第一波形的平均占空比为20%(即D=20%),第二波形的平均占空比为80%(即1-D=80%)。In the present application, the first waveform is a PWM waveform with an average duty ratio of D, and the second waveform is a PWM waveform with an average duty ratio of 1-D, 0≤D≤100%. For example, if one of the X first duty ratios has a first duty ratio of 20%, then the average duty ratio of the first waveform in the UPWM symbol corresponding to 20% is 20% (ie, D=20%) The average duty ratio of the second waveform is 80% (ie, 1-D=80%).
在一个示例中,第一波形可以包含连续的J1个第一子波形,该J1个第一子波形中的每个第一子波形为一个脉冲波形,即一个第一子波形为一个完整的PWM周期内的波形。该J1个第一子波形的占空比为均D;或者,该第一波形在任意第一预设时长Ti内包含的J2个第一子波形的平均占空比为D1,D1与D的差值的绝对值小于或者等于第一预设值,J2<J1。每一个第一子波形的PWM周期可以相同,也可以不同。In one example, the first waveform may include consecutive J1 first sub-waveforms, and each of the first sub-waveforms of the J1 first sub-waveforms is a pulse waveform, that is, a first sub-waveform is a complete PWM The waveform within the period. The duty ratio of the J1 first sub-waveforms is D; or the average duty ratio of the J2 first sub-waveforms included in the first waveform in any first preset duration Ti is D1, D1 and D The absolute value of the difference is less than or equal to the first preset value, J2 < J1. The PWM period of each of the first sub-waveforms may be the same or different.
在一个示例中,第二波形可以包含连续的J3个第二子波形,该J3个第二子波形中的每个第二子波形为一个脉冲波形,即一个第二子波形为一个完整的PWM周期内的波形。该J3个第二子波形的占空比为均1-D;或者,该第二波形在任意第一预设时长Ti内包含的J4个第二子波形的平均占空比为D2,D2与1-D的差值的绝对值小于或者等于第二预设值,J4<J3。每一个第二子波形的PWM周期可以相同,也可以不同。In one example, the second waveform may include consecutive J3 second sub-waveforms, each of the J3 second sub-waveforms being a pulse waveform, ie, a second sub-waveform is a complete PWM The waveform within the period. The duty ratio of the J3 second sub-waveforms is 1-D; or the average duty ratio of the J4 second sub-waveforms included in the second waveform in any first preset duration Ti is D2, D2 and The absolute value of the difference of 1-D is less than or equal to the second preset value, J4 < J3. The PWM period of each of the second sub-waveforms may be the same or different.
示例性的,以与X个第一占空比中的一个第一占空比对应的UPWM符号为例。假设,该第一占空比为20%,k=2,那么该对应的UPWM符号包括平均占空比为20%的2个第一波形,和平均占空比为80%的2个第二波形。该UPWM符号的时长为T,该UPWM符号的波形图可以如图5A-图5D所示。Illustratively, a UPWM symbol corresponding to one of the X first duty ratios is taken as an example. Assuming that the first duty cycle is 20%, k=2, then the corresponding UPWM symbol includes two first waveforms having an average duty ratio of 20%, and two seconds having an average duty ratio of 80%. Waveform. The UPWM symbol has a duration of T, and the waveform of the UPWM symbol can be as shown in FIGS. 5A-5D.
在图5A中,第一波形中的每个第一子波形的PWM周期相同,均为T1,每个第一子波形的占空比均为20%,第二波形中的每个第二子波形的PWM周期相同,均为T3,每个第二子波形的占空比均为80%。其中,T1可以与T3相等,也可以不相等。In FIG. 5A, each of the first sub-waveforms in the first waveform has the same PWM period, both of which are T1, and each of the first sub-waveforms has a duty ratio of 20%, and each second sub-waveform The PWM period of the waveform is the same, both are T3, and the duty ratio of each second sub-waveform is 80%. Among them, T1 may be equal to T3 or may not be equal.
在图5B中,第一波形中的各个第一子波形的PWM周期不相同,如图5B所示第1个第一子波形的PWM周期为T1,第2个第一子波形的PWM周期为T2,T1≠T2。每个第一子波形的占空比均为20%。第二波形中的各个第二子波形的PWM周期 不相同,如图5B所示第1个第二子波形的PWM周期为T3,第2个第二子波形的PWM周期为T4,T3≠T4。每个第一子波形的占空比均为80%。In FIG. 5B, the PWM periods of the first sub-waveforms in the first waveform are different, as shown in FIG. 5B, the PWM period of the first first sub-waveform is T1, and the PWM period of the second first sub-waveform is T2, T1 ≠ T2. The duty cycle of each first sub-waveform is 20%. PWM period of each second sub-waveform in the second waveform Different, as shown in FIG. 5B, the PWM period of the first second sub-waveform is T3, and the PWM period of the second second sub-waveform is T4, T3 ≠ T4. The duty cycle of each first sub-waveform is 80%.
在图5C中,第一预设值为3%,第二预设值为0。第一波形中的每个第一子波形的PWM周期相同,均为T1。第1个第一波形在第一个Ti内包含的3个第一子波形的平均占空比为D1,例如D1=21%,D1与20%的差值的绝对值小于3%,这3个第一子波形的占空比可以相同也可以不同。第二波形中的每个第二子波形的PWM周期相同,均为T3。第1个第二波形在第一个Ti内包含的4个第二子波形的平均占空比为D2,例如D2=80%,D2与80%的差值的绝对值等于0,这4个第二子波形的占空比可以相同也可以不同。其中,T1可以与T3相等,也可以不相等。In FIG. 5C, the first preset value is 3%, and the second preset value is 0. Each of the first sub-waveforms in the first waveform has the same PWM period, both of which are T1. The average duty ratio of the first first sub-waveforms included in the first first waveform is D1, for example, D1=21%, and the absolute value of the difference between D1 and 20% is less than 3%. The duty ratios of the first sub-waveforms may be the same or different. Each of the second sub-waveforms has the same PWM period, both of which are T3. The average duty ratio of the four second sub-waveforms contained in the first Ti in the first Ti is D2, for example, D2=80%, and the absolute value of the difference between D2 and 80% is equal to 0. The duty ratios of the second sub-waveforms may be the same or different. Among them, T1 may be equal to T3 or may not be equal.
在图5D中,第一预设值为0,第二预设值为4%。第一波形中的各个第一子波形的PWM周期不相同,如图5D所示,第1个第一子波形的PWM周期为T1,第2个第一子波形的PWM周期为T2,T1≠T2。第1个第一波形在第一个Ti内包含的3个第一子波形的平均占空比为D1,例如D1=20%,D1与20%的差值的绝对值等于0,这3个第一子波形的占空比可以相同也可以不同。第二波形中的各个第二子波形的PWM周期不相同,如图5D所示,第1个第二子波形的PWM周期为T3,第2个第二子波形的PWM周期为T4,T3≠T4。第1个第二波形在第一个Ti内包含的4个第二子波形的平均占空比为D2,例如D2=76%,D2与80%的差值的绝对值等于4%,这4个第二子波形的占空比可以相同也可以不同。In FIG. 5D, the first preset value is 0, and the second preset value is 4%. The PWM period of each first sub-waveform in the first waveform is different. As shown in FIG. 5D, the PWM period of the first first sub-waveform is T1, and the PWM period of the second first sub-waveform is T2, T1≠ T2. The average duty ratio of the first first sub-waveforms included in the first first waveform is D1, for example, D1=20%, and the absolute value of the difference between D1 and 20% is equal to 0. The duty ratios of the first sub-waveforms may be the same or different. The PWM period of each second sub-waveform in the second waveform is different. As shown in FIG. 5D, the PWM period of the first second sub-waveform is T3, and the PWM period of the second second sub-waveform is T4, T3≠ T4. The average duty ratio of the four second sub-waveforms included in the first Ti in the first Ti is D2, for example, D2=76%, and the absolute value of the difference between D2 and 80% is equal to 4%. The duty ratios of the second sub-waveforms may be the same or different.
在一个示例中,该X个UPWM符号、该Y个UPWM符号和该Z个UPWM符号中的每个UPWM符号均满足以下四个条件中的至少一个条件:In one example, the X UPWM symbols, the Y UPWM symbols, and each of the Z UPWM symbols meet at least one of the following four conditions:
一、该每个UPWM符号的时长均为T,T=1/Fc,Fc表示接收端的帧率。1. The duration of each UPWM symbol is T, T=1/Fc, and Fc represents the frame rate of the receiving end.
二、该k个第一波形的总时长为T/2。2. The total duration of the k first waveforms is T/2.
三、该第一波形和该第二波形的时长均小于或者等于第二预设时长。3. The duration of the first waveform and the second waveform are both less than or equal to the second preset duration.
四、该每段第一波形的时长和与该第一波形相邻的第二波形的时长之间的差值的绝对值小于等于第三预设值。4. The absolute value of the difference between the duration of the first waveform of each segment and the duration of the second waveform adjacent to the first waveform is less than or equal to a third predetermined value.
示例性的,以一个为30%的第二占空比对应的UPWM符号为例。当接收端帧率为50fps时,Fc=50Hz,因此T=0.02s。若k=10,D=30%,则在该UPWM符号内包括10个占空比为30%的PWM波形(第一波形),以及10个占空比为70%的PWM波形(第二波形)。Exemplarily, a UPWM symbol corresponding to a second duty ratio of 30% is taken as an example. When the frame rate at the receiving end is 50 fps, Fc = 50 Hz, so T = 0.02 s. If k=10, D=30%, then 10 PWM waveforms with a duty cycle of 30% (first waveform) and 10 PWM waveforms with a duty ratio of 70% (second waveform) are included in the UPWM symbol. ).
其中,该10个第一波形的总时长和10个第二波形的总时长均为0.01s。若第二预设时长为0.0012s,则发送端生成的第一波形和第二波形的时长均要小于0.0012s。若第三预设值为0.0002s,则相邻的任一第一波形和第二波形的时长差绝对值都要小 于等于0.0002s。The total duration of the 10 first waveforms and the total duration of the 10 second waveforms are both 0.01 s. If the second preset duration is 0.0012 s, the duration of the first waveform and the second waveform generated by the transmitting end are both less than 0.0012 s. If the third preset value is 0.0002 s, the absolute value of the duration difference between any adjacent first waveform and the second waveform is small. It is equal to 0.0002s.
可选的,任一第一波形和第二波形的时长均相等。Optionally, the durations of any of the first waveform and the second waveform are equal.
值得说明的是,通过本申请中设计的UPWM符号,能够使得发送端在发送UPWM符号的过程中,在保持平均功率不变的情况下,避免出现闪烁的问题。It should be noted that the UPWM symbol designed in the present application can enable the transmitting end to avoid the problem of flicker while maintaining the average power in the process of transmitting the UPWM symbol.
步骤403,发送端依次发送该X个UPWM符号、该Y个UPWM符号以及该Z个UPWM符号。Step 403: The transmitting end sequentially transmits the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
可以理解的是,发送端先发送X个UPWM符号,之后紧接着发送该Y个UPWM符号,最后紧接着发送该Z个UPWM符号。It can be understood that the transmitting end first transmits X UPWM symbols, then sends the Y UPWM symbols immediately, and finally sends the Z UPWM symbols.
需要说明的是,当本申请提供的信号传输方法应用于OCC通信系统中时,发送端可以通过LED灯发送上述UPWM符号。当本申请提供的信号传输方法应用于射频通信系统中时,发送端可以通过射频模块发送上述UPWM符号。It should be noted that when the signal transmission method provided by the present application is applied to an OCC communication system, the transmitting end may transmit the UPWM symbol by using an LED lamp. When the signal transmission method provided by the present application is applied to a radio frequency communication system, the transmitting end may send the UPWM symbol through the radio frequency module.
步骤404,接收端检测连续的X个第一信号,以获取第一参数。Step 404: The receiving end detects consecutive X first signals to obtain a first parameter.
其中,该X个第一信号包括L1个最小值、L2个最大值以及除该最小值和最大值以外的p个信号,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。Wherein, the X first signals include L1 minimum values, L2 maximum values, and p signals except the minimum value and the maximum value, L1≥0, L1 is an integer, L2≥0, L2 is an integer, p≥ 1, p is an integer.
该第一参数可以包括但不限于:该X个第一信号中的最大值、最小值、非线性数值序列和/或相位错误指示信息。The first parameter may include, but is not limited to, a maximum value, a minimum value, a non-linear value sequence, and/or phase error indication information among the X first signals.
下面结合示例一对步骤404进行示例性的说明。An exemplary illustration is made below in conjunction with an example pair of steps 404.
示例一,基于发送端在上述场景一中发送的UPWM信号,接收端接收到与各个UPWM信号分别对应的亮度信号或者幅值信号。下面以亮度信号为例:Example 1, based on the UPWM signal sent by the transmitting end in the above scenario 1, the receiving end receives the luminance signal or the amplitude signal respectively corresponding to each UPWM signal. Let's take the luminance signal as an example:
如果接收端从拍摄的视频帧中检测LED亮度信号(例如,RGB值)时,并未出现相位错误,那么假设接收端接检测到的亮度信号为{100,220,265,326,298,350,220,326,220,326,220,220,265,298,326,220,265,298,326,220,265}。接收端通过检测,确定350是所接收到的信号中的最大值,100为所接收到信号中的最小值。接收端可以根据最大值和最小值的位置确定连续接收到的{100,220,265,326,298,350}为6个第一信号。其中包括L1=1个最小值100,L2=1个最大值350,以及p=4个其他信号{220,265,326,298}。If the receiving end detects the LED luminance signal (for example, RGB value) from the captured video frame, there is no phase error, then it is assumed that the received luminance signal is {100, 220, 265, 326, 298, 350. , 220, 326, 220, 326, 220, 220, 265, 298, 326, 220, 265, 298, 326, 220, 265}. The receiving end determines by detection that 350 is the maximum value of the received signal and 100 is the minimum value of the received signal. The receiving end can determine that the continuously received {100, 220, 265, 326, 298, 350} is 6 first signals according to the positions of the maximum value and the minimum value. These include L1 = 1 minimum 100, L2 = 1 maximum 350, and p = 4 other signals {220, 265, 326, 298}.
接收端利用上述最大值和最小值对{100,220,265,326,298,350}进行归一化,得到{0%,48%,66%,90%,79%,100%}。然后通过将{0%,48%,66%,90%,79%,100%}进行从低到高的排列,得到非线性数值序列{0%,48%,66%,79%,90%,100%}。在该示例中,得到的非线性数值序列{0%,48%,66%,79%,90%,100%}可以通过Gamma曲线表示,该Gamma曲线的Gamma参数γ=2.2。The receiving end normalizes {100, 220, 265, 326, 298, 350} using the above maximum and minimum values to obtain {0%, 48%, 66%, 90%, 79%, 100%}. Then, by arranging {0%, 48%, 66%, 90%, 79%, 100%} from low to high, a sequence of nonlinear values is obtained {0%, 48%, 66%, 79%, 90% , 100%}. In this example, the resulting non-linear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%} can be represented by a Gamma curve with a Gamma parameter γ=2.2.
接收端利用得到的非线性数值序列{0%,48%,66%,79%,90%,100%},对{0%, 48%,66%,90%,79%,100%}进行非线性补偿,得到{0%,20%,40%,80%,60%,100%}。通过将{0%,20%,40%,80%,60%,100%}与上述表9对比,确定表9中存在与{0%,20%,40%,80%,60%,100%}中的6个百分比的排列顺序相同的序列。从而接收端即可确定不存在相位错误。The receiver uses the resulting non-linear value sequence {0%, 48%, 66%, 79%, 90%, 100%} for {0%, 48%, 66%, 90%, 79%, 100%} performed nonlinear compensation, resulting in {0%, 20%, 40%, 80%, 60%, 100%}. By comparing {0%, 20%, 40%, 80%, 60%, 100%} with Table 9 above, it is determined that the presence in Table 9 is {0%, 20%, 40%, 80%, 60%, 100 6 percent of the %7 sequences in the same order. Therefore, the receiving end can determine that there is no phase error.
那么,从6个第一信号{100,220,265,326,298,350}中,可以获得的第一参数为最大值350,最小值100,γ=2.2,第一相位错误指示参数(用于指示不存在相位错误)。Then, from the six first signals {100, 220, 265, 326, 298, 350}, the first parameter obtainable is a maximum value of 350, a minimum value of 100, γ = 2.2, and a first phase error indication parameter (using Indicates that there is no phase error).
可选的,如果接收端从拍摄的视频帧中检测亮度信号时,出现相位错误,那么假设接收端检测到的亮度信号为{350,326,298,220,265,100,326,220,326,220,326,326,298,265,220,326,298,265,220,326,298}。接收端通过检测,确定350是所接收到的信号中的最大值,100为所接收到信号中的最小值。接收端可以根据最大值和最小值的位置确定连续接收到的{350,326,298,220,265,100}为6个第一信号。其中包括L1=1个最小值100,L2=1个最大值350,以及p=4个其他信号{220,265,326,298}。Optionally, if the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be {350, 326, 298, 220, 265, 100, 326, 220, 326. , 220, 326, 326, 298, 265, 220, 326, 298, 265, 220, 326, 298}. The receiving end determines by detection that 350 is the maximum value of the received signal and 100 is the minimum value of the received signal. The receiving end can determine that the continuously received {350, 326, 298, 220, 265, 100} is 6 first signals according to the positions of the maximum value and the minimum value. These include L1 = 1 minimum 100, L2 = 1 maximum 350, and p = 4 other signals {220, 265, 326, 298}.
接收端利用上述最大值和最小值对{350,326,298,220,265,100}进行归一化,得到{100%,90%,79%,48%,66%,0%}。然后通过将{100%,90%,79%,48%,66%,0%}进行从低到高的排列,得到非线性数值序列{0%,48%,66%,79%,90%,100%}。The receiving end normalizes {350, 326, 298, 220, 265, 100} using the above maximum and minimum values to obtain {100%, 90%, 79%, 48%, 66%, 0%}. Then, by arranging {100%, 90%, 79%, 48%, 66%, 0%} from low to high, a sequence of nonlinear values is obtained {0%, 48%, 66%, 79%, 90% , 100%}.
利用得到的非线性数值序列{0%,48%,66%,79%,90%,100%}对{100%,90%,79%,48%,66%,0%}进行非线性补偿,得到{100%,80%,60%,20%,40%,0%}。通过将{100%,80%,60%,20%,40%,0%}与上述表9对比,确定表9中不存在与{100%,80%,60%,20%,40%,0%}中的6个百分比的排列顺序相同的序列。而将{100%,80%,60%,20%,40%,0%}中的每个百分比进行1-D操作之后,得到的{0%,20%,40%,80%,60%,100%},存在于表9中,从而接收端即可确定存在相位错误。Nonlinear compensation for {100%, 90%, 79%, 48%, 66%, 0%} using the resulting nonlinear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%} , get {100%, 80%, 60%, 20%, 40%, 0%}. By comparing {100%, 80%, 60%, 20%, 40%, 0%} with Table 9 above, it is determined that there is no {100%, 80%, 60%, 20%, 40%, 6 percent of the sequences in 0%} are in the same order. And after each 1-% operation in {100%, 80%, 60%, 20%, 40%, 0%}, the obtained {0%, 20%, 40%, 80%, 60% , 100%}, exists in Table 9, so that the receiving end can determine that there is a phase error.
那么,从6个第一信号{350,326,298,220,265,100}中,可以获得的第一参数为最大值350,最小值100,γ=2.2,第二相位错误指示参数(用于指示存在相位错误)。Then, from the six first signals {350, 326, 298, 220, 265, 100}, the first parameters that can be obtained are a maximum value of 350, a minimum value of 100, γ=2.2, and a second phase error indication parameter (using Indicates that there is a phase error).
步骤405,该接收端根据第二参数,检测连续的Y个第二信号。Step 405: The receiving end detects consecutive Y second signals according to the second parameter.
在本申请中,当Y>0时,接收端获取第二参数,以用于检测连续的Y个第二信号。该第二参数可以是预设的参数,或者是接收端根据X个第一百分比获取的。In the present application, when Y>0, the receiving end acquires a second parameter for detecting consecutive Y second signals. The second parameter may be a preset parameter, or the receiving end acquires according to the X first percentages.
其中,X个第一百分比是接收端对X个信号进行第一处理操作之后得到的。 The first percentage of X is obtained after the receiving end performs the first processing operation on the X signals.
在本申请中,第一处理操作可以包括非线性补偿、归一化和/或相位补偿操作。其中,相位补偿操作可以为1-D操作,例如,当D=20%时,在进行相位补偿操作后,该20%就变成了1-D=80%。In the present application, the first processing operation may include non-linear compensation, normalization, and/or phase compensation operations. The phase compensation operation may be a 1-D operation. For example, when D=20%, after performing the phase compensation operation, the 20% becomes 1-D=80%.
示例性的,结合上述示例一,接收端获取第一参数的过程中,对6个第一信号进行了非线性补偿、归一化以及相位补偿操作,得到6个第一百分比为{0%,20%,40%,80%,60%,100%}。Exemplarily, in combination with the above example 1, in the process of acquiring the first parameter by the receiving end, nonlinear compensation, normalization, and phase compensation operations are performed on the six first signals, and six first percentages are obtained as {0. %, 20%, 40%, 80%, 60%, 100%}.
接收端根据上述表9,确定{0%,20%,40%,80%,60%,100%}指示的第二参数为:Y=5,n=2,映射规则21。The receiving end determines, according to the above Table 9, that the second parameter indicated by {0%, 20%, 40%, 80%, 60%, 100%} is: Y=5, n=2, mapping rule 21.
接收端根据第二参数中的Y,确定该接收端接收到的6个第一信号之后连续的5个信号为第二信号。The receiving end determines, according to Y in the second parameter, that the five consecutive signals after the six first signals received by the receiving end are the second signals.
即在示例一中,当6个第一信号为{100,220,265,326,298,350}时,5个第二信号为{220,326,220,326,220}。当6个第一信号为{350,326,298,220,265,100}时,5个第二信号为{326,220,326,220,326}。That is, in the first example, when the six first signals are {100, 220, 265, 326, 298, 350}, the five second signals are {220, 326, 220, 326, 220}. When the six first signals are {350, 326, 298, 220, 265, 100}, the five second signals are {326, 220, 326, 220, 326}.
步骤406,该接收端根据该第一参数对该Y个第二信号进行第一处理操作,得到Y个第二百分比。Step 406: The receiving end performs a first processing operation on the Y second signals according to the first parameter, to obtain Y second percentages.
示例性的,基于示例一,当第一参数为最大值350,最小值100,非线性数值序列{0%,48%,66%,79%,90%,100%}以及第一相位错误指示参数;5个第二信号为{220,326,220,326,220}时,接收端根据该第一参数对该5个第二信号进行第一处理操作可以包括:接收端根据350和100对{220,326,220,326,220}进行归一化处理,得到{8%,90%,48%,90%,48%}。根据非线性数值序列{0%,48%,66%,79%,90%,100%}对{8%,90%,48%,90%,48%}进行非线性处理,得到5个第二百分比{20%,80%,20%,80%,20%}。根据第一相位错误指示参数确定没有相位错误,因此接收端无需进行相位补偿操作。Illustratively, based on Example 1, when the first parameter is a maximum value of 350, a minimum value of 100, a sequence of non-linear values {0%, 48%, 66%, 79%, 90%, 100%} and a first phase error indication The parameter; when the 5 second signals are {220, 326, 220, 326, 220}, the receiving end performing the first processing operation on the five second signals according to the first parameter may include: receiving, according to the 350 and 100 pairs {220, 326, 220, 326, 220} normalized to obtain {8%, 90%, 48%, 90%, 48%}. Non-linear processing of {8%, 90%, 48%, 90%, 48%} based on the nonlinear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%}, yielding 5 Two percentages {20%, 80%, 20%, 80%, 20%}. It is determined that there is no phase error according to the first phase error indication parameter, so the receiving end does not need to perform phase compensation operation.
或者,当第一参数为最大值350,最小值100,得到非线性数值序列{0%,48%,66%,79%,90%,100%},以及第二相位错误指示参数;5个第二信号为{326,220,326,220,326}时,接收端根据该第一参数对该5个第二信号进行第一处理操作可以包括:接收端根据350和100对{326,220,326,220,326}进行归一化处理,得到{90%,48%,90%,48%,90%}。根据非线性数值序列{0%,48%,66%,79%,90%,100%}对{90%,48%,90%,48%,90%}进行非线性处理,得到{80%,20%,80%,20%,80%}。根据第二相位错误指示参数确定存在相位错误,因此对{80%,20%,80%,20%,80%}进行相位补偿操作,得到5个第二百分比{20%,80%,20%,80%,20%}。Alternatively, when the first parameter is a maximum value of 350 and a minimum value of 100, a non-linear value sequence {0%, 48%, 66%, 79%, 90%, 100%}, and a second phase error indication parameter are obtained; When the second signal is {326, 220, 326, 220, 326}, the receiving end performing the first processing operation on the five second signals according to the first parameter may include: receiving, according to the 350 and 100 pairs, {326, 220 , 326, 220, 326} normalized to obtain {90%, 48%, 90%, 48%, 90%}. Non-linear processing of {90%, 48%, 90%, 48%, 90%} according to the nonlinear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%}, yielding {80% , 20%, 80%, 20%, 80%}. According to the second phase error indication parameter, it is determined that there is a phase error, so the phase compensation operation is performed on {80%, 20%, 80%, 20%, 80%}, and 5 second percentages are obtained {20%, 80%, 20%, 80%, 20%}.
步骤407,该接收端使用该第二参数对该Y个第二百分比进行解调,得到第一数据。 Step 407: The receiving end demodulates the Y second percentages by using the second parameter to obtain first data.
在一个示例中,当第二参数包括第一映射规则时,接收端可以根据该第一映射规则将该Y个第二百分比中的每个第二百分比映射为该第一数据中的N个比特。然后根据该Y个第二信号的接收顺序,将由每个第二百分比映射出N个比特进行排列,得到该第一数据。In an example, when the second parameter includes the first mapping rule, the receiving end may map each second percentage of the Y second percentages to the first data according to the first mapping rule. N bits. Then, according to the receiving order of the Y second signals, N bits are mapped by each second percentage to obtain the first data.
示例性的,基于示例一,接收端获取的第二参数中包括n=2,映射规则21。那么接收端可以根据映射规则21,将5个第二百分比{20%,80%,20%,80%,20%}分别映射为{0,1,0,1,0}。然后根据5个第二信号的接收顺序,将{0,1,0,1,0}进行排列,得到第一数据“01010”。Exemplarily, based on the first example, the second parameter acquired by the receiving end includes n=2, and the mapping rule 21 is included. Then, the receiving end can map 5 second percentages {20%, 80%, 20%, 80%, 20%} to {0, 1, 0, 1, 0} according to the mapping rule 21. Then, according to the order of reception of the five second signals, {0, 1, 0, 1, 0} is arranged to obtain the first data "01010".
步骤408,该接收端根据第三参数,检测连续的Z个第三信号。Step 408: The receiving end detects consecutive Z third signals according to the third parameter.
在本申请中,当Z>0时,接收端可以获取第三参数,以用于检测连续的Z个第三信号。该第三参数可以是预设的,也可以是接收端根据X个第一百分比获取的,还可以是接收端从第一数据中获取的。In the present application, when Z>0, the receiving end can acquire a third parameter for detecting consecutive Z third signals. The third parameter may be preset, or may be obtained by the receiving end according to the X first percentages, or may be obtained by the receiving end from the first data.
示例性的,基于示例一,接收端得到的6个第一百分比为{0%,20%,40%,80%,60%,100%}。Exemplarily, based on the first example, the six first percentages obtained by the receiving end are {0%, 20%, 40%, 80%, 60%, 100%}.
接收端根据上述表9,确定{0%,20%,40%,80%,60%,100%}指示的第三参数为:m=4,映射规则41。接收端得到的第一数据中也包括第三参数,即“01010”表示Z=10。即接收端从6个第一百分比,以及第一数据“01010”中获取的第三参数为:Z=10,m=4,映射规则41。The receiving end determines, according to the above Table 9, that the third parameter indicated by {0%, 20%, 40%, 80%, 60%, 100%} is: m=4, mapping rule 41. The third parameter obtained by the receiving end also includes a third parameter, that is, "01010" indicates Z=10. That is, the third parameter obtained by the receiving end from the six first percentages and the first data "01010" is: Z=10, m=4, mapping rule 41.
接收端根据第三参数中的Z,确定该接收端接收到的5个第二信号之后连续的10个信号为第三信号。The receiving end determines, according to Z in the third parameter, that the consecutive 10 signals are the third signal after the 5 second signals received by the receiving end.
即在示例一中,当5个第二信号为{220,326,220,326,220}时,10个第三信号为{220,265,298,326,220,265,298,326,220,265}。当5个第二信号为{326,220,326,220,326}时,10个第三信号为{326,298,265,220,326,298,265,220,326,298}。That is, in the first example, when the five second signals are {220, 326, 220, 326, 220}, the ten third signals are {220, 265, 298, 326, 220, 265, 298, 326, 220. , 265}. When the five second signals are {326, 220, 326, 220, 326}, the ten third signals are {326, 298, 265, 220, 326, 298, 265, 220, 326, 298}.
步骤409,该接收端使用该第一参数对该Z个第三信号进行该第一处理操作,得到Z个第三百分比。Step 409: The receiving end performs the first processing operation on the Z third signals by using the first parameter, to obtain Z third percentages.
示例性的,基于示例一,当第一参数为最大值350,最小值100,γ=2.2以及第一相位错误指示参数,10个第三信号为{220,265,298,326,220,265,298,326,220,265}时,接收端根据该第一参数对该10个第三信号进行第一处理操作可以包括:接收端根据350和100对{220,265,298,326,220,265,298,326,220,265}进行归一化处理,得到{48%,66%,79%,90%,48%,66%,79%,90%,48%,66%}。根据非线性数值序列{0%,48%,66%,79%,90%,100%}对{48%,66%,79%,90%,48%,66%,79%,90%,48%,66%}进行非线性处理,得到10个第三百分比{20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}。根据第一相位错误指示参 数确定没有相位错误,因此接收端无需进行相位补偿操作。Exemplarily, based on the first example, when the first parameter is the maximum value 350, the minimum value 100, the γ=2.2, and the first phase error indication parameter, the ten third signals are {220, 265, 298, 326, 220, 265. At 298, 326, 220, 265}, the receiving end performing the first processing operation on the ten third signals according to the first parameter may include: receiving, according to the 350 and 100 pairs, {220, 265, 298, 326, 220 , 265, 298, 326, 220, 265} normalized to give {48%, 66%, 79%, 90%, 48%, 66%, 79%, 90%, 48%, 66%}. According to the non-linear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%} for {48%, 66%, 79%, 90%, 48%, 66%, 79%, 90%, 48%, 66%} nonlinear processing, resulting in 10 third percentages {20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40%} . According to the first phase error indication The number determines that there is no phase error, so the receiver does not need to perform phase compensation.
或者,当第一参数为最大值350,最小值100,非线性数值序列{0%,48%,66%,79%,90%,100%},以及第二相位错误指示参数,10个第三信号为{326,298,265,220,326,298,265,220,326,298}时,接收端根据该第一参数对该10个第三信号进行第一处理操作可以包括:接收端根据350和100对{326,298,265,220,326,298,265,220,326,298}进行归一化处理,得到{90%,79%,66%,48%,90%,79%,66%,48%,90%,79%}。根据非线性数值序列{0%,48%,66%,79%,90%,100%}对{90%,79%,66%,48%,90%,79%,66%,48%,90%,79%}进行非线性处理,得到{80%,60%,40%,20%,80%,60%,40%,20%,80%,60%}。根据第二相位错误指示参数确定存在相位错误,因此对{80%,60%,40%,20%,80%,60%,40%,20%,80%,60%}进行相位补偿操作,得到10个第三百分比{20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}。Or, when the first parameter is the maximum value 350, the minimum value is 100, the non-linear value sequence {0%, 48%, 66%, 79%, 90%, 100%}, and the second phase error indication parameter, 10 When the three signals are {326, 298, 265, 220, 326, 298, 265, 220, 326, 298}, the receiving end performs the first processing operation on the ten third signals according to the first parameter, which may include: receiving end Normalized according to 350 and 100 for {326,298,265,220,326,298,265,220,326,298}, yielding {90%, 79%, 66%, 48%, 90%, 79 %, 66%, 48%, 90%, 79%}. According to the non-linear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%} for {90%, 79%, 66%, 48%, 90%, 79%, 66%, 48%, 90%, 79%} were subjected to nonlinear processing to obtain {80%, 60%, 40%, 20%, 80%, 60%, 40%, 20%, 80%, 60%}. According to the second phase error indication parameter, it is determined that there is a phase error, so phase compensation operation is performed on {80%, 60%, 40%, 20%, 80%, 60%, 40%, 20%, 80%, 60%}, Get 10 third percentages {20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40%}.
步骤410,该接收端使用该第三参数对该Z个第三百分比进行解调,得到第二数据。Step 410: The receiving end demodulates the Z third percentages by using the third parameter to obtain second data.
在一个示例中,当第三参数包括第二映射规则时,接收端可以根据该第二映射规则将该Z个第三百分比中的每个第三百分比映射为该第二数据中的M个比特。然后根据该Z个第三信号的接收顺序,将由每个第三百分比映射出M个比特进行排列,得到该第二数据。In an example, when the third parameter includes the second mapping rule, the receiving end may map each third percentage of the Z third percentages to the second data according to the second mapping rule. M bits. Then, according to the order of receiving the Z third signals, M bits are mapped out by each third percentage to obtain the second data.
示例性的,基于示例一,接收端获取的第二参数中包括:m=4,映射规则41。那么接收端可以根据映射规则41,将10个第三百分比{20%,40%,60%,80%,20%,40%,60%,80%,20%,40%}分别映射为{00,01,10,11,00,01,10,11,00,01}。然后根据10个第三信号的接收顺序,将{00,01,10,11,00,01,10,11,00,01}进行排列,得到第一数据“00011011000110110001”。Exemplarily, based on the first example, the second parameter acquired by the receiving end includes: m=4, mapping rule 41. Then the receiving end can map 10 third percentages {20%, 40%, 60%, 80%, 20%, 40%, 60%, 80%, 20%, 40%} according to the mapping rule 41, respectively. It is {00,01,10,11,00,01,10,11,00,01}. Then, according to the order of receiving the ten third signals, {00, 01, 10, 11, 00, 01, 10, 11, 00, 01} are arranged to obtain the first data "00011011000110110001".
可选的,X个第一百分比也可以用于指示第三数据。那么接收端还可以根据预设的第三映射规则和所述X个第一百分比中的t个互不相同的第一百分比的排列顺序,获取第三数据。Optionally, X first percentages may also be used to indicate the third data. The receiving end may further acquire the third data according to the preset third mapping rule and the first percentage order of the t first different percentages among the X first percentages.
下面结合示例二对接收端获取第三数据的方式进行示例性说明。The following is a description of the manner in which the receiving end acquires the third data in conjunction with the second example.
示例二,基于发送端在上述场景二中发送的UPWM信号,接收端接收到与各个UPWM信号分别对应的亮度信号。在该示例中,Y=0,Z=0,即接收端接收到的亮度信号均为第一信号。For example, based on the UPWM signal sent by the transmitting end in the above scenario 2, the receiving end receives the luminance signal corresponding to each UPWM signal. In this example, Y=0, Z=0, that is, the luminance signals received by the receiving end are all the first signals.
如果接收端从拍摄的视频帧中检测LED亮度信号时,并未出现相位错误,那么假设接收端接检测到的亮度信号为{100,265,220,326,298,350,100,298,220,265,326,350}。接收端通过检测,确定350是所接收到的信号中的最大值,100为 所接收到信号中的最小值。根据上述最大值与最小值的位置,可以确定连续接收到的{100,265,220,326,298,350}为一组6个第一信号,{100,298,220,265,326,350}为另一组6个第一信号。其中包括L1=1个最小值100,L2=1个最大值350,以及p=4个其他信号。If the receiving end detects the LED luminance signal from the captured video frame, there is no phase error, then it is assumed that the received luminance signal is {100, 265, 220, 326, 298, 350, 100, 298, 220 , 265, 326, 350}. The receiving end determines through detection that 350 is the maximum value of the received signal, 100 is The minimum value in the received signal. According to the positions of the above maximum and minimum values, it can be determined that the continuously received {100, 265, 220, 326, 298, 350} is a set of six first signals, {100, 298, 220, 265, 326, 350 } is another set of 6 first signals. These include L1 = 1 minimum 100, L2 = 1 maximum 350, and p = 4 other signals.
接收端根据最大值和最小值,对该这两组第一信号进行归一化,得到{0%,66%,48%,90%,79%,100%,0%},{79%,48%,66%,90%,100%}。The receiving end normalizes the two sets of first signals according to the maximum value and the minimum value, and obtains {0%, 66%, 48%, 90%, 79%, 100%, 0%}, {79%, 48%, 66%, 90%, 100%}.
选择两组6个第一信号中的任意一组第一信号经过归一化之后得到的序列确定非线性数值序列。例如,将{0%,66%,48%,90%,79%,100%}进行从低到高的排列,得到非线性数值序列{0%,48%,66%,79%,90%,100%}。A sequence obtained by normalizing any one of the two sets of the first signals is selected to determine a sequence of non-linear values. For example, {0%, 66%, 48%, 90%, 79%, 100%} is arranged from low to high, resulting in a sequence of nonlinear values {0%, 48%, 66%, 79%, 90% , 100%}.
利用得到的非线性数值序列{0%,48%,66%,79%,90%,100%}对{0%,66%,48%,90%,79%,100%},{0%,79%,48%,66%,90%,100%}进行非线性补偿,得到{0%,40%,20%,80%,60%,100%},{0%,60%,20%,40%,80%,100%}。通过将{0%,40%,20%,80%,60%,100%}与上述表10对比,确定表10中存在与{0%,40%,20%,80%,60%,100%}中的6个百分比的排列顺序相同的序列。从而接收端即可确定不存在相位错误。那么接收端即可确定{0%,40%,20%,80%,60%,100%}为一组第一百分比,并且根据表10可以确定这6个第一百分比映射的第三数据为“011”。{0%,60%,20%,40%,80%,100%}为一组第一百分比,并且根据表10可以确定这6个第一百分比映射的第三数据为“110”。接收端根据这两组第一信号的接收顺序,对“011”和“110”进行排列,得到发送端发送的数据“011110”。Using the resulting non-linear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%} versus {0%, 66%, 48%, 90%, 79%, 100%}, {0% , 79%, 48%, 66%, 90%, 100%} nonlinear compensation, get {0%, 40%, 20%, 80%, 60%, 100%}, {0%, 60%, 20 %, 40%, 80%, 100%}. By comparing {0%, 40%, 20%, 80%, 60%, 100%} with Table 10 above, it is determined that the presence in Table 10 is {0%, 40%, 20%, 80%, 60%, 100 6 percent of the %7 sequences in the same order. Therefore, the receiving end can determine that there is no phase error. Then the receiving end can determine that {0%, 40%, 20%, 80%, 60%, 100%} is a set of first percentages, and according to Table 10, the six first percentage maps can be determined. The third data is "011". {0%, 60%, 20%, 40%, 80%, 100%} is a set of first percentages, and according to Table 10, it can be determined that the third data of the six first percentage maps is "110" ". The receiving end arranges "011" and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
可选的,如果接收端从拍摄的视频帧中检测亮度信号时,出现相位错误,那么假设接收端检测到的亮度信号为{350,298,326,220,265,100,350,265,326,298,220,100}时,接收端通过检测,确定350是所接收到的信号中的最大值,100为所接收到信号中的最小值。根据上述最大值与最小值的位置,可以确定连续接收到的{350,298,326,220,265,100}为一组6个第一信号,{350,265,326,298,220,100}为另一组6个第一信号。其中包括L1=1个最小值100,L2=1个最大值350,以及p=4个其他信号。Optionally, if the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be {350, 298, 326, 220, 265, 100, 350, 265, 326. At 298, 220, 100}, the receiving end determines by detection that 350 is the maximum value of the received signal, and 100 is the minimum value of the received signal. According to the above positions of the maximum value and the minimum value, it can be determined that the continuously received {350, 298, 326, 220, 265, 100} is a set of 6 first signals, {350, 265, 326, 298, 220, 100 } is another set of 6 first signals. These include L1 = 1 minimum 100, L2 = 1 maximum 350, and p = 4 other signals.
接收端根据最大值和最小值,对该这两组第一信号进行归一化,得到{100%,79%,90%,48%,66%,0%},{79%,48%,66%,90%,100%}。选择两组6个第一信号中的任意一组第一信号经过归一化之后得到的序列确定非线性数值序列。例如,将{100%,79%,90%,48%,66%,0%}进行从低到高的排列,得到非线性数值序列{0%,48%,66%,79%,90%,100%}。The receiving end normalizes the two sets of first signals according to the maximum value and the minimum value, and obtains {100%, 79%, 90%, 48%, 66%, 0%}, {79%, 48%, 66%, 90%, 100%}. A sequence obtained by normalizing any one of the two sets of the first signals is selected to determine a sequence of non-linear values. For example, {100%, 79%, 90%, 48%, 66%, 0%} is arranged from low to high, resulting in a non-linear numerical sequence {0%, 48%, 66%, 79%, 90% , 100%}.
利用得到的非线性数值序列{0%,48%,66%,79%,90%,100%}对{100%,79%,90%,48%,66%,0%},{79%,48%,66%,90%,100%}进行非线性补偿,得到{100%,60%,80%,20%,40%,0%},{100%,40%,80%,60%,20%,0%}。通过将{100%, 60%,80%,20%,40%,0%}与上述表10对比,确定表10中不存在与{100%,60%,80%,20%,40%,0%}中的6个百分比的排列顺序相同的序列。而将{100%,60%,80%,20%,40%,0%},{100%,40%,80%,60%,20%,0%}中的每个百分比进行相位补偿操作之后,得到的{0%,20%,40%,80%,60%,100%},{100%,40%,80%,60%,20%,0%},均存在于表10中。从而接收端即可确定存在相位错误。那么接收端即可确定{0%,40%,20%,80%,60%,100%}为一组第一百分比,并且根据表10可以确定这6个第一百分比映射的第三数据为“011”。{0%,60%,20%,40%,80%,100%}为一组第一百分比,并且根据表10可以确定这6个第一百分比映射的第三数据为“110”。接收端根据这两组第一信号的接收顺序,对“011”和“110”进行排列,得到发送端发送的数据“011110”。Using the resulting non-linear numerical sequence {0%, 48%, 66%, 79%, 90%, 100%} versus {100%, 79%, 90%, 48%, 66%, 0%}, {79% , 48%, 66%, 90%, 100%} nonlinear compensation, get {100%, 60%, 80%, 20%, 40%, 0%}, {100%, 40%, 80%, 60 %, 20%, 0%}. By putting {100%, 60%, 80%, 20%, 40%, 0%} compared with Table 10 above, it is determined that there is no 6 in {100%, 60%, 80%, 20%, 40%, 0%} in Table 10. A sequence of identical percentages in the same order. Phase compensation operation for each percentage in {100%, 60%, 80%, 20%, 40%, 0%}, {100%, 40%, 80%, 60%, 20%, 0%} After that, the obtained {0%, 20%, 40%, 80%, 60%, 100%}, {100%, 40%, 80%, 60%, 20%, 0%} are all present in Table 10. . Therefore, the receiving end can determine that there is a phase error. Then the receiving end can determine that {0%, 40%, 20%, 80%, 60%, 100%} is a set of first percentages, and according to Table 10, the six first percentage maps can be determined. The third data is "011". {0%, 60%, 20%, 40%, 80%, 100%} is a set of first percentages, and according to Table 10, it can be determined that the third data of the six first percentage maps is "110" ". The receiving end arranges "011" and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
需要说明的是,在本申请中,对X个第一信号,Y个第二信号以及Z个第三信号进行第一处理操作时,可以分别执行,也可以串行执行,对此,本申请不做限制。It should be noted that, in the present application, when the first processing operations are performed on the X first signals, the Y second signals, and the Z third signals, the first processing may be performed separately or in series. No restrictions.
其中,串行执行可以是指接收端在获取第一参数的过程中,对接收端获取的所有信号同时执行第一处理操作,且在执行第一处理操作时为分别执行第一处理操作中的归一化、非线性补偿和/或相位补偿操作,得到所有信号对应的百分比,然后确定哪些是第二信号对应的第二百分比,哪些是第三信号对应的第三百分比。The serial execution may be performed by the receiving end performing the first processing operation on all the signals acquired by the receiving end in the process of acquiring the first parameter, and performing the first processing operation separately when performing the first processing operation. The normalized, nonlinearly compensated, and/or phase compensated operations obtain a percentage of all signals and then determine which are the second percentage corresponding to the second signal and which are the third percentages corresponding to the third signal.
下面结合以下两个示例,对串行执行的方式进行示例性的说明。The manner in which serial execution is performed is exemplified below in conjunction with the following two examples.
示例三,基于发送端在上述场景三中发送的UPWM信号,接收端接收到与各个UPWM信号分别对应的亮度信号或者幅值信号。下面以亮度信号为例:Example 3, based on the UPWM signal sent by the transmitting end in the above scenario 3, the receiving end receives the luminance signal or the amplitude signal corresponding to each UPWM signal respectively. Let's take the luminance signal as an example:
如果接收端从拍摄的视频帧中检测LED亮度信号时,并未出现相位错误,那么假设接收端接检测到的亮度信号为{220,249,310,292,272,326,220,326,220,326,220,220,265,298,326,220,265,298,326,220,265}。接收端通过检测,确定326是所接收到的信号中的最大值,220为所接收到信号中的最小值。根据上述最大值与最小值的位置,可以确定连续接收到的{220,249,310,292,272,326}为6个第一信号。If the receiving end detects the LED luminance signal from the captured video frame without phase error, it is assumed that the received luminance signal is {220, 249, 310, 292, 272, 326, 220, 326, 220. , 326, 220, 220, 265, 298, 326, 220, 265, 298, 326, 220, 265}. The receiving end determines by detection 326 that it is the maximum value of the received signal, and 220 is the minimum value of the received signal. According to the above positions of the maximum value and the minimum value, it can be determined that the continuously received {220, 249, 310, 292, 272, 326} is 6 first signals.
接收端根据最大值和最小值,对{220,249,310,292,272,326,220,326,220,326,220,220,265,298,326,220,265,298,326,220,265}进行归一化,得到{0%,27%,85%,68%,50%,100%,0%,100%,0%,100%,0%,0%,42%,74%,100%,0%,42%,74%,100%,0%,42%},并通过对6个第一信号归一化后的{0%,27%,85%,68%,50%,100%}进行从低到高的排列,得到非线性数值序列{0%,27%,50%,68%,85%,100%}。The receiving end is {220, 249, 310, 292, 272, 326, 220, 326, 220, 326, 220, 220, 265, 298, 326, 220, 265, 298, 326, 220 according to the maximum value and the minimum value. , 265} normalized to get {0%, 27%, 85%, 68%, 50%, 100%, 0%, 100%, 0%, 100%, 0%, 0%, 42%, 74 %, 100%, 0%, 42%, 74%, 100%, 0%, 42%}, and normalized by 6 first signals {0%, 27%, 85%, 68%, 50%, 100%} from low to high alignment, resulting in a non-linear numerical sequence {0%, 27%, 50%, 68%, 85%, 100%}.
接收端利用得到的非线性数值序列{0%,27%,50%,68%,85%,100%}对{0%,27%,85%,68%,50%,100%,0%,100%,0%,100%,0%,0%,42%,74%, 100%,0%,42%,74%,100%,0%,42%}进行非线性补偿,得到:{0%,20%,80%,60%,40%,100%,0%,100%,0%,100%,0%,0%,33%,67%,100%,0%,33%,67%,100%,0%,33%}。The receiving end uses the obtained nonlinear numerical sequence {0%, 27%, 50%, 68%, 85%, 100%} for {0%, 27%, 85%, 68%, 50%, 100%, 0% , 100%, 0%, 100%, 0%, 0%, 42%, 74%, 100%, 0%, 42%, 74%, 100%, 0%, 42%} nonlinear compensation, get: {0%, 20%, 80%, 60%, 40%, 100%, 0%, 100%, 0%, 100%, 0%, 0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33%}.
通过将6个第一信号经过归一化以及非线性补偿的序列{0%,20%,80%,60%,40%,100%}与如下表13对比,得到第二参数为:Y=5,n=2,映射规则21;第三参数为:m=4,映射规则41。因此可以确定不存在相位错误。从而接收端可以确定{0%,20%,80%,60%,40%,100%}为6个第一百分比。进而接收端可以确定,6个第一信号之后连续的5个信号为{220,326,220,326,220}为第二信号,相应的{0%,100%,0%,100%,0%}为5个第二百分比。By comparing the normalized and non-linearly compensated sequences of the six first signals {0%, 20%, 80%, 60%, 40%, 100%} with Table 13 below, the second parameter is: Y= 5, n=2, mapping rule 21; the third parameter is: m=4, mapping rule 41. It can therefore be determined that there is no phase error. Thus the receiving end can determine that {0%, 20%, 80%, 60%, 40%, 100%} is 6 first percentages. Furthermore, the receiving end can determine that the five consecutive signals after the six first signals are {220, 326, 220, 326, 220} as the second signal, corresponding to {0%, 100%, 0%, 100%, 0. %} is 5 second percentages.
根据映射规则21(如下表14所示)对{0%,100%,0%,100%,0%}进行映射,得到第一数据为:01010。第一数据表示Z=10,那么接收端可以确定5个第二信号{220,326,220,326,220}之后连续的10个信号{220,265,298,326,220,265,298,326,220,265}为第三信号,相应的{0%,33%,67%,100%,0%,33%,67%,100%,0%,33%}为10个第三百分比。根据映射规则41(如下表15)对{0%,33%,67%,100%,0%,33%,67%,100%,0%,33%}进行映射,得到第二数据为:00011011000110110001。According to mapping rule 21 (shown in Table 14 below), {0%, 100%, 0%, 100%, 0%} is mapped, and the first data is obtained as: 01010. The first data represents Z=10, then the receiving end can determine 10 consecutive signals {220, 265, 298, 326, 220, 265, 298 after 5 second signals {220, 326, 220, 326, 220}, 326,220,265} is the third signal, corresponding to {0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33%} for 10 third hundred The ratio. Mapping {0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33%} according to mapping rule 41 (see Table 15 below), the second data is: 00011011000110110001.
表13Table 13
Figure PCTCN2017085384-appb-000006
Figure PCTCN2017085384-appb-000006
表14Table 14
比特组合Bit combination 百分比percentage
0000 0%0%
0101 33%33%
1010 66%66%
1111 100%100%
表15Table 15
比特组合Bit combination 百分比percentage
00 0%0%
11 100%100%
需要说明的是,接收端中存储的第三映射规则(即表13)可以是发送端中存储的第三映射规则(即表11)归一化后的结果,即表13中的X个第一百分比为对表11中X个第一占空比进行归一化后得到的。接收端中存储的映射规则21(即表14)可以是发送端中存储的映射规则21(即表8)归一化后的结果,即表14中的百分比为对表8中占空比进行归一化后得到的。接收端中存储的映射规则41(即表15)可以是发送端中存储的映射规则41(即表1)归一化后的结果,即表15中的百分比为对表1中占空比进行归一化后得到的。It should be noted that the third mapping rule (ie, Table 13) stored in the receiving end may be a normalized result of the third mapping rule (ie, Table 11) stored in the sending end, that is, the X numbers in Table 13. A percentage is obtained by normalizing the X first duty cycles in Table 11. The mapping rule 21 (ie, Table 14) stored in the receiving end may be a normalized result of the mapping rule 21 (ie, Table 8) stored in the transmitting end, that is, the percentage in Table 14 is the duty ratio in Table 8. Obtained after normalization. The mapping rule 41 (ie, Table 15) stored in the receiving end may be a normalized result of the mapping rule 41 (ie, Table 1) stored in the transmitting end, that is, the percentage in Table 15 is the duty ratio in Table 1. Obtained after normalization.
可选的,如果接收端从拍摄的视频帧中检测亮度信号时,出现相位错误,那么假设接收端检测到的亮度信号为{326,310,249,272,292,220,326,220,326,220,326,326,298,265,220,326,298,265,220,326,298}。接收端通过检测,确定326是所接收到的信号中的最大值,220为所接收到信号中的最小值。根据上述最大值与最小值的位置,可以确定连续接收到的{326,310,249,272,292,220}为6个第一信号。Optionally, if the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be {326, 310, 249, 272, 292, 220, 326, 220, 326. , 220, 326, 326, 298, 265, 220, 326, 298, 265, 220, 326, 298}. The receiving end determines by detection 326 that it is the maximum value of the received signal, and 220 is the minimum value of the received signal. According to the positions of the above maximum and minimum values, it can be determined that the continuously received {326, 310, 249, 272, 292, 220} is 6 first signals.
接收端根据最大值和最小值,对该{326,310,249,272,292,220,326,220,326,220,326,326,298,265,220,326,298,265,220,326,298}进行归一化,得到{100%,85%,27%,50%,68%,0%,100%,0%,100%,0%,100%,100%,74%,42%,0%,100%,74%,42%,0%,100%,74%}。并通过对6个第一信号 归一化后的{100%,85%,27%,50%,68%,0%}进行从低到高的排列,得到非线性数值序列{0%,27%,50%,68%,85%,100%}。The receiving end is based on the maximum value and the minimum value, and is {326, 310, 249, 272, 292, 220, 326, 220, 326, 220, 326, 326, 298, 265, 220, 326, 298, 265, 220, 326,298} normalized to get {100%, 85%, 27%, 50%, 68%, 0%, 100%, 0%, 100%, 0%, 100%, 100%, 74%, 42%, 0%, 100%, 74%, 42%, 0%, 100%, 74%}. And pass the 6 first signals The normalized {100%, 85%, 27%, 50%, 68%, 0%} are arranged from low to high, resulting in a non-linear numerical sequence {0%, 27%, 50%, 68%, 85%, 100%}.
利用得到的非线性数值序列{0%,27%,50%,68%,85%,100%}对{100%,85%,27%,50%,68%,0%,100%,0%,100%,0%,100%,100%,74%,42%,0%,100%,74%,42%,0%,100%,74%}进行非线性补偿,得到{100%,80%,20%,40%,60%,0%,100%,0%,100%,0%,100%,100%,67%,33%,0%,100%,67%,33%,0%,100%,67%}。Using the obtained nonlinear numerical sequence {0%, 27%, 50%, 68%, 85%, 100%} for {100%, 85%, 27%, 50%, 68%, 0%, 100%, 0 %, 100%, 0%, 100%, 100%, 74%, 42%, 0%, 100%, 74%, 42%, 0%, 100%, 74%} nonlinear compensation, get {100% , 80%, 20%, 40%, 60%, 0%, 100%, 0%, 100%, 0%, 100%, 100%, 67%, 33%, 0%, 100%, 67%, 33 %, 0%, 100%, 67%}.
通过将6个第一信号经过归一化以及非线性补偿后得到的序列{100%,80%,20%,40%,60%,0%}与如下表13对比,确定表13中没有与{100%,80%,20%,40%,60%,0%}排列顺序相同的序列,那么,接收端对{100%,80%,20%,40%,60%,0%,100%,0%,100%,0%,100%,100%,67%,33%,0%,100%,67%,33%,0%,100%,67%}进行相位补偿操作,得到{0%,20%,80%,60%,40%,100%,0%,100%,0%,100%,0%,0%,33%,67%,100%,0%,33%,67%,100%,0%,33%}。The sequence {100%, 80%, 20%, 40%, 60%, 0%} obtained by normalizing and non-linearly compensating the six first signals is compared with Table 13 below, and it is determined that there is no {100%, 80%, 20%, 40%, 60%, 0%} The sequence is the same in order, then the receiving end is {100%, 80%, 20%, 40%, 60%, 0%, 100 %, 0%, 100%, 0%, 100%, 100%, 67%, 33%, 0%, 100%, 67%, 33%, 0%, 100%, 67%} for phase compensation operation, {0%, 20%, 80%, 60%, 40%, 100%, 0%, 100%, 0%, 100%, 0%, 0%, 33%, 67%, 100%, 0%, 33 %, 67%, 100%, 0%, 33%}.
接收端将相位补偿后得到的{0%,20%,80%,60%,40%,100%},与表13进行对比,得到第二参数为:Y=5,n=2,映射规则21;第三参数为:m=4,映射规则41。因此接收端可以确定存在相位错误。从而接收端可以确定经过相位补偿后的{0%,20%,80%,60%,40%,100%}为6个第一百分比。进而接收端可以确定,6个第一信号之后连续的5个信号为{326,220,326,220,326}为第二信号,相应的经过相位补偿后的{0%,100%,0%,100%,0%}为5个第二百分比。The receiving end compares {0%, 20%, 80%, 60%, 40%, 100%} obtained after phase compensation with Table 13, and obtains the second parameter: Y=5, n=2, mapping rule 21; The third parameter is: m=4, mapping rule 41. Therefore, the receiving end can determine that there is a phase error. Therefore, the receiving end can determine that the phase compensation is {0%, 20%, 80%, 60%, 40%, 100%} as 6 first percentages. Furthermore, the receiving end can determine that the five consecutive signals after the six first signals are {326, 220, 326, 220, 326} as the second signal, and the corresponding phase-compensated {0%, 100%, 0% , 100%, 0%} is 5 second percentages.
根据映射规则21对{0%,100%,0%,100%,0%}进行映射,得到第一数据为:01010。第一数据表示Z=10,那么接收端可以确定5个第二信号{326,220,326,220,326}之后连续的10个信号{326,298,265,220,326,298,265,220,326,298}为第三信号,相应的经过相位补偿后的{0%,33%,67%,100%,0%,33%,67%,100%,0%,33%}为10个第三百分比。根据映射规则41对{0%,33%,67%,100%,0%,33%,67%,100%,0%,33%}进行映射,得到第二数据为:00011011000110110001。According to the mapping rule 21, {0%, 100%, 0%, 100%, 0%} is mapped, and the first data is obtained as: 01010. The first data represents Z=10, then the receiving end can determine 10 consecutive signals {326, 298, 265, 220, 326, 298, 265 after 5 second signals {326, 220, 326, 220, 326}, 220, 326, 298} is the third signal, and the corresponding phase-compensated {0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33%} is 10 third percentages. Mapping {0%, 33%, 67%, 100%, 0%, 33%, 67%, 100%, 0%, 33%} according to the mapping rule 41, the second data is: 00011011000110110001.
示例四、基于发送端在上述场景四中发送的UPWM信号,接收端接收到与各个UPWM信号分别对应的亮度信号或者幅值信号。下面以亮度信号为例:Example 4: Based on the UPWM signal sent by the transmitting end in the above scenario 4, the receiving end receives the luminance signal or the amplitude signal corresponding to each UPWM signal respectively. Let's take the luminance signal as an example:
如果接收端从拍摄的视频帧中检测LED亮度信号时,并未出现相位错误,那么假设接收端接检测到的亮度信号为{220,272,249,310,292,326,220,272,310,292,249,326}时,确定326是所接收到的信号中的最大值,220为所接收到信号中的最小值。根据最大值和最小值的位置可以确定{220,272,249,310,292,326},{220,272,310,292,249,326}为两组第信号。If the receiving end detects the LED luminance signal from the captured video frame without phase error, it is assumed that the received luminance signal is {220, 272, 249, 310, 292, 326, 220, 272, 310. At 292, 249, 326}, the determination 326 is the maximum of the received signals and 220 is the minimum of the received signals. According to the positions of the maximum and minimum values, {220, 272, 249, 310, 292, 326}, {220, 272, 310, 292, 249, 326} can be determined as two sets of first signals.
利用最大值和最小值,对{220,272,249,310,292,326,220,272,310,292, 249,326}进行归一化,得到{0%,50%,27%,85%,68%,100%,0%,50%,85%,68%,27%,100%}。可以根据两组中的一组第一信号经过归一化后得到的序列确定非线性数值序列。例如,对{0%,50%,27%,85%,68%,100%}进行从低到高的排列,得到非线性数值序列{0%,27%,50%,68%,85%,100%}。然后利用得到的非线性数值序列{0%,27%,50%,68%,85%,100%}对{0%,40%,20%,80%,60%,100%,0%,40%,80%,60%,20%,100%}进行非线性补偿得到{0%,50%,27%,85%,68%,100%,0%,50%,85%,68%,27%,100%}。通过将{0%,40%,20%,80%,60%,100%}与上述表16对比,确定表16中存在与{0%,40%,20%,80%,60%,100%}中的6个百分比的排列顺序相同的序列。从而接收端即可确定不存在相位错误。那么接收端即可确定{0%,40%,20%,80%,60%,100%}为一组第一百分比,并且根据表16可以确定这6个第一百分比映射的第三数据为“011”。{0%,60%,20%,40%,80%,100%}为一组第一百分比,并且根据表16可以确定这6个第一百分比映射的第三数据为“110”。接收端根据这两组第一信号的接收顺序,对“011”和“110”进行排列,得到发送端发送的数据“011110”。 Using the maximum and minimum values, for {220, 272, 249, 310, 292, 326, 220, 272, 310, 292, 249,326} normalized to get {0%, 50%, 27%, 85%, 68%, 100%, 0%, 50%, 85%, 68%, 27%, 100%}. A sequence of non-linear values can be determined from a sequence obtained by normalizing a set of first signals in two sets. For example, a low-to-high alignment of {0%, 50%, 27%, 85%, 68%, 100%} yields a sequence of nonlinear values {0%, 27%, 50%, 68%, 85% , 100%}. Then use the resulting non-linear numerical sequence {0%, 27%, 50%, 68%, 85%, 100%} for {0%, 40%, 20%, 80%, 60%, 100%, 0%, 40%, 80%, 60%, 20%, 100%} nonlinear compensation is obtained {0%, 50%, 27%, 85%, 68%, 100%, 0%, 50%, 85%, 68% , 27%, 100%}. By comparing {0%, 40%, 20%, 80%, 60%, 100%} with Table 16 above, it is determined that the presence in Table 16 is {0%, 40%, 20%, 80%, 60%, 100 6 percent of the %7 sequences in the same order. Therefore, the receiving end can determine that there is no phase error. Then the receiving end can determine that {0%, 40%, 20%, 80%, 60%, 100%} is a set of first percentages, and according to Table 16, the six first percentage maps can be determined. The third data is "011". {0%, 60%, 20%, 40%, 80%, 100%} is a set of first percentages, and according to Table 16, it can be determined that the third data of the six first percentage maps is "110" ". The receiving end arranges "011" and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
表16Table 16
Figure PCTCN2017085384-appb-000007
Figure PCTCN2017085384-appb-000007
需要说明的是,接收端中存储的第三映射规则(即表16)可以是发送端中存储的第三映射规则(即表2)归一化后的结果,即表16中的X个第一百分比为对表12中X个第一占空比进行归一化后得到的。It should be noted that the third mapping rule (ie, Table 16) stored in the receiving end may be a normalized result of the third mapping rule (ie, Table 2) stored in the transmitting end, that is, the X numbers in Table 16. A percentage is obtained by normalizing the X first duty cycles in Table 12.
可选的,如果接收端从拍摄的视频帧中检测亮度信号时,出现相位错误,那么假设接收端检测到的亮度信号为{326,292,310,249,272,220,326,292,249,272,310,220}。接收端通过检测,确定326是所接收到的信号中的最大值,220为所接收到信号中的最小值。根据上述最大值与最小值的位置,可以确定连续接收到的{326,292,310,249,272,220}为一组6个第一信号,{326,292,249,272,310,220}为另一组6个第一信号。Optionally, if the receiving end detects a luminance signal from the captured video frame, a phase error occurs, and then the luminance signal detected by the receiving end is assumed to be {326, 292, 310, 249, 272, 220, 326, 292, 249 , 272, 310, 220}. The receiving end determines by detection 326 that it is the maximum value of the received signal, and 220 is the minimum value of the received signal. According to the positions of the above maximum and minimum values, it can be determined that the continuously received {326, 292, 310, 249, 272, 220} is a set of six first signals, {326, 292, 249, 272, 310, 220 } is another set of 6 first signals.
接收端根据最大值和最小值,对{326,292,310,249,272,220,326,292, 249,272,310,220}进行归一化,得到{100%,68%,85%,27%,50%,0%,100%,68%,27%,50%,85%,0%}。选择两组6个第一信号中的任意一组第一信号经过归一化之后得到的序列确定非线性数值序列。例如,将{100%,68%,85%,27%,50%,0%}进行从低到高的排列,得到非线性数值序列{0%,27%,50%,68%,85%,100%}。The receiving end is based on the maximum and minimum values, for {326, 292, 310, 249, 272, 220, 326, 292, 249, 272, 310, 220} normalized to get {100%, 68%, 85%, 27%, 50%, 0%, 100%, 68%, 27%, 50%, 85%, 0% }. A sequence obtained by normalizing any one of the two sets of the first signals is selected to determine a sequence of non-linear values. For example, {100%, 68%, 85%, 27%, 50%, 0%} is arranged from low to high, resulting in a sequence of nonlinear values {0%, 27%, 50%, 68%, 85% , 100%}.
利用得到的非线性数值序列{0%,27%,50%,68%,85%,100%}对{100%,68%,85%,27%,50%,0%,100%,68%,27%,50%,85%,0%}进行非线性补偿,得到{100%,60%,80%,20%,40%,0%,100%,60%,20%,40%,80%,0%}。通过将{100%,60%,80%,20%,40%,0%}与上述表16对比,确定表16中不存在与{100%,60%,80%,20%,40%,0%}中的6个百分比的排列顺序相同的序列。而将{100%,60%,80%,20%,40%,0%,100%,60%20%,40%,80%,0%}进行相位补偿操作之后,得到{0%,50%,27%,85%,68%,100%,0%,50%,85%,68%,27%,100%}。通过将{0%,40%,20%,80%,60%,100%}与上述表16对比,确定表16中存在与{0%,40%,20%,80%,60%,100%}中的6个百分比的排列顺序相同的序列。从而接收端即可确定不存在相位错误。那么接收端即可确定{0%,40%,20%,80%,60%,100%}为一组第一百分比,并且根据表16可以确定这6个第一百分比映射的第三数据为“011”。{0%,60%,20%,40%,80%,100%}为一组第一百分比,并且根据表16可以确定这6个第一百分比映射的第三数据为“110”。接收端根据这两组第一信号的接收顺序,对“011”和“110”进行排列,得到发送端发送的数据“011110”。Using the resulting non-linear numerical sequence {0%, 27%, 50%, 68%, 85%, 100%} versus {100%, 68%, 85%, 27%, 50%, 0%, 100%, 68 %, 27%, 50%, 85%, 0%} for nonlinear compensation, giving {100%, 60%, 80%, 20%, 40%, 0%, 100%, 60%, 20%, 40% , 80%, 0%}. By comparing {100%, 60%, 80%, 20%, 40%, 0%} with Table 16 above, it is determined that Table 16 does not exist with {100%, 60%, 80%, 20%, 40%, 6 percent of the sequences in 0%} are in the same order. After {100%, 60%, 80%, 20%, 40%, 0%, 100%, 60% 20%, 40%, 80%, 0%}, after phase compensation operation, get {0%, 50 %, 27%, 85%, 68%, 100%, 0%, 50%, 85%, 68%, 27%, 100%}. By comparing {0%, 40%, 20%, 80%, 60%, 100%} with Table 16 above, it is determined that the presence in Table 16 is {0%, 40%, 20%, 80%, 60%, 100 6 percent of the %7 sequences in the same order. Therefore, the receiving end can determine that there is no phase error. Then the receiving end can determine that {0%, 40%, 20%, 80%, 60%, 100%} is a set of first percentages, and according to Table 16, the six first percentage maps can be determined. The third data is "011". {0%, 60%, 20%, 40%, 80%, 100%} is a set of first percentages, and according to Table 16, it can be determined that the third data of the six first percentage maps is "110" ". The receiving end arranges "011" and "110" according to the receiving order of the two sets of first signals, and obtains data "011110" sent by the transmitting end.
从上述实施例可以看出,采用本申请提供的信号传输方法,通过将待发送的第一数据和第二数据映射为对应的占空比,使得每个占空比能够传递1个或1个以上比特信息。通过发送与占空比对应的UPWM信号来传输第一数据和第二数据。那么,当发送端和接收端进行相机通信时,接收端能够通过一帧图像采集到一个占空比信息,从而提取出该占空比对应的1个或1个以上比特信息,从而提高了数据传输效率。It can be seen from the foregoing embodiment that, by using the signal transmission method provided by the present application, by mapping the first data and the second data to be transmitted to corresponding duty ratios, each duty ratio can be transmitted by one or one. The above bit information. The first data and the second data are transmitted by transmitting a UPWM signal corresponding to the duty ratio. Then, when the transmitting end and the receiving end perform camera communication, the receiving end can collect one duty piece information through one frame image, thereby extracting one or more bit information corresponding to the duty ratio, thereby improving data. Transmission efficiency.
上述主要从各个网元之间交互的角度对本申请提供的方案进行了介绍。可以理解的是,各个网元,例如发送端、接收端等为了实现上述功能,其包含了执行各个功能相应的硬件结构和/或软件模块。本领域技术人员应该很容易意识到,结合本文中所公开的实施例描述的各示例的单元及算法步骤,本申请能够以硬件或硬件和计算机软件的结合形式来实现。某个功能究竟以硬件还是计算机软件驱动硬件的方式来执行,取决于技术方案的特定应用和设计约束条件。专业技术人员可以对每个特定的应用来使用不同方法来实现所描述的功能,但是这种实现不应认为超出本申请的范围。The foregoing provides a description of the solution provided by the present application from the perspective of interaction between the various network elements. It can be understood that each network element, such as a transmitting end, a receiving end, etc., in order to implement the above functions, includes hardware structures and/or software modules corresponding to performing respective functions. Those skilled in the art will readily appreciate that the present application can be implemented in a combination of hardware or hardware and computer software in combination with the elements and algorithm steps of the various examples described in the embodiments disclosed herein. Whether a function is implemented in hardware or computer software to drive hardware depends on the specific application and design constraints of the solution. A person skilled in the art can use different methods to implement the described functions for each particular application, but such implementation should not be considered to be beyond the scope of the present application.
本申请可以根据上述方法示例对发送端、接收端等进行功能模块的划分,例如,可以对应各个功能划分各个功能模块,也可以将两个或两个以上的功能集成在一个处 理模块中。上述集成的模块既可以采用硬件的形式实现,也可以采用软件功能模块的形式实现。需要说明的是,本申请中对模块的划分是示意性的,仅仅为一种逻辑功能划分,实际实现时可以有另外的划分方式。The application may divide the function module by the sending end, the receiving end, and the like according to the foregoing method example. For example, each function module may be divided according to each function, or two or more functions may be integrated into one place. In the module. The above integrated modules can be implemented in the form of hardware or in the form of software functional modules. It should be noted that the division of modules in the present application is schematic, and is only a logical function division, and may be further divided in actual implementation.
在采用对应各个功能划分各个功能模块的情况下,图6A示出了上述实施例中所涉及的发送端的一种可能的结构示意图,发送端包括:处理单元和发送单元。处理单元用于支持发送端执行图4中的步骤401-402;发送单元用于支持发送端执行图4中的步骤403。其中,上述方法实施例涉及的各步骤的所有相关内容均可以援引到对应功能模块的功能描述,在此不再赘述。FIG. 6A shows a possible structural diagram of a transmitting end involved in the foregoing embodiment, where the transmitting end includes: a processing unit and a sending unit. The processing unit is configured to support the transmitting end to perform steps 401-402 in FIG. 4; the sending unit is configured to support the transmitting end to perform step 403 in FIG. 4. All the related content of the steps involved in the foregoing method embodiments may be referred to the functional descriptions of the corresponding functional modules, and details are not described herein again.
在采用集成的单元的情况下,图6B示出了上述实施例中所涉及的发送端的一种可能的结构示意图。发送端包括:处理模块602和通信模块603。处理模块602用于对发送端的动作进行控制管理,例如,处理模块602用于支持发送端执行图4中的步骤401-403,和/或用于本文所描述的技术的其它过程。通信模块603用于支持发送端与其他网络实体的通信,例如与图1中示出的功能模块或网络实体之间的通信。发送端还可以包括存储模块601,用于存储发送端的程序代码和数据。In the case of employing an integrated unit, FIG. 6B shows a possible structural diagram of the transmitting end involved in the above embodiment. The transmitting end includes a processing module 602 and a communication module 603. The processing module 602 is configured to control and manage the actions of the sender. For example, the processing module 602 is configured to support the sender to perform steps 401-403 in FIG. 4, and/or other processes for the techniques described herein. The communication module 603 is configured to support communication between the sender and other network entities, such as communication with the functional modules or network entities shown in FIG. The sending end may further include a storage module 601 for storing program code and data of the transmitting end.
其中,处理模块602可以是处理器或控制器,例如可以是中央处理器(Central Processing Unit,CPU),通用处理器,数字信号处理器(Digital Signal Processor,DSP),专用集成电路(Application-Specific Integrated Circuit,ASIC),现场可编程门阵列(Field Programmable Gate Array,FPGA)或者其他可编程逻辑器件、晶体管逻辑器件、硬件部件或者其任意组合。其可以实现或执行结合本申请公开内容所描述的各种示例性的逻辑方框,模块和电路。所述处理器也可以是实现计算功能的组合,例如包含一个或多个微处理器组合,DSP和微处理器的组合等等。通信模块603可以是收发器、收发电路或通信接口等。存储模块601可以是存储器。The processing module 602 can be a processor or a controller, for example, a central processing unit (CPU), a general-purpose processor, a digital signal processor (DSP), and an application-specific integrated circuit (Application-Specific). Integrated Circuit (ASIC), Field Programmable Gate Array (FPGA) or other programmable logic device, transistor logic device, hardware component, or any combination thereof. It is possible to implement or carry out the various illustrative logical blocks, modules and circuits described in connection with the present disclosure. The processor may also be a combination of computing functions, for example, including one or more microprocessor combinations, a combination of a DSP and a microprocessor, and the like. The communication module 603 can be a transceiver, a transceiver circuit, a communication interface, or the like. The storage module 601 can be a memory.
当处理模块602为处理器,通信模块603为收发器,存储模块601为存储器时,本申请所涉及的发送端可以为图6C所示的发送端。When the processing module 602 is a processor, the communication module 603 is a transceiver, and the storage module 601 is a memory, the transmitting end of the present application may be the transmitting end shown in FIG. 6C.
参阅图6C所示,该发送端包括:处理器612、收发器613、存储器611以及总线614。其中,收发器613、处理器612以及存储器611通过总线614相互连接;总线614可以是外设部件互连标准(Peripheral Component Interconnect,PCI)总线或扩展工业标准结构(Extended Industry StandardArchitecture,EISA)总线等。所述总线可以分为地址总线、数据总线、控制总线等。为便于表示,图6C中仅用一条粗线表示,但并不表示仅有一根总线或一种类型的总线。Referring to FIG. 6C, the transmitting end includes a processor 612, a transceiver 613, a memory 611, and a bus 614. The transceiver 613, the processor 612, and the memory 611 are connected to each other through a bus 614. The bus 614 may be a Peripheral Component Interconnect (PCI) bus or an Extended Industry Standard Architecture (EISA) bus. . The bus can be divided into an address bus, a data bus, a control bus, and the like. For ease of representation, only one thick line is shown in Figure 6C, but it does not mean that there is only one bus or one type of bus.
在采用对应各个功能划分各个功能模块的情况下,图7A示出了上述实施例中所涉及的接收端的一种可能的结构示意图,接收端包括:处理单元。处理单元用于支持接收端执行图4中的步骤403-409。其中,上述方法实施例涉及的各步骤的所有相关 内容均可以援引到对应功能模块的功能描述,在此不再赘述。FIG. 7A shows a possible structural diagram of the receiving end involved in the foregoing embodiment, where the receiving end includes: a processing unit. The processing unit is configured to support the receiving end to perform steps 403-409 in FIG. Wherein all the steps involved in the above method embodiments are related The content can be referred to the functional description of the corresponding function module, and will not be described here.
在采用集成的单元的情况下,图7B示出了上述实施例中所涉及的接收端的一种可能的结构示意图。接收端包括:处理模块702和通信模块703。处理模块702用于对接收端的动作进行控制管理,例如,处理模块702用于支持接收端执行图4中的步骤403-409,和/或用于本文所描述的技术的其它过程。通信模块703用于支持接收端与其他网络实体的通信,例如与图1中示出的功能模块或网络实体之间的通信。接收端还可以包括存储模块701,用于存储接收端的程序代码和数据。In the case of employing an integrated unit, FIG. 7B shows a possible structural diagram of the receiving end involved in the above embodiment. The receiving end includes a processing module 702 and a communication module 703. The processing module 702 is configured to control and manage the actions of the receiving end. For example, the processing module 702 is configured to support the receiving end to perform steps 403-409 in FIG. 4, and/or other processes for the techniques described herein. The communication module 703 is configured to support communication between the receiving end and other network entities, such as communication with the functional modules or network entities shown in FIG. The receiving end may further include a storage module 701 for storing program codes and data of the receiving end.
其中,处理模块702可以是处理器或控制器,例如可以是CPU,通用处理器,DSP,ASIC,FPGA或者其他可编程逻辑器件、晶体管逻辑器件、硬件部件或者其任意组合。其可以实现或执行结合本申请公开内容所描述的各种示例性的逻辑方框,模块和电路。所述处理器也可以是实现计算功能的组合,例如包含一个或多个微处理器组合,DSP和微处理器的组合等等。通信模块703可以是收发器、收发电路或通信接口等。存储模块701可以是存储器。The processing module 702 can be a processor or a controller, such as a CPU, a general purpose processor, a DSP, an ASIC, an FPGA or other programmable logic device, a transistor logic device, a hardware component, or any combination thereof. It is possible to implement or carry out the various illustrative logical blocks, modules and circuits described in connection with the present disclosure. The processor may also be a combination of computing functions, for example, including one or more microprocessor combinations, a combination of a DSP and a microprocessor, and the like. The communication module 703 can be a transceiver, a transceiver circuit, a communication interface, or the like. The storage module 701 can be a memory.
当处理模块702为处理器,通信模块703为收发器,存储模块701为存储器时,本申请所涉及的接收端可以为图7C所示的接收端。When the processing module 702 is a processor, the communication module 703 is a transceiver, and the storage module 701 is a memory, the receiving end of the present application may be the receiving end shown in FIG. 7C.
参阅图7C所示,该接收端包括:处理器712、收发器713、存储器711以及总线714。其中,收发器713、处理器712以及存储器711通过总线714相互连接;总线714可以是PCI总线或EISA总线等。所述总线可以分为地址总线、数据总线、控制总线等。为便于表示,图7C中仅用一条粗线表示,但并不表示仅有一根总线或一种类型的总线。Referring to FIG. 7C, the receiving end includes a processor 712, a transceiver 713, a memory 711, and a bus 714. The transceiver 713, the processor 712, and the memory 711 are connected to each other through a bus 714; the bus 714 may be a PCI bus or an EISA bus or the like. The bus can be divided into an address bus, a data bus, a control bus, and the like. For ease of representation, only one thick line is shown in Figure 7C, but it does not mean that there is only one bus or one type of bus.
结合本申请公开内容所描述的方法或者算法的步骤可以硬件的方式来实现,也可以是由处理器执行软件指令的方式来实现。软件指令可以由相应的软件模块组成,软件模块可以被存放于随机存取存储器(RandomAccess Memory,RAM)、闪存、只读存储器(Read Only Memory,ROM)、可擦除可编程只读存储器(Erasable Programmable ROM,EPROM)、电可擦可编程只读存储器(Electrically EPROM,EEPROM)、寄存器、硬盘、移动硬盘、只读光盘(CD-ROM)或者本领域熟知的任何其它形式的存储介质中。一种示例性的存储介质耦合至处理器,从而使处理器能够从该存储介质读取信息,且可向该存储介质写入信息。当然,存储介质也可以是处理器的组成部分。处理器和存储介质可以位于ASIC中。另外,该ASIC可以位于核心网接口设备中。当然,处理器和存储介质也可以作为分立组件存在于核心网接口设备中。The steps of a method or algorithm described in connection with the present disclosure may be implemented in a hardware or may be implemented by a processor executing software instructions. The software instructions can be composed of corresponding software modules, which can be stored in random access memory (RAM), flash memory, read only memory (ROM), and erasable programmable read only memory (Erasable). Programmable ROM (EPROM), electrically erasable programmable read only memory (EEPROM), registers, hard disk, removable hard disk, compact disk read only (CD-ROM) or any other form of storage medium known in the art. An exemplary storage medium is coupled to the processor to enable the processor to read information from, and write information to, the storage medium. Of course, the storage medium can also be an integral part of the processor. The processor and the storage medium can be located in an ASIC. Additionally, the ASIC can be located in a core network interface device. Of course, the processor and the storage medium may also exist as discrete components in the core network interface device.
具体实现中,本发明还提供一种计算机存储介质,其中,该计算机存储介质可存储有程序,该程序执行时可包括本发明提供的资源调度方法的各实施例中的部分或全 部步骤。所述的存储介质可为磁碟、光盘、只读存储记忆体(英文:read-only memory,简称:ROM)或随机存储记忆体(英文:random access memory,简称:RAM)等。In a specific implementation, the present invention further provides a computer storage medium, wherein the computer storage medium may store a program, where the program may include part or all of the embodiments of the resource scheduling method provided by the present invention. Steps. The storage medium may be a magnetic disk, an optical disk, a read-only memory (English: read-only memory, abbreviated as: ROM) or a random access memory (English: random access memory, abbreviation: RAM).
如图1所示,本申请还提供一种通信系统,包括如图6A、6B或6C所示的发送端,和如图7A、7B或7C的接收端。As shown in FIG. 1, the present application also provides a communication system including a transmitting end as shown in FIG. 6A, 6B or 6C, and a receiving end as shown in FIG. 7A, 7B or 7C.
本领域的技术人员可以清楚地了解到本申请中的技术可借助软件加必需的通用硬件平台的方式来实现。基于这样的理解,本申请中的技术方案本质上或者说对现有技术做出贡献的部分可以以软件产品的形式体现出来,该计算机软件产品可以存储在存储介质中,如ROM/RAM、磁碟、光盘等,包括若干指令用以使得一台计算机设备(可以是个人计算机,服务器,或者接收端等)执行本发明各个实施例或者实施例的某些部分所述的方法。Those skilled in the art will clearly understand that the techniques in this application can be implemented by means of software plus the necessary general hardware platform. Based on such understanding, the technical solutions in the present application may be embodied in the form of software products in essence or in the form of software products, which may be stored in a storage medium such as ROM/RAM, magnetic Discs, optical discs, etc., include instructions for causing a computer device (which may be a personal computer, server, or receiving end, etc.) to perform the methods described in various embodiments of the present invention or portions of the embodiments.
本说明书中各个实施例之间相同相似的部分互相参见即可。尤其,对于装置实施例而言,由于其基本相似于方法实施例,所以描述的比较简单,相关之处参见方法实施例中的说明即可。The same and similar parts between the various embodiments in this specification can be referred to each other. In particular, for the device embodiment, since it is basically similar to the method embodiment, the description is relatively simple, and the relevant points can be referred to the description in the method embodiment.
以上所述的本发明实施方式并不构成对本发明保护范围的限定。 The embodiments of the invention described above are not intended to limit the scope of the invention.

Claims (42)

  1. 一种信号传输方法,其特征在于,包括:A signal transmission method, comprising:
    发送端获取X个第一占空比、Y个第二占空比和Z个第三占空比,所述Y个第二占空比中的每个第二占空比由待发送的第一数据中的N个比特映射所得,所述Z个第三占空比中的每个第三占空比由待发送的第二数据中的M个比特映射所得,X≥1,X为整数,Y≥0,Y为整数,Z≥0,Z为整数,N≥1,N为整数,M≥1,M为整数;The transmitting end acquires X first duty ratios, Y second duty ratios, and Z third duty ratios, and each of the Y second duty ratios is to be sent by the second Obtained by N bit mappings in a data, each of the Z duty ratios is obtained by mapping M bits in the second data to be transmitted, X≥1, and X is an integer. Y≥0, Y is an integer, Z≥0, Z is an integer, N≥1, N is an integer, M≥1, M is an integer;
    所述发送端生成与所述X个第一占空比对应的X个基于欠采样脉冲宽度调制UPWM符号、与所述Y个第二占空比对应的Y个UPWM符号,和与所述Z个第三占空比对应的Z个UPWM符号;The transmitting end generates X sub-sampled pulse width modulated UPWM symbols corresponding to the X first duty ratios, and Y UPWM symbols corresponding to the Y second duty ratios, and the Z Z UPWM symbols corresponding to the third duty cycle;
    所述发送端依次发送所述X个UPWM符号、所述Y个UPWM符号以及所述Z个UPWM符号。The transmitting end sequentially transmits the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
  2. 根据权利要求1所述的方法,其特征在于,当Y>0时,所述发送端获取Y个第二占空比,包括:The method according to claim 1, wherein when Y>0, the transmitting end acquires Y second duty ratios, including:
    所述发送端按照所述第一数据中比特的排列顺序,以N个比特为一组,将所述第一数据中的比特划分为Y组,N=log2n;The transmitting end divides the bits in the first data into Y groups according to the order of the bits in the first data, and divides the bits in the first data into N groups, N=log 2 n;
    所述发送端根据预设的第一映射规则,将所述Y组比特映射为所述Y个第二占空比,所述第一映射规则包括不同的n组比特与不同的n个占空比之间的一一对应关系。The transmitting end maps the Y group bits to the Y second duty ratios according to a preset first mapping rule, where the first mapping rule includes different n groups of bits and different n duty cycles One-to-one correspondence between the ratios.
  3. 根据权利要求1或2所述的方法,其特征在于,当Z>0时,所述发送端获取Z个第三占空比,包括:The method according to claim 1 or 2, wherein, when Z>0, the transmitting end acquires Z third duty ratios, including:
    所述发送端按照所述第二数据中比特的排列顺序,以M个比特为一组,将所述第二数据中的比特划分为Z组,M=log2m;The transmitting end divides the bits in the second data into Z groups according to the order of the bits in the second data, and divides the bits in the second data into M groups, M=log 2 m;
    所述发送端根据预设的第二映射规则,将所述Z组比特映射为所述Z个第三占空比,所述第二映射规则包括不同的m组比特与不同的m个占空比之间的一一对应关系。The transmitting end maps the Z group bits to the Z third duty ratios according to a preset second mapping rule, where the second mapping rule includes different m group bits and different m duty cycles One-to-one correspondence between the ratios.
  4. 根据权利要求1-3任一项所述的方法,其特征在于,所述X个UPWM符号、所述Y个UPWM符号和所述Z个UPWM符号中的每个UPWM符号,均包括k段第一波形和k段第二波形,所述第一波形为平均占空比为D的PWM波 形,所述第二波形为平均占空比为1-D的PWM波形,所述k段第一波形中的每段第一波形之后均与所述k段第二波形中的一段第二波形相邻,k≥1,k为整数,0≤D≤100%。The method according to any one of claims 1 to 3, wherein each of the X UPWM symbols, the Y UPWM symbols, and the U UPWM symbols includes a k segment a waveform and a k-segment second waveform, the first waveform being a PWM wave having an average duty ratio D Forming, the second waveform is a PWM waveform with an average duty ratio of 1-D, and each of the first waveforms of the k-segment first waveform is followed by a second waveform of the second waveform of the k-segment Adjacent, k≥1, k is an integer, and 0≤D≤100%.
  5. 根据权利要求4所述的方法,其特征在于,所述第一波形包含连续的J1个第一子波形,所述J1个第一子波形中的每个第一子波形为一个脉冲波形;The method according to claim 4, wherein the first waveform comprises consecutive J1 first sub-waveforms, and each of the first sub-waveforms of the J1 first sub-waveforms is a pulse waveform;
    所述J1个第一子波形的占空比为均D;或者,The duty ratios of the J1 first sub-waveforms are all D; or
    所述第一波形在任意第一预设时长Ti内包含的J2个第一子波形的平均占空比为D1,D1与D的差值的绝对值小于或者等于第一预设值,J2<J1。The average duty ratio of the J2 first sub-waveforms included in any first preset duration Ti is D1, and the absolute value of the difference between D1 and D is less than or equal to the first preset value, J2< J1.
  6. 根据权利要求4或5所述的方法,其特征在于,所述第二波形包含连续的J3个第二子波形,所述J3个第二子波形中的每个第二子波形为一个脉冲波形;The method according to claim 4 or 5, wherein the second waveform comprises consecutive J3 second sub-waveforms, and each of the J3 second sub-waveforms is a pulse waveform ;
    所述J3个第二子波形的占空比为均1-D;或者,The duty ratio of the J3 second sub-waveforms is 1-D; or
    所述第二波形在任意第一预设时长Ti内包含的J4个第二子波形的平均占空比为D2,D2与1-D的差值的绝对值小于或者等于第二预设值,J4<J3。The average duty ratio of the J4 second sub-waveforms included in the second waveform in any first preset duration Ti is D2, and the absolute value of the difference between D2 and 1-D is less than or equal to the second preset value. J4<J3.
  7. 根据权利要求4-6任一项所述的方法,其特征在于,所述每个UPWM符号均满足以下四个条件中的至少一个条件:The method according to any one of claims 4-6, wherein each of the UPWM symbols satisfies at least one of the following four conditions:
    一、所述每个UPWM符号的时长均为T,T=1/Fc,Fc表示接收端的帧率;1. The duration of each UPWM symbol is T, T=1/Fc, and Fc represents the frame rate of the receiving end;
    二、所述k个第一波形的总时长为T/2;2. The total duration of the k first waveforms is T/2;
    三、所述第一波形和所述第二波形的时长均小于或者等于第二预设时长;3. The durations of the first waveform and the second waveform are both less than or equal to a second preset duration;
    四、所述每段第一波形的时长和与该第一波形相邻的第二波形的时长之间的差值的绝对值小于等于第三预设值。4. The absolute value of the difference between the duration of the first waveform of each segment and the duration of the second waveform adjacent to the first waveform is less than or equal to a third predetermined value.
  8. 根据权利要求1-7任一项所述的方法,其特征在于,所述X个第一占空比包括p个占空比、L1个最小占空比以及L2个最大占空比,所述最小占空比和所述最大占空比均为预设的,所述最小占空比小于所述p个占空比中的任意一个占空比,所述最大占空比大于所述p个占空比中的任意一个占空比,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。The method according to any one of claims 1 to 7, wherein the X first duty ratios include p duty ratios, L1 minimum duty ratios, and L2 maximum duty ratios, a minimum duty ratio and a maximum duty ratio are preset, the minimum duty ratio being less than any one of the p duty ratios, the maximum duty ratio being greater than the p Any one of the duty ratios, L1 ≥ 0, L1 is an integer, L2 ≥ 0, L2 is an integer, p ≥ 1, and p is an integer.
  9. 根据权利要求8所述的方法,其特征在于,所述X个第一占空比指示第二参数、第三参数和/或第三数据;The method according to claim 8, wherein the X first duty ratios indicate a second parameter, a third parameter, and/or third data;
    其中,所述第二参数用于辅助接收端将所述Y个第二占空比还原为所述第 一数据;The second parameter is used by the auxiliary receiving end to restore the Y second duty ratios to the first One data
    所述第三参数用于辅助接收端将所述Z个第三占空比还原为所述第二数据。The third parameter is used to assist the receiving end to restore the Z third duty ratios to the second data.
  10. 根据权利要求8或9所述的方法,其特征在于,当所述X个第一占空比用于指示所述第三数据时,所述X个第一占空比中的t个互不相同的第一占空比的排列顺序根据所述第三数据和预设的第三映射规则得到,所述第三映射规则包括t个互不相同的第一占空比的y种不同的排列顺序与y组比特之间的一一对应关系,所述y组比特中的每一组比特包括R个比特,所述第三数据为所述y组比特中的一组,t≤X,t为整数,R≥1,R为整数。The method according to claim 8 or 9, wherein when the X first duty ratios are used to indicate the third data, t of the X first duty ratios are not mutually The order of arrangement of the same first duty ratio is obtained according to the third data and a preset third mapping rule, where the third mapping rule includes y different arrangements of t different first duty ratios a one-to-one correspondence between the sequence and the y group bits, each of the y group bits includes R bits, and the third data is a group of the y group bits, t ≤ X, t Is an integer, R ≥ 1, and R is an integer.
  11. 根据权利要求2所述的方法,其特征在于,所述第一数据包括第三参数,所述第三参数用于辅助接收端将所述Z个第三占空比还原为所述第二数据。The method according to claim 2, wherein the first data comprises a third parameter, the third parameter is used to assist the receiving end to restore the Z third duty ratios to the second data .
  12. 一种信号传输方法,其特征在于,包括:A signal transmission method, comprising:
    接收端检测连续的X个第一信号,以获取第一参数,X≥1,X为整数;The receiving end detects consecutive X first signals to obtain a first parameter, X≥1, and X is an integer;
    所述接收端根据第二参数,检测连续的Y个第二信号,Y≥0,Y为整数;The receiving end detects consecutive Y second signals according to the second parameter, Y≥0, and Y is an integer;
    所述接收端根据所述第一参数对所述Y个第二信号进行第一处理操作,得到Y个第二百分比;The receiving end performs a first processing operation on the Y second signals according to the first parameter, to obtain Y second percentages;
    所述接收端使用所述第二参数对所述Y个第二百分比进行解调,得到第一数据;The receiving end demodulates the Y second percentages by using the second parameter to obtain first data;
    所述接收端根据第三参数,检测连续的Z个第三信号,Z≥0,Z为整数;The receiving end detects consecutive Z third signals according to the third parameter, Z≥0, and Z is an integer;
    所述接收端使用所述第一参数对所述Z个第三信号进行所述第一处理操作,得到Z个第三百分比;The receiving end performs the first processing operation on the Z third signals by using the first parameter, to obtain Z third percentages;
    所述接收端使用所述第三参数对所述Z个第三百分比进行解调,得到第二数据。The receiving end demodulates the Z third percentages by using the third parameter to obtain second data.
  13. 根据权利要求12所述的方法,其特征在于,所述方法还包括:The method of claim 12, wherein the method further comprises:
    所述接收端对所述X个第一信号进行所述第一处理操作,得到X个第一百分比。The receiving end performs the first processing operation on the X first signals to obtain X first percentages.
  14. 根据权利要求13所述的方法,其特征在于,所述第二参数为预设的;或者,所述第二参数为所述接收端根据所述X个第一百分比获取的。 The method according to claim 13, wherein the second parameter is preset; or the second parameter is obtained by the receiving end according to the X first percentages.
  15. 根据权利要求14所述的方法,其特征在于,当Y>0时,所述第二参数包括第一映射规则,所述第一映射规则包括不同的n组比特与不同的n个百分比之间的一一对应关系,所述n组比特中的每一组比特中包括N个比特;The method according to claim 14, wherein when Y>0, the second parameter comprises a first mapping rule, the first mapping rule comprising between different n groups of bits and different n percentages a one-to-one correspondence, each of the n sets of bits includes N bits;
    所述接收端使用所述第二参数对所述Y个第二百分比进行解调,得到第一数据,包括:The receiving end demodulates the Y second percentages by using the second parameter to obtain first data, including:
    所述接收端根据所述第一映射规则将所述Y个第二百分比中的每个第二百分比映射为所述第一数据中的N个比特;The receiving end maps each second percentage of the Y second percentages to N bits in the first data according to the first mapping rule;
    所述接收端根据所述Y个第二信号的接收顺序,将由所述每个第二百分比映射出N个比特进行排列,得到所述第一数据。The receiving end maps the N bits by the second percentage according to the receiving order of the Y second signals to obtain the first data.
  16. 根据权利要求13所述的方法,其特征在于,所述第三参数为预设的;或者,所述第三参数为所述接收端根据所述X个第一百分比获取的。The method according to claim 13, wherein the third parameter is preset; or the third parameter is obtained by the receiving end according to the X first percentages.
  17. 根据权利要求16所述的方法,其特征在于,当Z>0时,所述第三参数包括第二映射规则,所述第二映射规则包括不同的m组比特与不同的m个百分比之间的一一对应关系,所述m组比特中的每一组比特中包括M个比特;The method according to claim 16, wherein when Z>0, the third parameter comprises a second mapping rule, the second mapping rule comprising between different m groups of bits and different m percentages a one-to-one correspondence, each of the m sets of bits includes M bits;
    所述接收端使用所述第三参数对所述Z个第三百分比进行解调,得到第二数据,包括:The receiving end demodulates the Z third percentages by using the third parameter to obtain second data, including:
    所述接收端根据所述第二映射规则将所述Z个第三百分比中的每个第三百分比映射为所述第二数据中的M个比特;The receiving end maps each third percentage of the Z third percentages to M bits in the second data according to the second mapping rule;
    所述接收端根据所述Z个第三信号的接收顺序,将由所述每个第三百分比映射出M个比特进行排列,得到所述第二数据。The receiving end maps the M bits by each of the third percentages according to the receiving order of the Z third signals to obtain the second data.
  18. 根据权利要求13所述的方法,其特征在于,所述接收端对所述X个第一信号进行所述第一处理操作,得到X个第一百分比之后,所述方法还包括:The method according to claim 13, wherein after the receiving end performs the first processing operation on the X first signals to obtain X first percentages, the method further includes:
    所述接收端根据预设的第三映射规则和所述X个第一百分比中的t个互不相同的第一百分比的排列顺序,获取第三数据;The receiving end acquires third data according to a preset third mapping rule and an arrangement order of t first different percentages among the X first percentages;
    其中,所述第三映射规则包括所述X个第一百分比中的t个互不相同的第一百分比的y种不同的排列顺序与y组比特之间的一一对应关系,所述y组比特中的每一组比特包括R个比特,所述第三数据为所述y组比特中的一组,t≤X,t为整数,R≥1,R为整数。The third mapping rule includes a one-to-one correspondence between the different first order of the t percentages of the X first percentages and the y group bits. Each of the y group bits includes R bits, the third data is a group of the y group bits, t ≤ X, t is an integer, R ≥ 1, and R is an integer.
  19. 根据权利要求12-18任一项所述的方法,其特征在于,所述X个第一信号包括L1个最小值、L2个最大值以及除所述最小值和最大值以外的p个信号, L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。The method according to any one of claims 12 to 18, wherein the X first signals comprise L1 minimum values, L2 maximum values, and p signals other than the minimum and maximum values, L1 ≥ 0, L1 is an integer, L2 ≥ 0, L2 is an integer, p ≥ 1, and p is an integer.
  20. 根据权利要求19所述的方法,其特征在于,所述第一参数包括:所述X个第一信号中的最大值、最小值、非线性数值序列和/或相位错误指示信息。The method of claim 19, wherein the first parameter comprises: a maximum value, a minimum value, a sequence of non-linear values, and/or phase error indication information of the X first signals.
  21. 根据权利要求20所述的方法,其特征在于,所述第一处理操作包括:非线性补偿、归一化和/或相位补偿操作。The method of claim 20 wherein said first processing operation comprises: nonlinear compensation, normalization, and/or phase compensation operations.
  22. 一种发送端,其特征在于,包括:A transmitting end, comprising:
    处理单元,用于获取X个第一占空比、Y个第二占空比和Z个第三占空比,所述Y个第二占空比中的每个第二占空比由待发送的第一数据中的N个比特映射所得,所述Z个第三占空比中的每个第三占空比由待发送的第二数据中的M个比特映射所得,X≥1,X为整数,Y≥0,Y为整数,Z≥0,Z为整数,N≥1,N为整数,M≥1,M为整数;a processing unit, configured to acquire X first duty ratios, Y second duty ratios, and Z third duty ratios, wherein each of the Y second duty ratios is to be processed Obtaining N bit mappings in the transmitted first data, each third duty ratio of the Z third duty ratios is obtained by mapping M bits in the second data to be transmitted, X≥1, X is an integer, Y≥0, Y is an integer, Z≥0, Z is an integer, N≥1, N is an integer, M≥1, M is an integer;
    所述处理单元,还用于生成与所述X个第一占空比对应的X个基于欠采样脉冲宽度调制UPWM符号、与所述Y个第二占空比对应的Y个UPWM符号,和与所述Z个第三占空比对应的Z个UPWM符号;The processing unit is further configured to generate X UPWM symbols based on the undersampled pulse width modulation UPWM symbols corresponding to the X first duty ratios and corresponding to the Y second duty ratios, and Z UPWM symbols corresponding to the Z third duty cycles;
    发送单元,用于依次发送所述X个UPWM符号、所述Y个UPWM符号以及所述Z个UPWM符号。And a sending unit, configured to sequentially send the X UPWM symbols, the Y UPWM symbols, and the Z UPWM symbols.
  23. 根据权利要求22所述的发送端,其特征在于,当Y>0时,所述处理单元获取Y个第二占空比,具体包括:The transmitting end according to claim 22, wherein, when Y>0, the processing unit acquires the Y second duty ratios, specifically:
    按照所述第一数据中比特的排列顺序,以N个比特为一组,将所述第一数据中的比特划分为Y组,N=log2n;According to the order of the bits in the first data, the bits in the first data are divided into Y groups by N bits, N=log 2 n;
    根据预设的第一映射规则,将所述Y组比特映射为所述Y个第二占空比,所述第一映射规则包括不同的n组比特与不同的n个占空比之间的一一对应关系。Mapping the Y group bits to the Y second duty ratios according to a preset first mapping rule, where the first mapping rule includes between different n groups of bits and different n duty ratios One-to-one correspondence.
  24. 根据权利要求22或23所述的发送端,其特征在于,当Z>0时,所述处理单元获取Z个第三占空比,具体包括:The transmitting end according to claim 22 or 23, wherein, when Z>0, the processing unit acquires Z third duty ratios, specifically including:
    按照所述第二数据中比特的排列顺序,以M个比特为一组,将所述第二数据中的比特划分为Z组,M=log2m;According to the order of the bits in the second data, the bits in the second data are divided into Z groups by M bits, M=log 2 m;
    根据预设的第二映射规则,将所述Z组比特映射为所述Z个第三占空比,所述第二映射规则包括不同的m组比特与不同的m个占空比之间的一一对应关 系。Mapping the Z group bits to the Z third duty ratios according to a preset second mapping rule, where the second mapping rule includes between different m groups of bits and different m duty ratios One-to-one correspondence system.
  25. 根据权利要求22-24任一项所述的发送端,其特征在于,所述X个UPWM符号、所述Y个UPWM符号和所述Z个UPWM符号中的每个UPWM符号,均包括k段第一波形和k段第二波形,所述第一波形为平均占空比为D的PWM波形,所述第二波形为平均占空比为1-D的PWM波形,所述k段第一波形中的每段第一波形之后均与所述k段第二波形中的一段第二波形相邻,k≥1,k为整数,0≤D≤100%。The transmitting end according to any one of claims 22-24, wherein each of the X UPWM symbols, the Y UPWM symbols, and the U UPWM symbols comprises a k segment a first waveform and a k-segment second waveform, wherein the first waveform is a PWM waveform having an average duty ratio D, and the second waveform is a PWM waveform having an average duty ratio of 1-D, the k-segment first Each of the first waveforms in the waveform is followed by a second waveform of the second waveform of the k-segment, k≥1, k is an integer, and 0≤D≤100%.
  26. 根据权利要求25所述的发送端,其特征在于,所述第一波形包含连续的J1个第一子波形,所述J1个第一子波形中的每个第一子波形为一个脉冲波形;The transmitting end according to claim 25, wherein the first waveform comprises consecutive J1 first sub-waveforms, and each of the first sub-waveforms of the J1 first sub-waveforms is a pulse waveform;
    所述J1个第一子波形的占空比为均D;或者,The duty ratios of the J1 first sub-waveforms are all D; or
    所述第一波形在任意第一预设时长Ti内包含的J2个第一子波形的平均占空比为D1,D1与D的差值的绝对值小于或者等于第一预设值,J2<J1。The average duty ratio of the J2 first sub-waveforms included in any first preset duration Ti is D1, and the absolute value of the difference between D1 and D is less than or equal to the first preset value, J2< J1.
  27. 根据权利要求25或26所述的发送端,其特征在于,所述第二波形包含连续的J3个第二子波形,所述J3个第二子波形中的每个第二子波形为一个脉冲波形;The transmitting end according to claim 25 or 26, wherein said second waveform comprises consecutive J3 second sub-waveforms, and each of said J3 second sub-waveforms is a pulse Waveform
    所述J3个第二子波形的占空比为均1-D;或者,The duty ratio of the J3 second sub-waveforms is 1-D; or
    所述第二波形在任意第一预设时长Ti内包含的J4个第二子波形的平均占空比为D2,D2与1-D的差值的绝对值小于或者等于第二预设值,J4<J3。The average duty ratio of the J4 second sub-waveforms included in the second waveform in any first preset duration Ti is D2, and the absolute value of the difference between D2 and 1-D is less than or equal to the second preset value. J4<J3.
  28. 根据权利要求25-27任一项所述的发送端,其特征在于,所述每个UPWM符号均满足以下四个条件中的至少一个条件:The transmitting end according to any one of claims 25-27, wherein each of the UPWM symbols satisfies at least one of the following four conditions:
    一、所述每个UPWM符号的时长均为T,T=1/Fc,Fc表示接收端的帧率;1. The duration of each UPWM symbol is T, T=1/Fc, and Fc represents the frame rate of the receiving end;
    二、所述k个第一波形的总时长为T/2;2. The total duration of the k first waveforms is T/2;
    三、所述第一波形和所述第二波形的时长均小于或者等于第二预设时长;3. The durations of the first waveform and the second waveform are both less than or equal to a second preset duration;
    四、所述每段第一波形的时长和与该第一波形相邻的第二波形的时长之间的差值的绝对值小于等于第三预设值。4. The absolute value of the difference between the duration of the first waveform of each segment and the duration of the second waveform adjacent to the first waveform is less than or equal to a third predetermined value.
  29. 根据权利要求22-28任一项所述的发送端,其特征在于,所述X个第一占空比包括p个占空比、L1个最小占空比以及L2个最大占空比,所述最小占空比和所述最大占空比均为预设的,所述最小占空比小于所述p个占空比中的任意 一个占空比,所述最大占空比大于所述p个占空比中的任意一个占空比,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。The transmitting end according to any one of claims 22-28, wherein the X first duty ratios include p duty ratios, L1 minimum duty ratios, and L2 maximum duty ratios. The minimum duty ratio and the maximum duty ratio are both preset, and the minimum duty ratio is less than any of the p duty ratios a duty ratio, the maximum duty ratio is greater than any one of the p duty ratios, L1 ≥ 0, L1 is an integer, L2 ≥ 0, L2 is an integer, p ≥ 1, p is an integer .
  30. 根据权利要求29所述的发送端,其特征在于,所述X个第一占空比指示第二参数、第三参数和/或第三数据;The transmitting end according to claim 29, wherein the X first duty ratios indicate a second parameter, a third parameter, and/or third data;
    其中,所述第二参数用于辅助接收端将所述Y个第二占空比还原为所述第一数据;The second parameter is used by the auxiliary receiving end to restore the Y second duty ratios to the first data;
    所述第三参数用于辅助接收端将所述Z个第三占空比还原为所述第二数据。The third parameter is used to assist the receiving end to restore the Z third duty ratios to the second data.
  31. 根据权利要求29或30所述的发送端,其特征在于,当所述X个第一占空比用于指示所述第三数据时,所述X个第一占空比中的t个互不相同的第一占空比的排列顺序根据所述第三数据和预设的第三映射规则得到,所述第三映射规则包括t个互不相同的第一占空比的y种不同的排列顺序与y组比特之间的一一对应关系,所述y组比特中的每一组比特包括R个比特,所述第三数据为所述y组比特中的一组,t≤X,t为整数,R≥1,R为整数。The transmitting end according to claim 29 or 30, wherein when the X first duty ratios are used to indicate the third data, t of the X first duty ratios are mutually The order of arrangement of the different first duty ratios is obtained according to the third data and a preset third mapping rule, where the third mapping rule includes t different y different first duty ratios a one-to-one correspondence between the arrangement order and the y group bits, each of the y group bits includes R bits, and the third data is a group of the y group bits, t ≤ X, t is an integer, R≥1, and R is an integer.
  32. 根据权利要求24所述的发送端,其特征在于,所述第一数据包括第三参数,所述第三参数用于辅助接收端将所述Z个第三占空比还原为所述第二数据。The transmitting end according to claim 24, wherein the first data comprises a third parameter, and the third parameter is used to assist the receiving end to restore the Z third duty ratios to the second data.
  33. 一种接收端,其特征在于,包括:A receiving end, comprising:
    处理单元,用于检测连续的X个第一信号,以获取第一参数,X≥1,X为整数;a processing unit, configured to detect consecutive X first signals to obtain a first parameter, X≥1, where X is an integer;
    所述处理单元,还用于根据第二参数,检测连续的Y个第二信号,Y≥0,Y为整数;The processing unit is further configured to detect consecutive Y second signals according to the second parameter, where Y≥0, and Y is an integer;
    所述处理单元,还用于根据所述第一参数对所述Y个第二信号进行第一处理操作,得到Y个第二百分比;The processing unit is further configured to perform a first processing operation on the Y second signals according to the first parameter, to obtain Y second percentages;
    所述处理单元,还用于使用所述第二参数对所述Y个第二百分比进行解调,得到第一数据;The processing unit is further configured to demodulate the Y second percentages by using the second parameter to obtain first data;
    所述处理单元,还用于根据第三参数,检测连续的Z个第三信号,Z≥0,Z为整数;The processing unit is further configured to detect consecutive Z third signals according to the third parameter, Z≥0, and Z is an integer;
    所述处理单元,还用于使用所述第一参数对所述Z个第三信号进行所述第一处理操作,得到Z个第三百分比;The processing unit is further configured to perform the first processing operation on the Z third signals by using the first parameter, to obtain Z third percentages;
    所述处理单元,还用于使用所述第三参数对所述Z个第三百分比进行解调, 得到第二数据。The processing unit is further configured to demodulate the Z third percentages by using the third parameter, Get the second data.
  34. 根据权利要求33所述的接收端,其特征在于,所述处理单元,还用于在对所述X个第一信号进行所述第一处理操作,得到X个第一百分比。The receiving end according to claim 33, wherein the processing unit is further configured to perform the first processing operation on the X first signals to obtain X first percentages.
  35. 根据权利要求34所述的接收端,其特征在于,所述第二参数为预设的;或者,所述第二参数为所述处理单元根据所述X个第一百分比获取的。The receiving end according to claim 34, wherein the second parameter is preset; or the second parameter is obtained by the processing unit according to the X first percentages.
  36. 根据权利要求35所述的接收端,其特征在于,当Y>0时,所述第二参数包括第一映射规则,所述第一映射规则包括不同的n组比特与不同的n个百分比之间的一一对应关系,所述n组比特中的每一组比特中包括N个比特;The receiving end according to claim 35, wherein when Y>0, the second parameter comprises a first mapping rule, the first mapping rule comprising different n groups of bits and different n percentages a one-to-one correspondence, wherein each of the n sets of bits includes N bits;
    所述处理单元使用所述第二参数对所述Y个第二百分比进行解调,得到第一数据,具体包括:根据所述第一映射规则将所述Y个第二百分比中的每个第二百分比映射为所述第一数据中的N个比特;根据所述Y个第二信号的接收顺序,将由所述每个第二百分比映射出N个比特进行排列,得到所述第一数据。The processing unit demodulates the Y second percentages by using the second parameter to obtain the first data, and specifically includes: performing the Y second percentages according to the first mapping rule Each second percentage is mapped to N bits in the first data; according to the receiving order of the Y second signals, N bits are mapped by each of the second percentages Obtaining the first data.
  37. 根据权利要求34所述的接收端,其特征在于,所述第三参数为预设的;或者,所述第三参数为所述处理单元根据所述X个第一百分比获取的。The receiving end according to claim 34, wherein the third parameter is preset; or the third parameter is obtained by the processing unit according to the X first percentages.
  38. 根据权利要求37所述的接收端,其特征在于,当Z>0时,所述第三参数包括第二映射规则,所述第二映射规则包括不同的m组比特与不同的m个百分比之间的一一对应关系,所述m组比特中的每一组比特中包括M个比特;The receiving end according to claim 37, wherein when Z>0, the third parameter comprises a second mapping rule, wherein the second mapping rule comprises different m groups of bits and different m percentages a one-to-one correspondence, wherein each of the m groups of bits includes M bits;
    所述处理单元使用所述第三参数对所述Z个第三百分比进行解调,得到第二数据,具体包括:根据所述第二映射规则将所述Z个第三百分比中的每个第三百分比映射为所述第二数据中的M个比特;根据所述Z个第三信号的接收顺序,将由所述每个第三百分比映射出M个比特进行排列,得到所述第二数据。And the processing unit demodulates the Z third percentages by using the third parameter to obtain the second data, specifically, the method includes: performing the Z third percentages according to the second mapping rule Each third percentage is mapped to M bits in the second data; according to the receiving order of the Z third signals, M bits are mapped by each of the third percentages And obtaining the second data.
  39. 根据权利要求34所述的接收端,其特征在于,The receiving end according to claim 34, characterized in that
    所述处理单元,还用于在对所述X个第一信号进行所述第一处理操作,得到X个第一百分比之后,根据预设的第三映射规则和所述X个第一百分比中的t个互不相同的第一百分比的排列顺序,获取第三数据;The processing unit is further configured to perform, after performing the first processing operation on the X first signals, to obtain X first percentages, according to a preset third mapping rule and the X first Obtaining the third data in the order of the first percentages of the t different percentages in the percentage;
    其中,所述第三映射规则包括所述X个第一百分比中的t个互不相同的第一百分比的y种不同的排列顺序与y组比特之间的一一对应关系,所述y组比特中 的每一组比特包括R个比特,所述第三数据为所述y组比特中的一组,t≤X,t为整数,R≥1,R为整数。The third mapping rule includes a one-to-one correspondence between the different first order of the t percentages of the X first percentages and the y group bits. The y group of bits Each set of bits includes R bits, the third data is a set of the y set of bits, t ≤ X, t is an integer, R ≥ 1, and R is an integer.
  40. 根据权利要求33-39任一项所述的接收端,其特征在于,所述X个第一信号包括L1个最小值、L2个最大值以及除所述最小值和最大值以外的p个信号,L1≥0,L1为整数,L2≥0,L2为整数,p≥1,p为整数。The receiving end according to any one of claims 33 to 39, wherein the X first signals comprise L1 minimum values, L2 maximum values, and p signals other than the minimum and maximum values L1≥0, L1 is an integer, L2≥0, L2 is an integer, p≥1, and p is an integer.
  41. 根据权利要求40所述的接收端,其特征在于,所述第一参数包括:所述X个第一信号中的最大值、最小值、非线性数值序列和/或相位错误指示信息。The receiving end according to claim 40, wherein the first parameter comprises: a maximum value, a minimum value, a non-linear value sequence and/or phase error indication information among the X first signals.
  42. 根据权利要求41所述的方法,其特征在于,所述第一处理操作包括:非线性补偿、归一化和/或相位补偿操作。 The method of claim 41 wherein said first processing operation comprises: nonlinear compensation, normalization, and/or phase compensation operations.
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