WO2014205689A1 - Method and device for transmitting discovery signal - Google Patents

Method and device for transmitting discovery signal Download PDF

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Publication number
WO2014205689A1
WO2014205689A1 PCT/CN2013/078026 CN2013078026W WO2014205689A1 WO 2014205689 A1 WO2014205689 A1 WO 2014205689A1 CN 2013078026 W CN2013078026 W CN 2013078026W WO 2014205689 A1 WO2014205689 A1 WO 2014205689A1
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WO
WIPO (PCT)
Prior art keywords
discovery frame
user equipment
frame
discovery
current
Prior art date
Application number
PCT/CN2013/078026
Other languages
French (fr)
Chinese (zh)
Inventor
张祺智
Original Assignee
华为技术有限公司
Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)
Filing date
Publication date
Application filed by 华为技术有限公司 filed Critical 华为技术有限公司
Priority to PCT/CN2013/078026 priority Critical patent/WO2014205689A1/en
Priority to CN201380035268.2A priority patent/CN104488321B/en
Publication of WO2014205689A1 publication Critical patent/WO2014205689A1/en

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Classifications

    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04WWIRELESS COMMUNICATION NETWORKS
    • H04W8/00Network data management
    • H04W8/005Discovery of network devices, e.g. terminals
    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04WWIRELESS COMMUNICATION NETWORKS
    • H04W48/00Access restriction; Network selection; Access point selection
    • H04W48/16Discovering, processing access restriction or access information

Definitions

  • the present invention relates to communication technologies, and more particularly to a method and apparatus for transmitting a Discovery signal. Background technique
  • D2D communication means that the user equipment (User Equipment, UE for short) communicates directly without forwarding through the base station.
  • UEs In order to achieve direct communication with each other, UEs must first discover each other. Each UE needs to send a Discovery signal to let other UEs discover themselves.
  • the Discovery signal sent by all UEs consists of some Discovery frames that occupy the same frequency band, have the same duration, and do not intersect in time.
  • Each Discovery frame is divided into m frequency units from the frequency, m is a positive integer, i is the number of frequency units, and i is a non-negative integer less than m; each Discovery frame is divided into n time units from time, n Is a positive integer, j is the number of the time unit, and j is a non-negative integer less than n.
  • the intersection of the i-th frequency unit and the j-th time unit is referred to as the (i, j)th frequency unit.
  • Each UE uses only one frequency time unit in one Discovery frame. Considering the large number of UEs, m is usually not greater than n.
  • Let the frequency unit used by a certain UE in the tth Discovery frame be (i t , j t ).
  • the sequence (i., j.) (ii, ji) (i 2 , j 2 ) ... (i t , jt) is a Discovery resource, called the first
  • the (it, jt) frequency time unit is the frequency time position of the Discovery resource in the tth frame.
  • the Discovery signal sent by each UE needs to be time-hopped, that is, the Discovery resource used by the UE has different frequency/time positions in each frame, so that each UE has a chance to be discovered by all other UEs.
  • the prior art provides a technical solution.
  • the UE When determining a transmission location of a UE in a current Discovery frame, the UE needs to know the frame number of the current Discovery frame and/or the initial frequency time position of the UE, according to the frame number of the current Discovery frame. / or the initial transmission location of the UE, determining the transmission location of the UE in the current frame. Therefore, it is necessary to design a special method for the UE to obtain the frame number of the current Discovery frame and/or the initial frequency time position of the Discovery resource. Thus, the complexity of the system is increased, and the burden on the UE to transmit and listen to the Discovery signal is increased. Summary of the invention
  • Embodiments of the present invention provide a method and apparatus for transmitting a Discovery signal, which is used to reduce the complexity of the system in the case of jumping the transmission position of the Discovery signal.
  • the embodiment of the present invention provides a method for transmitting a Discovery signal, including: determining, according to a transmission location of a first user equipment in a previous Discovery frame, a transmission location of the first UE in a current Discovery frame;
  • the transmission location of the first user equipment in the current Discovery frame is determined only according to the transmission location of the first user equipment in the previous Discovery frame; the transmission location of the first user equipment in the current Discovery frame is The first user equipment is on the previous
  • the transmission locations in the Discovery frame are different in frequency; if the transmission location of the first user equipment in the last Discovery frame is the same in time as the transmission location of the second user equipment in the previous Discovery frame, The transmission location of the first user equipment in the current Discovery frame is different in time from the transmission location of the second user equipment in the current Discovery frame;
  • the Discovery signal is transmitted at the transmission location of the first user equipment in the current Discovery frame.
  • determining a transmission location of the first user equipment in a current Discovery frame according to a transmission location of the first user equipment in the previous Discovery frame includes:
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • i is a non-negative integer less than m
  • i t represents the frequency unit of the first user equipment in the current Discovery frame
  • j is a non-negative integer less than n
  • j t represents the number of the time unit of the first user equipment in the current Discovery frame
  • the m and the n are both positive integers
  • the m is less than or equal to the n
  • the m is the total number of units of the Discovery frame on the frequency
  • the n is the total number of units of the Discovery frame in time
  • the mod is a modulo operation symbol
  • Where a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime.
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame.
  • j is a non-negative integer less than n
  • j t represents the number of the time unit of the Discovery resource in the current Discovery frame, indicating the first The number of the time unit of the user equipment in the last Discovery frame
  • the m and the n are both positive integers
  • the m is divisible by the n
  • the m is the total number of units of the Discovery frame in frequency
  • the n is The total number of units of the Discovery frame in time
  • the mod is a modulo operation symbol
  • a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, ad-kbc and m are prime, gp, a X dk X b X c And m mutual prime.
  • the method further includes:
  • an embodiment of the present invention provides an apparatus for transmitting a Discovery signal, including: a determining module, configured to determine a transmission location of a first user equipment in a current Discovery frame in a transmission location in a previous Discovery frame; The device is disposed inside the first user equipment or the device is the first user equipment;
  • the transmission location of the first user equipment in the current Discovery frame is determined only according to the transmission location of the first user equipment in the previous Discovery frame; the transmission location of the first user equipment in the current Discovery frame is The transmission location of the first user equipment in the previous Discovery frame is different in frequency; if the first user equipment is on The transmission location in a Discovery frame is the same in time as the transmission location of the second user equipment in the previous Discovery frame, and the transmission location of the first user equipment in the current Discovery frame and the second user equipment are currently in the current location.
  • the transmission locations in the Discovery frame are not the same in time;
  • a transmission module configured to transmit a Discovery signal on a transmission location of the first user equipment in a current Discovery frame.
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • i is a non-negative integer less than m
  • i t represents the frequency unit of the first user equipment in the current Discovery frame
  • j is a non-negative integer less than n
  • j t represents the number of the time unit of the first user equipment in the current Discovery frame
  • the number of the time unit of the Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame on the frequency, and the n is the Discovery frame at the time The total number of units on; the mod is a modulo operation symbol;
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame.
  • j is a non-negative integer less than n
  • j t represents the number of the time unit of the Discovery resource in the current Discovery frame, indicating the first The number of the time unit of the user equipment in the last Discovery frame
  • the m and the n are both positive integers
  • the m is divisible by the n
  • the m is the total number of units of the Discovery frame in frequency
  • the n is The total number of units of the Discovery frame in time
  • the mod is a modulo operation symbol
  • a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, Ad-kbc and m-co-prime, gp, a X dk X b X c and m.
  • the detecting module is configured to be used by the first user equipment In the transmission location in the Discovery frame, before the Discovery signal is transmitted, the N Discovery frames before the current Discovery frame are detected, and the transmission location corresponding to the first user equipment in the N Discovery frames before the current Discovery frame is determined. Not used, where N is a non-negative integer.
  • the embodiment of the present invention further provides a first user equipment, including:
  • a processor configured to determine, in a transmission location in a previous Discovery frame, a transmission location of the first user equipment in the current Discovery frame;
  • the transmission location of the first user equipment in the current Discovery frame is determined only according to the transmission location of the first user equipment in the previous Discovery frame; the transmission location of the first user equipment in the current Discovery frame is The transmission location of the first user equipment in the previous Discovery frame is different in frequency; if the transmission location of the first user equipment in the previous Discovery frame and the transmission of the second user equipment in the previous Discovery frame If the location is the same in time, the transmission location of the first user equipment in the current Discovery frame is not the same as the transmission location of the second user equipment in the current Discovery frame;
  • a transceiver configured to transmit a Discovery signal on a transmission location of the first user equipment in a current Discovery frame.
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • i is a non-negative integer less than m
  • i t represents the frequency unit of the first user equipment in the current Discovery frame
  • j is a non-negative integer less than n
  • j t represents the number of the time unit of the first user equipment in the current Discovery frame
  • the m and the n are both positive integers
  • the m is less than or equal to the n
  • the m is the total number of units of the Discovery frame on the frequency
  • the n The total number of units in the time of the Discovery frame
  • the mod is a modulo operation symbol; where a, c, d, e, f are integers, a and m are mutually prime, c and n are mutually prime, d and n are mutually prime .
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame.
  • j is a non-negative integer less than n
  • j t represents the number of the time unit of the Discovery resource in the current Discovery frame, indicating the first The number of the time unit of the user equipment in the last Discovery frame
  • the m and the n are both positive integers
  • the m is divisible by the n
  • the m is the total number of units of the Discovery frame in frequency
  • the n is The total number of units of the Discovery frame in time
  • the mod is a modulo operation symbol
  • a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, ad-kbc and m are prime, gp, a X dk X b X c And m mutual prime.
  • the processor is further configured to transmit, at the transmission location of the first user equipment in the current Discovery frame.
  • the N Discovery frames before the current Discovery frame are detected, and the transmission location corresponding to the first user equipment is not used in the N Discovery frames before the current Discovery frame, where N is not Negative integer.
  • the first UE is determined according to the transmission position of the first UE in the previous Discovery frame, and the first UE is only related to the frame number of the current Discovery frame and the initial transmission position of the Discovery resource. It is necessary to know that the first UE is in the transmission position of the previous Discovery frame, and it is not necessary to acquire the frame number of the current Discovery frame and the initial transmission position of the first UE, so that the transmission position of the first UE in the current frame can be determined, thereby reducing the system.
  • the complexity reduces the burden on the first UE to transmit and listen to the Discovery signal.
  • the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame
  • the transmission position of the first UE in the current Discovery frame is The second UE is different in time in the transmission position of the current Discovery frame
  • the transmission position of the first UE in the current Discovery frame is in the same position as the first UE.
  • the transmission locations in the previous Discovery frame are not the same in frequency, so that the Discovery signal sent by each UE can be found by other UEs.
  • FIG. 1 is a flowchart of a method for transmitting a Discovery signal according to an embodiment of the present invention
  • FIG. 2A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention
  • FIG. 2B is a schematic diagram of a Discovery frame pattern generated according to the method provided in FIG. 2A;
  • FIG. 3A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention
  • FIG. 3B is a schematic diagram of a Discovery frame pattern generated according to the method provided in FIG. 3A;
  • FIG. 4A is a schematic structural diagram of an apparatus for transmitting a Discovery signal according to an embodiment of the present invention
  • FIG. 4B is a schematic structural diagram of another apparatus for transmitting a Discovery signal according to an embodiment of the present invention.
  • FIG. 5 is a schematic structural diagram of a first user equipment according to an embodiment of the present invention. detailed description
  • FIG. 1 is a flowchart of a method for transmitting a Discovery signal according to an embodiment of the present invention.
  • the executor of this embodiment may be a UE that transmits a Discovery signal or a receiver.
  • the UE of the Discovery signal As shown in FIG. 1, the method provided in this embodiment includes:
  • Step 11 The first UE determines, according to the transmission location of the first Discovery frame of the first UE, the transmission location of the first UE in the current Discovery frame, where the first UE meets the following conditions in the current transmission location of the Discovery frame. a and 1).
  • the transmission location may be a frequency location of the first UE transmitting the Discovery signal.
  • the first UE is a UE that sends a Discovery signal
  • the Discovery signal may be sent at the transmission location of the current Discovery frame.
  • an idle transmission location of the previous Discovery frame It is called the transmission location of the first UE in the previous Discovery frame.
  • the technical solution provided by the embodiment of the present invention may be adopted.
  • a UE determines the transmission location of the first UE in the current Discovery frame at the transmission location of the previous Discovery frame.
  • the first UE may continue to detect whether the transmission position of the current UE in the current Discovery frame and the next Discovery frame is idle, and detect that the first UE is in the current Discovery frame and
  • the first UE may also adopt the technical solution provided by the embodiment of the present invention, and determine the transmission location of the first UE in the current Discovery frame according to the transmission position of the first UE in the previous Discovery frame. After detecting that the first UE is idle at the transmission location of the at least one Discovery frame, the first UE may send a Discovery signal.
  • the first UE may also use the technical solution provided by the embodiment of the present invention to determine the transmission location of the current Discovery frame of the UE according to the transmission position of the first UE in the previous Discovery frame.
  • the first UE transmits the location of the Discovery signal sent by the second UE on the previous Discovery frame in the transmission location of the previous Discovery frame, and the first UE transmits the location of the current Discovery frame.
  • a UE listens to the transmission location of the Discovery signal sent by other UEs in the current Discovery frame.
  • the first UE satisfies the conditions A and B at the frequency time position of each Discovery frame.
  • Condition A is that the transmission position of the first UE in the current Discovery frame is determined only according to the transmission position of the first UE in the previous Discovery frame.
  • Condition B is that if the transmission location of the first UE in the last Discovery frame is the same in time as the transmission location of the second UE in the previous Discovery frame, the transmission location of the first UE in the current Discovery frame and the second UE The transmission locations in the current Discovery frame are not the same in time. That is, if two UEs transmit the Discovery signal at the same transmission position of the last Discovery frame, the two UEs transmit the Discovery signal at different transmission positions at different times on the current Discovery frame.
  • the function f is expressed as: Z/mZ x Z/nZ ⁇ Z/tnZ x Z/; Where is the component of f in frequency and the component of f in time. m is the total number of units in the frequency of the Discovery frame, and n is the total number of units in the time of the Discovery frame.
  • Z/mZ f0, 1 in- indicates the total number of frequency units of the Discovery frame
  • Z/llZ ⁇ 0 , ,., , - indicates the total number of time units of the Discovery frame.
  • the symbol for the direct product operation represents the direct product of the set.
  • ZlnZ, iG ZIr. i denotes the number of the frequency unit of the UE in the Discovery frame
  • j denotes the number of the time unit of the UE in the Discovery frame.
  • the representation map f reflects each element in the open interval (0, 1) as its own square.
  • Step 12 The first UE transmits the Discovery signal at the frequency position of the current Discovery frame by the first UE.
  • the transmitting the Discovery signal includes transmitting a Discovery signal and/or receiving a Discovery signal.
  • the Discovery signal is sent by the first UE in the transmission position of the current Discovery frame. If the first UE is the UE that receives the Discovery signal, after determining that the first UE is in the frequency position of the current Discovery frame, the first UE receives the Discovery signal at the frequency position of the current Discovery frame.
  • the first UE further satisfies the following condition c in the transmission location of the current Discovery frame: the transmission location of the first UE in the current Discovery frame is the previous one with the first UE.
  • the transmission positions in the Discovery frame are different in frequency, and frequency diversity gain can be achieved.
  • the above function f satisfies: f ⁇ (i, j) ⁇ i, for all i,.
  • the first UE before the transmitting, by the first UE, in the current location of the Discovery frame, before detecting the Discovery signal, detecting N Discovery frames before the current Discovery frame, and determining N Discoverys before the current Discovery frame.
  • the transmission location corresponding to the first UE in the frame is not used, where N is a non-negative integer.
  • the first UE is in accordance with the first time in the frequency position of the current Discovery frame.
  • the UE determines in the transmission location of the previous Discovery frame that, regardless of the frame number of the current Discovery frame and the initial transmission location of the Discovery resource, the first UE only needs to know the transmission location of the first UE in the previous Discovery frame, and does not need to acquire the current Discovery.
  • the frame number of the frame and the initial transmission position of the first UE can determine the transmission position of the first UE in the current frame. Therefore, the complexity of the system is reduced, and the burden of transmitting and listening to the Discovery signal by the first UE is reduced.
  • the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame
  • the transmission position of the first UE in the current Discovery frame is The two UEs are different in time in the transmission position of the current Discovery frame, and the transmission position of the first UE in the current Discovery frame is different in frequency from the transmission position of the first UE in the previous Discovery frame, so that each UE The transmitted Discovery signal can be found by other UEs.
  • FIG. 2A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention.
  • the executor of this embodiment may be a UE that transmits a Discovery signal or a UE that receives a Discovery signal.
  • the method provided in this embodiment includes:
  • Step 21 The UE determines, according to the formula (i t , jt fit — hj "): ( ( ai "+e ) mod m, ( ci ⁇ +dj t-i+g ) mod n ), that the UE is in the current Discovery (discovery) ) The frequency position of the frame.
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • the i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the UE in the current Discovery frame
  • j is a non-negative integer less than n
  • j t represents a number of time units of the UE in the current Discovery frame
  • the m and the n are both positive integers, and the m is less than or equal to the n
  • m is the total number of units in the frequency of the Discovery frame
  • n is the total number of units in the time of the Discovery frame.
  • Mod is a modulo operation.
  • (it-hjt-L ⁇ UE is in the frequency position of the previous Discovery frame
  • (i t , j t ) is the frequency position of the UE in the current Discovery frame.
  • a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime.
  • the function f is a mapping function, f satisfies ifi ⁇ , then f 2 (i,j) ⁇ f 2 (i , for all js Z/nZ, ie ZlrJe Zlr.
  • at least one prime number p exists to achieve frequency diversity gain , p divides al, the p divides m, and the p does not divide e, so that f satisfies: fl(i, j) ⁇ i, for all i,.
  • each Discovery frame has 3 frequency units in frequency and 3 time units in time.
  • Each large box in Figure 2B represents a Discovery frame, which is listed as Time, frequency of behavior.
  • the first large box from the left in Figure 2B represents the 0th Discovery frame
  • the second largest box from the left in Figure 2B represents the first Discovery frame
  • the third largest box from the left in Figure 2B represents the second.
  • the Discovery frame, the fourth largest box from the left in Figure 2B represents the third Discovery frame.
  • the frequency position of UE0 in the 0th Discovery frame is (0, 0), and the frequency position of UE1 in the 0th Discovery frame is (0, 1), ...
  • the frequency position of UE8 in the 0th Discovery frame is (2, 2).
  • the frequency position of UE0 in the first Discovery frame is (1, 0)
  • the frequency position of UE1 in the first Discovery frame is (1, 1)
  • the frequency position of UE8 in the first Discovery frame is (0, 1). It can also be seen from 2B that different UEs have different frequency positions in each Discovery frame.
  • the time positions of UE0, UE3, and UE6 in the 0th Discovery frame are all in time unit 0, but on the first Discovery frame, the time position of UE0 is in time unit 0, and the time position of UE3 On time unit 1, the time position of UE6 is on time unit 2. Therefore, two or more UEs transmit the Discovery signal at the same time position of the previous Discovery frame, and on the current Discovery frame, at different time positions. Send a Discovery signal.
  • UE0 is on frequency unit 0 in the frequency domain of the 0th Discovery frame and on frequency unit 1 in the frequency domain of the first Discovery frame. Therefore, the frequency position of the current UE in the current Discovery frame is different from the frequency position of the second UE in the previous Discovery frame, so that the frequency diversity gain can be achieved.
  • Step 22 The UE transmits a Discovery signal at the frequency position of the current Discovery frame by the UE.
  • FIG. 3A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention.
  • the executor of this embodiment may be a UE that transmits a Discovery signal or a UE that receives a Discovery signal.
  • the method provided in this embodiment includes:
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • the i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the UE in the current Discovery frame
  • j is a non-negative integer less than n
  • j t represents a number of time units of the UE in the current Discovery frame
  • m is the total number of units in the frequency of the Discovery frame
  • n is the total number of units in the time of the Discovery frame.
  • Mod is the modulo operator symbol.
  • a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are mutually prime, ad-kbc and m are prime, gp, aXd-kXbXc and m are mutually Prime.
  • the function f is a mapping function, f satisfies ifi ⁇ , then f 2 (i,j) ⁇ f 2 (i, for all jGZ/nZ, i Zlr, ie Zlr.
  • f satisfies ifi ⁇ , then f 2 (i,j) ⁇ f 2 (i, for all jGZ/nZ, i Zlr, ie Zlr.
  • at least one prime number exists for achieving frequency diversity gain p
  • the p is divisible by b
  • the p is divisible by m
  • the p is divisible by al
  • the p is not Divide e such that f satisfies: fl(j, j) ⁇ i, for all .
  • each Discovery frame has 4 frequency units in frequency and 4 time units in time.
  • Each large box in Fig. 3B represents a Discovery frame, which is listed as time and behavior frequency.
  • Step 32 The UE transmits the Discovery signal at the frequency position of the current Discovery frame by the UE.
  • the UE determines the time-frequency location of each Discovery frame, and the UE does not need to obtain the frame number of the current Discovery frame and the initial frequency-time position of the UE. Therefore, the burden of transmitting the Discovery signal by the UE and the production cost of the UE are alleviated.
  • FIG. 4A is a schematic structural diagram of an apparatus for transmitting a Discovery signal according to an embodiment of the present invention.
  • the apparatus can be used to implement the above method, and therefore, the features in the above method embodiments can be applied to the present embodiment.
  • the device can be the first UE.
  • the apparatus provided in this embodiment includes: a determining module 41 and a transmitting module 42.
  • a determining module 41 configured to determine, in a transmission location in a previous Discovery frame, a transmission location of the first UE in a current Discovery frame; the apparatus is disposed in the first UE or the apparatus is the first UE .
  • the transmission location of the first UE in the current Discovery frame is determined only according to the transmission location of the first UE in the previous Discovery frame; if the transmission location of the first UE in the previous Discovery frame is The transmission positions of the two UEs in the previous Discovery frame are the same in time, and the transmission position of the first UE in the current Discovery frame is different in time from the transmission position of the second UE in the current Discovery frame.
  • the transmission position of the first UE in the current Discovery frame is The transmission positions of the first UE in the previous Discovery frame are different in frequency.
  • the transmitting module 42 is configured to transmit a Discovery signal on a transmission location of the first UE in a current Discovery frame.
  • the apparatus provided in FIG. 4A may further include: a detecting module 43.
  • a detecting module configured to detect N Discovery frames before the current Discovery frame before transmitting the Discovery signal in the transmission position of the first UE in the current Discovery frame, and determine N before the current Discovery frame The transmission location corresponding to the first UE in the Discovery frame is not used, where N is a non-negative integer.
  • the determining module is specifically used for a formula (( ai t _ 1+ e ) mod m, ( ci M+dj M+g ) mod n ) , determining a transmission location of the first UE in a current Discovery frame;
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • the i is a non-negative integer less than m
  • i t represents the frequency unit of the first UE in the current Discovery frame a number indicating a frequency unit of the first UE in the previous Discovery frame
  • j is a non-negative integer less than n
  • j t represents a number of the time unit of the first UE in the current Discovery frame, indicating the number a number of time units of a UE in a previous Discovery frame
  • the m and the n are both positive integers
  • the m is less than or equal to the n
  • the m is a total number of units of the Discovery frame in frequency
  • the n The total number of units in the time of the Discovery frame
  • the mod is a modulo operation symbol
  • a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime.
  • there is at least one prime number p the p is divisible by a, the p is divisible by m, and the p is not divisible by e, so that the condition is met: the first UE is in the frequency position of the current Discovery frame, and the first The two UEs have different frequency in the frequency position of the previous Discovery frame, and the frequency diversity gain is realized.
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • the i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame.
  • a number indicating a frequency unit of the first UE in the previous Discovery frame
  • j is a non-negative integer less than n
  • j t represents a number of the time unit of the Discovery resource in the current Discovery frame, indicating the first UE
  • the m and the n are both positive integers
  • the m is divisible by the n
  • the m is the total number of units of the Discovery frame in frequency
  • the n is a Discovery frame The total number of units in time
  • the mod is a modulo operation symbol
  • a, b, c, d, e, g are integers, a and k are mutually prime, d and k are mutually prime, c and m are mutually prime, ad-kbc and m are prime.
  • the frequency-time position of the first UE satisfies the following conditions:
  • the frequency-time position of the first UE in the current Discovery frame is different from the frequency-time position of the second UE in the previous Discovery frame, and the frequency diversity gain is implemented.
  • the first UE determines the frequency position of the current Discovery frame according to the transmission position of the first UE in the previous Discovery frame, and is independent of the initial transmission position of the frame number of the current Discovery frame and the Discovery resource, the first UE. It is only necessary to know the transmission position of the first UE in the previous Discovery frame, and it is not necessary to acquire the frame number of the current Discovery frame and the initial transmission position of the first UE, so that the transmission position of the first UE in the current frame can be determined, and therefore, the transmission is reduced.
  • the complexity of the system reduces the burden on the first UE to transmit and listen to the Discovery signal. In each Discovery frame, if the transmission location of the first UE in the previous Discovery frame is the same in time as the transmission location of the second UE in the previous Discovery frame, the first UE is currently
  • the transmission location of the Discovery frame is different in time from the transmission location of the current UE in the current Discovery frame, and the transmission location of the first UE in the current Discovery frame and the transmission location of the first UE in the previous Discovery frame are in frequency. Different, so that the Discovery signal sent by each UE can be found by other UEs.
  • FIG. 5 is a schematic structural diagram of a first UE according to an embodiment of the present invention.
  • the user equipment can be used to implement the foregoing method. Therefore, the features in the foregoing method embodiments can be applied to the implementation.
  • the first UE includes: a processor 51 and a transceiver 52.
  • the processor 51 is configured to determine, in a transmission location in the last Discovery frame, a transmission location of the first UE in the current Discovery frame;
  • the transmission location of the first UE in the current Discovery frame is determined only according to the transmission location of the first UE in the previous Discovery frame; the transmission location of the first UE in the current Discovery frame and the first The transmission position of a UE in the last Discovery frame is different in frequency; if the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame, And the transmission location of the first UE in the current Discovery frame and the second UE are currently
  • the transmission locations in the Discovery frame are not the same in time;
  • the transceiver 52 is configured to transmit a Discovery signal at a transmission location of the first UE in a current Discovery frame.
  • the processor is further configured to: before transmitting the Discovery signal in the transmission position of the first UE in the current Discovery frame, detecting N Discovery frames before the current Discovery frame, determining that The transmission location corresponding to the first UE in the N Discovery frames before the current Discovery frame is not used, where N is a non-negative integer.
  • the processor is specifically configured according to a formula (i t , ( ( ai t _ 1+ e ) mod m,
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • the i is a non-negative integer less than m
  • i t represents the frequency unit of the first UE in the current Discovery frame a number indicating a frequency unit of the first UE in the previous Discovery frame
  • j is a non-negative integer less than n
  • j t represents a number of the time unit of the first UE in the current Discovery frame, indicating the number a number of time units of a UE in a previous Discovery frame
  • the m and the n are both positive integers, and the m is less than or equal to the n
  • the m is a total number of units of the Discovery frame in frequency
  • the n The total number of units in the time of the Discovery frame
  • the mod is a modulo operation symbol
  • a, c, d, e, f are integers
  • a and m are prime
  • c and n are prime
  • t is the frame number of the current Discovery frame
  • t-1 is the frame number of the previous Discovery frame
  • the i is a non-negative integer less than m
  • i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame.
  • a number indicating a frequency unit of the first UE in the previous Discovery frame
  • j is a non-negative integer less than n
  • j t represents a number of the time unit of the Discovery resource in the current Discovery frame, indicating the first UE
  • the m and the n are both positive integers
  • the m is divisible by the n
  • the m is the total number of units of the Discovery frame in frequency
  • the n is a Discovery frame The total number of units in time
  • the mod is a modulo operation symbol
  • a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, ad-kbc and m are prime.
  • there is at least one prime p the p is divisible b, the p is divisible by m, the p is divisible by a-1, and the p is not divisible by e.
  • the first UE is determined according to the transmission position of the first UE in the previous Discovery frame, and the first UE is only related to the frame number of the current Discovery frame and the initial transmission position of the Discovery resource. It is necessary to know that the first UE is in the transmission position of the previous Discovery frame, and it is not necessary to acquire the frame number of the current Discovery frame and the initial transmission position of the first UE, so that the transmission position of the first UE in the current frame can be determined, thereby reducing the system.
  • the complexity reduces the burden on the first UE to transmit and listen to the Discovery signal.
  • the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame
  • the transmission position of the first UE in the current Discovery frame is The second UE is different in time in the transmission position of the current Discovery frame, and the transmission position of the first UE in the current Discovery frame is in the same position as the first UE.
  • the transmission locations in the previous Discovery frame are not the same in frequency, so that the Discovery signal sent by each UE can be found by other UEs.
  • the aforementioned program can be stored in a computer readable storage medium.
  • the program when executed, performs the steps including the above method embodiments; and the foregoing storage medium includes: a medium that can store program codes, such as a ROM, a RAM, a magnetic disk, or an optical disk.

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Abstract

Provided are a method and device for transmitting a discovery signal. The method comprises: according to a transmission position of a first user equipment in a previous discovery frame, determining a transmission position of the first UE in a current discovery frame, wherein the transmission position of the first user equipment in the current discovery frame is different from that of the first user equipment in the previous discovery frame in frequency; and if the transmission position of the first user equipment in the previous discovery frame is the same as that of a second user equipment in the previous discovery frame in time, then the transmission position of the first user equipment in the current discovery frame is different from that of the second user equipment in the current discovery frame in time; and transmitting a discovery signal on the transmission position of the first user equipment in the current discovery frame. The embodiments of the present invention alleviate the burden of the UE to transmit a discovery signal, thereby reducing the production cost of the UE.

Description

传输 Discovery信号的方法和装置  Method and apparatus for transmitting a Discovery signal
技术领域 Technical field
本发明涉及通信技术, 尤其涉及一种传输 Discovery信号的方法和装 置。 背景技术  The present invention relates to communication technologies, and more particularly to a method and apparatus for transmitting a Discovery signal. Background technique
设备至设备 (Device to Device, 简称 D2D) 通信是指用户设备 (User Equipment, 简称 UE) 之间直接进行通信, 不需要经过基站转发。 UE间 为了实现相互间的直接通信, 必须先相互发现。 每个 UE 需要发送 Discovery信号来让其它 UE发现自己。 所有 UE发送的发现 (Discovery) 信号由一些占用频带相同, 持续时间相同, 时间上不相交的 Discovery帧 组成。 每个 Discovery帧从频率上分为 m个频率单元, m为正整数, i为 频率单元的编号, i为小于 m的非负整数; 每个 Discovery帧从时间上分 为 n个时间单元, n为正整数, j为时间单元的编号, j为小于 n的非负整 数。 称第 i个频率单元与第 j个时间单元的交为第 (i, j)个频时单元。 每个 UE在一个 Discovery帧中仅使用 1个频时单元。 考虑到 UE数的众多, 通 常情况下, m不大于 n。 设某个 UE在第 t个 Discovery帧使用的频时单元 为 (it, jt)。称序列 (i。, j。) (ii, ji) (i2, j2) ... ... (it, jt) 为一个 Discovery资源,称第Device to Device (D2D) communication means that the user equipment (User Equipment, UE for short) communicates directly without forwarding through the base station. In order to achieve direct communication with each other, UEs must first discover each other. Each UE needs to send a Discovery signal to let other UEs discover themselves. The Discovery signal sent by all UEs consists of some Discovery frames that occupy the same frequency band, have the same duration, and do not intersect in time. Each Discovery frame is divided into m frequency units from the frequency, m is a positive integer, i is the number of frequency units, and i is a non-negative integer less than m; each Discovery frame is divided into n time units from time, n Is a positive integer, j is the number of the time unit, and j is a non-negative integer less than n. The intersection of the i-th frequency unit and the j-th time unit is referred to as the (i, j)th frequency unit. Each UE uses only one frequency time unit in one Discovery frame. Considering the large number of UEs, m is usually not greater than n. Let the frequency unit used by a certain UE in the tth Discovery frame be (i t , j t ). The sequence (i., j.) (ii, ji) (i 2 , j 2 ) ... (i t , jt) is a Discovery resource, called the first
(it, jt)个频时单元为该 Discovery资源在第 t帧的频时位置。 The (it, jt) frequency time unit is the frequency time position of the Discovery resource in the tth frame.
通常情况下, 需要对每个 UE发送的 Discovery信号进行频时跳转, 即该 UE使用的 Discovery资源在每帧内的频 /时位置不同, 使得每个 UE 都有机会被其它所有 UE发现。  Generally, the Discovery signal sent by each UE needs to be time-hopped, that is, the Discovery resource used by the UE has different frequency/time positions in each frame, so that each UE has a chance to be discovered by all other UEs.
现有技术提供了一种技术方案, 确定 UE在当前 Discovery帧内的传 输位置时, UE需要知道当前 Discovery帧的帧号和 /或该 UE的初始频时 位置,根据当前 Discovery帧的帧号和 /或该 UE的初始传输位置,确定 UE 在当前帧的传输位置。 因此, 需要设计专门的方法使 UE 获取当前 Discovery帧的帧号和 /或该 Discovery资源的初始频时位置。 因而, 增加 了系统的复杂度, 增加了 UE传输和侦听 Discovery信号的负担。 发明内容 The prior art provides a technical solution. When determining a transmission location of a UE in a current Discovery frame, the UE needs to know the frame number of the current Discovery frame and/or the initial frequency time position of the UE, according to the frame number of the current Discovery frame. / or the initial transmission location of the UE, determining the transmission location of the UE in the current frame. Therefore, it is necessary to design a special method for the UE to obtain the frame number of the current Discovery frame and/or the initial frequency time position of the Discovery resource. Thus, the complexity of the system is increased, and the burden on the UE to transmit and listen to the Discovery signal is increased. Summary of the invention
本发明实施例提供一种传输 Discovery信号的方法和装置, 用于在跳 转 Discovery信号的传输位置的情况下, 降低系统的复杂度。 第一方面, 本发明实施例提供一种传输 Discovery信号的方法, 包括: 根据第一用户设备在上一个 Discovery帧中的传输位置, 确定所述第 一 UE在当前 Discovery帧中的传输位置;  Embodiments of the present invention provide a method and apparatus for transmitting a Discovery signal, which is used to reduce the complexity of the system in the case of jumping the transmission position of the Discovery signal. In a first aspect, the embodiment of the present invention provides a method for transmitting a Discovery signal, including: determining, according to a transmission location of a first user equipment in a previous Discovery frame, a transmission location of the first UE in a current Discovery frame;
其中, 所述第一用户设备在当前 Discovery帧中的传输位置仅根据所 述第一用户设备在上一个 Discovery帧中的传输位置确定; 所述第一用户 设备在当前 Discovery帧中的传输位置与所述第一用户设备在上一个 The transmission location of the first user equipment in the current Discovery frame is determined only according to the transmission location of the first user equipment in the previous Discovery frame; the transmission location of the first user equipment in the current Discovery frame is The first user equipment is on the previous
Discovery帧中的传输位置在频率上不相同; 如果所述第一用户设备在上 一个 Discovery帧中的传输位置与第二用户设备在上一个 Discovery帧中的 传输位置在时间上相同, 则所述第一用户设备在当前 Discovery帧中的传 输位置与所述第二用户设备在当前 Discovery帧中的传输位置在时间上不 相同; The transmission locations in the Discovery frame are different in frequency; if the transmission location of the first user equipment in the last Discovery frame is the same in time as the transmission location of the second user equipment in the previous Discovery frame, The transmission location of the first user equipment in the current Discovery frame is different in time from the transmission location of the second user equipment in the current Discovery frame;
在所述第一用户设备在当前 Discovery帧中的传输位置上, 传输 Discovery信号。  The Discovery signal is transmitted at the transmission location of the first user equipment in the current Discovery frame.
结合第一方面, 在第一种可能的实现方式中, 根据第一用户设备在上 一个 Discovery帧中的传输位置, 确定所述第一用户设备在当前 Discovery 帧中的传输位置, 包括:  With reference to the first aspect, in a first possible implementation manner, determining a transmission location of the first user equipment in a current Discovery frame according to a transmission location of the first user equipment in the previous Discovery frame includes:
根据公式 (it, jt)=f(i t—i, j t—i)= ( ( ait-i+e ) mod m, ( ci t-i+dj t-i+g ) mod n ), 确定所述第一用户设备在当前 Discovery帧中的传输位置; Determining the above according to the formula (it, jt)=f(it—i, jt−i)=( ( ait-i+e ) mod m, ( ci t —i+dj t —i+g ) mod n ) a transmission location of the first user equipment in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一用户设备在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述第一用户设备在当前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n 为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; 其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 结合第一方面, 在第二种可能的实现方式中, 根据第一用户设备在上 一个 Discovery帧中的传输位置, 确定所述第一用户设备在当前 Discovery 帧中的传输位置, 包括: Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the frequency unit of the first user equipment in the current Discovery frame The number indicating the frequency unit of the first user equipment in the previous Discovery frame, j is a non-negative integer less than n, and j t represents the number of the time unit of the first user equipment in the current Discovery frame, The number of the time unit of the first user equipment in the previous Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame on the frequency The n is the total number of units of the Discovery frame in time; the mod is a modulo operation symbol; Where a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime. With reference to the first aspect, in a second possible implementation manner, determining a transmission location of the first user equipment in a current Discovery frame according to a transmission location of the first user equipment in a previous Discovery frame, including:
丰艮据公式 (it, jt)= f(i t-i, j t-i)= ( ( a it-i+b j t-i+e ) mod m, (kc it-i+d j t-i+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置; According to the formula (i t , j t )= f(i ti, j ti)= ( ( a it-i+bj t-i+e ) mod m, (kc i t -i+dj t -i+ g) mod n ) , determining a frequency time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当 前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame. Indicates the number of the frequency unit of the first user equipment in the previous Discovery frame, j is a non-negative integer less than n, and j t represents the number of the time unit of the Discovery resource in the current Discovery frame, indicating the first The number of the time unit of the user equipment in the last Discovery frame; the m and the n are both positive integers, and the m is divisible by the n; the m is the total number of units of the Discovery frame in frequency, and the n is The total number of units of the Discovery frame in time; the mod is a modulo operation symbol;
其中 a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, a X d-k X b X c和 m互素。  Where a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, ad-kbc and m are prime, gp, a X dk X b X c And m mutual prime.
结合第一方面, 或第一方面的第一种可能实现方式和第二种可能实现 方式,在第三种可能的实现方式中,在所述第一用户设备在当前 Discovery 帧中的传输位置上, 传输 Discovery信号之前, 所述方法还包括:  With reference to the first aspect, or the first possible implementation manner of the first aspect, and the second possible implementation manner, in a third possible implementation manner, where the first user equipment is in the current location of the Discovery frame Before transmitting the Discovery signal, the method further includes:
所述第一用户设备检测所述当前 Discovery帧之前的 N个 Discovery 帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所述第一用 户设备对应的传输位置未被使用, 其中, N为非负整数。 第二方面, 本发明实施例提供一种传输 Discovery信号的装置, 包括: 确定模块, 用于在上一个 Discovery帧中的传输位置, 确定第一用户 设备在当前 Discovery帧中的传输位置; 所述装置设置于所述第一用户设 备内部或所述装置为所述第一用户设备;  The first user equipment detects the N Discovery frames before the current Discovery frame, and determines that the transmission location corresponding to the first user equipment is not used in the N Discovery frames before the current Discovery frame, where, Is a non-negative integer. In a second aspect, an embodiment of the present invention provides an apparatus for transmitting a Discovery signal, including: a determining module, configured to determine a transmission location of a first user equipment in a current Discovery frame in a transmission location in a previous Discovery frame; The device is disposed inside the first user equipment or the device is the first user equipment;
其中, 所述第一用户设备在当前 Discovery帧中的传输位置仅根据所 述第一用户设备在上一个 Discovery帧中的传输位置确定; 所述第一用户 设备在当前 Discovery帧中的传输位置与所述第一用户设备在上一个 Discovery帧中的传输位置在频率上不相同; 如果所述第一用户设备在上 一个 Discovery帧中的传输位置与第二用户设备在上一个 Discovery帧中的 传输位置在时间上相同, 则所述第一用户设备在当前 Discovery帧中的传 输位置与所述第二用户设备在当前 Discovery帧中的传输位置在时间上不 相同; The transmission location of the first user equipment in the current Discovery frame is determined only according to the transmission location of the first user equipment in the previous Discovery frame; the transmission location of the first user equipment in the current Discovery frame is The transmission location of the first user equipment in the previous Discovery frame is different in frequency; if the first user equipment is on The transmission location in a Discovery frame is the same in time as the transmission location of the second user equipment in the previous Discovery frame, and the transmission location of the first user equipment in the current Discovery frame and the second user equipment are currently in the current location. The transmission locations in the Discovery frame are not the same in time;
传输模块, 用于在所述第一用户设备在当前 Discovery帧中的传输位 置上, 传输 Discovery信号。  And a transmission module, configured to transmit a Discovery signal on a transmission location of the first user equipment in a current Discovery frame.
结合第二方面, 在第一种可能的实现方式中, 所述确定模块, 具体用 于根据公式 (it, jt)=f(i ", j t—i)= ( ( ait_i+e ) mod m, ( ci "+dj "+g ) mod n ) , 确定所述第一用户设备在当前 Discovery帧中的传输位置; With reference to the second aspect, in a first possible implementation, the determining module is specifically configured to use the formula (i t , jt)=f(i “, jt—i)= ( ( ai t _i+e ) according to the formula Mod m, ( ci "+dj "+g ) mod n ) , determining a transmission location of the first user equipment in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一用户设备在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述第一用户设备在当前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the frequency unit of the first user equipment in the current Discovery frame The number indicating the frequency unit of the first user equipment in the previous Discovery frame, j is a non-negative integer less than n, and j t represents the number of the time unit of the first user equipment in the current Discovery frame, The first user equipment is on the previous
Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n 为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; The number of the time unit of the Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame on the frequency, and the n is the Discovery frame at the time The total number of units on; the mod is a modulo operation symbol;
其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 结合第二方面, 在第二种可能的实现方式中, 所述确定模块, 具体用 于根据公式 (it, jt)= f(i t_!, j t-1)= ( ( a it_!+b j t_!+e ) mod m, (kc it_!+d j t-1+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置; Where a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime. With reference to the second aspect, in a second possible implementation manner, the determining module is specifically configured to use the formula (i t , j t )= f(i t _!, j t-1 )= ( ( ai t _!+bj t _!+e ) mod m, (kc i t _!+dj t-1 +g ) mod n ) , determining the frequency time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当 前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame. Indicates the number of the frequency unit of the first user equipment in the previous Discovery frame, j is a non-negative integer less than n, and j t represents the number of the time unit of the Discovery resource in the current Discovery frame, indicating the first The number of the time unit of the user equipment in the last Discovery frame; the m and the n are both positive integers, and the m is divisible by the n; the m is the total number of units of the Discovery frame in frequency, and the n is The total number of units of the Discovery frame in time; the mod is a modulo operation symbol;
其中, a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, a X d-k X b X c和 m互素。 Where a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, Ad-kbc and m-co-prime, gp, a X dk X b X c and m.
结合第二方面, 或第二方面的第一种或能实现方式和第二种可能实现 方式, 在在第三种可能的实现方式中, 检测模块, 用于在所述第一用户设 备在当前 Discovery帧中的传输位置上, 传输 Discovery信号之前, 检测所 述当前 Discovery帧之前的 N个 Discovery帧, 确定在所述当前 Discovery 帧之前的 N个 Discovery帧中所述第一用户设备对应的传输位置未被使用, 其中, N为非负整数。  With reference to the second aspect, or the first or the implementation of the second aspect, and the second possible implementation manner, in a third possible implementation, the detecting module is configured to be used by the first user equipment In the transmission location in the Discovery frame, before the Discovery signal is transmitted, the N Discovery frames before the current Discovery frame are detected, and the transmission location corresponding to the first user equipment in the N Discovery frames before the current Discovery frame is determined. Not used, where N is a non-negative integer.
第三方面, 本发明实施例还提供一种第一用户设备, 包括:  In a third aspect, the embodiment of the present invention further provides a first user equipment, including:
处理器, 用于在上一个 Discovery帧中的传输位置, 确定第一用户设 备在当前 Discovery帧中的传输位置;  a processor, configured to determine, in a transmission location in a previous Discovery frame, a transmission location of the first user equipment in the current Discovery frame;
其中, 所述第一用户设备在当前 Discovery帧中的传输位置仅根据所 述第一用户设备在上一个 Discovery帧中的传输位置确定; 所述第一用户 设备在当前 Discovery帧中的传输位置与所述第一用户设备在上一个 Discovery帧中的传输位置在频率上不相同; 如果所述第一用户设备在上 一个 Discovery帧中的传输位置与第二用户设备在上一个 Discovery帧中的 传输位置在时间上相同, 则所述第一用户设备在当前 Discovery帧中的传 输位置与所述第二用户设备在当前 Discovery帧中的传输位置在时间上不 相同;  The transmission location of the first user equipment in the current Discovery frame is determined only according to the transmission location of the first user equipment in the previous Discovery frame; the transmission location of the first user equipment in the current Discovery frame is The transmission location of the first user equipment in the previous Discovery frame is different in frequency; if the transmission location of the first user equipment in the previous Discovery frame and the transmission of the second user equipment in the previous Discovery frame If the location is the same in time, the transmission location of the first user equipment in the current Discovery frame is not the same as the transmission location of the second user equipment in the current Discovery frame;
收发器, 用于在所述第一用户设备在当前 Discovery帧中的传输位置 上, 传输 Discovery信号。  And a transceiver, configured to transmit a Discovery signal on a transmission location of the first user equipment in a current Discovery frame.
结合第三方面, 在第一种可能实现方式中, 所述处理器, 具体用于根 据公式 (it, jt)=f(i t-i, j t-i)= ( ( ait-i+e ) mod m, ( ci "+dj "+g ) mod n ) , 确 定所述第一用户设备在当前 Discovery帧中的传输位置; With reference to the third aspect, in a first possible implementation, the processor is specifically configured to use the formula (i t , j t )=f(i ti, j ti)= ( ( ait-i+e ) mod according to the formula m, ( ci "+dj "+g ) mod n ) , determining a transmission location of the first user equipment in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一用户设备在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述第一用户设备在当前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n 为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; 其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 结合第三方面, 在第二种可能实现方式中, 所述处理器, 具体用于根 据公式 (it, jt)= f(i t-i, j t-i)= ( ( it-i+b j t-i+e ) mod m, (kc it-i+d j t-i+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the frequency unit of the first user equipment in the current Discovery frame The number indicating the frequency unit of the first user equipment in the previous Discovery frame, j is a non-negative integer less than n, and j t represents the number of the time unit of the first user equipment in the current Discovery frame, The number of the time unit of the first user equipment in the previous Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame on the frequency , the n The total number of units in the time of the Discovery frame; the mod is a modulo operation symbol; where a, c, d, e, f are integers, a and m are mutually prime, c and n are mutually prime, d and n are mutually prime . With reference to the third aspect, in a second possible implementation manner, the processor is specifically configured to use the formula (i t , j t )= f(i ti, j ti)= ( ( it-i+bj t - i+e ) mod m, (kc i t -i+dj t -i+g ) mod n ) , determining a frequency time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当 前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame. Indicates the number of the frequency unit of the first user equipment in the previous Discovery frame, j is a non-negative integer less than n, and j t represents the number of the time unit of the Discovery resource in the current Discovery frame, indicating the first The number of the time unit of the user equipment in the last Discovery frame; the m and the n are both positive integers, and the m is divisible by the n; the m is the total number of units of the Discovery frame in frequency, and the n is The total number of units of the Discovery frame in time; the mod is a modulo operation symbol;
其中 a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, a X d-k X b X c和 m互素。  Where a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, ad-kbc and m are prime, gp, a X dk X b X c And m mutual prime.
结合第三方面, 或第三方面的第一种和第二种或能的实现方式中, 所 述处理器, 还用于在所述第一用户设备在当前 Discovery帧中的传输位置 上, 传输 Discovery信号之前, 检测所述当前 Discovery帧之前的 N个 Discovery帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所 述第一用户设备对应的传输位置未被使用, 其中, N为非负整数。  In combination with the third aspect, or the first and the second or the second implementation of the third aspect, the processor is further configured to transmit, at the transmission location of the first user equipment in the current Discovery frame. Before the discovery signal, the N Discovery frames before the current Discovery frame are detected, and the transmission location corresponding to the first user equipment is not used in the N Discovery frames before the current Discovery frame, where N is not Negative integer.
上述技术方案中, 第一 UE在当前 Discovery帧的频时位置根据第一 UE在上一个 Discovery帧的传输位置确定, 与当前 Discovery帧的帧号和 Discovery资源的初始传输位置无关, 第一 UE只需知道第一 UE在上一个 Discovery帧的传输位置, 不需要获取当前 Discovery帧的帧号和第一 UE 的初始传输位置, 就可确定第一 UE在当前帧的传输位置, 因此, 降低了 系统的复杂度, 减轻了第一 UE传输及侦听 Discovery信号的负担。 由于 在每个 Discovery帧中, 如果第一 UE在上一个 Discovery帧的传输位置与 第二 UE在上一个 Discovery帧的传输位置在时间上相同,则第一 UE在当 前 Discovery帧的传输位置与第二 UE在当前 Discovery帧的传输位置在时 间上不相同,并且第一 UE在当前 Discovery帧中的传输位置与第一 UE在 上一个 Discovery帧中的传输位置在频率上不相同, 使得每个 UE发送的 Discovery信号都可以被其它 UE发现。 附图说明 In the foregoing technical solution, the first UE is determined according to the transmission position of the first UE in the previous Discovery frame, and the first UE is only related to the frame number of the current Discovery frame and the initial transmission position of the Discovery resource. It is necessary to know that the first UE is in the transmission position of the previous Discovery frame, and it is not necessary to acquire the frame number of the current Discovery frame and the initial transmission position of the first UE, so that the transmission position of the first UE in the current frame can be determined, thereby reducing the system. The complexity reduces the burden on the first UE to transmit and listen to the Discovery signal. In each of the Discovery frames, if the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame, the transmission position of the first UE in the current Discovery frame is The second UE is different in time in the transmission position of the current Discovery frame, and the transmission position of the first UE in the current Discovery frame is in the same position as the first UE. The transmission locations in the previous Discovery frame are not the same in frequency, so that the Discovery signal sent by each UE can be found by other UEs. DRAWINGS
图 1为本发明实施例提供的一种传输 Discovery信号的方法流程图; 图 2A为本发明实施例提供的另一种传输 Discovery信号的方法流程 图;  1 is a flowchart of a method for transmitting a Discovery signal according to an embodiment of the present invention; FIG. 2A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention;
图 2B为根据图 2A提供的方法产生的一种 Discovery帧图样示意图; 图 3A为本发明实施例提供的又一种传输 Discovery信号的方法流程 图;  2B is a schematic diagram of a Discovery frame pattern generated according to the method provided in FIG. 2A; FIG. 3A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention;
图 3B为根据图 3A提供的方法产生的一种 Discovery帧图样示意图; 图 4A为本发明实施例提供的一种传输 Discovery信号的装置结构示意 图;  FIG. 3B is a schematic diagram of a Discovery frame pattern generated according to the method provided in FIG. 3A; FIG. 4A is a schematic structural diagram of an apparatus for transmitting a Discovery signal according to an embodiment of the present invention;
图 4B为本发明实施例提供的另一种传输 Discovery信号的装置结构示 意图;  FIG. 4B is a schematic structural diagram of another apparatus for transmitting a Discovery signal according to an embodiment of the present invention; FIG.
图 5为本发明实施例提供的一种第一用户设备结构示意图。 具体实施方式  FIG. 5 is a schematic structural diagram of a first user equipment according to an embodiment of the present invention. detailed description
图 1为本发明实施例提供一种传输 Discovery信号的方法流程图。 本 实施例的执行主体可以是发送 Discovery信号的 UE也可以是接收  FIG. 1 is a flowchart of a method for transmitting a Discovery signal according to an embodiment of the present invention. The executor of this embodiment may be a UE that transmits a Discovery signal or a receiver.
Discovery信号的 UE。 如图 1所示, 本实施例提供的方法包括: The UE of the Discovery signal. As shown in FIG. 1, the method provided in this embodiment includes:
步骤 11 : 第一 UE根据第一 UE在上一个 Discovery帧的传输位置, 确定所述第一 UE在当前 Discovery帧的传输位置; 其中, 所述第一 UE在 当前 Discovery帧的传输位置满足以下条件 a和1)。其中,所述传输位置可 以是第一 UE传输 Discovery信号的频时位置。  Step 11: The first UE determines, according to the transmission location of the first Discovery frame of the first UE, the transmission location of the first UE in the current Discovery frame, where the first UE meets the following conditions in the current transmission location of the Discovery frame. a and 1). The transmission location may be a frequency location of the first UE transmitting the Discovery signal.
如果第一 UE是发送 Discovery信号的 UE, 在上一个 Discovery帧发 现某个传输位置空闲时, 可以在当前 Discovery帧的传输位置上发送 Discovery信号, 此时, 上一个 Discovery帧的某个空闲传输位置称为第一 UE在上一个 Discovery帧的传输位置。 第一 UE在当前 Discovery帧上发 送 Discovery信号之前, 可以采用本发明实施例提供的技术方案, 根据第 一 UE在上一个 Discovery帧的传输位置, 确定第一 UE在当前 Discovery 帧的传输位置。 另外, 第一 UE在上一个 Discovery帧发现某个传输位置 空闲后, 还可以继续检测第一 UE在当前 Discovery帧和下一个 Discovery 帧的传输位置是否空闲, 检测第一 UE在当前 Discovery帧和下一个 If the first UE is a UE that sends a Discovery signal, when the previous Discovery frame finds that a certain transmission location is idle, the Discovery signal may be sent at the transmission location of the current Discovery frame. At this time, an idle transmission location of the previous Discovery frame. It is called the transmission location of the first UE in the previous Discovery frame. Before the first UE sends the Discovery signal on the current Discovery frame, the technical solution provided by the embodiment of the present invention may be adopted. A UE determines the transmission location of the first UE in the current Discovery frame at the transmission location of the previous Discovery frame. In addition, after the first Discovery frame finds that a certain transmission location is idle, the first UE may continue to detect whether the transmission position of the current UE in the current Discovery frame and the next Discovery frame is idle, and detect that the first UE is in the current Discovery frame and One
Discovery帧的传输位置是否空闲时,第一 UE也可以采用本发明实施例提 供的技术方案, 根据第一 UE在上一个 Discovery帧的传输位置, 确定第 一 UE在当前 Discovery帧的传输位置。 在检测到第一 UE在至少一个 Discovery帧的传输位置空闲之后, 第一 UE可以发送 Discovery信号。 When the transmission location of the Discovery frame is idle, the first UE may also adopt the technical solution provided by the embodiment of the present invention, and determine the transmission location of the first UE in the current Discovery frame according to the transmission position of the first UE in the previous Discovery frame. After detecting that the first UE is idle at the transmission location of the at least one Discovery frame, the first UE may send a Discovery signal.
如果第一 UE是侦听 Discovery信号的 UE, 在上一个 Discovery帧的 某个传输上侦听到第二 UE发送的 Discovery信号, 为了在当前 Discovery 帧的传输位置上继续侦听 Discovery信号, 第一 UE也可以采用本发明实 施例提供的技术方案, 根据第一 UE在上一个 Discovery帧的传输位置, 确定该 UE在当前 Discovery帧的传输位置。 此时, 第一 UE在上一个 Discovery帧的传输位置为第一 UE在上一个 Discovery帧上侦听到第二 UE 发送的 Discovery信号的传输位置,第一 UE在当前 Discovery帧的传输位 置为第一 UE在当前 Discovery帧仙听其它 UE发送的 Discovery信号的传 输位置。 本发明实施例中, 第一 UE在每个 Discovery帧的频时位置均满 足条件 A和 B。条件 A 为第一 UE在当前 Discovery帧中的传输位置仅根 据第一 UE在上一个 Discovery帧中的传输位置确定。 条件 B为如果第一 UE在上一个 Discovery帧中的传输位置与第二 UE在上一个 Discovery帧 中的传输位置在时间上相同, 则第一 UE在当前 Discovery帧中的传输位 置与第二 UE在当前 Discovery帧中的传输位置在时间上不相同。 也就是 说, 如果两个 UE在上一个 Discovery帧的时间相同的传输位置上传输 Discovery信号, 则在当前 Discovery帧上, 这两个 UE在不同的时间不同 的传输位置上传输 Discovery信号。  If the first UE is the UE that listens to the Discovery signal, the Discovery signal sent by the second UE is detected on a certain transmission of the previous Discovery frame, in order to continue to listen for the Discovery signal in the transmission location of the current Discovery frame, the first The UE may also use the technical solution provided by the embodiment of the present invention to determine the transmission location of the current Discovery frame of the UE according to the transmission position of the first UE in the previous Discovery frame. At this time, the first UE transmits the location of the Discovery signal sent by the second UE on the previous Discovery frame in the transmission location of the previous Discovery frame, and the first UE transmits the location of the current Discovery frame. A UE listens to the transmission location of the Discovery signal sent by other UEs in the current Discovery frame. In the embodiment of the present invention, the first UE satisfies the conditions A and B at the frequency time position of each Discovery frame. Condition A is that the transmission position of the first UE in the current Discovery frame is determined only according to the transmission position of the first UE in the previous Discovery frame. Condition B is that if the transmission location of the first UE in the last Discovery frame is the same in time as the transmission location of the second UE in the previous Discovery frame, the transmission location of the first UE in the current Discovery frame and the second UE The transmission locations in the current Discovery frame are not the same in time. That is, if two UEs transmit the Discovery signal at the same transmission position of the last Discovery frame, the two UEs transmit the Discovery signal at different transmission positions at different times on the current Discovery frame.
为使第一 UE在每个 Discovery帧的频时位置满足上述条件, 可以设 计一一映射的函数 f。第一 UE
Figure imgf000010_0001
U (i", j"), (it-1, jt-1))、 或 (it, jt)=f ( i", j" )
Figure imgf000010_0002
f4(i", j"))计算自己在每个 Discovery帧的频 时位置。
In order for the first UE to satisfy the above conditions in the frequency-time position of each Discovery frame, a one-to-map function f can be designed. First UE
Figure imgf000010_0001
U (i", j"), (i t-1 , j t-1 )), or (i t , j t )=f ( i", j" )
Figure imgf000010_0002
f 4 (i", j")) Calculates the frequency position of each Discovery frame.
优选地, 函数 f表示为: Z/mZ x Z/nZ→ Z/tnZ x Z/; 其中, 是 f在频率上的分量, 是 f在时间上的分量。 m为 Discovery 帧在频率上的单元总数, n为 Discovery帧在时间上的单元总数, Preferably, the function f is expressed as: Z/mZ x Z/nZ→ Z/tnZ x Z/; Where is the component of f in frequency and the component of f in time. m is the total number of units in the frequency of the Discovery frame, and n is the total number of units in the time of the Discovery frame.
Z/mZ = f0, 1 in-表示 Discovery帧的频率单元的编号全体, Z/llZ = {0 , ,., , -表示 Discovery帧的时间单元的编号全体。 为直积 运算的符号表示集合的直积。 ZlnZ, iG ZIr。 i表示 UE在 Discovery帧 中的频率单元的编号, j表示 UE在 Discovery帧中的时间单元的编号。  Z/mZ = f0, 1 in- indicates the total number of frequency units of the Discovery frame, and Z/llZ = {0 , ,., , - indicates the total number of time units of the Discovery frame. The symbol for the direct product operation represents the direct product of the set. ZlnZ, iG ZIr. i denotes the number of the frequency unit of the UE in the Discovery frame, and j denotes the number of the time unit of the UE in the Discovery frame.
f、 if足: if i≠, then f2(i, j)≠ f2({, for all f, if foot: if i≠, then f 2 (i, j)≠ f 2 ({, for all
js Z/nZ, i Zlr,( Zlr。 相当于, 如果第一 UE在上一个 Discovery 帧中的传输位置与第二 UE在上一个 Discovery帧中的传输位置在时间上 相同,则所述第一 UE在当前 Discovery帧中的传输位置与第二 UE在当前 Discovery帧中的传输位置在时间上不相同。  Js Z/nZ, i Zlr, ( Zlr. Equivalent to, if the transmission position of the first UE in the last Discovery frame is the same in time as the transmission position of the second UE in the previous Discovery frame, then the first The transmission location of the UE in the current Discovery frame is different in time from the transmission location of the second UE in the current Discovery frame.
下面对函数 f 的表达式的含义进行说明:  The following describes the meaning of the expression of the function f:
设八、 B是两个非空集合, 如果存在一个法则 f, 使得对 A中的每个 元素 a, 按法则 f, 在 B中存在唯一确定的元素 f(a)与之对应, 则称 f为从 A到 B的映射, 记作:
Figure imgf000011_0001
Let 八, B be two non-empty sets, if there is a law f, so that for each element a in A, according to the law f, there is a uniquely determined element f(a) in B corresponding to it, then f For the mapping from A to B, it is recorded as:
Figure imgf000011_0001
例如:  E.g:
f φ. ΐ) ^ (Ό,Ι)  f φ. ΐ) ^ (Ό,Ι)
H- a  H- a
表示映射 f将开区间 (0,1)中的每个元素映为自己的平方。  The representation map f reflects each element in the open interval (0, 1) as its own square.
步骤 12: 第一 UE在第一 UE在当前 Discovery帧的频时位置上, 传 输 Discovery信号。 其中, 传输 Discovery信号包括发送 Discovery信号禾口 /或接收 Discovery信号。  Step 12: The first UE transmits the Discovery signal at the frequency position of the current Discovery frame by the first UE. The transmitting the Discovery signal includes transmitting a Discovery signal and/or receiving a Discovery signal.
如果第一 UE是发送 Discovery信号的 UE, 确定第一 UE在当前 Discovery帧的频时位置后,在第一 UE在当前 Discovery帧的传输位置上, 发送 Discovery信号。 如果第一 UE是接收 Discovery信号的 UE, 在确定 第一 UE在当前 Discovery帧的频时位置后, 在第一 UE在当前 Discovery 帧的频时位置上, 接收 Discovery信号。  If the first UE is the UE that sends the Discovery signal, after determining that the first UE is in the frequency position of the current Discovery frame, the Discovery signal is sent by the first UE in the transmission position of the current Discovery frame. If the first UE is the UE that receives the Discovery signal, after determining that the first UE is in the frequency position of the current Discovery frame, the first UE receives the Discovery signal at the frequency position of the current Discovery frame.
优选地, 所述第一 UE在当前 Discovery帧的传输位置还满足以下条 件 c: 第一 UE在当前 Discovery帧中的传输位置与第一 UE在上一个 Discovery帧中的传输位置在频率上不相同, 可以实现频率分集增益。 例 如, UE0到 UE3的信道在频率单元 i=0衰落很严重, 而 UE0还可以在其 它频率单元上发送 Discovery信号, UE3可以在其它频率单元上发现 UE0 发送的 Discovery信号。为满足频率分集低增益,上述函数 f满足: f\(i, j)≠i, for all i,。 Preferably, the first UE further satisfies the following condition c in the transmission location of the current Discovery frame: the transmission location of the first UE in the current Discovery frame is the previous one with the first UE. The transmission positions in the Discovery frame are different in frequency, and frequency diversity gain can be achieved. For example, the channel of UE0 to UE3 is very fading in frequency unit i=0, and UE0 can also transmit the Discovery signal on other frequency units, and UE3 can discover the Discovery signal sent by UE0 on other frequency units. In order to satisfy the frequency diversity low gain, the above function f satisfies: f\(i, j) ≠i, for all i,.
可选地, 在所述第一 UE在当前 Discovery帧中的传输位置上, 传输 Discovery信号之前,检测所述当前 Discovery帧之前的 N个 Discovery帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所述第一 UE对应 的传输位置未被使用, 其中, N为非负整数。  Optionally, before the transmitting, by the first UE, in the current location of the Discovery frame, before detecting the Discovery signal, detecting N Discovery frames before the current Discovery frame, and determining N Discoverys before the current Discovery frame. The transmission location corresponding to the first UE in the frame is not used, where N is a non-negative integer.
上述技术方案中, 第一 UE在当前 Discovery帧的频时位置根据第一 In the foregoing technical solution, the first UE is in accordance with the first time in the frequency position of the current Discovery frame.
UE在上一个 Discovery帧的传输位置确定, 与当前 Discovery帧的帧号和 Discovery资源的初始传输位置无关, 第一 UE只需知道第一 UE在上一个 Discovery帧的传输位置, 不需要获取当前 Discovery帧的帧号和第一 UE 的初始传输位置, 就可确定第一 UE在当前帧的传输位置, 因此, 降低了 系统的复杂度, 减轻了第一 UE传输及侦听 Discovery信号的负担。 由于 在每个 Discovery帧中, 如果第一 UE在上一个 Discovery帧的传输位置与 第二 UE在上一个 Discovery帧的传输位置在时间上相同,则第一 UE在当 前 Discovery帧的传输位置与第二 UE在当前 Discovery帧的传输位置在时 间上不相同,并且第一 UE在当前 Discovery帧中的传输位置与第一 UE在 上一个 Discovery帧中的传输位置在频率上不相同, 使得每个 UE发送的 Discovery信号都可以被其它 UE发现。 The UE determines in the transmission location of the previous Discovery frame that, regardless of the frame number of the current Discovery frame and the initial transmission location of the Discovery resource, the first UE only needs to know the transmission location of the first UE in the previous Discovery frame, and does not need to acquire the current Discovery. The frame number of the frame and the initial transmission position of the first UE can determine the transmission position of the first UE in the current frame. Therefore, the complexity of the system is reduced, and the burden of transmitting and listening to the Discovery signal by the first UE is reduced. In each of the Discovery frames, if the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame, the transmission position of the first UE in the current Discovery frame is The two UEs are different in time in the transmission position of the current Discovery frame, and the transmission position of the first UE in the current Discovery frame is different in frequency from the transmission position of the first UE in the previous Discovery frame, so that each UE The transmitted Discovery signal can be found by other UEs.
图 2A为本发明实施例提供的另一种传输 Discovery信号的方法流程 图。 本实施例的执行主体可以是发送 Discovery信号的 UE也可以是接收 Discovery信号的 UE。 如图 2A所示, 本实施例提供的方法包括:  2A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention. The executor of this embodiment may be a UE that transmits a Discovery signal or a UE that receives a Discovery signal. As shown in FIG. 2A, the method provided in this embodiment includes:
步骤 21 : UE根据公式 (it, jt f i t—h j "): ( ( ai"+e ) mod m, ( ci ^+dj t-i+g ) mod n ) , 确定该 UE在当前 Discovery (发现) 帧的频时位置。 Step 21: The UE determines, according to the formula (i t , jt fit — hj "): ( ( ai "+e ) mod m, ( ci ^+dj t-i+g ) mod n ), that the UE is in the current Discovery (discovery) ) The frequency position of the frame.
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示 UE在当前 Discovery帧的频率单元的 编号, j为小于 n的非负整数, jt表示 UE在当前 Discovery帧的时间单元 的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, n为 Discovery帧在时间上的单元 总数。 mod为取模运算。 Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the UE in the current Discovery frame, j is a non-negative integer less than n, j t represents a number of time units of the UE in the current Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; m is the total number of units in the frequency of the Discovery frame, and n is the total number of units in the time of the Discovery frame. Mod is a modulo operation.
其中, (it-hjt-L^ UE在上一个 Discovery帧的频时位置, (it,jt) 为 UE 在当前 Discovery帧的频时位置。 Where (it-hjt-L^ UE is in the frequency position of the previous Discovery frame, (i t , j t ) is the frequency position of the UE in the current Discovery frame.
其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 函数 f为 映射函数, f满足 ifi≠, then f2(i,j)≠ f2(i , for all js Z/nZ, ie ZlrJe Zlr。 优选地, 为实现频率分集增益, 至少存在一个 素数 p , 所述 p整除 a-l, 所述 p整除 m, 所述 p不整除 e, 使得 f满足: fl(i, j)≠ i, for all i,。 Where a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime. The function f is a mapping function, f satisfies ifi≠, then f 2 (i,j)≠ f 2 (i , for all js Z/nZ, ie ZlrJe Zlr. Preferably, at least one prime number p exists to achieve frequency diversity gain , p divides al, the p divides m, and the p does not divide e, so that f satisfies: fl(i, j)≠ i, for all i,.
举例来说,设 m=n=3, 令 a=c=d=e=l, g=0,则 (it, jt)=f(it j t-1)=( (i"+l ) mod 3, (it-1+jt-1)mod3)。以数字 0, 1, , 8表示的 9个 UE的 Discovery 资源在 Discovery帧 t=0, 1, 2, 3的频时位置如图 2B所示。 如图 2B所示, 每个 Discovery帧在频率上有 3个频率单元,在时间上也有 3个时间单元, 图 2B中每一个大方框表示一个 Discovery帧, 其中列为时间, 行为频率。 For example, let m=n=3, let a=c=d=e=l, g=0, then (i t , j t )=f(i t j t-1 )=( (i"+ l) mod 3, (i t-1 +j t-1 ) mod3). The Discovery resources of the 9 UEs represented by the numbers 0, 1, and 8 are in the frequency of the Discovery frame t=0, 1, 2, 3 The position is shown in Figure 2B. As shown in Figure 2B, each Discovery frame has 3 frequency units in frequency and 3 time units in time. Each large box in Figure 2B represents a Discovery frame, which is listed as Time, frequency of behavior.
图 2B中从左边数第一个大方框表示第 0个 Discovery帧, 图 2B中从 左边数第二个大方框表示第 1个 Discovery帧,图 2B中从左边数第三个大 方框表示第 2个 Discovery帧, 图 2B中从左边数第四个大方框表示第 3 个 Discovery帧。  The first large box from the left in Figure 2B represents the 0th Discovery frame, the second largest box from the left in Figure 2B represents the first Discovery frame, and the third largest box from the left in Figure 2B represents the second. The Discovery frame, the fourth largest box from the left in Figure 2B represents the third Discovery frame.
参见图 2B中从左边数第一个大方框, UE0在第 0个 Discovery帧的 频时位置为(0, 0), UE1在第 0个 Discovery帧的频时位置为(0, 1), …, UE8在第 0个 Discovery帧的频时位置为 (2, 2) 。 参见图 2B中左边第 二个大方框, UE0在第 1个 Discovery帧的频时位置为 (1, 0) , UE1在 第 1个 Discovery帧的频时位置为 (1, 1) , ..., UE8在第 1个 Discovery 帧的频时位置为 (0, 1) 。 从 2B还可以看出, 不同 UE在每个 Discovery 帧的频时位置上都不相同。  Referring to the first large block from the left in FIG. 2B, the frequency position of UE0 in the 0th Discovery frame is (0, 0), and the frequency position of UE1 in the 0th Discovery frame is (0, 1), ... The frequency position of UE8 in the 0th Discovery frame is (2, 2). Referring to the second big box on the left side of FIG. 2B, the frequency position of UE0 in the first Discovery frame is (1, 0), and the frequency position of UE1 in the first Discovery frame is (1, 1), ... The frequency position of UE8 in the first Discovery frame is (0, 1). It can also be seen from 2B that different UEs have different frequency positions in each Discovery frame.
如图 2B所示, UE0、 UE3和 UE6在第 0个 Discovery帧的时间位置 均在时间单元 0上, 但是在第 1个 Discovery帧上, UE0的时间位置在时 间单元 0上, UE3的时间位置在时间单元 1上, UE6的时间位置在时间单 元 2上。 因此, 两个或两个以上 UE在上一个 Discovery帧的相同时间位 置上发送 Discovery信号, 则在当前 Discovery帧上, 在不同的时间位置上 发送 Discovery信号。 As shown in FIG. 2B, the time positions of UE0, UE3, and UE6 in the 0th Discovery frame are all in time unit 0, but on the first Discovery frame, the time position of UE0 is in time unit 0, and the time position of UE3 On time unit 1, the time position of UE6 is on time unit 2. Therefore, two or more UEs transmit the Discovery signal at the same time position of the previous Discovery frame, and on the current Discovery frame, at different time positions. Send a Discovery signal.
如图 2B所示, UE0在第 0个 Discovery帧的频域位置在频率单元 0 上, 而在第 1个 Discovery帧的频域位置在频率单元 1上。 因此, 第一 UE 在当前 Discovery帧的频时位置, 与第二 UE在上一个 Discovery帧的频时 位置在频率上不相同, 从而可以实现频率分集增益。  As shown in FIG. 2B, UE0 is on frequency unit 0 in the frequency domain of the 0th Discovery frame and on frequency unit 1 in the frequency domain of the first Discovery frame. Therefore, the frequency position of the current UE in the current Discovery frame is different from the frequency position of the second UE in the previous Discovery frame, so that the frequency diversity gain can be achieved.
步骤 22: UE在该 UE在当前 Discovery帧的频时位置上,传输 Discovery 信号。  Step 22: The UE transmits a Discovery signal at the frequency position of the current Discovery frame by the UE.
本实施例中, 根据公式 (it,jt)=f(it-1, jt-1)= ( (ai^+e) modm, (ci^+dj t-i+g) modn) , 确定 UE在每个 Discovery帧的频时位置, UE不需要获 取当前 Discovery帧的帧号和该 UE的初始频时位置, 因此, 减轻了 UE传 输 Discovery信号的负担和 UE的生产成本。 In this embodiment, according to the formula (i t , j t )=f(i t-1 , j t-1 )= ( (ai^+e) modm, (ci^+dj t-i+g) modn) The UE determines the time-frequency location of each Discovery frame, and the UE does not need to obtain the frame number of the current Discovery frame and the initial frequency-time position of the UE. Therefore, the burden of transmitting the Discovery signal by the UE and the production cost of the UE are alleviated.
图 3A为本发明实施例提供的另一种传输 Discovery信号的方法流程 图。 本实施例的执行主体可以是发送 Discovery信号的 UE也可以是接收 Discovery信号的 UE。 如图 3A所示, 本实施例提供的方法包括:  FIG. 3A is a flow chart of another method for transmitting a Discovery signal according to an embodiment of the present invention. The executor of this embodiment may be a UE that transmits a Discovery signal or a UE that receives a Discovery signal. As shown in FIG. 3A, the method provided in this embodiment includes:
步骤 31: UE根据公式 (it,jt)=f(it—丄, 丄)= ( (ai"+bjt-1+e) modm,Step 31: The UE according to the formula (i t , j t )=f(i t —丄, 丄)= ( (ai"+bj t-1 +e) modm,
(kci"+djt-1+g) modn) , 确定该 UE在当前 Discovery (发现) 帧的频 时位置。 (kci"+dj t-1 +g) modn) , determines the frequency position of the UE in the current Discovery frame.
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 UE在当前 Discovery帧的频率单 元的编号, j为小于 n的非负整数, jt表示所述 UE在当前 Discovery帧的 时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n, m=nk; 所述 m为 Discovery帧在频率上的单元总数, n为 Discovery帧在 时间上的单元总数。 mod为取模运算符号。 Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the UE in the current Discovery frame, j is a non-negative integer less than n, j t represents a number of time units of the UE in the current Discovery frame; the m and the n are both positive integers, and the m divisible by the n, m=nk; m is the total number of units in the frequency of the Discovery frame, and n is the total number of units in the time of the Discovery frame. Mod is the modulo operator symbol.
其中, (it-hjt-L^ UE在上一个 Discovery帧的频时位置, (it,jt) 为 UE 在当前 Discovery帧的频时位置; Where (it-hjt-L^ UE is in the frequency position of the previous Discovery frame, (i t , j t ) is the frequency time position of the UE in the current Discovery frame;
其中, a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, aXd-kXbXc和 m互素。  Where a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are mutually prime, ad-kbc and m are prime, gp, aXd-kXbXc and m are mutually Prime.
函数 f为 映射函数, f满足 ifi≠, then f2(i,j)≠ f2(i, for all jGZ/nZ, i Zlr,ie Zlr。 优选地, 为实现频率分集增益, 至少存在一个 素数 p , 所述 p 整除 b, 所述 p 整除 m , 所述 p整除 a-l, 所述 p不 整除 e, 使得 f满足: fl(j, j)≠i, for all ,。 The function f is a mapping function, f satisfies ifi≠, then f 2 (i,j)≠ f 2 (i, for all jGZ/nZ, i Zlr, ie Zlr. Preferably, at least one prime number exists for achieving frequency diversity gain p, the p is divisible by b, the p is divisible by m, the p is divisible by al, and the p is not Divide e such that f satisfies: fl(j, j)≠i, for all .
举例来说, 设 m=n=4, k=l, 令 a=3, b=2, c=-l, d=e=g=l, (it, jt)= f(i j t_i)= ( ( 3 it-i+2 j t_i+l ) mod 4, ( it-i+ j t-i+1 ) mod 4 ) 。 For example, let m=n=4, k=l, let a=3, b=2, c=-l, d=e=g=l, (i t , j t )= f(ij t_i) = ( ( 3 it-i+2 j t _i+l ) mod 4, ( i t -i+ j t-i+1 ) mod 4 ) .
以十六进制数字 0, 1, ...... , f表示的 16个 UE的 Discovery资源在 Discovery帧 t=0, 1, 2, 3, 4的位置如图 3B所示。 如图 3B所示, 每个 Discovery帧在频率上有 4个频率单元, 在时间上也有 4个时间单元, 图 3B中每一个大方框表示一个 Discovery帧, 其中列为时间, 行为频率。  The Discovery resources of the 16 UEs represented by the hexadecimal digits 0, 1, ..., f are located at the position of the Discovery frame t = 0, 1, 2, 3, 4 as shown in Fig. 3B. As shown in Fig. 3B, each Discovery frame has 4 frequency units in frequency and 4 time units in time. Each large box in Fig. 3B represents a Discovery frame, which is listed as time and behavior frequency.
从图 3B可以看出,不同 UE资源在每个 Discovery帧的频时位置上都 不相同。 两个 UE在上一个 Discovery帧的相同时间位置上发送 Discovery 信号,则在当前 Discovery帧上,在不同的时间位置上发送 Discovery信号。 不同 UE在每个 Discovery帧的频时位置上都不相同。  As can be seen from Figure 3B, different UE resources are different in the frequency time position of each Discovery frame. The two UEs transmit the Discovery signal at the same time position of the previous Discovery frame, and then transmit the Discovery signal at different time positions on the current Discovery frame. Different UEs are different in the frequency position of each Discovery frame.
步骤 32: UE在该 UE在当前 Discovery帧的频时位置上,传输 Discovery 信号。  Step 32: The UE transmits the Discovery signal at the frequency position of the current Discovery frame by the UE.
本实施例中, 根据公式 2 : (it, jt)= f(i t-i, j t-i)= ( ( a i^+b j ^+e ) mod m, (kc i"+d j "+g ) mod n ) , 确定 UE在每个 Discovery帧的频时位置, UE 不需要获取当前 Discovery帧的帧号和 UE的初始频时位置, 因此, 减轻 了 UE传输 Discovery信号的负担和 UE的生产成本。 In this embodiment, according to the formula 2: (i t , jt)= f(i ti, j ti)= ( ( ai^+bj ^+e ) mod m, (kc i"+dj "+g ) mod n The UE determines the time-frequency location of each Discovery frame, and the UE does not need to obtain the frame number of the current Discovery frame and the initial frequency-time position of the UE. Therefore, the burden of transmitting the Discovery signal by the UE and the production cost of the UE are alleviated.
图 4A为本发明实施例提供的一种传输 Discovery信号的装置结构示意 图。 该装置可以用于实现上述方法, 因此, 上述方法实施例中的特征可以 应用到本实施例中。 该装置可以是第一 UE。 如图 4A所示, 本实施例提供 的装置包括: 确定模块 41和传输模块 42。  FIG. 4A is a schematic structural diagram of an apparatus for transmitting a Discovery signal according to an embodiment of the present invention. The apparatus can be used to implement the above method, and therefore, the features in the above method embodiments can be applied to the present embodiment. The device can be the first UE. As shown in FIG. 4A, the apparatus provided in this embodiment includes: a determining module 41 and a transmitting module 42.
确定模块 41,用于在上一个 Discovery帧中的传输位置,确定第一 UE 在当前 Discovery帧中的传输位置; 所述装置设置于所述第一 UE内部或 所述装置为所述第一 UE。  a determining module 41, configured to determine, in a transmission location in a previous Discovery frame, a transmission location of the first UE in a current Discovery frame; the apparatus is disposed in the first UE or the apparatus is the first UE .
其中, 所述第一 UE在当前 Discovery帧中的传输位置仅根据所述第 一 UE在上一个 Discovery帧中的传输位置确定;如果所述第一 UE在上一 个 Discovery帧中的传输位置与第二 UE在上一个 Discovery帧中的传输位 置在时间上相同, 则所述第一 UE在当前 Discovery帧中的传输位置与所 述第二 UE在当前 Discovery帧中的传输位置在时间上不相同。 优选地, 为实现频率分集增益, 所述第一 UE在当前 Discovery帧中的传输位置与 所述第一 UE在上一个 Discovery帧中的传输位置在频率上不相同。 The transmission location of the first UE in the current Discovery frame is determined only according to the transmission location of the first UE in the previous Discovery frame; if the transmission location of the first UE in the previous Discovery frame is The transmission positions of the two UEs in the previous Discovery frame are the same in time, and the transmission position of the first UE in the current Discovery frame is different in time from the transmission position of the second UE in the current Discovery frame. Preferably, to implement the frequency diversity gain, the transmission position of the first UE in the current Discovery frame is The transmission positions of the first UE in the previous Discovery frame are different in frequency.
传输模块 42, 用于在所述第一 UE在当前 Discovery帧中的传输位置 上, 传输 Discovery信号。  The transmitting module 42 is configured to transmit a Discovery signal on a transmission location of the first UE in a current Discovery frame.
可选地, 如图 4B所示, 图 4A提供的装置还可包括: 检测模块 43。 检测模块, 用于在所述第一 UE在当前 Discovery帧中的传输位置上, 传 输 Discovery信号之前, 检测所述当前 Discovery帧之前的 N个 Discovery 帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所述第一 UE 对应的传输位置未被使用, 其中, N为非负整数。  Optionally, as shown in FIG. 4B, the apparatus provided in FIG. 4A may further include: a detecting module 43. a detecting module, configured to detect N Discovery frames before the current Discovery frame before transmitting the Discovery signal in the transmission position of the first UE in the current Discovery frame, and determine N before the current Discovery frame The transmission location corresponding to the first UE in the Discovery frame is not used, where N is a non-negative integer.
本实施例可参见图 1对应实施例中描述, 在此不再赘述。  This embodiment can be described in the corresponding embodiment of FIG. 1, and details are not described herein again.
可选地, 还存在以下实施例:  Optionally, the following embodiments are also present:
所述确定模块, 具体用于公式
Figure imgf000016_0001
( ( ait_1+e ) mod m, ( ci M+dj M+g ) mod n ) , 确定所述第一 UE在当前 Discovery帧中的传输 位置;
The determining module is specifically used for a formula
Figure imgf000016_0001
(( ai t _ 1+ e ) mod m, ( ci M+dj M+g ) mod n ) , determining a transmission location of the first UE in a current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一 UE在当前 Discovery帧的频 率单元的编号, 表示所述第一 UE在上一个 Discovery帧的频率单元的 编号, j为小于 n的非负整数, jt表示所述第一 UE在当前 Discovery帧的 时间单元的编号, 表示所述第一 UE在上一个 Discovery帧的时间单元 的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数,所述 n为 Discovery帧在时间上的 单元总数; 所述 mod为取模运算符号; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the frequency unit of the first UE in the current Discovery frame a number indicating a frequency unit of the first UE in the previous Discovery frame, j is a non-negative integer less than n, and j t represents a number of the time unit of the first UE in the current Discovery frame, indicating the number a number of time units of a UE in a previous Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is a total number of units of the Discovery frame in frequency, the n The total number of units in the time of the Discovery frame; the mod is a modulo operation symbol;
其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 优选地, 至少存在一个素数 p, 所述 p整除 a-l, 所述 p整除 m, 所述 p不整除 e,从而满足条件:所述第一 UE在当前 Discovery帧的频时位置, 与所述第二 UE在上一个 Discovery帧的频时位置在频率上不相同, 实现 频率分集增益。  Where a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime. Preferably, there is at least one prime number p, the p is divisible by a, the p is divisible by m, and the p is not divisible by e, so that the condition is met: the first UE is in the frequency position of the current Discovery frame, and the first The two UEs have different frequency in the frequency position of the previous Discovery frame, and the frequency diversity gain is realized.
本实施例可参见图 2对应实施例中描述, 在此不再赘述。  This embodiment can be described in the corresponding embodiment of FIG. 2, and details are not described herein again.
可选地, 还可存在实施例:  Optionally, there may also be embodiments:
所述确定模块,具体用于根据公式 (it, jt)= f(i
Figure imgf000016_0002
( a it1+b j t1+e ) mod m, (kc it_i+d j t_i+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置。
The determining module is specifically configured to use the formula (i t , j t )= f(i
Figure imgf000016_0002
( ai t1+ bj t1+ e ) mod m, (kc i t _i+dj t _i+g ) mod n ) , determining that the Discovery resource is currently The frequency position of the Discovery frame.
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一 UE在上一个 Discovery帧的频率 单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当前 Discovery帧的时间单元的编号, 表示所述第一 UE在上一个 Discovery 帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所 述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery 帧在时间上的单元总数; 所述 mod为取模运算符号; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame. a number indicating a frequency unit of the first UE in the previous Discovery frame, where j is a non-negative integer less than n, and j t represents a number of the time unit of the Discovery resource in the current Discovery frame, indicating the first UE The number of the time unit of the previous Discovery frame; the m and the n are both positive integers, and the m is divisible by the n; the m is the total number of units of the Discovery frame in frequency, and the n is a Discovery frame The total number of units in time; the mod is a modulo operation symbol;
其中, a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc禾口 m互素。  Where a, b, c, d, e, g are integers, a and k are mutually prime, d and k are mutually prime, c and m are mutually prime, ad-kbc and m are prime.
优选地, 至少存在一个素数 p , 所述 p整除 b, 所述 p整除 m, 所述 p整除 a-1, 所述 p不整除 e, 从而使第一 UE的频时位置满足以下条件: 所述第一 UE在当前 Discovery帧的频时位置, 与所述第二 UE在上一个 Discovery帧的频时位置在频率上不相同, 实现频率分集增益。  Preferably, there is at least one prime number p, the p is divisible by b, the p is divisible by m, the p is divisible by a-1, and the p is not divisible by e, so that the frequency-time position of the first UE satisfies the following conditions: The frequency-time position of the first UE in the current Discovery frame is different from the frequency-time position of the second UE in the previous Discovery frame, and the frequency diversity gain is implemented.
本实施例可参见图 3A对应实施例中描述, 在此不再赘述。 上述技术 方案中, 第一 UE在当前 Discovery帧的频时位置根据第一 UE在上一个 Discovery帧的传输位置确定, 与当前 Discovery帧的帧号禾卩 Discovery资 源的初始传输位置无关, 第一 UE只需知道第一 UE在上一个 Discovery 帧的传输位置, 不需要获取当前 Discovery帧的帧号和第一 UE的初始传 输位置, 就可确定第一 UE在当前帧的传输位置, 因此, 降低了系统的复 杂度, 减轻了第一 UE传输及侦听 Discovery信号的负担。 由于在每个 Discovery帧中,如果第一 UE在上一个 Discovery帧的传输位置与第二 UE 在上一个 Discovery帧的传输位置在时间上相同, 则第一 UE在当前  This embodiment can be described in the corresponding embodiment of FIG. 3A, and details are not described herein again. In the foregoing technical solution, the first UE determines the frequency position of the current Discovery frame according to the transmission position of the first UE in the previous Discovery frame, and is independent of the initial transmission position of the frame number of the current Discovery frame and the Discovery resource, the first UE. It is only necessary to know the transmission position of the first UE in the previous Discovery frame, and it is not necessary to acquire the frame number of the current Discovery frame and the initial transmission position of the first UE, so that the transmission position of the first UE in the current frame can be determined, and therefore, the transmission is reduced. The complexity of the system reduces the burden on the first UE to transmit and listen to the Discovery signal. In each Discovery frame, if the transmission location of the first UE in the previous Discovery frame is the same in time as the transmission location of the second UE in the previous Discovery frame, the first UE is currently
Discovery帧的传输位置与第二 UE在当前 Discovery帧的传输位置在时间 上不相同,并且第一 UE在当前 Discovery帧中的传输位置与第一 UE在上 一个 Discovery帧中的传输位置在频率上不相同, 使得每个 UE发送的 Discovery信号都可以被其它 UE发现。 The transmission location of the Discovery frame is different in time from the transmission location of the current UE in the current Discovery frame, and the transmission location of the first UE in the current Discovery frame and the transmission location of the first UE in the previous Discovery frame are in frequency. Different, so that the Discovery signal sent by each UE can be found by other UEs.
图 5为本发明实施例提供的一种第一 UE结构示意图。 该用户设备可 以用于实现上述方法, 因此, 上述方法实施例中的特征可以应用到本实施 例中。 如图 5所示, 第一 UE包括: 处理器 51和收发器 52。 FIG. 5 is a schematic structural diagram of a first UE according to an embodiment of the present invention. The user equipment can be used to implement the foregoing method. Therefore, the features in the foregoing method embodiments can be applied to the implementation. In the example. As shown in FIG. 5, the first UE includes: a processor 51 and a transceiver 52.
处理器 51, 用于在上一个 Discovery帧中的传输位置, 确定第一 UE 在当前 Discovery帧中的传输位置;  The processor 51 is configured to determine, in a transmission location in the last Discovery frame, a transmission location of the first UE in the current Discovery frame;
其中, 所述第一 UE在当前 Discovery帧中的传输位置仅根据所述第 一 UE在上一个 Discovery帧中的传输位置确定; 所述第一 UE在当前 Discovery帧中的传输位置与所述第一 UE在上一个 Discovery帧中的传输 位置在频率上不相同; 如果所述第一 UE在上一个 Discovery帧中的传输 位置与第二 UE在上一个 Discovery帧中的传输位置在时间上相同, 则所 述第一 UE在当前 Discovery帧中的传输位置与所述第二 UE在当前  The transmission location of the first UE in the current Discovery frame is determined only according to the transmission location of the first UE in the previous Discovery frame; the transmission location of the first UE in the current Discovery frame and the first The transmission position of a UE in the last Discovery frame is different in frequency; if the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame, And the transmission location of the first UE in the current Discovery frame and the second UE are currently
Discovery帧中的传输位置在时间上不相同; The transmission locations in the Discovery frame are not the same in time;
收发器 52,用于在所述第一 UE在当前 Discovery帧中的传输位置上, 传输 Discovery信号。  The transceiver 52 is configured to transmit a Discovery signal at a transmission location of the first UE in a current Discovery frame.
可选地, 所述处理器, 还用于在所述第一 UE在当前 Discovery帧中 的传输位置上, 传输 Discovery信号之前, 检测所述当前 Discovery帧之前 的 N个 Discovery帧, 确定在所述当前 Discovery帧之前的 N个 Discovery 帧中所述第一 UE对应的传输位置未被使用, 其中, N为非负整数。  Optionally, the processor is further configured to: before transmitting the Discovery signal in the transmission position of the first UE in the current Discovery frame, detecting N Discovery frames before the current Discovery frame, determining that The transmission location corresponding to the first UE in the N Discovery frames before the current Discovery frame is not used, where N is a non-negative integer.
本实施例中处理器和收发器的具体功能可参见图 1对应实施例中描 述, 在此不再赘述。  The specific functions of the processor and the transceiver in this embodiment are described in the corresponding embodiment of FIG. 1, and details are not described herein again.
可选地, 还存在以下实施例:  Optionally, the following embodiments are also present:
所述处理器, 具体用于根据公式 (it,
Figure imgf000018_0001
( ( ait_1+e ) mod m,
The processor is specifically configured according to a formula (i t ,
Figure imgf000018_0001
( ( ai t _ 1+ e ) mod m,
( ci M+dj M+g ) mod n ) , 确定所述第一 UE在当前 Discovery帧中的传输 位置; (ci M+dj M+g ) mod n ) , determining a transmission location of the first UE in a current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一 UE在当前 Discovery帧的频 率单元的编号, 表示所述第一 UE在上一个 Discovery帧的频率单元的 编号, j为小于 n的非负整数, jt表示所述第一 UE在当前 Discovery帧的 时间单元的编号, 表示所述第一 UE在上一个 Discovery帧的时间单元 的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数,所述 n为 Discovery帧在时间上的 单元总数; 所述 mod为取模运算符号; 其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 优 选地, 至少存在一个素数 p , 所述 p整除 a-l, 所述 p整除 m, 所述 p不 整除 e。 Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the frequency unit of the first UE in the current Discovery frame a number indicating a frequency unit of the first UE in the previous Discovery frame, j is a non-negative integer less than n, and j t represents a number of the time unit of the first UE in the current Discovery frame, indicating the number a number of time units of a UE in a previous Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is a total number of units of the Discovery frame in frequency, the n The total number of units in the time of the Discovery frame; the mod is a modulo operation symbol; Where a, c, d, e, f are integers, a and m are prime, c and n are prime, d and n are prime. Preferably, there is at least one prime number p, the p is divisible by a, the p is divisible by m, and the p is not divisible by e.
例中描述, 在此不再赘述。 The description is in the example and will not be described here.
Figure imgf000019_0001
Figure imgf000019_0001
所述处理器, 具体用于根据公式 (it, jt)= f(i t-i, j t-i)= ( ( a it_1+b j t_1+e ) mod m, (kc it_i+d j t-i+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置; The processor is specifically configured according to the formula (i t , j t )= f(i t -i, j ti)= ( ( ai t _ 1+ bj t _ 1+ e ) mod m, (kc i t _i+dj t -i+g ) mod n ) , determining a frequency time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一 UE在上一个 Discovery帧的频率 单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当前 Discovery帧的时间单元的编号, 表示所述第一 UE在上一个 Discovery 帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所 述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery 帧在时间上的单元总数; 所述 mod为取模运算符号; Where t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, and i t represents the number of the frequency unit of the Discovery resource in the current Discovery frame. a number indicating a frequency unit of the first UE in the previous Discovery frame, where j is a non-negative integer less than n, and j t represents a number of the time unit of the Discovery resource in the current Discovery frame, indicating the first UE The number of the time unit of the previous Discovery frame; the m and the n are both positive integers, and the m is divisible by the n; the m is the total number of units of the Discovery frame in frequency, and the n is a Discovery frame The total number of units in time; the mod is a modulo operation symbol;
其中 a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素。 优选地, 至少存在一个素数 p , 所述 p整除 b, 所述 p 整除 m, 所述 p整除 a-1, 所述 p不整除 e。  Where a, b, c, d, e, g are integers, a and k are prime, d and k are prime, c and m are prime, ad-kbc and m are prime. Preferably, there is at least one prime p, the p is divisible b, the p is divisible by m, the p is divisible by a-1, and the p is not divisible by e.
本实施例可参见图 3A对应实施例中描述, 在此不再赘述。  This embodiment can be described in the corresponding embodiment of FIG. 3A, and details are not described herein again.
上述技术方案中, 第一 UE在当前 Discovery帧的频时位置根据第一 UE在上一个 Discovery帧的传输位置确定, 与当前 Discovery帧的帧号和 Discovery资源的初始传输位置无关, 第一 UE只需知道第一 UE在上一个 Discovery帧的传输位置, 不需要获取当前 Discovery帧的帧号和第一 UE 的初始传输位置, 就可确定第一 UE在当前帧的传输位置, 因此, 降低了 系统的复杂度, 减轻了第一 UE传输及侦听 Discovery信号的负担。 由于 在每个 Discovery帧中, 如果第一 UE在上一个 Discovery帧的传输位置与 第二 UE在上一个 Discovery帧的传输位置在时间上相同,则第一 UE在当 前 Discovery帧的传输位置与第二 UE在当前 Discovery帧的传输位置在时 间上不相同,并且第一 UE在当前 Discovery帧中的传输位置与第一 UE在 上一个 Discovery帧中的传输位置在频率上不相同, 使得每个 UE发送的 Discovery信号都可以被其它 UE发现。 In the foregoing technical solution, the first UE is determined according to the transmission position of the first UE in the previous Discovery frame, and the first UE is only related to the frame number of the current Discovery frame and the initial transmission position of the Discovery resource. It is necessary to know that the first UE is in the transmission position of the previous Discovery frame, and it is not necessary to acquire the frame number of the current Discovery frame and the initial transmission position of the first UE, so that the transmission position of the first UE in the current frame can be determined, thereby reducing the system. The complexity reduces the burden on the first UE to transmit and listen to the Discovery signal. In each of the Discovery frames, if the transmission position of the first UE in the previous Discovery frame is the same as the transmission position of the second UE in the previous Discovery frame, the transmission position of the first UE in the current Discovery frame is The second UE is different in time in the transmission position of the current Discovery frame, and the transmission position of the first UE in the current Discovery frame is in the same position as the first UE. The transmission locations in the previous Discovery frame are not the same in frequency, so that the Discovery signal sent by each UE can be found by other UEs.
本领域普通技术人员可以理解: 实现上述各方法实施例的全部或部分 步骤可以通过程序指令相关的硬件来完成。 前述的程序可以存储于一计算 机可读取存储介质中。 该程序在执行时, 执行包括上述各方法实施例的步 骤; 而前述的存储介质包括: ROM、 RAM, 磁碟或者光盘等各种可以存 储程序代码的介质。  One of ordinary skill in the art will appreciate that all or part of the steps to implement the various method embodiments described above can be accomplished by hardware associated with the program instructions. The aforementioned program can be stored in a computer readable storage medium. The program, when executed, performs the steps including the above method embodiments; and the foregoing storage medium includes: a medium that can store program codes, such as a ROM, a RAM, a magnetic disk, or an optical disk.
最后应说明的是: 以上各实施例仅用以说明本发明的技术方案, 而非 对其限制; 尽管参照前述各实施例对本发明进行了详细的说明, 本领域的 普通技术人员应当理解: 其依然可以对前述各实施例所记载的技术方案进 行修改, 或者对其中部分或者全部技术特征进行等同替换; 而这些修改或 者替换, 并不使相应技术方案的本质脱离本发明各实施例技术方案的范 围。  It should be noted that the above embodiments are merely illustrative of the technical solutions of the present invention, and are not intended to be limiting; although the present invention has been described in detail with reference to the foregoing embodiments, those of ordinary skill in the art The technical solutions described in the foregoing embodiments may be modified, or some or all of the technical features may be equivalently replaced; and the modifications or substitutions do not deviate from the technical solutions of the embodiments of the present invention. range.

Claims

权 利 要 求 书 claims
1、 一种传输发现 Discovery信号的方法, 其特征在于, 包括: 根据第一用户设备在上一个 Discovery帧中的传输位置, 确定所述第 一 UE在当前 Discovery帧中的传输位置; 1. A method of transmitting a Discovery signal, characterized by comprising: determining the transmission position of the first UE in the current Discovery frame based on the transmission position of the first user equipment in the previous Discovery frame;
其中, 所述第一用户设备在当前 Discovery帧中的传输位置仅根据所 述第一用户设备在上一个 Discovery帧中的传输位置确定; 所述第一用户 设备在当前 Discovery帧中的传输位置与所述第一用户设备在上一个 Wherein, the transmission position of the first user equipment in the current Discovery frame is determined only based on the transmission position of the first user equipment in the previous Discovery frame; the transmission position of the first user equipment in the current Discovery frame is the same as The first user equipment is in the previous
Discovery帧中的传输位置在频率上不相同; 如果所述第一用户设备在上 一个 Discovery帧中的传输位置与第二用户设备在上一个 Discovery帧中的 传输位置在时间上相同, 则所述第一用户设备在当前 Discovery帧中的传 输位置与所述第二用户设备在当前 Discovery帧中的传输位置在时间上不 相同; The transmission positions in the Discovery frame are different in frequency; if the transmission position of the first user equipment in the previous Discovery frame is the same in time as the transmission position of the second user equipment in the previous Discovery frame, then the The transmission position of the first user equipment in the current Discovery frame is different in time from the transmission position of the second user equipment in the current Discovery frame;
在所述第一用户设备在当前 Discovery帧中的传输位置上, 传输 At the transmission position of the first user equipment in the current Discovery frame, transmit
Discovery信号。 Discovery signal.
2、 根据权利要求 1所述的方法, 其特征在于, 根据第一用户设备在 上一个 Discovery帧中的传输位置, 确定所述第一用户设备在当前 2. The method according to claim 1, characterized in that, according to the transmission position of the first user equipment in the previous Discovery frame, it is determined that the first user equipment is in the current
Discovery帧中的传输位置, 包括: The transmission position in the Discovery frame, including:
根据公式 (it, jt)=f(i ", j t—i)= ( ( ait-i+e ) mod m, ( ci t-i+dj t-i+g ) mod n ), 确定所述第一用户设备在当前 Discovery帧中的传输位置; According to the formula (it, jt)=f(i ", jt—i)= ( ( ait-i+e ) mod m, ( ci t -i+dj t -i+g ) mod n ), determine the The transmission position of the user equipment in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t- 1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一用户设备在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述第一用户设备在当前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Among them, t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, i t represents the frequency unit of the first user equipment in the current Discovery frame represents the number of the first user equipment in the frequency unit of the previous Discovery frame, j is a non-negative integer less than n, j t represents the number of the first user equipment in the time unit of the current Discovery frame, represents The first user equipment is in the previous
Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n 为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; The number of the time unit of the Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame in frequency, and the n is the Discovery frame in time The total number of units on; The mod is the modulo operation symbol;
其中, a, c, d, e,f为整数, a和 m互素, c和 n互素, d和 n互素。 Among them, a, c, d, e, f are integers, a and m are relatively prime, c and n are relatively prime, and d and n are relatively prime.
3、 根据权利要求 2所述的方法, 其特征在于, 至少存在一个素数 p , 所述 p整除 a- l, 所述 p整除 m, 所述 p不整除 e。 3. The method according to claim 2, characterized in that there is at least one prime number p, said p divides a-l, said p divides m, and p does not divide e.
4、 根据权利要求 1所述的方法, 其特征在于, 根据第一用户设备在 上一个 Discovery帧中的传输位置, 确定所述第一用户设备在当前 4. The method according to claim 1, characterized in that, according to the transmission position of the first user equipment in the previous Discovery frame, it is determined that the first user equipment is in the current
Discovery帧中的传输位置, 包括: The transmission position in the Discovery frame, including:
根据公式 (it, jt)= f(i ", j ")= ( ( a it-i+b j t-i+e ) mod m, (kc i"+d j "+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置; According to the formula (i t , j t )= f(i ", j ")= ( ( a it-i+bj t-i+e ) mod m, (kc i"+dj "+g ) mod n ) , Determine the frequency and time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当 前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; Among them, t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, i t represents the frequency unit number of the Discovery resource in the current Discovery frame , represents the number of the first user equipment in the frequency unit of the previous Discovery frame, j is a non-negative integer less than n, j t represents the number of the Discovery resource in the time unit of the current Discovery frame, represents the first The number of the user equipment in the time unit of the previous Discovery frame; The m and the n are both positive integers, and the m is evenly divided by the n; The m is the total number of units in the frequency of the Discovery frame, and the n is The total number of units in time of the Discovery frame; the mod is the modulo operation symbol;
其中, a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, a X d-k X b X c和 m互素。 Among them, a, b, c, d, e, g are integers, a and k are mutually prime, d and k are mutually prime, c and m are mutually prime, ad-kbc and m are mutually prime, gp, a X d-k X b X c and m are relatively prime.
5、 根据权利要求 4所述的方法, 其特征在于, 至少存在一个素数 p , 所述 p整除 b, 所述 p整除 m, 所述 p整除 a-1, 所述 p不整除 e。 5. The method according to claim 4, characterized in that there is at least one prime number p, said p divides b, p divides m, p divides a-1, and p does not divide e.
6、 根据权利要求 1至 5中任一项所述的方法, 其特征在于, 在所述 第一用户设备在当前 Discovery帧中的传输位置上,传输 Discovery信号之 前, 所述方法还包括: 6. The method according to any one of claims 1 to 5, characterized in that, before transmitting the Discovery signal at the transmission position of the first user equipment in the current Discovery frame, the method further includes:
所述第一用户设备检测所述当前 Discovery帧之前的 N个 Discovery 帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所述第一用 户设备对应的传输位置未被使用, 其中, N为非负整数。 The first user equipment detects N Discovery frames before the current Discovery frame, and determines that the transmission location corresponding to the first user equipment in the N Discovery frames before the current Discovery frame is not used, where, N is a non-negative integer.
7、 一种传输 Discovery信号的装置, 其特征在于, 包括: 7. A device for transmitting Discovery signals, characterized by including:
确定模块, 用于在上一个 Discovery帧中的传输位置, 确定第一用户 设备在当前 Discovery帧中的传输位置; 所述装置设置于所述第一用户设 备内部或所述装置为所述第一用户设备; Determining module, used to determine the transmission position of the first user equipment in the current Discovery frame based on the transmission position in the previous Discovery frame; the device is provided inside the first user equipment or the device is the first user equipment User equipment;
其中, 所述第一用户设备在当前 Discovery帧中的传输位置仅根据所 述第一用户设备在上一个 Discovery帧中的传输位置确定; 所述第一用户 设备在当前 Discovery帧中的传输位置与所述第一用户设备在上一个 Discovery帧中的传输位置在频率上不相同; 如果所述第一用户设备在上 一个 Discovery帧中的传输位置与第二用户设备在上一个 Discovery帧中的 传输位置在时间上相同, 则所述第一用户设备在当前 Discovery帧中的传 输位置与所述第二用户设备在当前 Discovery帧中的传输位置在时间上不 相同; Wherein, the transmission position of the first user equipment in the current Discovery frame is determined only based on the transmission position of the first user equipment in the previous Discovery frame; the transmission position of the first user equipment in the current Discovery frame is the same as The first user equipment is in the previous The transmission positions in the Discovery frame are different in frequency; if the transmission position of the first user equipment in the previous Discovery frame is the same in time as the transmission position of the second user equipment in the previous Discovery frame, then the The transmission position of the first user equipment in the current Discovery frame is different in time from the transmission position of the second user equipment in the current Discovery frame;
传输模块, 用于在所述第一用户设备在当前 Discovery帧中的传输位 置上, 传输 Discovery信号。 A transmission module, configured to transmit a Discovery signal at the transmission position of the first user equipment in the current Discovery frame.
8、 根据权利要求 7所述的装置, 其特征在于, 所述确定模块, 具体 用于根据公式 (it, jt)=f(i ", j ")= ( ( ait-i+e ) mod m, ( ci t-i+dj t-i+g ) mod n ), 确定所述第一用户设备在当前 Discovery帧中的传输位置; 8. The device according to claim 7, characterized in that the determination module is specifically used to determine according to the formula (i t , j t )=f (i ", j ")= ( ( ait-i+e ) mod m, ( ci t -i+dj t -i+g ) mod n ), determine the transmission position of the first user equipment in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一用户设备在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述第一用户设备在当前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Among them, t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, i t represents the frequency unit of the first user equipment in the current Discovery frame represents the number of the first user equipment in the frequency unit of the previous Discovery frame, j is a non-negative integer less than n, j t represents the number of the first user equipment in the time unit of the current Discovery frame, represents The first user equipment is in the previous
Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n 为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; The number of the time unit of the Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame in frequency, and the n is the Discovery frame in time The total number of units on; The mod is the modulo operation symbol;
其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 Among them, a, c, d, e, f are integers, a and m are relatively prime, c and n are relatively prime, and d and n are relatively prime.
9、 根据权利要求 8所述的装置, 其特征在于, 至少存在一个素数 p , 所述 p整除 a-l, 所述 p整除 m, 所述 p不整除 e。 9. The device according to claim 8, characterized in that there is at least one prime number p, said p divides a-l, said p divides m, and p does not divide e.
10、 根据权利要求 7所述的装置, 其特征在于, 所述确定模块, 具体 用于卞艮据公式 (it, jt)= f(i t-i, j t-i)= ( ( a it_i+b j t_i+e ) mod m, (kc it_i+d j t_i+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位置; 10. The device according to claim 7, characterized in that the determination module is specifically used to determine according to the formula ( it , jt )=f(iti, jti)=(( ait_i + bj t _i+e ) mod m, (kc i t _i+dj t _i+g ) mod n ), determine the frequency-time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当 前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; Among them, t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, i t represents the frequency unit number of the Discovery resource in the current Discovery frame , represents the number of the first user equipment in the frequency unit of the previous Discovery frame, j is a non-negative integer less than n, j t represents the number of the Discovery resource in the time unit of the current Discovery frame, represents the first The number of the user equipment in the time unit of the previous Discovery frame; the m and the n are both positive integers, and the m is an even divisor of n; m is the total number of units in the Discovery frame in frequency, and n is the total number of units in the Discovery frame in time; the mod is the modulo operation sign;
其中, a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, a X d-k X b X c和 m互素。 Among them, a, b, c, d, e, g are integers, a and k are mutually prime, d and k are mutually prime, c and m are mutually prime, ad-kbc and m are mutually prime, gp, a X d-k X b X c and m are relatively prime.
11、根据权利要求 10所述的装置,其特征在于,至少存在一个素数 p , 所述 p整除 b, 所述 p整除 m, 所述 p整除 a-1, 所述 p不整除 e。 11. The device according to claim 10, characterized in that there is at least one prime number p, the p is an integer division of b, the p is an integer division of m, the p is an integer division of a-1, and the p is not an integer division of e.
12、 根据权利要求 8至 11任一项所述的装置, 其特征在于, 还包括: 检测模块, 用于在所述第一用户设备在当前 Discovery帧中的传输位 置上, 传输 Discovery信号之前, 检测所述当前 Discovery帧之前的 N个 Discovery帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所 述第一用户设备对应的传输位置未被使用, 其中, N为非负整数。 12. The device according to any one of claims 8 to 11, further comprising: a detection module, configured to transmit a Discovery signal at the transmission position of the first user equipment in the current Discovery frame, Detect N Discovery frames before the current Discovery frame, and determine that the transmission location corresponding to the first user equipment in the N Discovery frames before the current Discovery frame is not used, where N is a non-negative integer.
13、 一种第一用户设备, 其特征在于, 包括: 13. A first user equipment, characterized by: including:
处理器, 用于在上一个 Discovery帧中的传输位置, 确定第一用户设 备在当前 Discovery帧中的传输位置; The processor is used to determine the transmission position of the first user equipment in the current Discovery frame based on the transmission position in the previous Discovery frame;
其中, 所述第一用户设备在当前 Discovery帧中的传输位置仅根据所 述第一用户设备在上一个 Discovery帧中的传输位置确定; 所述第一用户 设备在当前 Discovery帧中的传输位置与所述第一用户设备在上一个 Wherein, the transmission position of the first user equipment in the current Discovery frame is determined only based on the transmission position of the first user equipment in the previous Discovery frame; the transmission position of the first user equipment in the current Discovery frame is the same as The first user equipment is in the previous
Discovery帧中的传输位置在频率上不相同; 如果所述第一用户设备在上 一个 Discovery帧中的传输位置与第二用户设备在上一个 Discovery帧中的 传输位置在时间上相同, 则所述第一用户设备在当前 Discovery帧中的传 输位置与所述第二用户设备在当前 Discovery帧中的传输位置在时间上不 相同; The transmission positions in the Discovery frame are different in frequency; if the transmission position of the first user equipment in the previous Discovery frame is the same in time as the transmission position of the second user equipment in the previous Discovery frame, then the The transmission position of the first user equipment in the current Discovery frame is different in time from the transmission position of the second user equipment in the current Discovery frame;
收发器, 用于在所述第一用户设备在当前 Discovery帧中的传输位置 上, 传输 Discovery信号。 A transceiver, configured to transmit a Discovery signal at the transmission position of the first user equipment in the current Discovery frame.
14、 根据权利要求 13所述的用户设备, 其特征在于, 所述处理器, 具体用于根据公式 (it, jt)=f(i ", j ")= ( ( ait_!+e ) mod m, ( ci M+dj ^+g ) mod n ) , 确定所述第一用户设备在当前 Discovery帧中的传输位置; 14. The user equipment according to claim 13, characterized in that the processor is specifically configured to perform according to the formula (i t , j t )=f (i ", j ")= ( ( ai t _!+ e) mod m, (ci M+dj ^+g) mod n), determine the transmission position of the first user equipment in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述第一用户设备在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述第一用户设备在当前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Among them, t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, i t represents the frequency unit of the first user equipment in the current Discovery frame The number represents the number of the first user equipment in the previous Discovery frame. The number of the frequency unit, j is a non-negative integer less than n, j t represents the number of the first user equipment in the time unit of the current Discovery frame, represents the number of the first user equipment in the previous
Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m小于等于所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n 为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; The number of the time unit of the Discovery frame; the m and the n are both positive integers, and the m is less than or equal to the n; the m is the total number of units of the Discovery frame in frequency, and the n is the Discovery frame in time The total number of units on; The mod is the modulo operation symbol;
其中, a, c, d, e, f为整数, a和 m互素, c和 n互素, d和 n互素。 Among them, a, c, d, e, f are integers, a and m are relatively prime, c and n are relatively prime, and d and n are relatively prime.
15、 根据权利要求 14所述的用户设备, 其特征在于, 至少存在一个 素数 p , 所述 p整除 a-1, 所述 p整除 m, 所述 p不整除 e。 15. The user equipment according to claim 14, characterized in that there is at least one prime number p, the p is an integer divisor of a-1, the p is an integer divisor of m, and the p is not an integer divisor of e.
16、 根据权利要求 13所述的用户设备, 其特征在于, 所述处理器, 具体用于卞艮据公式 (it, jt)= f(i t-i, j t-i)= ( ( a it_i+b j t_i+e ) mod m, (kc it_i+d j t-i+g ) mod n ) , 确定所述 Discovery资源在当前 Discovery帧的频时位 直; 16. The user equipment according to claim 13, characterized in that the processor is specifically configured to calculate (i t , j t ) = f (i t , j t ) = ( ( ai t _i +bj t _i+e ) mod m, (kc i t _i+dj t-i+g ) mod n ) , determine the frequency and time position of the Discovery resource in the current Discovery frame;
其中, t为当前 Discovery帧的帧号, t-1为上一个 Discovery帧的帧号, 所述 i为小于 m的非负整数, it表示所述 Discovery资源在当前 Discovery 帧的频率单元的编号, 表示所述第一用户设备在上一个 Discovery帧的 频率单元的编号, j为小于 n的非负整数, jt表示所述 Discovery资源在当 前 Discovery帧的时间单元的编号, 表示所述第一用户设备在上一个 Discovery帧的时间单元的编号; 所述 m和所述 n均是正整数, 并且所述 m整除所述 n; 所述 m为 Discovery帧在频率上的单元总数, 所述 n为 Discovery帧在时间上的单元总数; 所述 mod为取模运算符号; Among them, t is the frame number of the current Discovery frame, t-1 is the frame number of the previous Discovery frame, the i is a non-negative integer less than m, i t represents the frequency unit number of the Discovery resource in the current Discovery frame , represents the number of the first user equipment in the frequency unit of the previous Discovery frame, j is a non-negative integer less than n, j t represents the number of the Discovery resource in the time unit of the current Discovery frame, represents the first The number of the user equipment in the time unit of the previous Discovery frame; The m and the n are both positive integers, and the m is evenly divided by the n; The m is the total number of units in the frequency of the Discovery frame, and the n is The total number of units in time of the Discovery frame; the mod is the modulo operation symbol;
其中, a, b, c, d, e, g为整数, a和 k 互素, d和 k 互素, c和 m互素, ad-kbc和 m互素, gp, aX d-k X b X c和 m互素。 Among them, a, b, c, d, e, g are integers, a and k are mutually prime, d and k are mutually prime, c and m are mutually prime, ad-kbc and m are mutually prime, gp, aX d-k X b X c and m are relatively prime.
17、 根据权利要求 16所述的用户设备, 其特征在于, 至少存在一个 素数 p , 所述 p整除 b, 所述 p整除 m, 所述 p整除 a-1, 所述 p不整除 e。 17. The user equipment according to claim 16, characterized in that there is at least one prime number p, the p is evenly divided by b, the p is divided by m, the p is divided by a-1, and the p is not divided by e.
18、 根据权利要求 13至 17任一项所述的用户设备, 其特征在于, 所 述处理器, 还用于在所述第一用户设备在当前 Discovery帧中的传输位置 上, 传输 Discovery 信号之前, 检测所述当前 Discovery 帧之前的 N个 Discovery帧, 确定在所述当前 Discovery帧之前的 N个 Discovery帧中所 述第一用户设备对应的传输位置未被使用, 其中, N为非负整数。 18. The user equipment according to any one of claims 13 to 17, characterized in that the processor is further configured to transmit a Discovery signal at the transmission position of the first user equipment in the current Discovery frame. , detect N Discovery frames before the current Discovery frame, and determine that the transmission position corresponding to the first user equipment in the N Discovery frames before the current Discovery frame is not used, where N is a non-negative integer.
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