CN101854727B - Method for configuring resource mapping indication information - Google Patents

Method for configuring resource mapping indication information Download PDF

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CN101854727B
CN101854727B CN 200910130558 CN200910130558A CN101854727B CN 101854727 B CN101854727 B CN 101854727B CN 200910130558 CN200910130558 CN 200910130558 CN 200910130558 A CN200910130558 A CN 200910130558A CN 101854727 B CN101854727 B CN 101854727B
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pru
fps
parameter
indication
bandwidth
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CN101854727A (en
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宁丁
关艳峰
刘向宇
刘颖
方惠英
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ZTE Corp
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Priority to PCT/CN2009/073922 priority patent/WO2010111859A1/en
Priority to PCT/CN2010/071554 priority patent/WO2010111974A1/en
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    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04W—WIRELESS COMMUNICATION NETWORKS
    • H04W28/00—Network traffic management; Network resource management
    • H04W28/02—Traffic management, e.g. flow control or congestion control
    • H04W28/06—Optimizing the usage of the radio link, e.g. header compression, information sizing, discarding information
    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04W—WIRELESS COMMUNICATION NETWORKS
    • H04W72/00—Local resource management

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Abstract

The invention discloses a method for configuring resource mapping indication information. The method comprises the following steps of: indicating at least one parameter for the resource mapping; and determining the bit number needed for indicating the parameter according to bandwidth, wherein the bit number needed for indicating the parameter is partially same or completely different as for a plurality of different bandwidths. By means of the technical scheme of the invention, the bit number needed for indicating the parameter is configured for each bandwidth supported by a system and the bit number for indicating the same parameter at different bandwidths are partially same or completely different, so the bit number used by a physical resource mapping indication signaling can flexibly change along with the bandwidth used by the system so as to reduce the bit number transmitted; therefore, the method avoids the problem of high expenditure of a control channel in the related technology and saves the downlink control expenditure on the premise of not affecting the normal running of the system so as to improve the working efficiency of the system.

Description

The collocation method of resource mapping indication information
Technical field
The present invention relates to the communications field, relate in particular to a kind of collocation method of resource mapping indication information.
Background technology
In wireless communication system, the base station typically refers in certain area of radio coverage can believe the radio station by the transceiving that mobile switching center and the terminal information of carrying out is transmitted.In actual applications, the base station can communicate by Uplink/Downlink and terminal, and wherein, down link refers to that the base station arrives the transmission direction of terminal, and up link refers to the transmission direction of terminal to base station.And a plurality of terminals can send data simultaneously by up link to the base station, also can be by down link simultaneously from the base station receive data.In addition, between base station and terminal, can carry out relaying to the data of transmission by relay station.
In the wireless communication system that adopts the control of base station realization wireless resource scheduling, the dispatching distribution of system wireless resource is finished by the base station.Ascending resource assignment information of required use when employed downlink resource assignment information and terminal are carried out uplink in the time of for example, can being provided this base station and carried out downlink transfer by the base station etc.
In the wireless communication system of commercialization, the base station is when the Radio Resource that scheduling is eated dishes without rice or wine, usually with a radio frames as a dispatching cycle, and Radio Resource (for example is divided into several radio resource unit, can be with a time slot or a code word as a Resource Unit) to dispatch, the base station provides data or multimedia service by the scheduling radio resource unit to the terminal of its covering.
Particularly, at second generation wireless communication system (for example, at global system for mobile communications (Global System for Mobile communication, referred to as GSM)) in, the base station is divided into time division multiple access (Time Division Multiple Address take 4.615ms as the cycle with the Radio Resource on each frequency, referred to as TDMA) radio frames, each radio frames comprises 8 time slots, time slot transmits the speech channel of a full rate or transmits the speech channel of two half rates, and also can be used for realizing the data service of low speed; In 2.5 generation wireless communication systems (for example, at wireless universal packet service (General Packet RadioService is referred to as GPRS)), by introducing the speed that can continue to promote data service based on the fixing packet switching of time slot; And at the third generation wireless communication system (for example, at TD SDMA (Time-Division Synchronous Code DivisionMultiple Address, referred to as TD-SCDMA) in, the base station can be divided into the Radio Resource of eating dishes without rice or wine the radio frames take 10ms as the cycle equally, each 10ms comprises 14 conventional time slots and 6 special time slots, conventional time slot is used for transmitting concrete business and signaling, on each conventional time slot, the user is distinguished by different code words in the base station.
Can find out by above description, GSM and TD-SCDMA system mainly adopt TDMA and/or CDMA technology, and these technology are based on time slot and code word is carried out resource mapping and resource is distributed, and processing procedure is fairly simple.
Based on OFDM (Orthogonal Frequency DivisionMultiplexing, referred to as OFDM) and OFDM (Orthogonal FrequencyDivision Multiple Address, referred to as OFDMA) in the communication system of technology, for example, at Long Term Evolution (Long Term Evolution, referred to as LTE), Ultra-Mobile Broadband (Ultra Mobile Broadband, referred to as UMB) and the wireless communication system such as IEEE 802.16m in, although Radio Resource also is divided framing and manages, but each OFDMA symbol comprises a plurality of mutually orthogonal subcarriers, and terminal takies the parton carrier wave usually, thereby can adopt partial frequency multiplexing (Fractional Frequency Reuse, referred to as FFR) etc. technology reduce interference, improve to cover; Secondly, because wireless channel environment changes frequent, the base station is in order to obtain the gain of frequency diversity gain and frequency selective scheduling, the available physical subcarrier is divided into physical resource unit (Physical Resource Unit, referred to as PRU), and then physical resource unit is mapped as continuous resource unit (ContiguousResource Unit, referred to as CRU) and distributed resource unit (Distributed ResourceUnit, referred to as DRU), to improve transmission performance, wherein, the subcarrier in the continuous resource unit is all continuous, and the subcarrier in the distributed resource unit is fully discontinuous or not exclusively continuous; In addition, along with frequency resource is day by day rare, multiple different bandwidth (for example, 5MHz, 10MHz or 20MHz) or multi-carrier operation need to be supported in the base station, to utilize different frequency resources and to satisfy the demand of different operators.
Owing to above reason, resource mapping process based on the wireless communication system of OFDM or OFDMA technology is more complicated, thereby cause for the indication signaling expense of control resource mapping process greatlyr, the resource allocation information of terminal parses base station increases with the complexity of process of determining its reception and sending the physical resource position of data.
As seen, consider that the resource mapping process of the wireless communication system of OFDM or OFDMA technology will more complicated, in order to reduce the indication signaling expense of its resource mapping, optimization system information management and transmission method need to carry out rational resource mapping.In order to ensure the efficient of wireless communication system, the base station is mapped as the Radio Resource of physics the Radio Resource of logic usually, for example, physical sub-carrier is mapped as the logical resource unit, and the scheduling of Radio Resource is realized in the base station by the scheduling logic Resource Unit.
Particularly, for the wireless communication system based on OFDM or OFDMA, frame structure and the resource structures of its this wireless communication system of Radio Resource mapping Main Basis, frame structure is described the control structure of Radio Resource on time domain, and resource structures has been described the control structure of Radio Resource on frequency domain.Frame structure is divided into Radio Resource the unit of different brackets in time domain, such as superframe (Superframe), frame (Frame), subframe (Subframe) and symbol (Symbol), by being set, different control channel (for example, broadcast channel, clean culture and Multicast Channel etc.) realizes scheduling controlling.
For example, shown in Figure 1, Radio Resource is divided into superframe in time domain, each superframe comprises 4 frames, each frame comprises 8 subframes, subframe is comprised of 6 basic OFDMA symbols, and what OFDMA symbols are the factors such as circulating prefix-length of the bandwidth that real system is supported as required and/or OFDMA symbol determine specifically to comprise in each grade unit in the frame structure; In addition, system arranges broadcast channel (owing to being positioned at the superframe head, being also referred to as superframe head (Superframe Header)) in can first descending sub frame in superframe and sends the system information such as resource mapping; And system can also arrange clean culture and/control channel of multicast character sends the scheduling control informations such as resource distribution.
According to factors such as networking technology, interference mitigation technology and types of service, resource structures is divided into a plurality of frequency partition (Frequency Partition with bandwidth available on the frequency domain, referred to as FP), and then the frequency resource in the frequency partition is divided into the continuous resource unit and/or the distributed resource unit is dispatched.For example, as shown in Figure 2, the available physical subcarrier of a subframe is divided into 3 frequency partition, and each frequency partition is divided into continuous resource and distributed resource unit, and the continuous resource unit is used for frequency selective scheduling, and the distributed resource unit is used for frequency diversity scheduling.
Method for mapping resource generally need to be supported 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz system bandwidth (abbreviation bandwidth); and when the part when not considering multi-carrier operation protection subcarrier is used for mapping PRU under 5MHz, 7MHz, 8.75MHz, 10MHz and the 20MHz bandwidth; the number of corresponding PRU is 24,48,48,48 and 96; therefore, the indication parameter of resource mapping is different under the different system bandwidths.For example, bandwidth is that each subframe of system of 5MHz has 24 PRU, when 4 PRU form a subband (Subband), 6 Subband are then arranged at most, and there are 48 PRU in 7MHz, 8.75MHz or 10MHz system, 12 Subband are then arranged at most, just need to carry out different settings to pare down expenses to subband allotment (Subband Allocation Count is referred to as SAC) to these two kinds of systems.The following execution mode that illustrates resource mapping according to different bandwidth:
The downlink resource mapping process generally includes: sub-band division (Subband Partitioning), little band displacement (Miniband Permutation), frequency partition is divided (FrequencyPartitioning), continuous resource unit/distributed resource unit distributes (Contigous ResourceUnit/Distributed Resource Unit Allocation, referred to as CRU/DRUAllocation) and subcarrier permutation (Subcarrier Permutation), comprise in the ascending resource mapping process: sub-band division, little band displacement, frequency partition is divided, continuous resource unit/distributed resource unit distributes and Tile displacement (Tile Permutation).Subband is comprised of N1 continuous PRU, N1=4 for example, and Miniband is comprised of N2 continuous PRU, for example N2=1.Concrete, as shown in Figure 3, the descending sub frame resource mapping process of 5MHz OFDMA system has been described.Wherein, it is 512 that the FFT of 5MHz system counts, usable data subcarriers is 432 in the subframe, be divided into into N=24 PRU, each PRU size is 18x6, and namely frequency domain is 18 carrier waves, and time domain is 6 symbols, owing to reasons such as Cyclic Prefix, superframe head, transfer point and control channels, the number of symbols on the time domain may be 5 or 7.
Below introduce each step of physical resource mapping as an example of the 5MHz bandwidth example:
(1) sub-band division, that is, a part of PRU is mapped as Subband as unit extracts take a Subband.The number of descending Subband and the number of up Subband are respectively by descending allocation of subbands number (Downlink Subband Allocation Count, referred to as DSAC) and two parameter indications of up allocation of subbands number (Uplink Subband Allocation Count is referred to as USAC).Take descending 5MHz bandwidth as example, when the number of the descending Subband of DSAC indication was 3,12 PRU were mapped as 3 Subband.As shown in Figure 4, Subband Partitioning is indicated with DSAC in the base station, obtains PRU SB(the shadow-free part among the figure) is Miniband with remaining part mapping, such as the PRU among the figure MB(dash area among the figure).
(2) little band displacement, that is, the PRU that will not be mapped as Subband is mapped as Miniband.Number at the descending Subband of 5MHz bandwidth and DSAC indication is 3 o'clock, has 12 PRU to be mapped as Miniband, as shown in Figure 5, these 12 PRU is replaced.This step does not need the additional parameter indication, finishes according to DSAC.Up little band displacement USAC finishes.
(3) frequency partition is divided, and, the Subband that divided and Miniband after the displacement is divided into each frequency partition that is.This step needs two parameters, a parameter is used to indicate each frequency partition number, size and/or ratio, descending and up respectively by downstream frequency zone configuration (Downlink Frequency Partition Configuration, referred to as DFPC) and upstream frequency zone configuration (Uplink Frequency PartitionConfiguration is referred to as UFPC) indication; Another parameter then be used to indicate except first frequency partition (be FP 0) number of Subband in addition the frequency partition, descending and up respectively by downstream frequency subregion allocation of subbands number (Downlink FrequencyPartition Subband Count, referred to as DFPSC) and upstream frequency subregion allocation of subbands number (Uplink Frequency Partition Subband Count is referred to as DFPSC) indication.Under the size of the frequency partition except first frequency partition condition equal with the number of the Subband that comprises, can carry out certain redundancy and delete, remove some impossible values.Fig. 6 shows that removing first frequency partition size is 24 PRU, and other frequency partition size is 0, and other frequency partition Subband number is 0 frequency partition dividing condition.
(4) continuous resource unit/distributed resource unit distributes, that is, each frequency partition is carried out separately continuous resource unit/distributed resource unit distribute.The downstream frequency subregion is by descending continuous resource unit allocated size (Downlink CRU Allocation Size, referred to as DCAS) indication, the upstream frequency subregion is indicated by up continuous resource unit allocated size (UplinkCRU Allocation Size is referred to as UCAS).If the size of some frequency partition is 0, then can not need to carry this parameter.As shown in Figure 3, when system bandwidth is 5MHz, the number of the descending Subband of DSAC indication is 3, first frequency partition size is 24 PRU, other frequency partition size is 0, first frequency partition CRU number is 12 schematic diagram, and wherein last shadow-free that lists partly represents CRU, and shade then represents DRU.Need to prove, can indicate whether Subband acquiescence as CRU by 1 bit, and with the Miniband acquiescence as DRU, at this moment, can not send DCAS or UCAS further pares down expenses.
(5) subcarrier permutation or Tile displacement, that is, and to carrying out the displacement of subcarrier for the PRU that is mapped as DRU in each frequency partition in the descending sub frame, to carrying out the Tile displacement for the PRU that is mapped as DRU in each frequency partition in the sub-frame of uplink.
Fig. 3 illustrates the instantiation that the 5MHz bandwidth situation has been descended the physical resource mapping, comprising the process of continuous resource unit/distributed resource unit distribution.Fig. 4 to Fig. 6 shows the processing procedure that sub-band division is divided to frequency partition, for the situation of resource mapping under other bandwidth more clearly is described, concrete mapping situation when Fig. 7 shows 10MHz (also comprising 7MHz, 8.75MHz) bandwidth, wherein, the Subband number is 5, and has 4 frequency partition, and each frequency partition size is 12 PRU, first frequency partition comprises 8 CRU and 4 DRU, and other frequency partition all comprises 4 CRU and 8 DRU.Fig. 8 shows the situation of the concrete mapping under the 20M bandwidth situation.
In Fig. 3～Fig. 8, PRU SBRefer to the PRU for Subband, PRU MBRefer to the PRU for Miniband, PPRU MBRefer to the PRU through Miniband Permutation.
Can find out by above description, in the resource mapping process, after bandwidth is determined, still need to determine other some parameters (for example, the Subband on the definite Subband number of needs, frequency partition number, each frequency partition and CRU number etc.).
In communication system, resource mapping indication information is all given terminal by the base station by broadcast channel or superframe hair, and terminal determines to receive and/or send the resource location of data according to resource mapping indication information and resource allocation information.Resource mapping indication information has been indicated division and the mapping of frequency resource, specifically can comprise following information: the number of the number of the number that the number that descending allocation of subbands number, up allocation of subbands number, downstream frequency zone configuration, upstream frequency zone configuration, downstream frequency subregion allocation of subbands number, upstream frequency subregion allocation of subbands number, descending continuous resource unit distribute, up continuous resource unit distribute, descending continuous resource unit based on Miniband, up continuous resource unit based on Miniband.
Because concrete resource mapping process is a lot, therefore, the setting of above-mentioned indication parameter has stronger flexibility, and needed bit number increases but this can cause indicating these parameters simultaneously, and then increased in the control channel expense of transmission during these bit waste large volumes of channels resource.Channel overhead for parameter indication and transmission in the correlation technique is large, the problem of waste system resource, not yet proposes at present effective solution.
Summary of the invention
Consider exist in the correlation technique problem and the present invention is proposed, for this reason, main purpose of the present invention is to provide a kind of collocation method of resource mapping indication information, to address the above problem.
A kind of collocation method of resource mapping indication information is provided according to an aspect of the present invention.
Collocation method according to resource mapping indication information of the present invention comprises: at least one parameter of indexed resource mapping, determine the required bit number of the described parameter of indication according to bandwidth; Wherein, for a plurality of different bandwidth, indicate the each other identical or complete difference of part of the required bit number of described parameter.
By means of above-mentioned at least one technical scheme of the present invention, the required bit number of each band width configuration indication parameter for system's support, and the bit number part of the indicating identical or complete difference of same parameters under different bandwidth, so that the bit number that physical resource mapping indication signaling uses can change flexibly according to the bandwidth that system uses, reduce as much as possible the bit number of transmission, avoided the large problem of control channel expense in the correlation technique, under the prerequisite that does not affect the normal running of system, save descending control overhead, thereby improve the operating efficiency of system.
Description of drawings
Accompanying drawing is used to provide a further understanding of the present invention, and consists of the part of specification, is used for together with embodiments of the present invention explaining the present invention, is not construed as limiting the invention.In the accompanying drawings:
Fig. 1 is the frame structure schematic diagram according to the wireless communication system of correlation technique;
Fig. 2 is the resource structures schematic diagram according to the wireless communication system of correlation technique;
Fig. 3 is the resource mapping process schematic diagram according to wireless communication system under the 5MHz bandwidth situation of correlation technique;
Fig. 4 is the sub-band division process schematic diagram according to wireless communication system under the 5MHz bandwidth situation of correlation technique;
Fig. 5 is the little band replacement process schematic diagram according to wireless communication system under the 5MHz bandwidth situation of correlation technique;
Fig. 6 divides schematic diagram according to the frequency partition of wireless communication system under the 5MHz bandwidth situation of correlation technique;
Fig. 7 is the resource mapping process schematic diagram of wireless communication system under 10MHz (can be 7MHz or the 8.75MHz) bandwidth situation according to correlation technique;
Fig. 8 is the schematic diagram according to the resource mapping process of wireless communication system under the 20MHz bandwidth situation of correlation technique;
Fig. 9 is the flow chart according to the collocation method of the resource mapping indication information of the embodiment of the invention;
Figure 10 is the collocation method of the resource mapping indication information of the embodiment of the invention of application schematic diagram signaling DSAC when adopting the bit indication parameter of varying number according to to(for) the 5MHz system bandwidth;
Figure 11 is the collocation method of the resource mapping indication information of the embodiment of the invention of application schematic diagram signaling USAC when adopting the bit indication parameter of varying number according to to(for) the 5MHz system bandwidth;
Figure 12 is the collocation method of the resource mapping indication information of the embodiment of the invention of application schematic diagram signaling DFPC when adopting the bit indication parameter of varying number according to to(for) the 10MHz system bandwidth;
Figure 13 is the collocation method of the resource mapping indication information of the embodiment of the invention of application schematic diagram signaling UFPC when adopting the bit indication parameter of varying number according to to(for) the 10MHz system bandwidth;
Figure 14 is the collocation method of the resource mapping indication information of the embodiment of the invention of application schematic diagram signaling DFPSC when adopting the bit indication parameter of varying number according to to(for) the 10MHz system bandwidth;
Figure 15 is the collocation method of the resource mapping indication information of the embodiment of the invention of application schematic diagram signaling UFPSC when adopting the bit indication parameter of varying number according to to(for) the 10MHz system bandwidth;
Figure 16 is signaling DCAS when adopting the bit indication parameter of varying number according to the collocation method of the resource mapping indication information of the embodiment of the invention for the 10MHz system bandwidth SBiThe application schematic diagram.
Figure 17 is signaling UCAS when adopting the bit indication parameter of varying number according to the collocation method of the resource mapping indication information of the embodiment of the invention for the 10MHz system bandwidth SBiThe application schematic diagram.
Figure 18 is signaling DCAS when adopting the bit indication parameter of varying number according to the collocation method of the resource mapping indication information of the embodiment of the invention for the 5MHz system bandwidth MB ÷The application schematic diagram.
Figure 19 is signaling UCAS when adopting the bit indication parameter of varying number according to the collocation method of the resource mapping indication information of the embodiment of the invention for the 5MHz system bandwidth MBThe application schematic diagram.
Embodiment
Functional overview
Consider in the correlation technique that the control channel expense is excessive in the parameter indication process, the problem of parameter indication bit transmission waste system resource, the present invention proposes: for the required bit number of each band width configuration indication parameter of system's support, and the bit number part of the indicating identical or complete difference of same parameters under different bandwidth, so that little bandwidth situation is lacked when carrying out bit that parameter when indication adopt than large bandwidth as far as possible, reduced the bit number that physical resource mapping indication signaling uses, can be under the prerequisite that does not affect the normal running of system, the bandwidth of using according to system changes the descending control overhead of saving flexibly.Namely the IE of transmission resource mapping indication information or message or attached bag are determined according to system bandwidth in broadcast channel or superframe head, thereby improve the operating efficiency of system.
Describe the present invention in detail below in conjunction with accompanying drawing.
Embodiment of the method
According to the embodiment of the invention, provide a kind of collocation method of resource mapping indication information.
Fig. 9 is the flow chart according to the collocation method of the resource mapping indication information of the embodiment of the invention, and as shown in Figure 9, the method may further comprise the steps:
Step S902, at least one parameter of indexed resource mapping process, determine the required bit number of the described parameter of indication according to bandwidth, this parameter comprise following one of at least: descending allocation of subbands number, up allocation of subbands number, the downstream frequency zone configuration, the upstream frequency zone configuration, downstream frequency subregion allocation of subbands number, upstream frequency subregion allocation of subbands number, the number that descending continuous resource unit distributes, the number that up continuous resource unit distributes, the number of descending continuous resource unit based on Miniband, the number of up continuous resource unit based on Miniband;
Step S904, a plurality of different bandwidths (this bandwidth can be system bandwidth) for system's support, the required bit number of indication parameter is the identical or complete difference of part each other, particularly, these a plurality of bandwidth can comprise the first bandwidth, the second bandwidth and the 3rd bandwidth, wherein, for a parameter in the resource indication information: corresponding to the first bandwidth, indicating the required bit number of this parameter is M; Corresponding to the second bandwidth, indicating the required bit number of this parameter is N; Corresponding to the 3rd bandwidth, indicating the required bit number of this parameter is P, and, the value of M, N, P is the identical or complete difference of part each other, preferably, above-mentioned the first bandwidth comprises: 5MHz, it is one of following that the second bandwidth comprises: 7MHz, 8.75MHz, 10MHz, the 3rd bandwidth comprise: 20MHz.Need to prove, for the bandwidth of 40MHz, the required bit number of indication parameter can be X, X+1 or X+2, and wherein, X is bandwidth needed bit number of indication parameter when being 20MHz.
Wherein, part is identical each other refers to for the value of M, N, P: N=M+1 and P=M+1; Perhaps, N=M+2 and P=M+2; Perhaps, N=M and P=M+1; Perhaps, N=M and P=M+2, wherein M is the integer greater than 0, preferably, the value of M is 1 or 2 or 3 or 4.
Wherein, the value of M, N, P each other fully difference refer to: N=M+1 and P=M+2; Perhaps N=M+2 and P=M+3, perhaps N=M+1 and P=M+3, wherein M is the integer greater than 0, preferably, the value of M is 1 or 2 or 3 or 4.
And concrete indicating mode is as follows: the bit number at the descending allocation of subbands number of system's indicating is identical or different with the bit number of the up allocation of subbands number of indication; Bit number in system's indicating downstream frequency zone configuration is identical or different with the bit number of indication upstream frequency zone configuration; Bit number at system's indicating downstream frequency subregion allocation of subbands number is identical or different with the bit number of indication upstream frequency subregion allocation of subbands number; The bit number of the number that distributes in the descending continuous resource of system's indicating unit is identical or different with the bit number of the number that the up continuous resource of indication unit distributes.
The technical scheme that provides by the embodiment of the invention, the bit number that physical resource mapping indication signaling is used can change flexibly according to the bandwidth that system uses, reduce as much as possible the bit number of transmission, under the prerequisite that does not affect the normal running of system, save descending control overhead, thereby improve the operating efficiency of system.
Describe in detail corresponding to the various examples of indicating the bit number of same parameters in the different bandwidth situation below in conjunction with concrete example.
In describing hereinafter, show the concrete corresponding situation of the indicated physical meaning of a plurality of indication signaling values of parameter (for example, DSAC, USAC etc.) and indication signaling value (for example, hereinafter described sub band number) by a plurality of forms.Should be appreciated that each form that occurs below, according to actual needs, the corresponding relation of the physical meaning that the indication signaling value of parameter and indication signaling value are indicated can change, and is not limited to the corresponding relation shown in the form.
In the shown form of description hereinafter, specifically provided the situation of carrying out the parameter indication by 1bit, 2bits, 3bits and four kinds of modes of 4bits.In the situation that adopt 1bit and optional value to have two, indication signaling is with 0 in binary " 0 " expression form, with 1 in binary " 1 " expression form; In the situation that employing 2bit carries out the parameter indication and optional value has 4, indication signaling represents 0 in the form with binary " 00 ", with 1 in binary " 01 " expression form, with 2 in binary " 10 " expression form, with 3 in binary " 11 " expression form; In the situation that employing 3bits carries out the parameter indication and optional value has 8, indication signaling represents 0 in the form with binary " 000 ", with 1 in binary " 001 " expression form, with 2 in binary " 010 " expression form, with 3 in binary " 011 " expression form, with 4 in binary " 100 " expression form, with 5 in binary " 101 " expression form, with 6 in binary " 110 " expression form, with 7 in binary " 111 " expression form; In the situation that employing 4bits carries out the parameter indication and optional value has 16, indication signaling represents 0 in the form with binary " 0000 ", with 1 in binary " 0001 " expression form, with 2 in binary " 0010 " expression form, with 3 in binary " 0011 " expression form, with 4 in binary " 0100 " expression form, with 5 in binary " 0101 " expression form, with 6 in binary " 0110 " expression form, with 7 in binary " 0111 " expression form, with 8 in binary " 1000 " expression form, with 9 in binary " 1001 " expression form, with 10 in binary " 1010 " expression form, with 11 in binary " 1011 " expression form, with 12 in binary " 1100 " expression form, with 13 in binary " 1101 " expression form, with 14 in binary " 1110 " expression form, with 15 in binary " 1111 " expression form.
The collocation method of descending allocation of subbands number
Example 1
Such as Figure 10, shown in DSAC value not simultaneously (that is, the number of the descending Subband of DSAC indication is not simultaneously), descending Subband Partitioning process is different.The below take system bandwidth as 5MHz, 10MHz (also can be 7MHz or 8.75MHz), 20MHz be example, the bandwidth that is divided three classes describes the configuring condition of DSAC, the first kind is 5MHz, and Equations of The Second Kind is 10MHz or 7MHz or 8.75MHz, and the 3rd class is 20MHz.
The first kind: when system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 2bits;
For 5MHz, the possible value set of Subband number is A DSAC={ 0,1,2,3,4,5,6}.1.1 having described system bandwidth to table 1.8 is 5MHz, and indication DSAC required bit number is when being 2bits, the corresponding relation of the value of DSAC and Subband number.2bits represents 4 kinds of different Subband numbers, the possible value set A of Subband number when these 4 kinds different Subband numbers are taken from 5MHz DSAC, be total to C 7 4=35 kinds of combinations.For example, table 1.1 has got that { 0,1,2,3}, other combination is enumerated no longer one by one.Need explanation, from m different elements, get n unduplicated element and form a subset, and do not consider the order of its element, be called from m, get n without recombination, the sum C of all possible combination m nExpression.
Table 1.1
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 2
1 1 3 3
Table 1.2
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 2
1 1 3 4
Table 1.3
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 3
1 1 3 4
Table 1.4
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 3
1 1 3 5
Table 1.5
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 3
1 2 3 4
Table 1.6
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 4
1 3 3 5
Table 1.7
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 2 4
1 3 3 6
Table 1.8
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 1 2 3
1 2 3 4
Perhaps: when system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 3bits.3bits represents 8 kinds of different Subband numbers, and these 8 kinds different Subband numbers can represent set A DSACIn all elements.Shown in table 1.9.
Table 1.9
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 Keep
Equations of The Second Kind: when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication DSAC parameter was 3bits;
For 7MHz or 8.75MHz or 10MHz, the possible value set of Subband number is B DSAC={ 0,1,2,3,4,5,6,7,8,9,10,11,12}.It is 10MHz (also can be 7MHz or 8.75MHz) that table 1.10 to table 1.22 has been described system bandwidth, and indication DSAC required bit number is in the situation of 3bits, the corresponding relation of the value of DSAC and Subband number.3bits represents 8 kinds of different Subband numbers, the possible value set B of Subband number when these 8 kinds different Subband numbers are taken from 10MHz (also can be 7MHz or 8.75MHz) DSAC, be total to C 13 8=1287 kinds of combinations.For example, table 1.10 has got that { 0,1,2,3,4,5,6,7} is except table 1.10 to other combination the table 1.22 is enumerated no longer one by one.
Table 1.10
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 1.11
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 7
3 3 7 9
Table 1.12
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 5
1 1 5 7
2 2 6 9
3 3 7 11
Table 1.13
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 1.14
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 7
3 3 7 8
Table 1.15
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 1.16
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 9
Table 1.17
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 10
3 3 7 12
Table 1.18
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 9
Table 1.19
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 9
3 3 7 12
Table 1.20
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 1.21
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 6
1 1 5 8
2 2 6 10
3 4 7 12
Table 1.22
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 6
1 2 5 8
2 3 6 10
3 4 7 12
Perhaps: when system bandwidth was 10MHz or 7MHz or 8.75MHz, the required bit number of indication DSAC parameter was 4bits.4bits represents 16 kinds of different Subband numbers, and these 16 kinds different Subband numbers enough represent set B DSACIn all elements.Shown in table 1.23.
Table 1.23
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 Keep
6 6 14 Keep
7 7 15 Keep
The 3rd bandwidth: when system bandwidth was 20MHz, the required bit number of indication DSAC parameter was 3bits.
For 20MHz, the possible value set of Subband number is C DSAC={ 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}.It is 20MHz that table 1.24 to table 1.27 has been described system bandwidth, and indication DSAC required bit number is in the situation of 3bits, the corresponding relation of the value of DSAC and Subband number.3bits represents 8 kinds of different Subband numbers, the possible value set C of Subband number when these 8 kinds different Subband numbers are taken from 20MHz DSAC, be total to C 25 8=1081575 kinds of combinations, for example, table 1.24 has got that { 0,2,3,4,6,8,9,12} is except table 1.24 to other combination the table 1.27 is enumerated no longer one by one.
Table 1.24
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 6
1 2 5 8
2 3 6 9
3 4 7 12
Table 1.25
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 6
1 1 5 9
2 2 6 12
3 3 7 15
Table 1.26
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 9
1 3 5 12
2 4 6 15
3 6 7 18
Table 1.27
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 4 12
1 3 5 15
2 6 6 18
3 9 7 21
Perhaps, when system bandwidth was 20MHz, the required bit number of indication DSAC parameter was 4bits.4bits represents 16 kinds of different Subband numbers, and these 16 kinds different Subband numbers are taken from set C DSAC, be total to C 25 16=2042975 kinds of combinations, for example, table 1.28 has been got { 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, }, except table 1.28 to other combination the table 1.40 is enumerated no longer one by one.
Table 1.28
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 1.29
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 14
6 6 14 16
7 7 15 18
Table 1.30
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 15
6 6 14 16
7 7 15 18
Table 1.31
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 15
6 6 14 18
7 7 15 20
Table 1.32
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 16
6 6 14 18
7 7 15 20
Table 1.33
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 15
6 6 14 18
7 7 15 21
Table 1.34
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 13
5 5 13 15
6 6 14 18
7 7 15 21
Table 1.35
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 15
5 5 13 18
6 6 14 21
7 7 15 24
Table 1.36
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 15
6 6 14 18
7 7 15 24
Table 1.37
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 15
5 5 13 18
6 6 14 21
7 7 15 24
Table 1.38
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 8 10
1 3 9 11
2 4 10 12
3 5 11 13
4 6 12 14
5 7 13 15
6 8 14 16
7 9 15 18
Table 1.39
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 3 8 11
1 4 9 12
2 5 10 13
3 6 11 14
4 7 12 15
5 8 13 16
6 9 14 18
7 10 15 21
Table 1.40
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 1 8 9
1 2 9 10
2 3 10 11
3 4 11 12
4 5 12 13
5 6 13 14
6 7 14 15
7 8 15 16
Perhaps, when system bandwidth was 20MHz, the required bit number of indication DSAC parameter was 5bits.5bits represents 32 kinds of different Subband numbers, and these 32 kinds different Subband numbers enough represent to gather C DSACIn all elements.Shown in table 1.41.
Table 1.41
DSAC Corresponding Subband number DSAC Corresponding Subband number
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 Keep
10 10 26 Keep
11 11 27 Keep
12 12 28 Keep
13 13 29 Keep
14 14 30 Keep
15 15 31 Keep
Can from said method, determine for the required bit number of indication DSAC parameter under each bandwidth, but for a plurality of different bandwidths that system supports, indicate the each other identical or complete difference of part of the required bit number of described parameter.For example,
When system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
Need to prove, in this method, even two different bandwidth have been used identical bit number indication DSAC parameter, but corresponding form also can be different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits, but corresponding form is table 1.15; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits, but corresponding form is table 1.25.
Because system bandwidth is 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be 10MHz (can be 7MHz or 8.75MHz) adopts identical DSAC when being 20MHz with system bandwidth value and corresponding relation with system bandwidth, thereby so that device fabrication is simpler, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth.For example, when system bandwidth was 5MHz, the required bit number of indication DSAC parameter was 2bits; And system bandwidth is when being 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and indicating the required bit number of this parameter is 4bits.
It is 5MHz that the same table 1.1 to table 1.8 has been described system bandwidth, and indication DSAC required bit number is in the situation of 2bits, and the corresponding relation of the value of DSAC and Subband number repeats no more here.
When system bandwidth is 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and the required bit number of indication DSAC is in the situation of 4bits, can all adopt one of 20MHz form when needing 4bits indication DSAC, for example, all be to determine the value of DSAC and the corresponding relation of Subband number with table 1.28 or table 1.33.
By above-mentioned example 1, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, the bit number of indication DSAC needs respectively 2bits, 3bits, 4bits, perhaps need respectively 2bits, 4bits, 4bits, perhaps need respectively 3bits, 4bits, 4bits, when perhaps needing respectively 3bits, 4bits, 5bits etc. to make up, in the situation that the possible value of DSAC reduces, deleted redundant and unnecessary information indication, save overhead bit, and guaranteed certain flexibility.
The collocation method of up allocation of subbands number
Example 2
As shown in figure 11, the USAC value is (that is, the number of the descending Subband of USAC indication is not simultaneously) not simultaneously, and descending Subband Partitioning process is different.The below (also can be 7MHz or 8.75MHz) take system bandwidth (or referred to as bandwidth) as 5MHz, 10MHz, 20MHz is as example, the bandwidth that is divided three classes describes the configuring condition of USAC, the first kind is 5MHz, Equations of The Second Kind is 10MHz or 7MHz or 8.75MHz, and the 3rd class is 20MHz.
The first kind: when system bandwidth was 5MHz, the required bit number of indication USAC parameter was 2bits;
For 5MHz, the possible value set of Subband number is A USAC={ 0,1,2,3,4,5,6}.1.1 having described system bandwidth to table 2.8 is 5MHz, and indication USAC required bit number is when being 2bits, the corresponding relation of the value of USAC and Subband number.2bits represents 4 kinds of different Subband numbers, the possible value set A of Subband number when these 4 kinds different Subband numbers are taken from 5MHz USAC, be total to C 7 4=35 kinds of combinations.For example, table 2.1 has got that { 0,1,2,3}, other combination is enumerated no longer one by one.Need explanation, from m different elements, get n unduplicated element and form a subset, and do not consider the order of its element, be called from m, get n without recombination, the sum C of all possible combination m nExpression.
Table 2.1
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 2
1 1 3 3
Table 2.2
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 2
1 1 3 4
Table 2.3
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 3
1 1 3 4
Table 2.4
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 3
1 1 3 5
Table 2.5
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 3
1 2 3 4
Table 2.6
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 4
1 3 3 5
Table 2.7
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 2 4
1 3 3 6
Table 2.8
USAC Corresponding Subband number USAC Corresponding Subband number
0 1 2 3
1 2 3 4
Perhaps: when system bandwidth was 5MHz, the required bit number of indication USAC parameter was 3bits.3bits represents 8 kinds of different Subband numbers, and these 8 kinds different Subband numbers can represent set A USACIn all elements.Shown in table 2.9.
Table 2.9
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 Keep
Equations of The Second Kind: when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication USAC parameter was 3bits;
For 7MHz or 8.75MHz or 10MHz, the possible value set of Subband number is B USAC={ 0,1,2,3,4,5,6,7,8,9,10,11,12}.It is 10MHz (also can be 7MHz or 8.75MHz) that table 2.10 to table 2.22 has been described system bandwidth, and indication USAC required bit number is in the situation of 3bits, the corresponding relation of the value of USAC and Subband number.3bits represents 8 kinds of different Subband numbers, the possible value set B of Subband number when these 8 kinds different Subband numbers are taken from 10MHz (also can be 7MHz or 8.75MHz) USAC, be total to C 13 8=1287 kinds of combinations.For example, table 2.10 has got that { 0,1,2,3,4,5,6,7} is except table 2.10 to other combination the table 2.22 is enumerated no longer one by one.
Table 2.10
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 2.11
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 7
3 3 7 9
Table 2.12
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 5
1 1 5 7
2 2 6 9
3 3 7 11
Table 2.13
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 2.14
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 7
3 3 7 8
Table 2.15
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 2.16
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 9
Table 2.17
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 10
3 3 7 12
Table 2.18
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 9
Table 2.19
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 4
1 1 5 6
2 2 6 9
3 3 7 12
Table 2.20
USAC Corresponding Subband number USAC Corresponding Subband number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 2.21
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 6
1 1 5 8
2 2 6 10
3 4 7 12
Table 2.22
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 6
1 2 5 8
2 3 6 10
3 4 7 12
Perhaps: when system bandwidth was 10MHz or 7MHz or 8.75MHz, the required bit number of indication USAC parameter was 4bits.4bits represents 16 kinds of different Subband numbers, and these 16 kinds different Subband numbers enough represent set B USACIn all elements.Shown in table 2.23.
Table 2.23
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 Keep
6 6 14 Keep
7 7 15 Keep
The 3rd bandwidth: when system bandwidth was 20MHz, the required bit number of indication USAC parameter was 3bits.
For 20MHz, the possible value set of Subband number is C USAC={ 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}.It is 20MHz that table 2.24 to table 2.27 has been described system bandwidth, and indication USAC required bit number is in the situation of 3bits, the corresponding relation of the value of USAC and Subband number.3bits represents 8 kinds of different Subband numbers, the possible value set C of Subband number when these 8 kinds different Subband numbers are taken from 20MHz USAC, be total to C 25 8=1081575 kinds of combinations, for example, table 2.24 has got that { 0,2,3,4,6,8,9,12} is except table 2.24 to other combination the table 2.27 is enumerated no longer one by one.
Table 2.24
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 6
1 2 5 8
2 3 6 9
3 4 7 12
Table 2.25
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 6
1 1 5 9
2 2 6 12
3 3 7 15
Table 2.26
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 9
1 3 5 12
2 4 6 15
3 6 7 18
Table 2.27
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 4 12
1 3 5 15
2 6 6 18
3 9 7 21
Perhaps, when system bandwidth was 20MHz, the required bit number of indication USAC parameter was 4bits.4bits represents 16 kinds of different Subband numbers, and these 16 kinds different Subband numbers are taken from set C USAC, be total to C 25 16=2042975 kinds of combinations, for example, table 2.28 has been got { 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, }, except table 2.28 to other combination the table 2.40 is enumerated no longer one by one.
Table 2.28
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 2.29
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 14
6 6 14 16
7 7 15 18
Table 2.30
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 15
6 6 14 16
7 7 15 18
Table 2.31
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 15
6 6 14 18
7 7 15 20
Table 2.32
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 16
6 6 14 18
7 7 15 20
Table 2.33
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 14
5 5 13 15
6 6 14 18
7 7 15 21
Table 2.34
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 13
5 5 13 15
6 6 14 18
7 7 15 21
Table 2.35
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 15
5 5 13 18
6 6 14 21
7 7 15 24
Table 2.36
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 15
6 6 14 18
7 7 15 24
Table 2.37
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 12
4 4 12 15
5 5 13 18
6 6 14 21
7 7 15 24
Table 2.38
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 8 10
1 3 9 11
2 4 10 12
3 5 11 13
4 6 12 14
5 7 13 15
6 8 14 16
7 9 15 18
Table 2.39
USAC Corresponding Subband number USAC Corresponding Subband number
0 3 8 11
1 4 9 12
2 5 10 13
3 6 11 14
4 7 12 15
5 8 13 16
6 9 14 18
7 10 15 21
Table 2.40
USAC Corresponding Subband number USAC Corresponding Subband number
0 1 8 9
1 2 9 10
2 3 10 11
3 4 11 12
4 5 12 13
5 6 13 14
6 7 14 15
7 8 15 16
Perhaps, when system bandwidth was 20MHz, the required bit number of indication USAC parameter was 5bits.5bits represents 32 kinds of different Subband numbers, and these 32 kinds different Subband numbers enough represent to gather C USACIn all elements.Shown in table 2.41.
Table 2.41
USAC Corresponding Subband number USAC Corresponding Subband number
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 Keep
10 10 26 Keep
11 11 27 Keep
12 12 28 Keep
13 13 29 Keep
14 14 30 Keep
15 15 31 Keep
Can from said method, determine for the required bit number of indication USAC parameter under each bandwidth, but for a plurality of different bandwidths that system supports, indicate the each other identical or complete difference of part of the required bit number of described parameter.For example,
When system bandwidth was 5MHz, the required bit number of indication USAC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication USAC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication USAC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication USAC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication USAC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
Need to prove, in this method, even two different bandwidth have been used identical bit number indication USAC parameter, but corresponding form also can be different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits, but corresponding form is table 2.15; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits, but corresponding form is table 2.25.
Because system bandwidth is 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be 10MHz (can be 7MHz or 8.75MHz) adopts identical USAC when being 20MHz with system bandwidth value and corresponding relation with system bandwidth, thereby so that device fabrication is simpler, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth.For example, when system bandwidth was 5MHz, the required bit number of indication USAC parameter was 2bits; And system bandwidth is when being 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and indicating the required bit number of this parameter is 4bits.
It is 5MHz that the same table 2.1 to table 2.8 has been described system bandwidth, and indication USAC required bit number is in the situation of 2bits, and the corresponding relation of the value of USAC and Subband number repeats no more here.
When system bandwidth is 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and the required bit number of indication USAC is in the situation of 4bits, can all adopt one of 20MHz form when needing 4bits indication USAC, for example, all be to determine the value of USAC and the corresponding relation of Subband number with table 2.28 or table 2.33.
By above-mentioned example 2, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, the bit number of indication USAC needs respectively 2bits, 3bits, 4bits, perhaps need respectively 2bits, 4bits, 4bits, perhaps need respectively 3bits, 4bits, 4bits, when perhaps needing respectively 3bits, 4bits, 5bits etc. to make up, in the situation that the possible value of USAC reduces, deleted redundant and unnecessary information indication, save overhead bit, and guaranteed certain flexibility.
The collocation method of downstream frequency zone configuration (DFPC)
DFPC has indicated size and the number of the frequency partition in the descending sub frame.When DFPC got different value, descending Frequency Partitioning process was different, as shown in figure 12.The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and is divided into three class bandwidth the configuring condition of DFPC is described, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.Wherein, NPRU is a PRU number on the subframe, and generally speaking, the NPRU that 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz are corresponding is respectively 24,48,48,48 and 96, but this method is not limited.And, each frequency partition ratio in following each form (FP0: FP1: FP2: FP3) in the expression formula, for the N1 that occurs: N2: N3: N4, wherein, N1 to N4 can represent the actual number of frequency partition, also can represent the proportionate relationship between each frequency partition.
Example 3
The first kind: when system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits.
For 5MHz, the set of the possible configuration of DFPC is A DFPC:
{ (1 frequency partition, the size of frequency partition are N PRU),
(3 frequency partition, the size of each frequency partition are N PRU* 1/3),
(4 frequency partition, and FPS 0=N PRU* 3/24, FPS 1=FPS 2=FPS 3=N PRU* 7/24),
(4 frequency partition, and FPS 0=N PRU* 6/24, FPS 1=FPS 2=FPS 3=N PRU* 1/4),
(4 frequency partition, and FPS 0=N PRU* 9/24, FPS 1=FPS 2=FPS 3=N PRU* 5/24),
(4 frequency partition, and FPS 0=N PRU* 1/2, FPS 1=FPS 2=FPS 3=N PRU* 1/6),
(4 frequency partition, and FPS 0=N PRU* 15/24, FPS 1=FPS 2=FPS 3=N PRU* 1/8),
(4 frequency partition, and FPS 0=N PRU* 18/24, FPS 1=FPS 2=FPS 3=N PRU* 1/12),
(4 frequency partition are arranged, and FPS 0=N PRU* 21/24, FPS 1=FPS 2=FPS 3=N PRU* 1/24) }.
2bits represents 4 kinds of different frequency partition numbers and frequency partition size, and these 4 kinds different frequency partition numbers and frequency partition size are taken from set A DFPC, be total to C 9 4=126 kinds of combinations.For example, table 3.1～table 3.3 has been described the corresponding relation of value and frequency partition number and the frequency partition size of DFPC, and other combination is enumerated no longer one by one.
Table 3.1
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
3 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
Table 3.2
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
3 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
Table 3.3
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
3 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
Perhaps, when system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 3bits.3bits represents 8 kinds of different frequency partition numbers and frequency partition size, and these 8 kinds different frequency partition numbers and frequency partition size are taken from A DFPC, be total to C 9 8=9 kinds of combinations.For example, table 3.4～table 3.5 has been described the corresponding relation of value and frequency partition number and the frequency partition size of DFPC, and other combination is enumerated no longer one by one.
Table 3.4
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 15∶3∶3∶3 4 N PRU*5/8 N PRU*1/8
7 18∶2∶2∶2 4 N PRU*3/4 N PRU*1/12
Table 3.5
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 15∶3∶3∶3 4 N PRU*5/8 N PRU*1/8
7 21∶1∶1∶1 4 N PRU*7/8 N PRU*1/24
Perhaps, although 3bits can represent 8 kinds of different frequency partition numbers and frequency partition size, owing to some frequency partition size can not be used substantially, so can be from A DFPC, the frequency partition size of selecting frequent usefulness represents, such as, 5 kinds, 6 kinds or 7 kinds, C altogether 9 5=126 kinds of combinations, C 9 6=84 kinds of combinations, C 9 7=36 kinds of combinations.For example, shown in table 3.6～table 3.8, other combination is enumerated no longer one by one.
Table 3.6
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
3 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
4 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
5 Keep Keep Keep Keep
6 Keep Keep Keep Keep
7 Keep Keep Keep Keep
Table 3.7
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 Keep Keep Keep Keep
7 Keep Keep Keep Keep
Table 3.8
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 15∶3∶3∶3 4 N PRU*5/8 N PRU*1/8
7 Keep Keep Keep Keep
Equations of The Second Kind: when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication DFPC parameter was 3bits.
The set of the possible configuration of DFPC is B DFPC=
{ (1 frequency partition, the size of frequency partition are N PRU),
(3 frequency partition, and the size of each frequency partition is N PRU* 1/3),
(4 frequency partition, and FPS 0=N PRU* 3/48, FPS 1=FPS 2=FPS 3=N PRU* 5/16),
(4 frequency partition, and FPS 0=N PRU* 6/48, FPS 1=FPS 2=FPS 3=N PRU* 7/24),
(4 frequency partition, and FPS 0=N PRU* 9/48, FPS 1=FPS 2=FPS 3=N PRU* 13/48),
(4 frequency partition, and FPS 0=N PRU* 12/48, FPS 1=FPS 2=FPS 3=N PRU* 1/4),
(4 frequency partition, and FPS 0=N PRU* 15/48, FPS 1=FPS 2=FPS 3=N PRU* 11/48),
(4 frequency partition, and FPS 0=N PRU* 18/48, FPS 1=FPS 2=FPS 3=N PRU* 5/24),
(4 frequency partition, and FPS 0=N PRU* 21/48, FPS 1=FPS 2=FPS 3=N PRU* 3/16),
(4 frequency partition, and FPS 0=N PRU* 24/48, FPS 1=FPS 2=FPS 3=N PRU* 1/6),
(4 frequency partition, and FPS 0=N PRU* 27/48, FPS 1=FPS 2=FPS 3=N PRU* 7/48),
(4 frequency partition, and FPS 0=N PRU* 30/48, FPS 1=FPS 2=FPS 3=N PRU* 1/8),
(4 frequency partition, and FPS 0=N PRU* 33/48, FPS 1=FPS 2=FPS 3=N PRU* 5/48),
(4 frequency partition, and FPS 0=N PRU* 36/48, FPS 1=FPS 2=FPS 3=N PRU* 1/12),
(4 frequency partition, and FPS 0=N PRU* 39/48, FPS 1=FPS 2=FPS 3=N PRU* 1/16),
(4 frequency partition, and FPS 0=N PRU* 42/48, FPS 1=FPS 2=FPS 3=N PRU* 1/24),
(4 frequency partition, and FPS 0=N PRU* 45/48, FPS 1=FPS 2=FPS 3=N PRU* 1/48) }.
3bits represents 8 kinds of different frequency partition numbers and frequency partition size, and these 8 kinds different frequency partition numbers and frequency partition size are taken from set B DFPC, be total to C 17 8=24310 kinds of combinations.For example, shown in table 3.9～table 3.11, other is enumerated no longer one by one.
Table 3.9
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*1/3
2 12∶12∶12∶12 4 N PRU*1/4 N PRU*1/4
3 24∶8∶8∶8 4 N PRU*3/6 N PRU*1/6
4 30∶6∶6∶6 4 N PRU*5/8 N PRU*1/8
5 6∶14∶14∶14 4 N PRU*3/24 N PRU*7/24
6 18∶10∶10∶10 4 N PRU*9/24 N PRU*5/24
7 21∶9∶9∶9 4 N PRU*7/16 N PRU*3/16
Table 3.10
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
FP 3)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*1/3
2 6∶14∶14∶14 4 N PRU*3/24 N PRU*7/24
3 12∶12∶12∶12 4 N PRU*1/4 N PRU*1/4
4 18∶10∶10∶10 4 N PRU*9/24 N PRU*5/24
5 21∶9∶9∶9 4 N PRU*7/16 N PRU*3/16
6 24∶8∶8∶8 4 N PRU*3/6 N PRU*1/6
7 30∶6∶6∶6 4 N PRU*5/8 N PRU*1/8
Table 3.11
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*1/3
2 9∶13∶13∶13 4 N PRU*3/16 N PRU*13/48
3 12∶12∶12∶12 4 N PRU*1/4 N PRU*1/4
4 18∶10∶10∶10 4 N PRU*9/24 N PRU*5/24
5 21∶9∶9∶9 4 N PRU*7/16 N PRU*3/16
6 24∶8∶8∶8 4 N PRU*3/6 N PRU*1/6
7 30∶6∶6∶6 4 N PRU*5/8 N PRU*1/8
Perhaps, when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication DFPC parameter was 4bits.4bits represents 16 kinds of different frequency partition numbers and frequency partition size, and these 16 kinds different frequency partition numbers and frequency partition size are taken from set B DFPC, be total to C 17 16=17 kinds of combinations.For example, shown in table 3.12～table 3.13.
Table 3.12
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*16/48
2 3∶15∶15∶15 4 N PRU*3/48 N PRU*15/48
3 6∶14∶14∶14 4 N PRU*6/48 N PRU*14/48
4 9∶13∶13∶13 4 N PRU*9/48 N PRU*13/48
5 12∶12∶12∶12 4 N PRU*12/48 N PRU*12/48
6 15∶11∶11∶11 4 N PRU*15/48 N PRU*11/48
7 18∶10∶10∶10 4 N PRU*18/48 N PRU*10/48
8 21∶9∶9∶9 4 N PRU*21/48 N PRU*9/48
9 24∶8∶8∶8 4 N PRU*24/48 N PRU*8/48
10 27∶7∶7∶7 4 N PRU*27/48 N PRU*7/48
11 30∶6∶6∶6 4 N PRU*30/48 N PRU*6/48
12 33∶5∶5∶5 4 N PRU*33/48 N PRU*5/48
13 36∶4∶4∶4 4 N PRU*36/48 N PRU*4/48
14 39∶3∶3∶3 4 N PRU*39/48 N PRU*3/48
15 42∶2∶2∶2 4 N PRU*42/48 N PRU*2/48
Table 3.13
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 0∶16∶16∶16 3 0 N PRU*16/48
1 3∶15∶15∶15 4 N PRU*3/48 N PRU*15/48
2 6∶14∶14∶14 4 N PRU*6/48 N PRU*14/48
3 9∶13∶13∶13 4 N PRU*9/48 N PRU*13/48
4 12∶12∶12∶12 4 N PRU*12/48 N PRU*12/48
5 15∶11∶11∶11 4 N PRU*15/48 N PRU*11/48
6 18∶10∶10∶10 4 N PRU*18/48 N PRU*10/48
7 21∶9∶9∶9 4 N PRU*21/48 N PRU*9/48
8 24∶8∶8∶8 4 N PRU*24/48 N PRU*8/48
9 27∶7∶7∶7 4 N PRU*27/48 N PRU*7/48
10 30∶6∶6∶6 4 N PRU*30/48 N PRU*6/48
11 33∶5∶5∶5 4 N PRU*33/48 N PRU*5/48
12 36∶4∶4∶4 4 N PRU*36/48 N PRU*4/48
13 39∶3∶3∶3 4 N PRU*39/48 N PRU*3/48
14 42∶2∶2∶2 4 N PRU*42/48 N PRU*2/48
15 48∶0∶0∶0 1 N PRU 0
Perhaps, although 4bits can represent 16 kinds of different frequency partition numbers and frequency partition size, owing to some frequency partition size can not be used substantially, so can be from B DFPCThe frequency partition size of selecting frequent usefulness represents, such as, 12 kinds, 13 kinds, 14 kinds or 15 kinds, C altogether 17 12=6188 kinds of combinations, C 17 13=2380 kinds of combinations, C 17 14=680 kinds of combinations, C 17 15=136 kinds of combinations.For example, shown in the table 3.14, other combination is enumerated no longer one by one.
Table 3.14
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*16/48
2 6∶14∶14∶14 4 N PRU*6/48 N PRU*14/48
3 9∶13∶13∶13 4 N PRU*9/48 N PRU*13/48
4 12∶12∶12∶12 4 N PRU*12/48 N PRU*12/48
5 15∶11∶11∶11 4 N PRU*15/48 N PRU*11/48
6 18∶10∶10∶10 4 N PRU*18/48 N PRU*10/48
7 21∶9∶9∶9 4 N PRU*21/48 N PRU*9/48
8 24∶8∶8∶8 4 N PRU*24/48 N PRU*8/48
9 27∶7∶7∶7 4 N PRU*27/48 N PRU*7/48
10 30∶6∶6∶6 4 N PRU*30/48 N PRU*6/48
11 33∶5∶5∶5 4 N PRU*33/48 N PRU*5/48
12 36∶4∶4∶4 4 N PRU*36/48 N PRU*4/48
13 Keep Keep Keep Keep
14 Keep Keep Keep Keep
15 Keep Keep Keep Keep
The 3rd class: when system bandwidth was 20MHz, the required bit number of indication DFPC parameter was 3bits.
The set of the possible configuration of DFPC is C DFPC=
{ (1 frequency partition, the size of frequency partition are N PRU),
(3 frequency partition, and the size of each frequency partition is N PRU* 1/3),
(4 frequency partition, and FPS 0=N PRU* 3/96, FPS 1=FPS 2=FPS 3=N PRU* 31/96),
(4 frequency partition, and FPS 0=N PRU* 6/96, FPS 1=FPS 2=FPS 3=N PRU* 30/96),
(4 frequency partition, and FPS 0=N PRU* 9/96, FPS 1=FPS 2=FPS 3=N PRU* 29/96),
(4 frequency partition, and FPS 0=N PRU* 12/96, FPS 1=FPS 2=FPS 3=N PRU* 28/96),
(4 frequency partition, and FPS 0=N PRU* 15/96, FPS 1=FPS 2=FPS 3=N PRU* 27/96),
(4 frequency partition, and FPS 0=N PRU* 18/96, FPS 1=FPS 2=FPS 3=N PRU* 26/96),
(4 frequency partition, and FPS 0=N PRU* 21/96, FPS 1=FPS 2=FPS 3=N PRU* 25/96),
(4 frequency partition, and FPS 0=N PRU* 24/96, FPS 1=FPS 2=FPS 3=N PRU* 24/96),
(4 frequency partition, and FPS 0=N PRU* 27/96, FPS 1=FPS 2=FPS 3=N PRU* 23/96),
(4 frequency partition, and FPS 0=N PRU* 30/96, FPS 1=FPS 2=FPS 3=N PRU* 22/96),
(4 frequency partition, and FPS 0=N PRU* 33/96, FPS 1=FPS 2=FPS 3=N PRU* 21/96),
(4 frequency partition, and FPS 0=N PRU* 36/96, FPS 1=FPS 2=FPS 3=N PRU* 20/96),
(4 frequency partition, and FPS 0=N PRU* 39/96, FPS 1=FPS 2=FPS 3=N PRU* 19/96),
(4 frequency partition, and FPS 0=N PRU* 42/96, FPS 1=FPS 2=FPS 3=N PRU* 18/96),
(4 frequency partition, and FPS 0=N PRU* 45/96, FPS 1=FPS 2=FPS 3=N PRU* 17/96),
(4 frequency partition, and FPS 0=N PRU* 48/96, FPS 1=FPS 2=FPS 3=N PRU* 16/96),
(4 frequency partition, and FPS 0=N PRU* 51/96, FPS 1=FPS 2=FPS 3=N PRU* 15/96),
(4 frequency partition, and FPS 0=N PRU* 54/96, FPS 1=FPS 2=FPS 3=N PRU* 14/96),
(4 frequency partition, and FPS 0=N PRU* 57/96, FPS 1=FPS 2=FPS 3=N PRU* 13/96),
(4 frequency partition, and FPS 0=N PRU* 60/96, FPS 1=FPS 2=FPS 3=N PRU* 12/96),
(4 frequency partition, and FPS 0=N PRU* 63/96, FPS 1=FPS 2=FPS 3=N PRU* 11/96),
(4 frequency partition, and FPS 0=N PRU* 66/96, FPS 1=FPS 2=FPS 3=N PRU* 10/96),
(4 frequency partition, and FPS 0=N PRU* 69/96, FPS 1=FPS 2=FPS 3=N PRU* 9/96),
(4 frequency partition, and FPS 0=N PRU* 72/96, FPS 1=FPS 2=FPS 3=N PRU* 8/96),
(4 frequency partition, and FPS 0=N PRU* 75/96, FPS 1=FPS 2=FPS 3=N PRU* 7/96),
(4 frequency partition, and FPS 0=N PRU* 78/96, FPS 1=FPS 2=FPS 3=N PRU* 6/96),
(4 frequency partition, and FPS 0=N PRU* 81/96, FPS 1=FPS 2=FPS 3=N PRU* 5/96),
(4 frequency partition, and FPS 0=N PRU* 84/96, FPS 1=FPS 2=FPS 3=N PRU* 4/96),
(4 frequency partition, and FPS 0=N PRU* 87/96, FPS 1=FPS 2=FPS 3=N PRU* 3/96),
(4 frequency partition, and FPS 0=N PRU* 90/96, FPS 1=FPS 2=FPS 3=N PRU* 2/96),
(4 frequency partition, and FPS 0=N PRU* 93/96, FPS 1=FPS 2=FPS 3=N PRU* 1/96) }.
3bits represents 8 kinds of different frequency partition numbers and frequency partition size, and these 8 kinds different frequency partition numbers and frequency partition size are taken from set C DFPC, be total to C 33 8=13884156 kinds of combinations.Can adopt the corresponding relation of any one combination indication DFPC value and frequency partition number and frequency partition size, for example, shown in table 3.15～3.17, other combination is not enumerated one by one.
Table 3.15
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
4 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
5 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
6 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
7 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
Table 3.16
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
4 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
5 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
6 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
7 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
Table 3.17
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
4 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
5 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
6 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
7 72∶8∶8∶8 4 N PRU*66/96 N PRU*8/96
Perhaps, when system bandwidth was 20MHz, the required bit number of indication DFPC parameter was 4bits.4bits represents 16 kinds of different frequency partition numbers and frequency partition size, and these 16 kinds different frequency partition numbers and frequency partition size are taken from set C DFPC, be total to C 33 16=1166803110 kinds of combinations.Can adopt the corresponding relation of any one combination indication DFPC value and frequency partition number and frequency partition size, for example, shown in table 3.18～table 19, other combination is not enumerated one by one.
Table 3.18
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
4 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
5 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
6 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
7 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
8 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
9 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
10 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
11 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
12 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
13 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
14 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
15 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
Table 3.19
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
3 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
4 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
5 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
6 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
7 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
8 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
9 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
10 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
11 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
12 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
13 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
14 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
15 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
Perhaps, when system bandwidth was 20MHz, the required bit number of indication DFPC parameter was 5bits.5bits represents 32 kinds of different frequency partition numbers and frequency partition size, and these 32 kinds different frequency partition numbers and frequency partition size are taken from set C DFPC, be total to C 33 32=33 kinds of combinations.Can adopt the corresponding relation of any one combination indication DFPC value and frequency partition number and frequency partition size, for example, shown in the table 3.20, other combination is not enumerated one by one.
Table 3.20
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 3∶31∶31∶31 4 N PRU*3/96 N PRU*31/96
3 6∶30∶30∶30 4 N PRU*6/96 N PRU*30/96
4 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
5 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
6 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
7 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
8 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
9 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
10 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
11 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
12 33∶21∶21∶21 4 N PRU*33/96 N PRU*21/96
13 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
14 39∶19∶19∶19 4 N PRU*39/96 N PRU*19/96
15 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
16 45∶17∶17∶17 4 N PRU*45/96 N PRU*17/96
17 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
18 51∶15∶15∶15 4 N PRU*51/96 N PRU*15/96
19 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
20 57∶13∶13∶13 4 N PRU*57/96 N PRU*13/96
21 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
22 63∶11∶11∶11 4 N PRU*63/96 N PRU*11/96
23 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
24 69∶9∶9∶9 4 N PRU*69/96 N PRU*9/96
25 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
26 75∶7∶7∶7 4 N PRU*75/96 N PRU*7/96
27 78∶6∶6∶6 4 N PRU*78/96 N PRU*6/96
28 81∶5∶5∶5 4 N PRU*81/96 N PRU*5/96
29 84∶4∶4∶4 4 N PRU*84/96 N PRU*4/96
30 87∶3∶3∶3 4 N PRU*87/96 N PRU*3/96
31 90∶2∶2∶2 4 N PRU*90/96 N PRU*2/96
Perhaps, although 5bits can represent 32 kinds of different frequency partition numbers and frequency partition size, owing to some frequency partition size can not be used substantially, so can be from C DFPCThe frequency partition size of selecting frequent usefulness represents, such as, M (1＜M＜32) plants, altogether C 33 MCombination.For example, shown in table 3.21～table 3.22, other combination is enumerated no longer one by one.
Table 3.21
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU *32/96
2 6∶30∶30∶30 4 N PRU*6/96 N PRU*30/96
3 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
4 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
5 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
6 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
7 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
8 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
9 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
10 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
11 33∶21∶21∶21 4 N PRU*33/96 N PRU*21/96
12 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
13 39∶19∶19∶19 4 N PRU*39/96 N PRU*19/96
14 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
15 45∶17∶17∶17 4 N PRU*45/96 N PRU*17/96
16 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
17 51∶15∶15∶15 4 N PRU*51/96 N PRU*15/96
18 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
19 57∶13∶13∶13 4 N PRU*57/96 N PRU*13/96
20 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
21 63∶11∶11∶11 4 N PRU*63/96 N PRU*11/96
22 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
23 69∶9∶9∶9 4 N PRU*69/96 N PRU*9/96
24 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
25 75∶7∶7∶7 4 N PRU*75/96 N PRU*7/96
26 78∶6∶6∶6 4 N PRU*78/96 N PRU*6/96
27 Keep Keep Keep Keep
28 Keep Keep Keep Keep
29 Keep Keep Keep Keep
30 Keep Keep Keep Keep
31 Keep Keep Keep Keep
Table 3.22
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
3 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
4 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
5 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
6 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
7 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
8 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
9 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
10 33∶21∶21∶21 4 N PRU*33/96 N PRU*21/96
11 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
12 39∶19∶19∶19 4 N PRU*39/96 N PRU*19/96
13 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
14 45∶17∶17∶17 4 N PRU*45/96 N PRU*17/96
15 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
16 51∶15∶15∶15 4 N PRU*51/96 N PRU*15/96
17 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
18 57∶13∶13∶13 4 N PRU*57/96 N PRU*13/96
19 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
20 63∶11∶11∶11 4 N PRU*63/96 N PRU*11/96
21 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
22 69∶9∶9∶9 4 N PRU*69/96 N PRU*9/96
23 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
24 Keep Keep Keep Keep
25 Keep Keep Keep Keep
26 Keep Keep Keep Keep
27 Keep Keep Keep Keep
28 Keep Keep Keep Keep
29 Keep Keep Keep Keep
30 Keep Keep Keep Keep
31 Keep Keep Keep Keep
Can from said method, determine for the required bit number of indication DFPC parameter under each bandwidth, but for different bandwidth, indicate the each other identical or complete difference of part of the required bit number of DFPC parameter.For example,
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
It may be noted that: in the collocation method of above-mentioned DFPC, used identical bit number indication DFPC parameter when two different bandwidth, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits, but corresponding form is table 3.13; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits, but corresponding form is table 3.18.
Because system bandwidth is 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be 10MHz (can be 7MHz or 8.75MHz) adopts identical DFPC when being 20MHz with system bandwidth value and corresponding relation with system bandwidth, thereby so that device fabrication is simpler, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth.For example, when system bandwidth was 5MHz, the required bit number of indication DFPC parameter was 2bits; And system bandwidth is when being 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and indicating the required bit number of this parameter is 4bits.
It is 5MHz that the same table 3.1 to table 3.3 has been described system bandwidth, and indication DFPC required bit number is the collocation method in the situation of 2bits, repeats no more here.
When system bandwidth is 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and the required bit number of indication DFPC is in the situation of 4bits, can all adopt one of 20MHz form when needing 4bits indication DFPC, for example, all be with table 3.13 or table 3.18.
In addition; it may be noted that: in the collocation method of above-mentioned DFPC; for each table; relation in the middle of the meaning of the value of DFPC and the indication of the value of DFPC can change; each form all is embodiment; the meaning that needs only the value indication of the DFPC that comprises in the table is identical, all is regarded as identical table, all within protection range.For example, table 3.23 all is considered as identical table with table 3.15, is identical because the value of the DFPC that comprises in two tables is the meaning of indication.
Table 3.23
DFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 0∶32∶32∶32 3 0 N PRU*32/96
1 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
2 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
3 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
4 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
5 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
6 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
7 96∶0∶0∶0 1 N PRU 0
Other form of listing and produced into by said method but all do not follow this principle at this form of listing.
By above-mentioned example 3, can find out, system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), during the 20MHz system, the bit number of indication DFPC needs respectively 2bits, 3bits, 4bits, perhaps need respectively 2bits, 4bits, 4bits, perhaps need respectively 3bits, 4bits, 4bits, perhaps need respectively 3bits, 4bits, when the combination such as 5bits or other combination, in the situation that the possible value of DFPC reduces, deleted redundant and unnecessary information indication, save overhead bit, and guaranteed certain flexibility.
The collocation method of upstream frequency zone configuration (UFPC)
UFPC has indicated size and the number of the frequency partition in the sub-frame of uplink.When UFPC got different value, up Frequency Partitioning process was different, as shown in figure 13.The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and is divided into three class bandwidth the configuring condition of UFPC is described, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.Wherein, N PRUA PRU number on the subframe, generally speaking, the N that 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz are corresponding PRUBe respectively 24,48,48,48 and 96, but this method is not limited.
Example 4
The first kind: when system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits.
For 5MHz, the set of the possible configuration of UFPC is A UFPC:
{ (1 frequency partition, the size of frequency partition are N PRU),
(3 frequency partition, the size of each frequency partition are N PRU* 1/3),
(4 frequency partition, and FPS 0=N PRU* 3/24, FPS 1=FPS 2=FPS 3=N PRU* 7/24),
(4 frequency partition, and FPS 0=N PRU* 6/24, FPS 1=FPS 2=FPS 3=N PRU* 1/4),
(4 frequency partition, and FPS 0=N PRU* 9/24, FPS 1=FPS 2=FPS 3=N PRU* 5/24),
(4 frequency partition, and FPS 0=N PRU* 1/2, FPS 1=FPS 2=FPS 3=N PRU* 1/6),
(4 frequency partition, and FPS 0=N PRU* 15/24, FPS 1=FPS 2=FPS 3=N PRU* 1/8),
(4 frequency partition, and FPS 0=N PRU* 18/24, FPS 1=FPS 2=FPS 3=N PRU* 1/12),
(4 frequency partition are arranged, and FPS 0=N PRU* 21/24, FPS 1=FPS 2=FPS 3=N PRU* 1/24) }.
2bits represents 4 kinds of different frequency partition numbers and frequency partition size, and these 4 kinds different frequency partition numbers and frequency partition size are taken from set A UFPC, be total to C 9 4=126 kinds of combinations.For example, table 4.1～table 4.3 has been described the corresponding relation of value and frequency partition number and the frequency partition size of UFPC, and other combination is enumerated no longer one by one.
Table 4.1
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
3 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
Table 4.2
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
3 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
Table 4.3
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
3 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
Perhaps, when system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 3bits.3bits represents 8 kinds of different frequency partition numbers and frequency partition size, and these 8 kinds different frequency partition numbers and frequency partition size are taken from A UFPC, be total to C 9 8=9 kinds of combinations.For example, table 4.4～table 4.5 has been described the corresponding relation of value and frequency partition number and the frequency partition size of UFPC, and other combination is enumerated no longer one by one.
Table 4.4
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 15∶3∶3∶3 4 N PRU*5/8 N PRU*1/8
7 18∶2∶2∶2 4 N PRU*3/4 N PRU*1/12
Table 4.5
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 15∶3∶3∶3 4 N PRU*5/8 N PRU*1/8
7 21∶1∶1∶1 4 N PRU*7/8 N PRU*1/24
Perhaps, although 3bits can represent 8 kinds of different frequency partition numbers and frequency partition size, owing to some frequency partition size can not be used substantially, so can be from A UFPC, the frequency partition size of selecting frequent usefulness represents, such as, 5 kinds, 6 kinds or 7 kinds, C altogether 9 5=126 kinds of combinations, C 9 6=84 kinds of combinations, C 9 7=36 kinds of combinations.For example, shown in table 4.6～table 4.8, other combination is enumerated no longer one by one.
Table 4.6
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
3 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
4 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
5 Keep Keep Keep Keep
6 Keep Keep Keep Keep
7 Keep Keep Keep Keep
Table 4.7
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 Keep Keep Keep Keep
7 Keep Keep Keep Keep
Table 4.8
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 24∶0∶0∶0 1 N PRU 0
1 0∶8∶8∶8 3 0 N PRU*1/3
2 3∶7∶7∶7 4 N PRU*1/8 N PRU*7/24
3 6∶6∶6∶6 4 N PRU*1/4 N PRU*1/4
4 9∶5∶5∶5 4 N PRU*3/8 N PRU*5/24
5 12∶4∶4∶4 4 N PRU*1/2 N PRU*1/6
6 15∶3∶3∶3 4 N PRU*5/8 N PRU*1/8
7 Keep Keep Keep Keep
Equations of The Second Kind: when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication UFPC parameter was 3bits.
The set of the possible configuration of UFPC is B UFPC=
{ (1 frequency partition, the size of frequency partition are N PRU),
(3 frequency partition, and the size of each frequency partition is N PRU* 1/3),
(4 frequency partition, and FPS 0=N PRU* 3/48, FPS 1=FPS 2=FPS 3=N PRU* 5/16),
(4 frequency partition, and FPS 0=N PRU* 6/48, FPS 1=FPS 2=FPS 3=N PRU* 7/24),
(4 frequency partition, and FPS 0=N PRU* 9/48, FPS 1=FPS 2=FPS 3=N PRU* 13/48),
(4 frequency partition, and FPS 0=N PRU* 12/48, FPS 1=FPS 2=FPS 3=N PRU* 1/4),
(4 frequency partition, and FPS 0=N PRU* 15/48, FPS 1=FPS 2=FPS 3=N PRU* 11/48),
(4 frequency partition, and FPS 0=N PRU* 18/48, FPS 1=FPS 2=FPS 3=N PRU* 5/24),
(4 frequency partition, and FPS 0=N PRU* 21/48, FPS 1=FPS 2=FPS 3=N PRU* 3/16),
(4 frequency partition, and FPS 0=N PRU* 24/48, FPS 1=FPS 2=FPS 3=N PRU* 1/6),
(4 frequency partition, and FPS 0=N PRU* 27/48, FPS 1=FPS 2=FPS 3=N PRU* 7/48),
(4 frequency partition, and FPS 0=N PRU* 30/48, FPS 1=FPS 2=FPS 3=N PRU* 1/8),
(4 frequency partition, and FPS 0=N PRU* 33/48, FPS 1=FPS 2=FPS 3=N PRU* 5/48),
(4 frequency partition, and FPS 0=N PRU* 36/48, FPS 1=FPS 2=FPS 3=N PRU* 1/12),
(4 frequency partition, and FPS 0=N PRU* 39/48, FPS 1=FPS 2=FPS 3=N PRU* 1/16),
(4 frequency partition, and FPS 0=N PRU* 42/48, FPS 1=FPS 2=FPS 3=N PRU* 1/24),
(4 frequency partition, and FPS 0=N PRU* 45/48, FPS 1=FPS 2=FPS 3=N PRU* 1/48) }.
3bits represents 8 kinds of different frequency partition numbers and frequency partition size, and these 8 kinds different frequency partition numbers and frequency partition size are taken from set B UFPC, be total to C 17 8=24310 kinds of combinations.For example, shown in table 4.9～table 4.11, other is enumerated no longer one by one.
Table 4.9
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*1/3
2 12∶12∶12∶12 4 N PRU*1/4 N PRU*1/4
3 24∶8∶8∶8 4 N PRU*3/6 N PRU*1/6
4 30∶6∶6∶6 4 N PRU*5/8 N PRU*1/8
5 6∶14∶14∶14 4 N PRU*3/24 N PRU*7/24
6 18∶10∶10∶10 4 N PRU*9/24 N PRU*5/24
7 21∶9∶9∶9 4 N PRU*7/16 N PRU*3/16
Table 4.10
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*1/3
2 6∶14∶14∶14 4 N PRU*3/24 N PRU*7/24
3 12∶12∶12∶12 4 N PRU*1/4 N PRU*1/4
4 18∶10∶10∶10 4 N PRU*9/24 N PRU*5/24
5 21∶9∶9∶9 4 N PRU*7/16 N PRU*3/16
6 24∶8∶8∶8 4 N PRU*3/6 N PRU*1/6
7 30∶6∶6∶6 4 N PRU*5/8 N PRU*1/8
Table 4.11
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*1/3
2 9∶13∶13∶13 4 N PRU*3/16 N PRU*13/48
3 12∶12∶12∶12 4 N PRU*1/4 N PRU*1/4
4 18∶10∶10∶10 4 N PRU*9/24 N PRU*5/24
5 21∶9∶9∶9 4 N PRU*7/16 N PRU*3/16
6 24∶8∶8∶8 4 N PRU*3/6 N PRU*1/6
7 30∶6∶6∶6 4 N PRU*5/8 N PRU*1/8
Perhaps, when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication UFPC parameter was 4bits.4bits represents 16 kinds of different frequency partition numbers and frequency partition size, and these 16 kinds different frequency partition numbers and frequency partition size are taken from set B UFPC, be total to C 17 16=17 kinds of combinations.For example, shown in table 4.12～table 4.13.
Table 4.12
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*16/48
2 3∶15∶15∶15 4 N PRU*3/48 N PRU*15/48
3 6∶14∶14∶14 4 N PRU*6/48 N PRU*14/48
4 9∶13∶13∶13 4 N PRU*9/48 N PRU*13/48
5 12∶12∶12∶12 4 N PRU*12/48 N PRU*12/48
6 15∶11∶11∶11 4 N PRU*15/48 N PRU*11/48
7 18∶10∶10∶10 4 N PRU*18/48 N PRU*10/48
8 21∶9∶9∶9 4 N PRU*21/48 N PRU*9/48
9 24∶8∶8∶8 4 N PRU*24/48 N PRU*8/48
10 27∶7∶7∶7 4 N PRU*27/48 N PRU*7/48
11 30∶6∶6∶6 4 N PRU*30/48 N PRU*6/48
12 33∶5∶5∶5 4 N PRU*33/48 N PRU*5/48
13 36∶4∶4∶4 4 N PRU*36/48 N PRU*4/48
14 39∶3∶3∶3 4 N PRU*39/48 N PRU*3/48
15 42∶2∶2∶2 4 N PRU*42/48 N PRU*2/48
Table 4.13
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 0∶16∶16∶16 3 0 N PRU*16/48
1 3∶15∶15∶15 4 N PRU*3/48 N PRU*15/48
2 6∶14∶14∶14 4 N PRU*6/48 N PRU*14/48
3 9∶13∶13∶13 4 N PRU*9/48 N PRU*13/48
4 12∶12∶12∶12 4 N PRU*12/48 N PRU*12/48
5 15∶11∶11∶11 4 N PRU*15/48 N PRU*11/48
6 18∶10∶10∶10 4 N PRU*18/48 N PRU*10/48
7 21∶9∶9∶9 4 N PRU*21/48 N PRU*9/48
8 24∶8∶8∶8 4 N PRU*24/48 N PRU*8/48
9 27∶7∶7∶7 4 N PRU*27/48 N PRU*7/48
10 30∶6∶6∶6 4 N PRU*30/48 N PRU*6/48
11 33∶5∶5∶5 4 N PRU*33/48 N PRU*5/48
12 36∶4∶4∶4 4 N PRU*36/48 N PRU*4/48
13 39∶3∶3∶3 4 N PRU*39/48 N PRU*3/48
14 42∶2∶2∶2 4 N PRU*42/48 N PRU*2/48
15 48∶0∶0∶0 1 N PRU 0
Perhaps, although 4bits can represent 16 kinds of different frequency partition numbers and frequency partition size, owing to some frequency partition size can not be used substantially, so can be from B UFPCThe frequency partition size of selecting frequent usefulness represents, such as, 12 kinds, 13 kinds, 14 kinds or 15 kinds, C altogether 17 12=6188 kinds of combinations, C 17 13=2380 kinds of combinations, C 17 14=680 kinds of combinations, C 17 15=136 kinds of combinations.For example, shown in the table 4.14, other combination is enumerated no longer one by one.
Table 4.14
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 48∶0∶0∶0 1 N PRU 0
1 0∶16∶16∶16 3 0 N PRU*16/48
2 6∶14∶14∶14 4 N PRU*6/48 N PRU*14/48
3 9∶13∶13∶13 4 N PRU*9/48 N PRU*13/48
4 12∶12∶12∶12 4 N PRU*12/48 N PRU*12/48
5 15∶11∶11∶11 4 N PRU*15/48 N PRU*11/48
6 18∶10∶10∶10 4 N PRU*18/48 N PRU*10/48
7 21∶9∶9∶9 4 N PRU*21/48 N PRU*9/48
8 24∶8∶8∶8 4 N PRU*24/48 N PRU*8/48
9 27∶7∶7∶7 4 N PRU*27/48 N PRU*7/48
10 30∶6∶6∶6 4 N PRU*30/48 N PRU*6/48
11 33∶5∶5∶5 4 N PRU*33/48 N PRU*5/48
12 36∶4∶4∶4 4 N PRU*36/48 N PRU*4/48
13 Keep Keep Keep Keep
14 Keep Keep Keep Keep
15 Keep Keep Keep Keep
The 3rd class: when system bandwidth was 20MHz, the required bit number of indication UFPC parameter was 3bits.
The set of the possible configuration of UFPC is C UFPC=
{ (1 frequency partition, the size of frequency partition are N PRU),
(3 frequency partition, and the size of each frequency partition is N PRU* 1/3),
(4 frequency partition, and FPS 0=N PRU* 3/96, FPS 1=FPS 2=FPS 3=N PRU* 31/96),
(4 frequency partition, and FPS 0=N PRU* 6/96, FPS 1=FPS 2=FPS 3=N PRU* 30/96),
(4 frequency partition, and FPS 0=N PRU* 9/96, FPS 1=FPS 2=FPS 3=N PRU* 29/96),
(4 frequency partition, and FPS 0=N PRU* 12/96, FPS 1=FPS 2=FPS 3=N PRU* 28/96),
(4 frequency partition, and FPS 0=N PRU* 15/96, FPS 1=FPS 2=FPS 3=N PRU* 27/96),
(4 frequency partition, and FPS 0=N PRU* 18/96, FPS 1=FPS 2=FPS 3=N PRU* 26/96),
(4 frequency partition, and FPS 0=N PRU* 21/96, FPS 1=FPS 2=FPS 3=N PRU* 25/96),
(4 frequency partition, and FPS 0=N PRU* 24/96, FPS 1=FPS 2=FPS 3=N PRU* 24/96),
(4 frequency partition, and FPS 0=N PRU* 27/96, FPS 1=FPS 2=FPS 3=N PRU* 23/96),
(4 frequency partition, and FPS 0=N PRU* 30/96, FPS 1=FPS 2=FPS 3=N PRU* 22/96),
(4 frequency partition, and FPS 0=N PRU* 33/96, FPS 1=FPS 2=FPS 3=N PRU* 21/96),
(4 frequency partition, and FPS 0=N PRU* 36/96, FPS 1=FPS 2=FPS 3=N PRU* 20/96),
(4 frequency partition, and FPS 0=N PRU* 39/96, FPS 1=FPS 2=FPS 3=N PRU* 19/96),
(4 frequency partition, and FPS 0=N PRU* 42/96, FPS 1=FPS 2=FPS 3=N PRU* 18/96),
(4 frequency partition, and FPS 0=N PRU* 45/96, FPS 1=FPS 2=FPS 3=N PRU* 17/96),
(4 frequency partition, and FPS 0=N PRU* 48/96, FPS 1=FPS 2=FPS 3=N PRU* 16/96),
(4 frequency partition, and FPS 0=N PRU* 51/96, FPS 1=FPS 2=FPS 3=N PRU* 15/96),
(4 frequency partition, and FPS 0=N PRU* 54/96, FPS 1=FPS 2=FPS 3=N PRU* 14/96),
(4 frequency partition, and FPS 0=N PRU* 57/96, FPS 1=FPS 2=FPS 3=N PRU* 13/96),
(4 frequency partition, and FPS 0=N PRU* 60/96, FPS 1=FPS 2=FPS 3=N PRU* 12/96),
(4 frequency partition, and FPS 0=N PRU* 63/96, FPS 1=FPS 2=FPS 3=N PRU* 11/96),
(4 frequency partition, and FPS 0=N PRU* 66/96, FPS 1=FPS 2=FPS 3=N PRU* 10/96),
(4 frequency partition, and FPS 0=N PRU* 69/96, FPS 1=FPS 2=FPS 3=N PRU* 9/96),
(4 frequency partition, and FPS 0=N PRU* 72/96, FPS 1=FPS 2=FPS 3=N PRU* 8/96),
(4 frequency partition, and FPS 0=N PRU* 75/96, FPS 1=FPS 2=FPS 3=N PRU* 7/96),
(4 frequency partition, and FPS 0=N PRU* 78/96, FPS 1=FPS 2=FPS 3=N PRU* 6/96),
(4 frequency partition, and FPS 0=N PRU* 81/96, FPS 1=FPS 2=FPS 3=N PRU* 5/96),
(4 frequency partition, and FPS 0=N PRU* 84/96, FPS 1=FPS 2=FPS 3=N PRU* 4/96),
(4 frequency partition, and FPS 0=N PRU* 87/96, FPS 1=FPS 2=FPS 3=N PRU* 3/96),
(4 frequency partition, and FPS 0=N PRU* 90/96, FPS 1=FPS 2=FPS 3=N PRU* 2/96),
(4 frequency partition, and FPS 0=N PRU* 93/96, FPS 1=FPS 2=FPS 3=N PRU* 1/96) }.
3bits represents 8 kinds of different frequency partition numbers and frequency partition size, and these 8 kinds different frequency partition numbers and frequency partition size are taken from set C UFPC, be total to C 33 8=13884156 kinds of combinations.Can adopt the corresponding relation of any one combination indication UFPC value and frequency partition number and frequency partition size, for example, shown in table 4.15～3.17, other combination is not enumerated one by one.
Table 4.15
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
4 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
5 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
6 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
7 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
Table 4.16
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
4 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
5 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
6 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
7 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
Table 4.17
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
4 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
5 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
6 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
7 72∶8∶8∶8 4 N PRU*66/96 N PRU*8/96
Perhaps, when system bandwidth was 20MHz, the required bit number of indication UFPC parameter was 4bits.4bits represents 16 kinds of different frequency partition numbers and frequency partition size, and these 16 kinds different frequency partition numbers and frequency partition size are taken from set C UFPC, be total to C 33 16=1166803110 kinds of combinations.Can adopt the corresponding relation of any one combination indication UFPC value and frequency partition number and frequency partition size, for example, shown in table 4.18～table 19, other combination is not enumerated one by one.
Table 4.18
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
3 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
4 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
5 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
6 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
7 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
8 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
9 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
10 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
11 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
12 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
13 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
14 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
15 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
Table 4.19
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
3 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
4 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
5 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
6 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
7 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
8 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
9 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
10 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
11 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
12 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
13 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
14 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
15 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
Perhaps, when system bandwidth was 20MHz, the required bit number of indication UFPC parameter was 5bits.5bits represents 32 kinds of different frequency partition numbers and frequency partition size, and these 32 kinds different frequency partition numbers and frequency partition size are taken from set C UFPC, be total to C 33 32=33 kinds of combinations.Can adopt the corresponding relation of any one combination indication UFPC value and frequency partition number and frequency partition size, for example, shown in the table 4.20, other combination is not enumerated one by one.
Table 4.20
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 3∶31∶31∶31 4 N PRU*3/96 N PRU*31/96
3 6∶30∶30∶30 4 N PRU*6/96 N PRU*30/96
4 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
5 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
6 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
7 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
8 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
9 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
10 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
11 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
12 33∶21∶21∶21 4 N PRU*33/96 N PRU*21/96
13 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
14 39∶19∶19∶19 4 N PRU*39/96 N PRU*19/96
15 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
16 45∶17∶17∶17 4 N PRU*45/96 N PRU*17/96
17 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
18 51∶15∶15∶15 4 N PRU*51/96 N PRU*15/96
19 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
20 57∶13∶13∶13 4 N PRU*57/96 N PRU*13/96
21 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
22 63∶11∶11∶11 4 N PRU*63/96 N PRU*11/96
23 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
24 69∶9∶9∶9 4 N PRU*69/96 N PRU*9/96
25 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
26 75∶7∶7∶7 4 N PRU*75/96 N PRU*7/96
27 78∶6∶6∶6 4 N PRU*78/96 N PRU*6/96
28 81∶5∶5∶5 4 N PRU*81/96 N PRU*5/96
29 84∶4∶4∶4 4 N PRU*84/96 N PRU*4/96
30 87∶3∶3∶3 4 N PRU*87/96 N PRU*3/96
31 90∶2∶2∶2 4 N PRU*90/96 N PRU*2/96
Perhaps, although 5bits can represent 32 kinds of different frequency partition numbers and frequency partition size, owing to some frequency partition size can not be used substantially, so can be from C UFPCThe frequency partition size of selecting frequent usefulness represents, such as, M (1＜M＜32) plants, altogether C 33 MCombination.For example, shown in table 4.21～table 4.22, other combination is enumerated no longer one by one.
Table 4.21
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 6∶30∶30∶30 4 N PRU*6/96 N PRU*30/96
3 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
4 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
5 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
6 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
7 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
8 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
9 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
10 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
11 33∶21∶21∶21 4 N PRU*33/96 N PRU*21/96
12 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
13 39∶19∶19∶19 4 N PRU*39/96 N PRU*19/96
14 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
15 45∶17∶17∶17 4 N PRU*45/96 N PRU*17/96
16 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
17 51∶15∶15∶15 4 N PRU*51/96 N PRU*15/96
18 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
19 57∶13∶13∶13 4 N PRU*57/96 N PRU*13/96
20 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
21 63∶11∶11∶11 4 N PRU*63/96 N PRU*11/96
22 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
23 69∶9∶9∶9 4 N PRU*69/96 N PRU*9/96
24 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
25 75∶7∶7∶7 4 N PRU*75/96 N PRU*7/96
26 78∶6∶6∶6 4 N PRU*78/96 N PRU*6/96
27 Keep Keep Keep Keep
28 Keep Keep Keep Keep
29 Keep Keep Keep Keep
30 Keep Keep Keep Keep
31 Keep Keep Keep Keep
Table 4.22
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 96∶0∶0∶0 1 N PRU 0
1 0∶32∶32∶32 3 0 N PRU*32/96
2 9∶29∶29∶29 4 N PRU*9/96 N PRU*29/96
3 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
4 15∶27∶27∶27 4 N PRU*15/96 N PRU*27/96
5 18∶26∶26∶26 4 N PRU*18/96 N PRU*26/96
6 21∶25∶25∶25 4 N PRU*21/96 N PRU*25/96
7 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
8 27∶23∶23∶23 4 N PRU*27/96 N PRU*23/96
9 30∶22∶22∶22 4 N PRU*30/96 N PRU*22/96
10 33∶21∶21∶21 4 N PRU*33/96 N PRU*21/96
11 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
12 39∶19∶19∶19 4 N PRU*39/96 N PRU*19/96
13 42∶18∶18∶18 4 N PRU*42/96 N PRU*18/96
14 45∶17∶17∶17 4 N PRU*45/96 N PRU*17/96
15 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
16 51∶15∶15∶15 4 N PRU*51/96 N PRU*15/96
17 54∶14∶14∶14 4 N PRU*54/96 N PRU*14/96
18 57∶13∶13∶13 4 N PRU*57/96 N PRU*13/96
19 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
20 63∶11∶11∶11 4 N PRU*63/96 N PRU*11/96
21 66∶10∶10∶10 4 N PRU*66/96 N PRU*10/96
22 69∶9∶9∶9 4 N PRU*69/96 N PRU*9/96
23 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
24 Keep Keep Keep Keep
25 Keep Keep Keep Keep
26 Keep Keep Keep Keep
27 Keep Keep Keep Keep
28 Keep Keep Keep Keep
29 Keep Keep Keep Keep
30 Keep Keep Keep Keep
31 Keep Keep Keep Keep
Can from said method, determine for the required bit number of indication UFPC parameter under each bandwidth, but for different bandwidth, indicate the each other identical or complete difference of part of the required bit number of UFPC parameter.For example,
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
It may be noted that: in the collocation method of above-mentioned UFPC, used identical bit number indication UFPC parameter when two different bandwidth, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits, but corresponding form is table 4.13; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits, but corresponding form is table 4.18.
Because system bandwidth is 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be 10MHz (can be 7MHz or 8.75MHz) adopts identical UFPC when being 20MHz with system bandwidth value and corresponding relation with system bandwidth, thereby so that device fabrication is simpler, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth.For example, when system bandwidth was 5MHz, the required bit number of indication UFPC parameter was 2bits; And system bandwidth is when being 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and indicating the required bit number of this parameter is 4bits.
It is 5MHz that the same table 4.1 to table 4.3 has been described system bandwidth, and indication UFPC required bit number is the collocation method in the situation of 2bits, repeats no more here.
When system bandwidth is 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and the required bit number of indication UFPC is in the situation of 4bits, can all adopt one of 20MHz form when needing 4bits indication UFPC, for example, all be with table 4.13 or table 4.18.
In addition; it may be noted that: in the collocation method of above-mentioned UFPC; for each table; relation in the middle of the meaning of the value of UFPC and the indication of the value of UFPC can change; each form all is embodiment; the value that needs only the UFPC that comprises in the table is that the meaning of indication is identical, all is regarded as identical table, all within protection range.For example, table 4.23 all is considered as identical table with table 4.15, is identical because the value of the UFPC that comprises in two tables is the meaning of indication.
Table 4.23
UFPC Each frequency partition ratio (FP 0∶FP 1∶FP 2∶ FP 3) Effective frequency number of partitions FPCT FP 0Big or small FPS 0 Other partition size FPS i(i＞0)
0 0∶32∶32∶32 3 0 N PRU*32/96
1 12∶28∶28∶28 4 N PRU*12/96 N PRU*28/96
2 24∶24∶24∶24 4 N PRU*24/96 N PRU*24/96
3 36∶20∶20∶20 4 N PRU*36/96 N PRU*20/96
4 48∶16∶16∶16 4 N PRU*48/96 N PRU*16/96
5 60∶12∶12∶12 4 N PRU*60/96 N PRU*12/96
6 72∶8∶8∶8 4 N PRU*72/96 N PRU*8/96
7 96∶0∶0∶0 1 N PRU 0
Other form of listing and produced into by said method but all do not follow this principle at this form of listing.
By above-mentioned example 4, can find out, system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), during the 20MHz system, the bit number of indication UFPC needs respectively 2bits, 3bits, 4bits, perhaps need respectively 2bits, 4bits, 4bits, perhaps need respectively 3bits, 4bits, 4bits, perhaps need respectively 3bits, 4bits, when the combination such as 5bits or other combination, in the situation that the possible value of UFPC reduces, deleted redundant and unnecessary information indication, save overhead bit, and guaranteed certain flexibility.
The collocation method of downstream frequency subregion Subband number (DFPSC)
When DFPSC got different values, descending Frequency Partitioning process was different, such as Figure 14.The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and is divided into three class bandwidth the configuring condition of DFPSC is described, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.
Example 5
The first kind: when system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits.
For 5MHz, the set of the number of the Subband that may indicate of DFPSC is: A DFPSC1={ 0,1,2}.1bits represents the number of 2 kinds of different Subband, and the number of these 2 kinds of different Subband is taken from set A DFPSC1, be total to C 3 2=3 kinds of combinations.For example, shown in table 5.1～table 5.3.
Table 5.1
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 1 1
Table 5.2
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 1 1 2
Table 5.3
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 1 2
Perhaps, when system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 2bits.Shown in table 5.4.
Table 5.4
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 2
1 1 3 Keep
Equations of The Second Kind: when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication DFPSC parameter was 2bits.
For 7MHz or 8.75MHz or 10MHz, the set of the number of the Subband that may indicate of DFPSC is: A DFPSC2={ 0,1,2,3,4}.2bits represents the number of 4 kinds of different Subband, and the number of these 4 kinds of different Subband is taken from set A DFPSC2, be total to C 5 4=5 kinds of combinations.For example, shown in table 5.5～table 5.9.
Table 5.5
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 2
1 1 3 3
Table 5.6
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 2
1 1 3 4
Table 5.7
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 3
1 1 3 4
Table 5.8
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 1 2 3
1 2 3 4
Table 5.9
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 3
1 2 3 4
Perhaps, when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication DFPSC parameter was 3bits.The corresponding relation of the Subband number of the frequency partition that the value of DFPSC is corresponding with it.For example, shown in the table 5.10.
Table 5.10
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 4 4
1 1 5 Keep
2 2 6 Keep
3 3 7 Keep
The 3rd class: when system bandwidth was 20MHz, the required bit number of indication DFPSC parameter was 2bits.
For 20MHz, the set of the number of the Subband that may indicate of DFPSC is: A DFPSC3={ 0,1,2,3,4,5,6,7,8}.2bits represents the number of 4 kinds of different Subband, and the number of these 4 kinds of different Subband is taken from set A DFPSC3, be total to C 9 4=126 kinds of combinations.For example, when table 5.11～table 5.14 has been described system bandwidth and is 20MHz, and the required bit number of indication DFPSC is in the situation of 2bits, the corresponding relation of the Subband number of the frequency partition that the value of DFPSC is corresponding with it, and other combination is not enumerated one by one.
Table 5.11
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 2 2
1 1 3 3
Table 5.12
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 2 2
1 1 3 4
Table 5.13
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 2 4
1 2 3 6
Table 5.14
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 1 2 3
1 2 3 4
Perhaps, when system bandwidth was 20MHz, the required bit number of indication DFPSC parameter was 3bits.3bits represents the number of 8 kinds of different Subband, and the number of these 4 kinds of different Subband is taken from set A DFPSC3, be total to C 9 8=9 kinds of combinations.For example, when table 5.15～table 5.17 has been described bit number and has been 3bits, the corresponding relation of the Subband number of the frequency partition that the value of DFPSC is corresponding with it, other combination is not enumerated one by one.
Table 5.15
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 5.16
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 5.17
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Perhaps, when system bandwidth was 20MHz, the required bit number of indication DFPSC parameter was 4bits.For example, when table 5.18 has been described bit number and has been 4bits, the corresponding relation of the Subband number of the frequency partition that the value of DFPSC is corresponding with it.
Table 5.18
DFPSC The corresponding Subband number of FPi (i＞0) DFPSC The corresponding Subband number of FPi (i＞0)
0 0 8 8
1 1 9 Keep
2 2 10 Keep
3 3 11 Keep
4 4 12 Keep
5 5 13 Keep
6 6 14 Keep
7 7 15 Keep
Can from said method, determine for the required bit number of indication DFPSC parameter under each bandwidth, but for different bandwidth, indicate the each other identical or complete difference of part of the required bit number of DFPSC parameter.For example,
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 2bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication DFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
It may be noted that: in the collocation method of above-mentioned DFPSC, used identical bit number indication DFPSC parameter when two different bandwidth, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits, but corresponding form is table 5.10; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits, but corresponding form is table 5.15.
Again for example, when system bandwidth is 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and indication DFPSC required bit number is in the situation of 3bits, can all adopt one of 20MHz form when needing 3bits indication DFPSC, for example, all be with table 5.15.
In addition; it may be noted that: in the collocation method of above-mentioned DFPSC; for each table; relation in the middle of the meaning of the value of DFPSC and the indication of the value of DFPSC can change; each form all is embodiment; the value that needs only the DFPSC that comprises in the table is that the meaning of indication is identical, all is regarded as identical table, all within protection range.For example, table 5.19 all is considered as identical table with table 5.15, is identical because the value of the DFPSC that comprises in two tables is the meaning of indication.
Table 5.19
DFPSC FP i(i＞0) corresponding Subband number DFPSC FP i(i＞0) corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 7
3 3 7 6
Other form of listing and produced into by said method but all do not follow this principle at this form of listing.
By above-mentioned example 5, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, the bit number of indication DFPSC needs respectively 1bits, 2bits, 2bits, perhaps need respectively 2bits, 3bits, 3bits, when perhaps needing respectively 2bits, 3bits, 4bits or other combination, in the situation that the possible value of DFPSC reduces, deleted redundant and unnecessary information indication, save overhead bit, and guaranteed certain flexibility.
The collocation method of upstream frequency subregion Subband number (UFPSC)
Example 6
When UFPSC got different values, up Frequency Partitioning process also was different, such as Figure 15.The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and is divided into three class bandwidth the configuring condition of UFPSC is described, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.
The first kind: when system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits.
For 5MHz, the set of the number of the Subband that may indicate of UFPSC is: A UFPSC={ 0,1,2}.1bits represents the number of 2 kinds of different Subband, and the number of these 2 kinds of different Subband is taken from set A UFPSC, be total to C 3 2=3 kinds of combinations.For example, shown in table 6.1～table 6.3.
Table 6.1
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 1 1
Table 6.2
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 1 1 2
Table 6.3
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 1 2
Perhaps, when system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 2bits.Shown in table 6.4.
Table 6.4
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 2
1 1 3 Keep
Equations of The Second Kind: when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication UFPSC parameter was 2bits.
For 7MHz or 8.75MHz or 10MHz, the set of the number of the Subband that may indicate of UFPSC is: B UFPSC={ 0,1,2,3,4}.2bits represents the number of 4 kinds of different Subband, and the number of these 4 kinds of different Subband is taken from set B UFPSC, be total to C 5 4=5 kinds of combinations.For example, shown in table 6.5～table 6.9.
Table 6.5
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 2
1 1 3 3
Table 6.6
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 2
1 1 3 4
Table 6.7
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 3
1 1 3 4
Table 6.8
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 1 2 3
1 2 3 4
Table 6.9
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 2 3
1 2 3 4
Perhaps, when system bandwidth was 7MHz or 8.75MHz or 10MHz, the required bit number of indication UFPSC parameter was 3bits.The corresponding relation of the Subband number of the frequency partition that the value of UFPSC is corresponding with it.For example, shown in the table 6.10.
Table 6.10
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 4 4
1 1 5 Keep
2 2 6 Keep
3 3 7 Keep
The 3rd class: when system bandwidth was 20MHz, the required bit number of indication UFPSC parameter was 2bits.
For 20MHz, the set of the number of the Subband that may indicate of UFPSC is: C UFPSC={ 0,1,2,3,4,5,6,7,8}.2bits represents the number of 4 kinds of different Subband, and the number of these 4 kinds of different Subband is taken from set C UFPSC, be total to C 9 4=126 kinds of combinations.For example, when table 6.11～table 6.14 has been described system bandwidth and is 20MHz, and the required bit number of indication UFPSC is in the situation of 2bits, the corresponding relation of the Subband number of the frequency partition that the value of UFPSC is corresponding with it, and other combination is not enumerated one by one.
Table 6.11
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 2 2
1 1 3 3
Table 6.12
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 2 2
1 1 3 4
Table 6.13
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 2 4
1 2 3 6
Table 6.14
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 1 2 3
1 2 3 4
Perhaps, when system bandwidth was 20MHz, the required bit number of indication UFPSC parameter was 3bits.3bits represents the number of 8 kinds of different Subband, and the number of these 4 kinds of different Subband is taken from set C UFPSC, be total to C 9 8=9 kinds of combinations.For example, when table 6.15～table 6.17 has been described bit number and has been 3bits, the corresponding relation of the Subband number of the frequency partition that the value of UFPSC is corresponding with it, other combination is not enumerated one by one.
Table 6.15
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 6.16
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 6.17
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Perhaps, when system bandwidth was 20MHz, the required bit number of indication UFPSC parameter was 4bits.For example, when table 6.18 has been described bit number and has been 4bits, the corresponding relation of the Subband number of the frequency partition that the value of UFPSC is corresponding with it.
Table 6.18
UFPSC The corresponding Subband number of FPi (i＞0) UFPSC The corresponding Subband number of FPi (i＞0)
0 0 8 8
1 1 9 Keep
2 2 10 Keep
3 3 11 Keep
4 4 12 Keep
5 5 13 Keep
6 6 14 Keep
7 7 15 Keep
Can from said method, determine for the required bit number of indication UFPSC parameter under each bandwidth, but for different bandwidth, indicate the each other identical or complete difference of part of the required bit number of UFPSC parameter.For example,
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 2bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth was 5MHz, the required bit number of indication UFPSC parameter was 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
It may be noted that: in the collocation method of above-mentioned UFPSC, used identical bit number indication UFPSC parameter when two different bandwidth, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits, but corresponding form is table 6.10; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits, but corresponding form is table 6.15.
Again for example, when system bandwidth is 10MHz (also can be 7MHz or 8.75MHz) and 20MHz, and indication UFPSC required bit number is in the situation of 3bits, can all adopt one of 20MHz form when needing 3bits indication UFPSC, for example, all be with table 6.15.
In addition; it may be noted that: in the collocation method of above-mentioned UFPSC; for each table; relation in the middle of the meaning of the value of UFPSC and the indication of the value of UFPSC can change; each form all is embodiment; the value that needs only the UFPSC that comprises in the table is that the meaning of indication is identical, all is regarded as identical table, all within protection range.For example, table 6.19 all is considered as identical table with table 6.15, is identical because the value of the UFPSC that comprises in two tables is the meaning of indication.
Table 6.19
UFPSC FP i(i＞0) corresponding Subband number UFPSC FP i(i＞0) corresponding Subband number
0 0 4 4
1 1 5 5
2 2 6 7
3 3 7 6
Other form of listing and produced into by said method but all do not follow this principle at this form of listing.
By above-mentioned example 6, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, the bit number of indication UFPSC needs respectively 1bits, 2bits, 2bits, perhaps need respectively 2bits, 3bits, 3bits, when perhaps needing respectively 2bits, 3bits, 4bits or other combination, in the situation that the possible value of UFPSC reduces, deleted redundant and unnecessary information indication, save overhead bit, and guaranteed certain flexibility.
Descending CRU allotment (DCAS based on Subband SB) collocation method
Example 7
DCAS SBiIndicated the number of CRU in the individual frequency partition of i (i 〉=0) and/or DRU take Subband as unit.DCAS SBiWhen getting different values, descending CRU/DRUAllocation process is different, as shown in figure 16.
The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and be divided into three class bandwidth to DCAS SBiConfiguring condition describe, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.
The first kind: when system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 2bits.
For 5MHz, DCAS SBiIndicated take Subband as unit the possible number set of the number of CRU in i the frequency partition and/or DRU as: A D CAS SBi = { 0,1,2,3 , 4,5,6 } . 2bits represents 4 different numbers, and these 4 different numbers are taken from set
Figure G2009101305580D00993
Be total to C 7 4=35 kinds of combinations.DCAS SBiFor example can adopt any one combination, a kind of in shown in table 7.1～table 7.6, other is similar, enumerates no longer one by one.
Table 7.1
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 3
Table 7.2
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 4
1 2 3 6
Table 7.3
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 3
1 2 3 4
Table 7.4
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 1 2 3
1 2 3 4
Table 7.5
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 6
Table 7.6
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 4
Perhaps, when system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 3bits.3bits represents 8 different numbers, can represent set
Figure G2009101305580D01001
In all numerical value.Shown in 7.7.
Table 7.7
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 Keep
Equations of The Second Kind: when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication DCAS SBiThe required bit number of parameter is 2bits.
For 7MHz or 8.75MHz or 10MHz, DCAS SBiIndicated take Subband as unit the possible number set of the number of CRU in i the frequency partition and/or DRU as: B DCAS SBi = { 0,1,2,3,4,5,6,7,8,9,10,11,12 } . 2bits represents 4 different numbers, and these 4 different numbers are taken from set
Figure G2009101305580D01012
Be total to C 13 4=715 kinds of combinations.DCAS SBiCan adopt any one combination.For example, a kind of in shown in table 7.8～table 7.9, other is similar, enumerates no longer one by one.
Table 7.8
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 4
Table 7.9
DCAS SBi FP i(i 〉=0) corresponding CRU number DCAS FP i(i 〉=0) corresponding CRU number
0 0 2 4
1 2 3 6
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication DCAS SBiThe required bit number of parameter is 3bits.3bits represents 8 different numbers, and these 8 different numbers are taken from set Be total to C 13 8=1287 kinds of combinations.DCAS SBiFor example can adopt any one combination, a kind of in shown in table 7.10～table 7.13, other is similar, enumerates no longer one by one.
Table 7.10
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 7.11
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 7.12
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 7.13
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 12
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication DCAS SBiThe required bit number of parameter is 4bits.4bits represents 16 different numbers, can represent set
Figure G2009101305580D01021
In all numerical value.For example, shown in table 7.14.
Table 7.14
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 Keep
6 6 14 Keep
7 7 15 Keep
The 3rd class: when system bandwidth is 20MHz, indication DCAS SBiThe required bit number of parameter is 3bits.
For 20MHz, DCAS SBiIndicated take Subband as unit the possible number set of the number of CRU in i the frequency partition and/or DRU as: C D CAS SBi = { 0,1,2,3 , 4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24 } . 3bits represents 8 different numbers, and these 8 different numbers are taken from set
Figure G2009101305580D01033
Be total to C 25 8=1081575 kinds of combinations.DCAS SBiCan adopt any one combination.For example, a kind of in shown in table 7.15～table 7.21, other is similar, enumerates no longer one by one.
Table 7.15
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 7.16
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 10
Table 7.17
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 6
1 1 5 8
2 2 6 10
3 4 7 12
Table 7.18
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 7.18
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 7.19
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 24
Table 7.20
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 7.21
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 1 4 6
1 2 5 8
2 3 6 10
3 4 7 12
Perhaps, when system bandwidth is 20MHz, indication DCAS SBiThe required bit number of parameter is 4bits.4bits represents 16 different numbers, and these 16 different numbers are taken from set
Figure G2009101305580D01051
Be total to C 25 16=2042975 kinds of combinations.DCAS SBiCan adopt any one combination.For example, a kind of in shown in table 7.22～table 7.25, other is similar, enumerates no longer one by one.
Table 7.22
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 8 10
1 1 9 12
2 2 10 14
3 3 11 16
4 4 12 18
5 5 13 20
6 6 14 22
7 8 15 24
Table 7.23
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 7.24
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 1 8 9
1 2 9 10
2 3 10 11
3 4 11 12
4 5 12 13
5 6 13 14
6 7 14 15
7 8 15 16
Table 7.25
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 1 8 10
1 2 9 12
2 3 10 14
3 4 11 16
4 5 12 18
5 6 13 20
6 7 14 22
7 8 15 24
Perhaps, when system bandwidth is 20MHz, indication DCAS SBiThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers can represent set
Figure G2009101305580D01061
In all numerical value.For example, shown in the table 7.26.
Table 7.26
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 1 16 17
1 2 17 18
2 3 18 19
3 4 19 20
4 5 20 21
5 6 21 22
6 7 22 23
7 8 23 24
8 9 24 Keep
9 10 25 Keep
10 11 26 Keep
11 12 27 Keep
12 13 28 Keep
13 14 29 Keep
14 15 30 Keep
15 16 31 Keep
For indicating DCAS under each bandwidth SBiThe required bit number of parameter can be determined from said method, but for different bandwidth, indicate DCAS SBiThe required bit number of parameter is the identical or complete difference of part each other.For example,
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 2bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication DCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
It may be noted that: at above-mentioned DCAS SBiCollocation method in, when two different bandwidth have been used identical bit number indication DCAS SBiParameter, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits, but corresponding form is table 7.14; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits, but corresponding form is table 7.23.
Identical form refers to: owing to system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be that 10MHz (can be 7MHz or 8.75MHz) adopts identical DCAS when being 20MHz with system bandwidth with system bandwidth SBiValue and corresponding relation, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth, for example, can adopt in table 7.22～table 7.25, the collocation method during perhaps according to 20MHz produces.Perhaps, produce according to the methods below:
Table 7.27
DCAS Account for the umber of PRU sum The corresponding CRU number of 10MHz (can be 7MHz or 8.75MHz) FPi (i＞0) The corresponding CRU number of 20MHzFPi (i＞0)
0 0/48 0 0
1 1/48 1 2
2 2/48 2 4
3 3/48 3 6
4 4/48 4 8
5 5/48 5 10
6 6/48 6 12
7 7/48 7 14
8 8/48 8 16
9 9/48 9 18
10 10/48 10 20
11 11/48 11 22
12 12/48 12 24
13 13/48 13 26
14 14/48 14 28
15 15/48 15 30
In addition, 5MHz can all adopt 2 bits or 3 bits with 10MHz (also can be 7MHz or 8.75MHz).
In addition, it may be noted that: at above-mentioned DCAS SBiCollocation method in, for each table, DCAS SBiValue and DCAS SBiThe meaning of value indication in the middle of relation can change, each form all is embodiment, as long as the DCAS that comprises in table SBiValue be the indication meaning be identical, all be regarded as identical table, all within protection range.
By above-mentioned example 7, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, indication DCAS SBiBit number need respectively 2bits, 3bits, 3bits, at DCAS SBiPossible value situation about reducing under, deleted that redundant and unnecessary information indicates, saved overhead bit, and guaranteed certain flexibility.
UCAS SBiIndicated the number of CRU in i the frequency partition and/or DRU take Subband as unit, as shown in figure 11, UCAS SBiWhen getting different values, descending CRU/DRU Allocation process is different.
Up CRU allotment (UCAS based on Subband SB) collocation method
Example 8
UCAS SBiIndicated the number of CRU in the individual frequency partition of i (i 〉=0) and/or DRU take Subband as unit.UCAS SBiWhen getting different values, up CRU/DRUAllocation process is different, as shown in figure 17.
The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and be divided into three class bandwidth to UCAS SBiConfiguring condition describe, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.
The first kind: when system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 2bits.
For 5MHz, UCAS SBiIndicated take Subband as unit the possible number set of the number of CRU in i the frequency partition and/or DRU as: A UCAS SBi = { 0,1,2 , 3,4,5,6 } . 2bits represents 4 different numbers, and these 4 different numbers are taken from set Be total to C 7 4=35 kinds of combinations.UCAS SBiFor example can adopt any one combination, a kind of in shown in table 8.1～table 8.6, other is similar, enumerates no longer one by one.
Table 8.1
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 3
Table 8.2
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 2 4
1 2 3 6
Table 8.3
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 2 3
1 2 3 4
Table 8.4
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 1 2 3
1 2 3 4
Table 8.5
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 6
Table 8.6
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 4
Perhaps, when system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 3bits.3bits represents 8 different numbers, can represent set
Figure G2009101305580D01121
In all numerical value.Shown in 8.7.
Table 8.7
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 Keep
Equations of The Second Kind: when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication UCAS SBiThe required bit number of parameter is 2bits.
For 7MHz or 8.75MHz or 10MHz, UCAS SBiIndicated take Subband as unit the possible number set of the number of CRU in i the frequency partition and/or DRU as: B UCA S SBi = { 0,1,2,3,4,5,6,7,8,9,10,11,12 } . 2bits represents 4 different numbers, and these 4 different numbers are taken from set
Figure G2009101305580D01123
Be total to C 13 4=715 kinds of combinations.UCAS SbiCan adopt any one combination.For example, a kind of in shown in table 8.8～table 8.9, other is similar, enumerates no longer one by one.
Table 8.8
UCAS SBi FP i(i 〉=0) corresponding CRU number UCAS FP i(i 〉=0) corresponding CRU number
0 0 2 2
1 1 3 4
Table 8.9
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication UCAS SBiThe required bit number of parameter is 3bits.3bits represents 8 different numbers, and these 8 different numbers are taken from set
Figure G2009101305580D01131
Be total to C 13 8=1287 kinds of combinations.UCAS SBiFor example can adopt any one combination, a kind of in shown in table 8.10～table 8.13, other is similar, enumerates no longer one by one.
Table 8.10
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 8.11
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 8
Table 8.12
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 8.13
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 12
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication UCAS SBiThe required bit number of parameter is 4bits.4bits represents 16 different numbers, can represent set
Figure G2009101305580D01141
In all numerical value.For example, shown in table 8.14.
Table 8.14
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 Keep
6 6 14 Keep
7 7 15 Keep
The 3rd class: when system bandwidth is 20MHz, indication UCAS SBiThe required bit number of parameter is 3bits.
For 20MHz, UCAS SBiIndicated take Subband as unit the possible number set of the number of CRU in i the frequency partition and/or DRU as: C UCA S SBi = { 0,1,2,3 , 4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24 } . 3bits represents 8 different numbers, and these 8 different numbers are taken from set
Figure G2009101305580D01144
Be total to C 25 8=1081575 kinds of combinations.UCAS SBiCan adopt any one combination.For example, a kind of in shown in table 8.15～table 8.21, other is similar, enumerates no longer one by one.
Table 8.15
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 8.16
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 10
Table 8.17
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 6
1 1 5 8
2 2 6 10
3 4 7 12
Table 8.18
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 8.18
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 8.19
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 24
Table 8.20
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 8.21
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 1 4 6
1 2 5 8
2 3 6 10
3 4 7 12
Perhaps, when system bandwidth is 20MHz, indication UCAS SBiThe required bit number of parameter is 4bits.4bits represents 16 different numbers, and these 16 different numbers are taken from set Be total to C 25 16=2042975 kinds of combinations.UCAS SBiCan adopt any one combination.For example, a kind of in shown in table 8.22～table 8.25, other is similar, enumerates no longer one by one.
Table 8.22
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 8 10
1 1 9 12
2 2 10 14
3 3 11 16
4 4 12 18
5 5 13 20
6 6 14 22
7 8 15 24
Table 8.23
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 8.24
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 1 8 9
1 2 9 10
2 3 10 11
3 4 11 12
4 5 12 13
5 6 13 14
6 7 14 15
7 8 15 16
Table 8.25
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 1 8 10
1 2 9 12
2 3 10 14
3 4 11 16
4 5 12 18
5 6 13 20
6 7 14 22
7 8 15 24
Perhaps, when system bandwidth is 20MHz, indication UCAS SBiThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers can represent set In all numerical value.For example, shown in the table 8.26.
Table 8.26
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 1 16 17
1 2 17 18
2 3 18 19
3 4 19 20
4 5 20 21
5 6 21 22
6 7 22 23
7 8 23 24
8 9 24 Keep
9 10 25 Keep
10 11 26 Keep
11 12 27 Keep
12 13 28 Keep
13 14 29 Keep
14 15 30 Keep
15 16 31 Keep
For indicating UCAS under each bandwidth SBiThe required bit number of parameter can be determined from said method, but for different bandwidth, indicate UCAS SBiThe required bit number of parameter is the identical or complete difference of part each other.For example,
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 2bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits;
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 1bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 2bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 3bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 3bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication UCAS SBiThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
It may be noted that: at above-mentioned UCAS SBiCollocation method in, when two different bandwidth have been used identical bit number indication UCAS SBiParameter, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits, but corresponding form is table 8.14; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits, but corresponding form is table 8.23.
Identical form refers to: owing to system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be that 10MHz (can be 7MHz or 8.75MHz) adopts identical UCAS when being 20MHz with system bandwidth with system bandwidth SBiValue and corresponding relation, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth, for example, can adopt in table 8.22～table 8.25, the collocation method during perhaps according to 20MHz produces.Perhaps, produce according to the methods below:
Table 8.27
UCAS Account for the umber of PRU sum The corresponding CRU number of 10MHz (can be 7MHz or 8.75MHz) FPi (i＞0) The corresponding CRU number of 20MHzFPi (i＞0)
0 0/48 0 0
1 1/48 1 2
2 2/48 2 4
3 3/48 3 6
4 4/48 4 8
5 5/48 5 10
6 6/48 6 12
7 7/48 7 14
8 8/48 8 16
9 9/48 9 18
10 10/48 10 20
11 11/48 11 22
12 12/48 12 24
13 13/48 13 26
14 14/48 14 28
15 15/48 15 30
In addition, 5MHz can all adopt 2 bits or 3 bits with 10MHz (also can be 7MHz or 8.75MHz).
In addition, it may be noted that: at above-mentioned UCAS SBiCollocation method in, for each table, UCAS SBiValue and UCAS SBiThe meaning of value indication in the middle of relation can change, each form all is embodiment, as long as the UCAS that comprises in table SBiValue be the indication meaning be identical, all be regarded as identical table, all within protection range.
By above-mentioned example 8, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, indication UCAS SBiBit number need respectively 2bits, 3bits, 3bits, at UCAS SBiPossible value situation about reducing under, deleted that redundant and unnecessary information indicates, saved overhead bit, and guaranteed certain flexibility.
Descending CRU allotment DCAS based on Miniband MBCollocation method
DCAS MBIndicated in the 0th frequency partition number based on the CRU of Miniband take Miniband as unit.
Example 9
The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and be divided into three class bandwidth to DCAS MBConfiguring condition describe, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.
The first kind: indication DCAS MBThe required bit number of parameter is 2bits.
For 5MHz, DCAS MBIndicated take Miniband as unit in the 0th frequency partition based on the possible number set of the number of the CRU of Miniband as:
Figure G2009101305580D01221
Figure G2009101305580D01222
2bits represents 4 different numbers, and these 4 different numbers are taken from set
Figure G2009101305580D01223
Be total to C 25 4=12650 kinds of combinations.DCAS MBCan adopt any one combination, for example, a kind of in shown in table 9.1～table 9.3, other is similar, enumerates no longer one by one.
Table 9.1
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 4
1 2 5 6
Table 9.2
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 6
1 4 5 8
Table 9.3
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 4 5 12
Perhaps, when system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 3bits.3bits represents 8 different numbers, and these 8 different numbers are taken from set
Figure G2009101305580D01231
Be total to C 25 8=1081575 kinds of combinations.DCAS MBCan adopt any one combination, for example, a kind of in shown in table 9.4～table 9.10, other is similar, enumerates no longer one by one.
Table 9.4
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 9.5
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 10
Table 9.6
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 6
1 1 5 8
2 2 6 10
3 4 7 12
Table 9.7
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 9.8
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 9.9
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 24
Table 9.10
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 9.11
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 1 4 6
1 2 5 8
2 3 6 10
3 4 7 12
Perhaps, when system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 4bits.4bits represents 16 different numbers, and these 16 different numbers are taken from set Be total to C 25 16=2042975 kinds of combinations.DCAS MBCan adopt any one combination, for example, a kind of in shown in table 9.12～table 9.15, other is similar, enumerates no longer one by one.
Table 9.12
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 10
1 1 9 12
2 2 10 14
3 3 11 16
4 4 12 18
5 5 13 20
6 6 14 22
7 8 15 24
Table 9.13
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 9.14
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 1 8 9
1 2 9 10
2 3 10 11
3 4 11 12
4 5 12 13
5 6 13 14
6 7 14 15
7 8 15 16
Table 9.15
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 1 8 10
1 2 9 12
2 3 10 14
3 4 11 16
4 5 12 18
5 6 13 20
6 7 14 22
7 8 15 24
Perhaps, when system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers can represent set
Figure G2009101305580D01261
In all numerical value.For example, shown in the table 9.16.
Table 9.16
DCAS SBi FP i(i＞0) corresponding CRU number DCAS SBi FP i(i＞0) corresponding CRU number
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 Keep
10 10 26 Keep
11 11 27 Keep
12 12 28 Keep
13 13 29 Keep
14 14 30 Keep
15 15 31 Keep
Equations of The Second Kind: when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication DCAS MBThe required bit number of parameter is 4bits.
For 7MHz or 8.75MHz or 10MHz, DCAS MBIndicated take Miniband as unit in the 0th frequency partition based on the possible number set of the number of the CRU of Miniband as: 4bits represents 16 different numbers, and these 16 different numbers are taken from set Be total to C 49 16=3348108992991 kinds of combinations.DCAS MBCan adopt any one combination, for example, a kind of in shown in table 9.17～table 9.20, other is similar, enumerates no longer one by one.
Table 9.17
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 9.18
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 16
Table 9.19
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 10
1 1 9 12
2 2 10 14
3 3 11 16
4 4 12 18
5 5 13 20
6 6 14 22
7 8 15 24
Table 9.20
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 20
1 1 9 24
2 2 10 28
3 3 11 32
4 4 12 36
5 8 13 40
6 12 14 44
7 16 15 48
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication DCAS MBThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers are taken from set
Figure G2009101305580D01291
Be total to C 49 32=6499270398159 kinds of combinations.DCAS MBCan adopt any one combination, for example, a kind of in shown in the table 9.21, other is similar, enumerates no longer one by one.
Table 9.21
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 25
10 10 26 26
11 11 27 27
12 12 28 28
13 13 29 29
14 14 30 30
15 15 31 31
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication DCAS SBiThe required bit number of parameter is 6bits.6bits represents 64 different numbers, can represent set
Figure G2009101305580D01292
In all numerical value.For example, shown in table 9.22.
Table 9.22
DCAS MB FP 0In based on Miniband DCAS MB FP 0In based on Miniband
The number of CRU The number of CRU
0 0 32 32
1 1 33 33
2 2 34 34
3 3 35 35
4 4 36 36
5 5 37 37
6 6 38 38
7 7 39 39
8 8 40 40
9 9 41 41
10 10 42 42
11 11 43 43
12 12 44 44
13 13 45 45
14 14 46 46
15 15 47 47
16 16 48 48
17 17 49 Keep
18 18 50 Keep
19 19 51 Keep
20 20 52 Keep
21 21 53 Keep
22 22 54 Keep
23 23 55 Keep
24 24 56 Keep
25 25 57 Keep
26 26 58 Keep
27 27 59 Keep
28 28 60 Keep
29 29 61 Keep
30 30 62 Keep
31 31 63 Keep
The 3rd class: for 20MHz, indication DCAS MBThe required bit number of parameter is 4bits.
DCAS MBIndicated take Miniband as unit in the 0th frequency partition based on the possible number set of the number of the CRU of Miniband as:
Figure G2009101305580D01311
Figure G2009101305580D01312
4bits represents 16 different numbers, and these 16 different numbers are taken from set Be total to C 97 16=793067310934425856 kinds of combinations.DCAS MBCan adopt any one combination, for example, a kind of in shown in table 9.17～table 9.20, other is similar, enumerates no longer one by one.
Perhaps, when system bandwidth is 20MHz, indication DCAS MBThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers are taken from set
Figure G2009101305580D01314
Be total to C 97 32Plant combination.DCAS MBCan adopt any one combination.For example, a kind of in shown in table 9.23～table 9.25, other is similar, enumerates no longer one by one.
Table 9.23
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 25
10 10 26 26
11 11 27 27
12 12 28 28
13 13 29 29
14 14 30 30
15 15 31 31
Table 9.24
DCAS MB FP 0In based on Miniband DCAS MB FP 0In based on Miniband
The number of CRU The number of CRU
0 0 16 16
1 1 17 18
2 2 18 20
3 3 19 22
4 4 20 24
5 5 21 26
6 6 22 28
7 7 23 30
8 8 24 32
9 9 25 34
10 10 26 36
11 11 27 38
12 12 28 40
13 13 29 42
14 14 30 46
15 15 31 48
Table 9.25
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 18
1 1 17 20
2 2 18 22
3 3 19 24
4 4 20 26
5 5 21 28
6 6 22 30
7 7 23 32
8 8 24 36
9 9 25 40
10 10 26 44
11 11 27 48
12 12 28 52
13 13 29 56
14 14 30 60
15 16 31 64
Perhaps, when system bandwidth is 20MHz, indication DCAS MBThe required bit number of parameter is 6bits.6bits represents 64 different numbers, and these 64 different numbers can represent set In all numerical value.For example, shown in the table 9.26.
Table 9.26
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0 32 32
1 1 33 33
2 2 34 34
3 3 35 35
4 4 36 36
5 5 37 37
6 6 38 38
7 7 39 39
8 8 40 40
9 9 41 41
10 10 42 42
11 11 43 43
12 12 44 44
13 13 45 45
14 14 46 46
15 15 47 47
16 16 48 48
17 17 49 49
18 18 50 50
19 19 51 51
20 20 52 52
21 21 53 53
22 22 54 54
23 23 55 55
24 24 56 56
25 25 57 57
26 26 58 58
27 27 59 59
28 28 60 60
29 29 61 61
30 30 62 62
31 31 63 63
Perhaps, when system bandwidth is 20MHz, indication DCAS MBThe required bit number of parameter is 7bits.7bits represents 128 different numbers, and these 128 different numbers can represent set
Figure G2009101305580D01341
In except 0 or 1 or 95 or 96 all numerical value.
For indicating DCAS under each bandwidth MBThe required bit number of parameter can be determined from said method, but for different bandwidth, indicate DCAS MBThe required bit number of parameter is the identical or complete difference of part each other.For example,
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 6bits; Perhaps,
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 5bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 6bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 6bits;
When system bandwidth is 5MHz, indication DCAS MBThe required bit number of parameter is 5bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 6bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 7bits; Perhaps,
It may be noted that: at above-mentioned DCAS MBCollocation method in, when two different bandwidth have been used identical bit number indication DCAS MBParameter, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits, but corresponding form is table 9.22; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits, but corresponding form is table 9.23.
Identical form refers to: owing to system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be that 10MHz (can be 7MHz or 8.75MHz) adopts identical DCAS when being 20MHz with system bandwidth with system bandwidth MBValue and corresponding relation, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth, for example, can adopt in table 9.23～table 9.25, the collocation method during perhaps according to 20MHz produces.Perhaps, produce according to the methods below:
Table 9.27
DCAS MB FP 0In based on the number of the CRU of Miniband DCAS MB FP 0In based on the number of the CRU of Miniband
0 0/48 0 0
1 1/48 1 2
2 2/48 2 4
3 3/48 3 6
4 4/48 4 8
5 5/48 5 10
6 6/48 6 12
7 7/48 7 14
8 8/48 8 16
9 9/48 9 18
10 10/48 10 20
11 11/48 11 22
12 12/48 12 24
13 13/48 13 26
14 14/48 14 28
15 15/48 15 30
In addition, 5MHz can all adopt 3 bits or 4 bits or 5 bits with 10MHz (also can be 7MHz or 8.75MHz).
In addition, it may be noted that: at above-mentioned DCAS MBCollocation method in, for each table, DCAS MBValue and DCAS MBThe meaning of value indication in the middle of relation can change, each form all is embodiment, as long as the DCAS that comprises in table MBValue be the indication meaning be identical, all be regarded as identical table, all within protection range.
By above-mentioned example 9, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, indication DCAS MBBit number need respectively 4bits, 5bits, 6bits, at DCAS MBPossible value situation about reducing under, deleted that redundant and unnecessary information indicates, saved overhead bit, and guaranteed certain flexibility.
Up CRU allotment UCAS based on Miniband MBCollocation method
UCAS MBIndicated in the 0th frequency partition number based on the CRU of Miniband take Miniband as unit.
Example 10
The below take system bandwidth as 5MHz, 7MHz, 8.75MHz, 10MHz and 20MHz be example, and be divided into three class bandwidth to UCAS MBConfiguring condition describe, the first kind is 5MHz, Equations of The Second Kind is 7MHz or 8.75MHz or 10MHz, the 3rd class is 20MHz.
The first kind: indication UCAS MBThe required bit number of parameter is 2bits.
For 5MHz, UCAS MBIndicated take Miniband as unit in the 0th frequency partition based on the possible number set of the number of the CRU of Miniband as: 2bits represents 4 different numbers, and these 4 different numbers are taken from set
Figure G2009101305580D01373
Be total to C 25 4=12650 kinds of combinations.UCAS MBCan adopt any one combination, for example, a kind of in shown in table 10.1～table 10.3, other is similar, enumerates no longer one by one.
Table 10.1
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 4
1 2 5 6
Table 10.2
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 6
1 4 5 8
Table 10.3
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 4 5 12
Perhaps, when system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 3bits.3bits represents 8 different numbers, and these 8 different numbers are taken from set Be total to C 25 8=1081575 kinds of combinations.UCAS MBCan adopt any one combination, for example, a kind of in shown in table 10.4～table 10.10, other is similar, enumerates no longer one by one.
Table 10.4
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 4
1 1 5 5
2 2 6 6
3 3 7 7
Table 10.5
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 4
1 1 5 6
2 2 6 8
3 3 7 10
Table 10.6
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 6
1 1 5 8
2 2 6 10
3 4 7 12
Table 10.7
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 10.8
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 14
Table 10.9
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 4 8
1 2 5 10
2 4 6 12
3 6 7 24
Table 10.10
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 1 4 5
1 2 5 6
2 3 6 7
3 4 7 8
Table 10.11
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 1 4 6
1 2 5 8
2 3 6 10
3 4 7 12
Perhaps, when system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 4bits.4bits represents 16 different numbers, and these 16 different numbers are taken from set
Figure G2009101305580D01401
Be total to C 25 16=2042975 kinds of combinations.UCAS MBCan adopt any one combination, for example, a kind of in shown in table 10.12～table 10.15, other is similar, enumerates no longer one by one.
Table 10.12
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 10
1 1 9 12
2 2 10 14
3 3 11 16
4 4 12 18
5 5 13 20
6 6 14 22
7 8 15 24
Table 10.13
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 10.14
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 1 8 9
1 2 9 10
2 3 10 11
3 4 11 12
4 5 12 13
5 6 13 14
6 7 14 15
7 8 15 16
Table 10.15
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 1 8 10
1 2 9 12
2 3 10 14
3 4 11 16
4 5 12 18
5 6 13 20
6 7 14 22
7 8 15 24
Perhaps, when system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers can represent set
Figure G2009101305580D01421
In all numerical value.For example, shown in the table 10.16.
Table 10.16
UCAS SBi FP i(i＞0) corresponding CRU number UCAS SBi FP i(i＞0) corresponding CRU number
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 Keep
10 10 26 Keep
11 11 27 Keep
12 12 28 Keep
13 13 29 Keep
14 14 30 Keep
15 15 31 Keep
Equations of The Second Kind: when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication UCAS MBThe required bit number of parameter is 4bits.
For 7MHz or 8.75MHz or 10MHz, UcAS MBIndicated take Miniband as unit in the 0th frequency partition based on the possible number set of the number of the CRU of Miniband as:
Figure G2009101305580D01422
4bits represents 16 different numbers, and these 16 different numbers are taken from set
Figure G2009101305580D01431
Be total to C 49 16=3348108992991 kinds of combinations.UCAS MBCan adopt any one combination, for example, a kind of in shown in table 10.17～table 10.20, other is similar, enumerates no longer one by one.
Table 10.17
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 15
Table 10.18
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 8
1 1 9 9
2 2 10 10
3 3 11 11
4 4 12 12
5 5 13 13
6 6 14 14
7 7 15 16
Table 10.19
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 10
1 1 9 12
2 2 10 14
3 3 11 16
4 4 12 18
5 5 13 20
6 6 14 22
7 8 15 24
Table 10.20
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 8 20
1 1 9 24
2 2 10 28
3 3 11 32
4 4 12 36
5 8 13 40
6 12 14 44
7 16 15 48
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication UCAS MBThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers are taken from set
Figure G2009101305580D01441
Be total to C 49 32=6499270398159 kinds of combinations.UCAS MBCan adopt any one combination, for example, a kind of in shown in the table 10.21, other is similar, enumerates no longer one by one.
Table 10.21
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 25
10 10 26 26
11 11 27 27
12 12 28 28
13 13 29 29
14 14 30 30
15 15 31 31
Perhaps, when system bandwidth is 7MHz or 8.75MHz or 10MHz, indication UCAS SBiThe required bit number of parameter is 6bits.6bits represents 64 different numbers, can represent set
Figure G2009101305580D01451
In all numerical value.For example, shown in table 10.22.
Table 10.22
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 32 32
1 1 33 33
2 2 34 34
3 3 35 35
4 4 36 36
5 5 37 37
6 6 38 38
7 7 39 39
8 8 40 40
9 9 41 41
10 10 42 42
11 11 43 43
12 12 44 44
13 13 45 45
14 14 46 46
15 15 47 47
16 16 48 48
17 17 49 Keep
18 18 50 Keep
19 19 51 Keep
20 20 52 Keep
21 21 53 Keep
22 22 54 Keep
23 23 55 Keep
24 24 56 Keep
25 25 57 Keep
26 26 58 Keep
27 27 59 Keep
28 28 60 Keep
29 29 61 Keep
30 30 62 Keep
31 31 63 Keep
The 3rd class: for 20MHz, indication UCAS MBThe required bit number of parameter is 4bits.UCAS MBIndicated take Miniband as unit in the 0th frequency partition based on the possible number set of the number of the CRU of Miniband as:
Figure G2009101305580D01461
Figure G2009101305580D01462
4bits represents 16 different numbers, and these 16 different numbers are taken from set Be total to C 97 16=793067310934425856 kinds of combinations.UCAS MBCan adopt any one combination, for example, a kind of in shown in table 10.17～table 10.20, other is similar, enumerates no longer one by one.
Perhaps, when system bandwidth is 20MHz, indication UCAS MBThe required bit number of parameter is 5bits.5bits represents 32 different numbers, and these 32 different numbers are taken from set
Figure G2009101305580D01464
Be total to C 97 32Plant combination.UCAS MBCan adopt any one combination.For example, a kind of in shown in table 10.23～table 10.25, other is similar, enumerates no longer one by one.
Table 10.23
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 16
1 1 17 17
2 2 18 18
3 3 19 19
4 4 20 20
5 5 21 21
6 6 22 22
7 7 23 23
8 8 24 24
9 9 25 25
10 10 26 26
11 11 27 27
12 12 28 28
13 13 29 29
14 14 30 30
15 15 31 31
Table 10.24
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 16
1 1 17 18
2 2 18 20
3 3 19 22
4 4 20 24
5 5 21 26
6 6 22 28
7 7 23 30
8 8 24 32
9 9 25 34
10 10 26 36
11 11 27 38
12 12 28 40
13 13 29 42
14 14 30 46
15 15 31 48
Table 10.25
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 16 18
1 1 17 20
2 2 18 22
3 3 19 24
4 4 20 26
5 5 21 28
6 6 22 30
7 7 23 32
8 8 24 36
9 9 25 40
10 10 26 44
11 11 27 48
12 12 28 52
13 13 29 56
14 14 30 60
15 16 31 64
Perhaps, when system bandwidth is 20MHz, indication UCAS MBThe required bit number of parameter is 6bits.6bits represents 64 different numbers, and these 64 different numbers can represent set
Figure G2009101305580D01481
In all numerical value.For example, shown in the table 10.26.
Table 10.26
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0 32 32
1 1 33 33
2 2 34 34
3 3 35 35
4 4 36 36
5 5 37 37
6 6 38 38
7 7 39 39
8 8 40 40
9 9 41 41
10 10 42 42
11 11 43 43
12 12 44 44
13 13 45 45
14 14 46 46
15 15 47 47
16 16 48 48
17 17 49 49
18 18 50 50
19 19 51 51
20 20 52 52
21 21 53 53
22 22 54 54
23 23 55 55
24 24 56 56
25 25 57 57
26 26 58 58
27 27 59 59
28 28 60 60
29 29 6l 6l
30 30 62 62
31 3 1 63 63
Perhaps, when system bandwidth is 20MHz, indication UCAS MBThe required bit number of parameter is 7bits.7bits represents 128 different numbers, and these 128 different numbers can represent set
Figure G2009101305580D01491
In except 0 or 1 or 95 or 96 all numerical value.
For indicating UCAS under each bandwidth MBThe required bit number of parameter can be determined from said method, but for different bandwidth, indicate UCAS MBThe required bit number of parameter is the identical or complete difference of part each other.For example,
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 2bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 4bits;
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 3bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 4bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits;
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 4bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 6bits; Perhaps,
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 5bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 6bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 6bits;
When system bandwidth is 5MHz, indication UCAS MBThe required bit number of parameter is 5bits; When system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 6bits; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 7bits; Perhaps,
It may be noted that: at above-mentioned UCAS MBCollocation method in, when two different bandwidth have been used identical bit number indication UCAS MBParameter, corresponding form can be identical or different.For example, when system bandwidth was 10MHz (also can be 7MHz or 8.75MHz), indicating the required bit number of this parameter was 5bits, but corresponding form is table 10.22; When system bandwidth was 20MHz, indicating the required bit number of this parameter was 5bits, but corresponding form is table 10.23.
Identical form refers to: owing to system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) is the more similar characteristics unification that can consider 10MHz (can be 7MHz or 8.75MHz) and 20MHz of situation of 20MHz with system bandwidth, can be that 10MHz (can be 7MHz or 8.75MHz) adopts identical UCAS when being 20MHz with system bandwidth with system bandwidth MBValue and corresponding relation, that is, system bandwidth is that 10MHz (can be 7MHz or 8.75MHz) adopts identical form when being 20MHz with system bandwidth, for example, can adopt in table 10.23～table 10.25, the collocation method during perhaps according to 20MHz produces.Perhaps, produce according to the methods below:
Table 10.27
UCAS MB FP 0In based on the number of the CRU of Miniband UCAS MB FP 0In based on the number of the CRU of Miniband
0 0/48 0 0
1 1/48 1 2
2 2/48 2 4
3 3/48 3 6
4 4/48 4 8
5 5/48 5 10
6 6/48 6 12
7 7/48 7 14
8 8/48 8 16
9 9/48 9 18
10 10/48 10 20
11 11/48 11 22
12 12/48 12 24
13 13/48 13 26
14 14/48 14 28
15 15/48 15 30
In addition, 5MHz can all adopt 3 bits or 4 bits or 5 bits with 10MHz (also can be 7MHz or 8.75MHz).
In addition, it may be noted that: at above-mentioned UCAS MBCollocation method in, for each table, UCAS MBValue and UCAS MBThe meaning of value indication in the middle of relation can change, each form all is embodiment, as long as the UCAS that comprises in table MBValue be the indication meaning be identical, all be regarded as identical table, all within protection range.
By above-mentioned example 9, can find out, when system bandwidth is respectively 5MHz, 10MHz (can be 7MHz or 8.75MHz), 20MHz system, indication UCAS MBBit number need respectively 4bits, 5bits, 6bits, at UCAS MBPossible value situation about reducing under, deleted that redundant and unnecessary information indicates, saved overhead bit, and guaranteed certain flexibility.
As mentioned above, collocation method by means of resource mapping indication information provided by the invention, the required bit number of each band width configuration indication parameter for system's support, and the bit number part of the indicating identical or complete difference of same parameters under different bandwidth, so that the bit number that physical resource mapping indication signaling uses can change flexibly according to the bandwidth that system uses, reduce as much as possible the bit number of transmission, avoided the large problem of control channel expense in the correlation technique, under the prerequisite that does not affect the normal running of system, save descending control overhead, thereby improve the operating efficiency of system.
The above is the preferred embodiments of the present invention only, is not limited to the present invention, and for a person skilled in the art, the present invention can have various modifications and variations.Within the spirit and principles in the present invention all, any modification of doing, be equal to replacement, improvement etc., all should be included within protection scope of the present invention.

Claims (9)

1. the collocation method of a resource mapping indication information, it is characterized in that, parameter in the described resource mapping indication information comprise following one of at least: descending allocation of subbands number, up allocation of subbands number, the downstream frequency zone configuration, the upstream frequency zone configuration, downstream frequency subregion allocation of subbands number, upstream frequency subregion allocation of subbands number, the number that descending continuous resource unit distributes, the number that up continuous resource unit distributes, the number of descending continuous resource unit based on Miniband, the number of up continuous resource unit based on Miniband, described method comprises:
At least one parameter of indexed resource mapping is determined the required bit number of the described parameter of indication according to bandwidth;
Wherein, for a plurality of different bandwidth, indicate the each other identical or complete difference of part of the required bit number of described parameter.
2. method according to claim 1, it is characterized in that, described resource mapping comprises downlink resource mapping and/or ascending resource mapping, wherein, the mapping of described downlink resource one of comprises the steps or combination: sub-band division (Subband Partitioning), little band displacement (Miniband Permutation), frequency partition is divided (Frequency Partitioning), continuous resource unit/distributed resource unit distributes (Contigous Resource Unit/Distributed Resource Unit Allocation, referred to as CRU/DRU Allocation) and subcarrier permutation (Subcarder Permutation), described ascending resource mapping one of comprises the steps or combination: sub-band division, little band displacement, frequency partition is divided, continuous resource unit/distributed resource unit distributes and Tile displacement (Tile Permutation).
3. method according to claim 1, it is characterized in that, described a plurality of bandwidth comprise the first bandwidth, the second bandwidth and the 3rd bandwidth, wherein, for a parameter in the described resource mapping indication information: corresponding to described the first bandwidth, indicating the required bit number of this parameter is M; Corresponding to described the second bandwidth, indicating the required bit number of this parameter is N; Corresponding to described the 3rd bandwidth, indicating the required bit number of this parameter is P, and the value of M, N, P is the identical or complete difference of part each other.
4. method according to claim 3 is characterized in that, part is identical each other refers to for the value of described M, N, P:
N=M+1 and P=M+1; Perhaps, N=M+2 and P=M+2; Perhaps, N=M and P=M+1; Perhaps, N=M and P=M+2, wherein M is the integer greater than 0.
5. method according to claim 3 is characterized in that, the value of described M, N, P each other complete difference refers to:
N=M+1 and P=M+2; Perhaps N=M+2 and P=M+3, perhaps N=M+1 and P=M+3, wherein M is the integer greater than 0.
6. each described method in 5 according to claim 3 is characterized in that, the value of M is 1 or 2 or 3 or 4.
7. each described method in 5 according to claim 3 is characterized in that,
Described the first bandwidth comprises: 5MHz, it is one of following that described the second bandwidth comprises: 7MHz, 8.75MHz, 10MHz, described the 3rd bandwidth comprise: 20MHz.
8. method according to claim 1 is characterized in that,
Bit number at the descending allocation of subbands number of system's indicating is same or different from the bit number of the up allocation of subbands number of indication;
Bit number in described system indicating downstream frequency zone configuration is identical or different with the bit number of indication upstream frequency zone configuration;
Bit number at described system indicating downstream frequency subregion allocation of subbands number is identical or different with the bit number of indication upstream frequency subregion allocation of subbands number;
The bit number of the number that distributes in the described system descending continuous resource of indicating unit is identical or different with the bit number of the number that the up continuous resource of indication unit distributes.
9. each described method in 5 according to claim 1, it is characterized in that, described descending allocation of subbands number and/or up allocation of subbands number refer to: the number of subband, downstream frequency zone configuration and/or upstream frequency zone configuration refer to that frequency partition is divided the number of medium frequency subregion and/or size or ratio, downstream frequency subregion allocation of subbands number and/or the upstream frequency subregion allocation of subbands number average of each frequency partition is except first frequency partition (FP in the indication frequency partition in the sub-band division 0) number, number that descending continuous resource unit distributes and/or the up continuous resource unit of subband distributes in addition the frequency partition number all refers to first frequency partition of number indicating downlink (FP of the continuous resource unit distributes in each frequency partition number, descending continuous resource unit based on Miniband 0) in based on up first frequency partition (FP of number indication of the number of the continuous resource unit of Miniband, up continuous resource unit based on Miniband 0) in based on the number of the continuous resource unit of Miniband, wherein, the unit of number is subband or little band or physical resource unit.
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