WO2010145549A1 - Method and system for scheduling uplink double data flow transmission - Google Patents

Method and system for scheduling uplink double data flow transmission Download PDF

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Publication number
WO2010145549A1
WO2010145549A1 PCT/CN2010/074020 CN2010074020W WO2010145549A1 WO 2010145549 A1 WO2010145549 A1 WO 2010145549A1 CN 2010074020 W CN2010074020 W CN 2010074020W WO 2010145549 A1 WO2010145549 A1 WO 2010145549A1
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Prior art keywords
bac
uplink
user terminal
bits
control information
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PCT/CN2010/074020
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French (fr)
Chinese (zh)
Inventor
袁明
毕峰
梁枫
杨瑾
吴栓栓
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中兴通讯股份有限公司
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Publication of WO2010145549A1 publication Critical patent/WO2010145549A1/en

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    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04WWIRELESS COMMUNICATION NETWORKS
    • H04W28/00Network traffic management; Network resource management
    • H04W28/02Traffic management, e.g. flow control or congestion control
    • H04W28/10Flow control between communication endpoints

Definitions

  • This relates to high- and long-term (TE, o g e Evo o Adva ced) wood, especially in the TEA system (R, Reay ode) or user terminal (E, se Eq p e) uplink method and system.
  • the long-term (TE, oge Evo o) system's downlink scheme is based on the OF wood force, in front of the OF (CP, Cyc c pe) to reduce the same, and support the downlink, that is, standing in the 1 meson can be the same 2 (T , TTa spo oc ).
  • the uplink scheme uses the FAN (Feq e cy vso pex gAcces) wood force, which can only support the uplink sheep, that is, E can be the same in the meson.
  • the downlink control information (C, ow Co ooao ) in the TE is in the upper and lower directions of E, and the information of the S (Syse oao) / ( ag ) / RA (Ra do access) phase.
  • the TE system has several different downlink control information formats (C oa ).
  • the God format is better than the solution (CS, od aoa dCod gSche e).
  • R is relative to E Edong
  • R is relative to e.
  • TEA supports the same as the uplink sheep user. That is to say, the user is on the line, and the 1 meson can be the same as 2, that is, up. Therefore, it is impossible to rule out that R can also support the uplink on the uplink.
  • the main purpose of this is to provide an uplink method and system, and the uplink in the TEA system.
  • the wood plan is
  • the medium-allied or user terminal receives the uplink downlink control information format and uplinks according to the received downlink control information format.
  • the user terminal of the version 9/10 of the high-long-term TEA system of the user terminal includes, resource allocation and resources, reference RS shift, uplink index, downlink configuration index A, Indication C, resolution scheme CS, Vietnamese indication, and version residual RV, power control TPC, AR, to flow, and upstream phase information.
  • the required bits of the uplink index are configured for the uplink and downlink, and the uplink bac a has more information than the downlink bac a.
  • the bits required by A are both configured for uplink and downlink, and the uplinks are the same or different.
  • the bit of AR is equal to AR.
  • ARQ is equal to the last upstream and the same configured upstream.
  • Upstream system including, and user terminals, wherein
  • the medium or user terminal is in the format of the received downlink control information.
  • the user terminal of the version 9/10 of the high-long-term TEA system of the user terminal The wood scheme provided in this book can be seen in the TEA system, due to In the uplink downlink control information format, downhole R or E, so that R or version 9/1 E, according to the received downlink control information format, is uplinked. 1 Explanation of the TEA system of R
  • the second 5 F mode of the configuration scheme of the uplink and downlink back sub-array In the 3F mode, in the 4th F mode of the configuration scheme of the uplink and downlink bac a sub-arrangement, the second 5 F mode of the configuration scheme of the uplink and downlink back sub-array, the AR ⁇ 6 F mode, ARQ
  • the method of the upstream method includes
  • step 200 the user terminal receives the downlink control information format of the uplink.
  • a new downlink control information format (C a ) is added, which is uplinked on R or E, and E is the version 9/10 of the TEA system. E of (Re 9/ 0). Add a new downlink control information format from the base R or E
  • the new C a content for the uplink is included.
  • Resource allocation and resource allocation R or TEARe 9/ E, uplink 0 can be a resource allocation method, or a non-resource allocation method
  • the shift of the RS is 3 bits. 1 No RS shift, that is, the same configuration as C a in the existing TE system, that is, only the RS shift is configured, but the physical hybrid AR indication is configured twice (C , hysca yb dARQ dcao Cha e ) 2 Existing configuration 5 Configure each RS shift separately, that is, configure 2 RS shifts.
  • the bits required for the hang are configured for the upstream and downstream bac ha, and only the upstream bac ha is more than the downstream bac ha.
  • the bit required for A hang has the configuration of the uplink and downlink bac h sub, and only the downlink bac ha sub is more than the upstream bac ha sub A information CQ 1 bit.
  • TPC Power control
  • T and T 2 are mapped to the above. Its 0 means that T is mapped to codewod and T 2 is mapped to codewod.
  • T is mapped to codewod and T 2 is mapped to codewod
  • step 201 the user terminal uplinks according to the received downlink control information format.
  • R or E is the composition and the conclusion of the channel downlink control information format. R or E only needs to complete the uplink in the manner specified by the downlink control information format.
  • the uplink method which provides an uplink system, is shown in the composition 1, including R and E, where, in the uplink downlink control information format, R or E R or E, in accordance with the received downlink control information format, uplink
  • the following combines the method of the downlink control information format
  • C o a contains the following information
  • the resource allocation mode of the non-bit is the bit, which is shown in Table 1.
  • This R configures the RS shift, and the 2 P C index.
  • the P C index is obtained in the wood of the field, which is no longer used.
  • the first indication of the configuration scheme of the upper and lower bac ha sub because 3, the square indicates the sub-, the downward-pointed square indicates the lower stroke, and the upward-pointed square indicates the upper stroke
  • Dex 10 representing bac h a e#6 dex 01 , representing backha S a e#7 dex 11 , indicating the same as bac ha S bfTa e#6 and bac ha S bf a e#7
  • Bac ha S b a e#2 or bac ha S b a e#3 phase AC / AC this requires 2 bits of A to indicate bac ha S a e
  • Bac ha S ae AC AC that is, the bit required for A is the same as the upper leg of the same leg. Hugh, can A 10, AC AC A 01 indicating bac a S ae#2 indicates AC AC of bac ha S ae #3, and A 11 indicates AC AC of bac ha S bfTa e#2 and bac ha S bae#3
  • CQ 1 bit.
  • 0 can be R or TE e 9/ 0 E CQ, where 1 means R or TEARe 9/ E requires C. In this, send a completely different C.
  • ARQ ARQ
  • bac ha sub-configured with H ARQ equal to 1 upstream and 1st configured upstream bac ha.
  • the ARQ's first indication because 5, the square indicates the sub-, the downward-pointing square indicates the lower stroke, the upward-pointed square indicates the upper stroke, the bac ha sub-configuration period s, It includes 10 mesons (0 to 9).
  • Configure bac ha and 1 bac ha in 1 as shown in 5, s a e#2 configure bac ha , s a e #6 configure bac ha .
  • 1 is on s a e#6, which is on the s a e#6 of the following, and they are not configured with the uplink bac ha, therefore, only 1 uplink ARQ, indicating ARQ.
  • the second indication of the ARQ because of the square, the square indicates the sub-segment, the downward-pointing square indicates the lower stroke, and the upward-pointed square indicates the upper stroke, the bac ha sub-configuration period s, Configure 2 bac ha and 2 bac ha, as shown in 6, sae#2 and sae#3 configure bac ha, sae#6 and sae#7 configure bac ha. So, 1 up in sbae#6, its waiter On sae#6, and they are configured with 1 uplink bac a, therefore, there are 2 uplink ARs, and 1 bit indicates ARQ. It is easy to conclude that the 1 upstream bac ha and the 1st upstream bac ha are configured with 2 uplink bac ha, that, there are 3 uplink ARQs, and 2 bits indicate ARQ.
  • the second indication of the ARQ because of the square, the square indicates the sub-head, the downward-pointing square indicates the lower leg, the upward-pointed square indicates the upper leg, and the bac ha sub-configuration period is 4 s. As shown in 7, it is easy to see that there is only 1 uplink ARQ, which in turn indicates ARQ.
  • T and T 2 are mapped to respectively. Its 0 means that T is mapped to codewod, T 2 is mapped to codewod, 1 means T is mapped to codewod, and T 2 is mapped to codewod.
  • Phase information Bits are taught, 2 2 bits, 4 requires 4 bits, and up to 5 bits. In the T way.
  • dex 3 bits.
  • the bit required for the hang has a configuration of the upper and lower bac ha, and only the number of upstream bac ha is more than the downstream bac ha.
  • the 6 sub-configuration scheme in 2 T mode which represents the uplink sub-symbol, indicating the downlink sub-.
  • bac ha S ae #4 can be scheduled at the same time for 3 different bac ha S ae, which is the sbae #8 and the lower Bae#2 and sbae#3, bac a S bae#9 can be the same as 3 different bac ha S bae, ie sbae#3 sbae#7 and sbae#8
  • the bac ha S bae #10 of the bac ha S bae #4 indicates that the bac ha S bae #110 of the present bac ha S bae #8 and the bac ha S bae #2 010 of the lower element
  • the bac ha S bfTa e #2 of the lower yuan is based on this. Among them, 1 means child, 0 means child.
  • the second indication of the configuration scheme of the upper and lower bac ha sub because of the square, the square indicates the sub-head, the downward-pointed square indicates the descending process, and the upward-pointed square indicates the uplink.
  • 9 shows that 2 bac ha S bae #2 can be the same as 1 bac haS bfTa e#3 and bac ha S bfTa e#4 phase / non (AC / AC), this requires 2 bits of A to indicate bac Ha S bae #2 is on
  • CQ 1 bit. It can be 0 or TEARe 9/ 0 EC, where 1 means R or TE e 9/1 E requires C. In this case, send the exact same CQ.
  • ARQ bac a sub-configuration, the value method F is the same, no longer.
  • Phase information Bits are taught, 2 2 bits, 4 requires 4 bits, and up to 5 bits.

Abstract

A method and a system for scheduling uplink double data flow transmission are provided by the present invention. In an LTE-A system, a base station (BS) sets the downlink control information (DCI) format for scheduling uplink double data flow transmission, and delivers the DCI format to a relay node (RN) or a user equipment (UE), which enables the RN or the UE of version 9/10 to transmit uplink double data flow according to the received downlink control information format.

Description

上行 的方法及 統 木領域  Upward method and the field of wood
本 涉及高 長期 ( TE , o g e Evo o Adva ced) 木, 尤指 TEA 統中 中 (R , Reay ode)或用戶終端 ( E, se Eq p e )上行 的方法及 統。 背景 木 This relates to high- and long-term (TE, o g e Evo o Adva ced) wood, especially in the TEA system (R, Reay ode) or user terminal (E, se Eq p e) uplink method and system. Background
TEA以及高 通信 木( T dva ced, e a o a ob e Teeco ca o Adva ce )等高 的 通信 統, 都是以 The communication systems of TEA and T dva ced, e a o a ob e Teeco ca o Adva ce are
(OF , OThogo a Feq e cy vso pex g) 木力 的。 長期 ( TE, o g e Evo o ) 統的下行 方案是以 統 的OF 木力 , 在OF 的 上 前 (CP,Cyc c pe ) 以減少 同的 , 而支持下行 的 , 即 站在 1介子 內可以同 2 (T , TTa spo oc )。但是, 在 TE 統中, 上行 方案以羊載波的 多 (F A, Feq e cy vso pex gAcces ) 木力 , 只能支持上行羊 的 , 也就是 , E在 介子 內可以同 1 。 (OF, OThogo a Feq e cy vso pex g) Mu Li. The long-term (TE, oge Evo o) system's downlink scheme is based on the OF wood force, in front of the OF (CP, Cyc c pe) to reduce the same, and support the downlink, that is, standing in the 1 meson can be the same 2 (T , TTa spo oc ). However, in the TE system, the uplink scheme uses the FAN (Feq e cy vso pex gAcces) wood force, which can only support the uplink sheep, that is, E can be the same in the meson.
TE中的下行控制信息 ( C , ow Co o o a o ) 于 E的上、 下行並各 , 以及 統消息 (S , Syse o a o ) / ( ag ) / 接 (RA, Ra do access) 相 的信息。 不 同的 模式, TE 統中 了若干神不同的下行控制信息格式 ( C o a ), 神格式除了包含 些最基本信息, 比 制解 方案 ( CS, od a o a dCod gSche e), 資源" 力配 (R , Reso ceAssg e ) 外,近包含 神格式各 特有的信息,比 , C a / A/ / C/ 表示 于 度下行羊 C a 2/2A表示 于 度下行 , C a 表示 于 上行 。 其中, TE 統中用于 上行 的 C a 的 休內容包括 The downlink control information (C, ow Co ooao ) in the TE is in the upper and lower directions of E, and the information of the S (Syse oao) / ( ag ) / RA (Ra do access) phase. In different modes, the TE system has several different downlink control information formats (C oa ). In addition to the most basic information, the God format is better than the solution (CS, od aoa dCod gSche e). , Reso ceAssg e ), except for the unique information of the god format, than, C a / A / / C / It is indicated that the degree of downhill sheep C a 2/2A is indicated by the degree of decline, and C a is expressed by the upward direction. Among them, the rest of the TE system used for uplink C a includes
用于 a 和 o a 的 (fag) 比特。 其 0 表示 a , 1表示 o a (fag) bits for a and o a. Its 0 means a and 1 means o a
( opp gfag) 1比特。 其 0表示 E的物理上行共 享 (P SC , hysca p S e Cha e ) 不需要 , 其  ( opp gfag) 1 bit. 0 means that the physical uplink sharing of E (P SC , hysca p S e Cha e ) is not required,
1表示 E的P SC 需要 1 indicates that the P SC of E needs
資源分配以及 資源分配 o9 +)/2 比特, 其中, 下的上行R 的 , 向上  Resource allocation and resource allocation o9 +)/2 bits, where, under the up R, up
制解 方案 ( CS) 以及版本 余 (RV Ted da cyVe so ) 5比 特  Solution (CS) and version (RV Ted da cyVe so) 5 bits
新教 指示 ( , ew aa dcaoT) b特 Protestant instruction ( , ew aa dcaoT) b
P SC 的 功率控制命令 (TPC Ta s owe Co o ) 2 比特  P SC power control command (TPC Ta s owe Co o ) 2 bits
參考 ( RS, e od a o Te比 e ceSg a )的 移位 3 比特  Reference (RS, e od a o Te to e ceSg a ) shift 3 bits
上行索引 ( dex) 2比特,只有T 模式的配置0才有 信息 下行配置索引 ( A , ow Assg e de ) 2比特,只有T 模式的配置 1~6才有 信息  Uplink index (dex) 2 bits, only T mode configuration 0 has information Downlink configuration index (A, ow Assg e de) 2 bits, only T mode configuration 1~6 only information
指示 (CQ , Cha e Q a y dcao 比特。  Indication (CQ, Cha e Q a y dcao bit.
未東元 通信或 統要求增 覆蓋 , 支持更高速率 , 通信 木提出了新的 同 , 統建造和 的 用 更 突出。 看 速率及通信 的增 , 的 也 得突出, 而 且未 的元 通信將 更高頻率, 由此造成的路往損耗 將更 。 了增 高教 速率、 、 部署的覆蓋 , 提高小 緣的吞 量, 以及 蜂窩 統的覆蓋漏洞內的用戶提供 各, 通 信 統中引 了中 ( ea ) 木, 因此, 中 木 視力 4G的 項 木。No TECO communication or system requirements increase coverage, support higher rates, communication wood proposes new, unified construction and use more prominent. Looking at the increase in speed and communication, it is also important, and the non-meta communication will be higher frequency, and the resulting road loss will be even more . Increased teaching rate, coverage of deployment, increased throughput of small margins, and user coverage within the coverage gap of the cellular system. The communication system has introduced medium (ea) wood. Therefore, Zhongmu vision 4G is the item.
1 引 R 的 TEA 統的示意 , 1所示, 在引 了 R 的 TEA 統中, R 同的 程 ( ac a , 也 中 )分力上行bac a 和下行bac ha R 其下 的 E 同的 接 ( ccess ) 分力上行access 和下行access 。 e 東說, R 就相 于 E E東說, R 就相 于e 目前, TEA 支持上行羊用戶的 同 , 也就是說, 用戶在 上 行 , 1介子 內可以同 2 , 即上行 。 因 此, 也就不能排除R 在 上行 也支持上行 的可能, 然而, 目前近沒有提供 于 上行 的 方案。  1 indicates the TEA system of R, as shown in 1 , in the TEA system with R, the same process of R (ac a , also in the middle) and the lower bac a and the lower bac ha R ( ccess ) Divided uplink access and downlink access. e East said, R is relative to E Edong, R is relative to e. Currently, TEA supports the same as the uplink sheep user. That is to say, the user is on the line, and the 1 meson can be the same as 2, that is, up. Therefore, it is impossible to rule out that R can also support the uplink on the uplink. However, there is currently no solution for the uplink.
內容  Content
有 于此, 本 的主要目的在于提供 上行 的 方法及 統, TEA 統中的上行 的 。  In this case, the main purpose of this is to provide an uplink method and system, and the uplink in the TEA system.
到上 目的, 本 的 木方案是 的  For the purpose of the purpose, the wood plan is
上行 的方法, 包括  Upward method, including
中牲 或用戶終端接收 于 上行 的下行控制信息 格式, 按照接收到的下行控制信息格式, 上行 。  The medium-allied or user terminal receives the uplink downlink control information format and uplinks according to the received downlink control information format.
用戶終端 高 長期 TEA 統中的版本9/10的用戶終端。 下行控制信息格式包含有 , 資源分配以及 資源 , 參考 RS的 移位, 上行索引 , 下行配置索引 A , 指示C , 制解 方案 CS、 新教 指示 以及版本 余 RV, 功率控制TPC, AR , 到 流特 的 , 上 行 相 的信息。The user terminal of the version 9/10 of the high-long-term TEA system of the user terminal. The downlink control information format includes, resource allocation and resources, reference RS shift, uplink index, downlink configuration index A, Indication C, resolution scheme CS, Protestant indication, and version residual RV, power control TPC, AR, to flow, and upstream phase information.
RS的 移位的 所述中 或用戶終端配置1 RS 移位, 以及2 P C 索引  The shift of the RS or the user terminal configuration 1 RS shift, and 2 P C index
或者, 分別 配置各 的 RS 移位, RS 移位得到2 P C 索引 。  Or, configure each RS shift separately, and RS shift to get 2 P C index.
上行索引 所需的比特 上下行 程子 的配置有 , 且 上行 程bac a 子 的 多于下行bac a 子 存在所 上行索引 的信息 The required bits of the uplink index are configured for the uplink and downlink, and the uplink bac a has more information than the downlink bac a.
A所需的比特 均上下行 程子 的配置有 ,且 上行 程 相同或不同的 C  The bits required by A are both configured for uplink and downlink, and the uplinks are the same or different.
相同或不同的 CS Same or different CS
AR 的比特 等于 AR The bit of AR is equal to AR.
ARQ 等于第 次上行 和第 次 同配置的上 行 程子 的 。  ARQ is equal to the last upstream and the same configured upstream.
上行 的 統, 包括 , 中 和用戶終端, 其中,  Upstream system, including, and user terminals, wherein
, 于 于 上行 的下行控制信息格式, 井 下 中 或用戶終端。  , in the uplink control information format, downhole or user terminal.
中 或用戶終端, 于按照接收到的下行控制信息格式, 上 行 。  The medium or user terminal is in the format of the received downlink control information.
用戶終端 高 長期 TEA 統中的版本9/10的用戶終端。 上 本 提供的 木方案可以看出, 在 TEA 統中, 由于 于 上行 的下行控制信息格式,井下 R 或 E, 使得R 或版本9/1 的 E, 通 按照接收到的下行控制信息格式, 了 上行 的 。 1 引 R 的 TEA 統的示意囤 The user terminal of the version 9/10 of the high-long-term TEA system of the user terminal. The wood scheme provided in this book can be seen in the TEA system, due to In the uplink downlink control information format, downhole R or E, so that R or version 9/1 E, according to the received downlink control information format, is uplinked. 1 Explanation of the TEA system of R
囤 2力本 上行 的方法的 程 囤 2 The method of the method of the uplink
3 F 方式下,上下行bac a 子 的配置方案的第 的 4 F 方式下, 上下行back 子 的配置方案的第二 的 5 F 方式下, AR 的第 的示意囤 囤 6 F 方式下, ARQ 的第二 的示意囤 囤 7 F 方式下, ARQ 的第三 的示意囤 囤 8 T 方式下, 上下行back 子 的配置方案的第 的 g T 方式下, 上下行back 子 的配置方案的第二 的  In the 3F mode, in the 4th F mode of the configuration scheme of the uplink and downlink bac a sub-arrangement, the second 5 F mode of the configuration scheme of the uplink and downlink back sub-array, the AR 的6 F mode, ARQ The second schematic 囤囤7 F mode, the third AR8 T mode of the ARQ, the second g T mode of the configuration scheme of the uplink and downlink back sub, and the second configuration scheme of the uplink and downlink back sub of
休 方式Hugh way
2力本 上行 的方法的 程 , 2所示, 包括  2 The method of the upstream method, shown in 2, includes
步驟200 中 或用戶終端接收 于 上行 的下行 控制信息格式。  In step 200, the user terminal receives the downlink control information format of the uplink.
在 TEA 統中, 新增下行控制信息格式 ( C a ), 于R 或 E 上行 , E是 TEA 統中的版本 9/10 (Re 9/ 0) 的 E。 新增下行控制信息格式由基 好 R 或 E In the TEA system, a new downlink control information format (C a ) is added, which is uplinked on R or E, and E is the version 9/10 of the TEA system. E of (Re 9/ 0). Add a new downlink control information format from the base R or E
本步驟中, 新增用于 上行 的 C a 休內容包 括  In this step, the new C a content for the uplink is included.
5 ( opp gfa ) 1比特。 于表示是否執行 操作, 其5 ( opp gfa ) 1 bit. Indicates whether to perform an operation,
0表示不執行 操作 1表示執行 操作。 特別地, 于 TEA Re 9/1 E, 具有 于R , 可以有, 也可以沒有  0 means no operation. 1 means execution. In particular, in TEA Re 9/1 E, with R, there may or may not be
資源分配以及 資源分配 R 或 TEARe 9/ E, 上行0 可以是 的資源分配方式, 也可以是非 的資源分配方式 Resource allocation and resource allocation R or TEARe 9/ E, uplink 0 can be a resource allocation method, or a non-resource allocation method
RS的 移位 3比特。 1 不 RS 移位, 即 現 有 TE 統中 C a 相同的方式 配置, 也就是 于 只 配置 RS 移位, 但是配置 2倍的物理混合 AR 指示 ( C , hysca yb dARQ dcao Cha e ) 2 現有配置5 方式分別 配置各 的 RS 移位,即 配置2 RS 移位 The shift of the RS is 3 bits. 1 No RS shift, that is, the same configuration as C a in the existing TE system, that is, only the RS shift is configured, but the physical hybrid AR indication is configured twice (C , hysca yb dARQ dcao Cha e ) 2 Existing configuration 5 Configure each RS shift separately, that is, configure 2 RS shifts.
dex 2比特或3比特。 休所需的比特 上下行bac ha 子 的配置有 , 且只有 上行bac ha 子 的 多于下行bac ha 子 存在 dex的信息 Dex 2 bits or 3 bits. The bits required for the hang are configured for the upstream and downstream bac ha, and only the upstream bac ha is more than the downstream bac ha.
A 休所需的比特 上下行bac h 子 的配置有 , 且只有 下行bac ha 子 的 多于上行bac ha 子 存在 A 信息 CQ 1比特。 于 R 或 TEARe 9/1 E 非周期的 CQ。 其中, 可以 相同或不同的CQ The bit required for A hang has the configuration of the uplink and downlink bac h sub, and only the downlink bac ha sub is more than the upstream bac ha sub A information CQ 1 bit. A non-periodic CQ for R or TEARe 9/1 E. Among them, can be the same or different CQ
CS, 以及RV 除了 1 的 CS, 和RV以外, 近5 2 的 CS, 和RV。 其中, 由于bac ha 1 的信 較好, 因此可以 減少 CS的神 。 此外, 可以 相同或 不同的 CS CS, and RV except for CS of 1 and RV, nearly 5 2 of CS, and RV. Among them, due to the letter of bac ha 1 Better, so you can reduce the god of CS. In addition, the same or different CS
功率控制 (TPC) R 上行 的 , 可以 不同的 功率, 因此, 需要分別指明 的TPC  Power control (TPC) R upstream, can have different power, therefore, need to specify the TPC separately
混合 ( AR ) 由于 TE的上行 是同步的 且非 的, 因此,在 C o a 中元 指明 的 。但是 本 存在R 的 TEA 統東說,上行 可以是非同步的且 的, 因此需要指明 休的 ARQ ,其中,配置 上下行bac a 子 的配置方案有  Hybrid (AR) Since the uplink of TE is synchronous and non-existent, it is specified in C o a . However, the TEA that has R is said that the uplink can be asynchronous, so it is necessary to specify the ARQ of Hugh, where the configuration scheme for configuring the uplink and downlink bac a sub
到 流特 ( odewod Sw ) 的 1比特。 于指示 1介 子 內同 的2 即T 和T 2分別映射到 上。 其 0表示T 映射到 codewod 上, T 2映射到 codewod 上 1 bit to the stream (odewod Sw). The same 2 in the indicator 1 meson, T and T 2 are mapped to the above. Its 0 means that T is mapped to codewod and T 2 is mapped to codewod.
1表示T 映射到 codewod 上, T 2映射到 codewod 上  1 means T is mapped to codewod and T 2 is mapped to codewod
上行 相 的信息 比特 口教有 , 2  Upstream information, bitch, 2
需要2比特, 4 需要4比特, 最多不 5比特。 目前 于上 行 需要多少 本近未 , 但是 不 32 。 It takes 2 bits, 4 requires 4 bits, and at most 5 bits. At present, how many of them are currently on the line, but not 32.
步驟201 中 或用戶終端按照接收到的下行控制信息格式, 上行 。  In step 201, the user terminal uplinks according to the received downlink control information format.
R 或 E是 道下行控制信息格式 包含的 的組成及結 的,R 或 E只需按照下行控制信息格式所指定的方式完成上行 的 即可。  R or E is the composition and the conclusion of the channel downlink control information format. R or E only needs to complete the uplink in the manner specified by the downlink control information format.
本 上行 的方法, 近提供 上行 的 統, 其組成 1所示, 包括 , R 和 E, 其中, , 于 于 上行 的下行控制信息格式, R 或 E R 或 E, 于按照接收到的下行控制信息格式, 上行The uplink method, which provides an uplink system, is shown in the composition 1, including R and E, where, in the uplink downlink control information format, R or E R or E, in accordance with the received downlink control information format, uplink
/ 。/ .
E TEARe 9/1 E  E TEARe 9/1 E
下面結合 , 本 下行控制信息格式的 方法  The following combines the method of the downlink control information format
。 以F 方式 。  . In the F way.
統帝 2 z, 即 00, F 上行  Confucius 2 z, ie 00, F up
的 C o a 包含 下信息 C o a contains the following information
1) : 1比特, "0"代表不執行 操作 "1"代表執行 操 作。 R , 可以保留 1比特的信息 TEA Re 9/ E, 需要 1 比特的信息。  1) : 1 bit, "0" means no operation. "1" stands for the operation. R, can retain 1 bit of information TEA Re 9/ E, requires 1 bit of information.
2) 資源分配以及 資源分配。  2) Resource allocation and resource allocation.
果 的資源分配方式 , 則
Figure imgf000010_0001
Method of resource allocation,
Figure imgf000010_0001
果 非 的資源分配方式 則 比特, 其中 的取 表1所示。  The resource allocation mode of the non-bit is the bit, which is shown in Table 1.
統帝  Confucius
<10  <10
11 26 2  11 26 2
27 63  27 63
64 110 4 64 110 4
1  1
3) RS的 移位, 3比特。本 中 R 配置 RS 移位, 以及2 P C 索引 。 于P C 索引 的得到 于本領 域 木 慣用 木手段, 里不再 。  3) RS shift, 3 bits. This R configures the RS shift, and the 2 P C index. The P C index is obtained in the wood of the field, which is no longer used.
4) dex, 2比特, 休所需的比特 上下行bac ha 子 的 置有 , 且只有 上行bac a 子 的 多于下行bac a 子 才 有 信息。4) dex, 2 bits, the bits required for the hang up and down bac ha There is, and only the upstream bac a is more than the downstream bac a.
3 F 方式下,上下行bac ha 子 的配置方案的第 的 示意 , 因 3 中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指方格表示上行 程子 , 3所示, 子 2 (s a e #2) 配置 下行健路 程子 ( bac ha S b a ), s a e #6和 s a e #7 配置 上行健路 程子 ( bac ha S a e ) bac ha S a e #2可以同 2 不同的 bac ha S a e #6和 bac ha S a e#7,此 ,需要2比特的 dex 出上行bac ha 在 bac ha S b a e , 即上行索引 所需的比特 下行 程子 同 度的不同的上行 程子 的 。 休 , 可以 In the 3 F mode, the first indication of the configuration scheme of the upper and lower bac ha sub, because 3, the square indicates the sub-, the downward-pointed square indicates the lower stroke, and the upward-pointed square indicates the upper stroke, 3 Show, sub 2 (sae #2) configures the downlink health path ( bac ha S ba ), sae #6 and sae #7 configures the uplink health path ( bac ha S ae ) bac ha S ae #2 can be different from 2 Bac ha S ae #6 and bac ha S ae#7, this requires 2 bits of dex to output the upstream bac ha in bac ha S bae, that is, the bit required for the uplink index is different from the upper leg of the same leg . Hugh, can
dex 10 , 表示 bac h a e#6 dex 01 , 表示 backha S a e#7 dex 11 , 表示同 bac ha S bfTa e#6和 bac ha S bf a e#7 Dex 10 , representing bac h a e#6 dex 01 , representing backha S a e#7 dex 11 , indicating the same as bac ha S bfTa e#6 and bac ha S bf a e#7
5) A, 休所需的比特 上下行bac ha 子 的配置有 , 且 只有 下行bac ha 子 的 多于上行bac ha 子 才有 信息。 5) A, the bits required for the hang The configuration of the upstream and downstream bac ha sub-has, and only the downstream bac ha sub-information is more than the upstream bac ha sub-.
4 F 方式下,上下行bac ha 子 的配置方案的第二 的 示意 , 因 4 中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指方格表示上行 程子 , 4所示, s a e # 2和S a e #3 配置 bac ha S b a , S b a e #7 配置 bac ha S a e。 那 , 在 bac h S a e #7上就要  In the 4 F mode, the second indication of the configuration scheme of the upper and lower bac ha sub, because 4, the square indicates the sub-head, the downward-pointed square indicates the lower stroke, and the upward-pointed square indicates the upper stroke, 4 As shown, sae # 2 and Sae #3 configure bac ha S ba , S bae #7 to configure bac ha S ae. Then, on bac h S a e #7
bac ha S b a e#2或者 bac ha S b a e#3相 的AC / AC , 此 需要2比特的 A 指示 bac ha S a e 的是 Bac ha S b a e#2 or bac ha S b a e#3 phase AC / AC , this requires 2 bits of A to indicate bac ha S a e
bac ha S a e的AC AC , 即 A 所需的比特 下行 程子 同 度的不同的上行 程子 的 。 休 , 可以 A 10 ,表示 bac a S a e#2 的AC AC A 01 表示 bac ha S a e #3 的AC AC , A 11 表示 bac ha S bfTa e#2和 bac ha S b a e#3 的AC ACBac ha S ae AC AC, that is, the bit required for A is the same as the upper leg of the same leg. Hugh, can A 10, AC AC A 01 indicating bac a S ae#2 indicates AC AC of bac ha S ae #3, and A 11 indicates AC AC of bac ha S bfTa e#2 and bac ha S bae#3
6)CQ ,1比特。可以 0表示R 或者 TE e 9/ 0 E CQ,其 1表示R 或者 TEARe 9/ E需要 C 。 本 中, 送完全不同的C 。  6) CQ, 1 bit. 0 can be R or TE e 9/ 0 E CQ, where 1 means R or TEARe 9/ E requires C. In this, send a completely different C.
7) ARQ , bac ha 子 配置有 , 休 , ARQ 等于 1 上行 和第 1 同配置的上行bac ha 子 的 。 7) ARQ, bac ha sub-configured with H, ARQ equal to 1 upstream and 1st configured upstream bac ha.
5 F 方式下, ARQ 的第 的示意 , 因 5中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指 方格表示上行 程子 , bac ha 子 的配置周期 s, 介 中包括10介子 ( 0至9)。 在1 中配置 bac ha 子 和1 bac ha 子 , 5所示,s a e#2 配置 bac ha 子 , s a e #6 配置 bac ha 子 。 此 , 1 上行 在 s a e#6, 其 侍在下 介 的 s a e#6上, 且它們 同沒有配 置上行bac ha 子 , 因此, 只有 1 上行 ARQ , 指示 ARQ 。 In the 5 F mode, the ARQ's first indication, because 5, the square indicates the sub-, the downward-pointing square indicates the lower stroke, the upward-pointed square indicates the upper stroke, the bac ha sub-configuration period s, It includes 10 mesons (0 to 9). Configure bac ha and 1 bac ha in 1 , as shown in 5, s a e#2 configure bac ha , s a e #6 configure bac ha . Thus, 1 is on s a e#6, which is on the s a e#6 of the following, and they are not configured with the uplink bac ha, therefore, only 1 uplink ARQ, indicating ARQ.
6 F 方式下, ARQ 的第二 的示意 , 因 6中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指 方格表示上行 程子 , bac ha 子 的配置周期 s, 配置2 bac ha 子 和2 bac ha 子 , 6所示, s a e#2和 s a e#3 配置 bac ha 子 , s a e#6和s a e#7 配置 bac ha 子 。 此 , 1 上行 在s b a e#6, 其 侍在下 的 s a e#6上, 且它們 同配置了 1 上行bac a 子 , 因 此,有2 上行 AR , 1比特 指示 ARQ 。容易得出 果在 1 上行bac ha 和第 1 上行bac ha 同配置了 2 上行bac ha 子 , 那 , 有3 上行 ARQ , 2比特 指示 ARQ 。In the 6 F mode, the second indication of the ARQ, because of the square, the square indicates the sub-segment, the downward-pointing square indicates the lower stroke, and the upward-pointed square indicates the upper stroke, the bac ha sub-configuration period s, Configure 2 bac ha and 2 bac ha, as shown in 6, sae#2 and sae#3 configure bac ha, sae#6 and sae#7 configure bac ha. So, 1 up in sbae#6, its waiter On sae#6, and they are configured with 1 uplink bac a, therefore, there are 2 uplink ARs, and 1 bit indicates ARQ. It is easy to conclude that the 1 upstream bac ha and the 1st upstream bac ha are configured with 2 uplink bac ha, that, there are 3 uplink ARQs, and 2 bits indicate ARQ.
7 F 方式下, ARQ 的第二 的示意 , 因 7中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指 方格表示上行 程子 , bac ha 子 的配置周期 4 s, 7所示, 容易看出只有 1 上行 ARQ , 依次 指示 ARQ 。 In the 7 F mode, the second indication of the ARQ, because of the square, the square indicates the sub-head, the downward-pointing square indicates the lower leg, the upward-pointed square indicates the upper leg, and the bac ha sub-configuration period is 4 s. As shown in 7, it is easy to see that there is only 1 uplink ARQ, which in turn indicates ARQ.
8) 到 流特 ( odewodSw )的 , 1比特。 于指示T 和T 2分別映射到 上。其 0表示T 映射到codewod 上, T 2映射到codewod 上 1表示T 映射到codewod 上,T 2 映射到 codewod 上。  8) to the stream (odewodSw), 1 bit. The indications T and T 2 are mapped to respectively. Its 0 means that T is mapped to codewod, T 2 is mapped to codewod, 1 means T is mapped to codewod, and T 2 is mapped to codewod.
9) 于T , CS 5比特, 1比特, RV 2比特, TPC2比特。 于T 2, CS 5比特, 1比特, RV 2比特, TPC 2比特。 其中, R 東說, 由于bac h 1 的信 很好, 因此可以 減少 CS的神 , 例 , 4比特 指示較高的 16 CS等。 本 中, 的 CS是完全不同的。  9) at T, CS 5 bits, 1 bit, RV 2 bits, TPC 2 bits. At T 2, CS 5 bits, 1 bit, RV 2 bits, TPC 2 bits. Among them, R Dong said that since the letter of bac h 1 is very good, it can reduce the god of CS, for example, 4 bits indicate higher 16 CS and so on. In this, the CS is completely different.
10) 相 的信息 比特 口教有 , 2 2 比特, 4 需要4比特, 最多不 5比特。 以 T 方式 。  10) Phase information Bits are taught, 2 2 bits, 4 requires 4 bits, and up to 5 bits. In the T way.
統 2 z, 即W W 00, 于T 且 aye ShH模式 下上行 的 C a 包含 下信息 System 2 z, ie W W 00, C a in the T and aye ShH mode, the following information is included
1) : 1比特, "0"代表不執行 操作 "1"代表執行 操 作。 R , 可以保留 1比特的信息 TEARe 9/ E, 需要 1 比特的信息。 資源分配以及 資源分配 1) : 1 bit, "0" means that the operation "1" is not performed. R, can retain 1 bit of information TEARe 9/ E, need 1 Bit information. Resource allocation and resource allocation
2) 資源分配以及 資源分配。  2) Resource allocation and resource allocation.
果 的資源分配方式 , 則 o9 3+ /2 og (1 X( 00 )/2 13比特  If the resource is allocated, then o9 3+ /2 og (1 X( 00 )/2 13 bits
果 非 的資源分配方式 則 / 比特, 其中 的取 表1所示。  If the resource allocation method is not / bit, which is shown in Table 1.
3) RS的 移位,3比特。本 中 配置2 RS 移位, 以便 RS得到 2 P C 索引 。 于P C 索引 的得到 于本領域 木 慣用 木手段, 里不再贅述。  3) RS shift, 3 bits. In this case, configure 2 RS shift so that RS gets 2 P C index. The P C index is obtained in the field of wood, and will not be described again.
4) dex, 3比特。 休所需的比特 上下行bac ha 子 的配 置有夫, 且只有 上行bac ha 子 的 數多于下行bac ha 子 有 信息。 4) dex, 3 bits. The bit required for the hang has a configuration of the upper and lower bac ha, and only the number of upstream bac ha is more than the downstream bac ha.
2 T 模式下的 6 子 配置方案, 其中 表示上行子 , 表 示下行子 。 上下 配置 下行到上行特 的周期 子  The 6 sub-configuration scheme in 2 T mode, which represents the uplink sub-symbol, indicating the downlink sub-. Up and down configuration down to uplink period
0 2 4 6 7 9 0 2 4 6 7 9
0 5 S S 0 5 S S
5 s S S  5 s S S
2 5 s S S  2 5 s S S
10 s S  10 s S
4 10 s S  4 10 s S
10 s S  10 s S
6 5 s S S 6 5 s S S
2  2
以 T J下行配置 1 , 因 8力 T 方式下, 上下行bac a 子 的配置方案的第 的示意 , 因 8 中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指方格表示上行 程子 。 8 所示, bac ha S a e #4可以同時調度 的 3 不同的 bac ha S a e, 即本 中的 s b a e #8 以及下 介 中的 b a e#2和s b a e#3 同 , bac a S b a e#9可以同 的3 不同的 bac ha S b a e, 即下 元 的 s b a e#3 s b a e#7和s b a e#8 With TJ downlink configuration 1, due to the 8 force T mode, the first indication of the configuration scheme of the upstream and downstream bac a sub, because 8 is the square, the lower indicated square indicates the lower stroke, the upward pointing The grid indicates the upper leg. As shown in Figure 8, bac ha S ae #4 can be scheduled at the same time for 3 different bac ha S ae, which is the sbae #8 and the lower Bae#2 and sbae#3, bac a S bae#9 can be the same as 3 different bac ha S bae, ie sbae#3 sbae#7 and sbae#8
此 , bac ha S b a e #4的上行 中 3比特的 dex 休含意可以 00表示 本 的 bac ha S b a e# 110表示 了本 的 bac ha S b a e #8以及下 元 的 bac ha S b a e #2 010表示 度下 元 的 bac ha S bfTa e #2 依此 。 其中, 1表示子 , 0表示子 未 。  In this case, the bac ha S bae #10 of the bac ha S bae #4 indicates that the bac ha S bae #110 of the present bac ha S bae #8 and the bac ha S bae #2 010 of the lower element The bac ha S bfTa e #2 of the lower yuan is based on this. Among them, 1 means child, 0 means child.
5) A, 休所需的比特 上下行bac ha 子 的配置有 , 且 只有 下行bac ha 子 的 多于上行bac ha 子 才有 信息。 5) A, the bits required for the hang The configuration of the upstream and downstream bac ha sub-has, and only the downstream bac ha sub-information is more than the upstream bac ha sub-.
g T 方式下,上下行bac ha 子 的配置方案的第二 的 示意 , 因 9 中, 方格表示子 , 向下 所指方格表示下行 程子 , 向上 所指方格表示上行 程子 。 9所示, 2 的 bac ha S b a e #2上可以同 1 的 bac haS bfTa e#3和 bac ha S bfTa e#4相 的 /非 (AC / AC ), 此 , 需要2比特的 A 指示 bac ha S b a e #2上 的是 In the g T mode, the second indication of the configuration scheme of the upper and lower bac ha sub, because of the square, the square indicates the sub-head, the downward-pointed square indicates the descending process, and the upward-pointed square indicates the uplink. 9 shows that 2 bac ha S bae #2 can be the same as 1 bac haS bfTa e#3 and bac ha S bfTa e#4 phase / non (AC / AC), this requires 2 bits of A to indicate bac Ha S bae #2 is on
bac ha S b a e的AC / AC 。 休 , 可以 A 10 ,表示 bac ha S b a e#3 的AC / AC , A 01 表示 bac ha S b a e #4 的AC AC , A 11 , 表示 bac ha S b a e #3 和 bac ha s b a e#4 的AC / AC 。 3 的 bac ha S b a e#7 上也有 似的情況, 里不再 。  Bac ha S b a e AC / AC. Hugh, can A 10 , AC / AC for bac ha S bae #3 , A 01 for AC AC of bac ha S bae #4 , A 11 , AC for bac ha S bae #3 and bac ha sbae #4 / AC. 3 bac ha S b a e#7 also has a similar situation, no longer.
6)CQ ,1比特。可以 0表示 或者 TEARe 9/ 0 E C ,其 1表示R 或者 TE e 9/1 E需要 C 。 本 中, 送完全相同的CQ。 ) ARQ , bac a 子 配置有 , 值方法 F 方式下的 相同, 里不再 。 6) CQ, 1 bit. It can be 0 or TEARe 9/ 0 EC, where 1 means R or TE e 9/1 E requires C. In this case, send the exact same CQ. ) ARQ, bac a sub-configuration, the value method F is the same, no longer.
8) 到 流特 ( odewodSw )的 , 1比特。 于指示T 和T 2分別映射到 上。 值方法 F 方式下的 相同, 里不再 。  8) to the stream (odewodSw), 1 bit. The indications T and T 2 are mapped to respectively. The value method is the same in F mode, no longer in it.
9) 于T , CS 5比特, 1比特, RV 2比特, TPC 2比特。 于T 2, CS 5比特, 1比特, RV 2比特, TPC 2比特。 其中, R 東說, 由于bac 的信 很好, 因此可以 減少 CS的神 , 例 , 4比特 指示較高的 16 CS等。 本 中, 的 CS是完全相同的。  9) On T, CS 5 bits, 1 bit, RV 2 bits, TPC 2 bits. At T 2, CS 5 bits, 1 bit, RV 2 bits, TPC 2 bits. Among them, R Dong said that because bac's letter is very good, it can reduce the god of CS, for example, 4 bits indicate higher 16 CS and so on. In this, the CS is exactly the same.
10) 相 的信息 比特 口教有 , 2 2 比特, 4 需要4比特, 最多不 5比特。  10) Phase information Bits are taught, 2 2 bits, 4 requires 4 bits, and up to 5 bits.
以上 , 力本 的較佳 而已, 非 于限定本 的保 , 凡在本 的精神和原則 內所作的任何修 、 等同替換和 等, 包含在本 的保 內。  The above is better than the limit, and any repairs, equivalents, and the like made in the spirit and principles of this article are included in this warranty.

Claims

要求 Claim
1、 上行 的方法, 其特 在于, 包括  1. The method of uplink, which is characterized by
中牲 或用戶終端接收 于 上行 的下行控制信息 格式, 按照接收到的下行控制信息格式, 上行 。 The medium-allied or user terminal receives the uplink downlink control information format and uplinks according to the received downlink control information format.
2、 要求1 的方法, 其特 在于, 用戶終端 高 長 期 TEA 統中的版本9/10的用戶終端。  2. The method of claim 1, which is characterized in that the user terminal has a version 9/10 user terminal in the high-end TEA system.
3、 要求2 的方法, 其特 在于, 下行控制信息格式 包含有 , 資源分配以及 資源分配, 參考 RS的 移位, 上行索引 , 下行配置索引 A, 指示C ,  3. The method of claim 2, wherein the downlink control information format includes, resource allocation, and resource allocation, reference RS shift, uplink index, downlink configuration index A, indication C,
方案 CS、 新教 指示 以及版本 余RV, 功率控制TPC, 混 合 ARQ , 到 流特 的 , 上行 相 的信息。  Scheme CS, Protestant Indications, and version RV, Power Control TPC, Mixed ARQ, Upstream, Upstream Information.
4、 要求3 的方法, 其特 在于, RS的 移位的 中 或用戶終端配置 RS 移位, 以及2 混合ARQ指示 P C 索引  4. The method of claim 3, which is characterized in that the RS shifts or the user terminal configures the RS shift, and the 2 hybrid ARQ indicates the P C index
或者, 分別 配置各 的 RS 移位, RS 移位得到2 P C 索引 。  Or, configure each RS shift separately, and RS shift to get 2 P C index.
5、 要求3 的方法, 其特 在于, 上行索引 所需的 比特 上下行 程子 的配置有 ,且 上行 程bac a 子 的 多于下行bac ha 子 存在所 上行索引 的信息 5. The method of claim 3, wherein the uplink and downlink parameters required for the uplink index are configured, and the uplink bac a sub-has more than the downlink bac ha sub-existing index information.
A所需的比特 上下行 程子 的配置有 ,且 上行 程 bac ha 子 的 多于下行bac ha 子 存在所 A的信息。  The bit required for A is configured for the uplink and downlink, and the information of the A is more than the downstream bac ha of the uplink bac ha.
6、 要求3 的方法, 其特 在于, 相 同或不同的 CQ 。 6. The method of claim 3, which is characterized by the same or different CQ.
7、 要求3 的方法, 其特 在于, 相 同或不同的 CS 7. The method of claim 3, which is characterized by Same or different CS
8、 要求3 的方法, 其特 在于, AR 的 特 等于 ARQ 8. The method of claim 3, characterized in that the AR is equal to the ARQ.
ARQ 等于第 次上行 和第 次 同配置的上 程子 的 。  ARQ is equal to the last upstream and the same configured upstream.
9、 上行 的 統, 其特 在于, 包括 , 中 和用戶終端, 其中,  9. The uplink system, which includes, and includes, a user terminal, wherein
, 于 于 上行 的下行控制信息格式, 井 下 中 或用戶終端。  , in the uplink control information format, downhole or user terminal.
中 或用戶終端, 于按照接收到的下行控制信息格式, 上 行 。  The medium or user terminal is in the format of the received downlink control information.
10、 要求 9 的 統, 其特 在于, 用戶終端 高 長期 TEA 統中的版本9/10的用戶終端。  10. The system of requirement 9 is characterized in that the user terminal has a version 9/10 user terminal in the high-long-term TEA system.
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