WO2010148939A1 - Method and device for data mapping, and method and device for data demapping - Google Patents

Method and device for data mapping, and method and device for data demapping Download PDF

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Publication number
WO2010148939A1
WO2010148939A1 PCT/CN2010/073634 CN2010073634W WO2010148939A1 WO 2010148939 A1 WO2010148939 A1 WO 2010148939A1 CN 2010073634 W CN2010073634 W CN 2010073634W WO 2010148939 A1 WO2010148939 A1 WO 2010148939A1
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Prior art keywords
valid
mapping
rate
received
receive
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PCT/CN2010/073634
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French (fr)
Chinese (zh)
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覃尉
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中兴通讯股份有限公司
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Publication of WO2010148939A1 publication Critical patent/WO2010148939A1/en

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    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04JMULTIPLEX COMMUNICATION
    • H04J3/00Time-division multiplex systems
    • H04J3/16Time-division multiplex systems in which the time allocation to individual channels within a transmission cycle is variable, e.g. to accommodate varying complexity of signals, to vary number of channels transmitted
    • H04J3/1605Fixed allocated frame structures
    • H04J3/1652Optical Transport Network [OTN]
    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04LTRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION
    • H04L49/00Packet switching elements
    • H04L49/35Switches specially adapted for specific applications
    • H04L49/356Switches specially adapted for specific applications for storage area networks
    • H04L49/357Fibre channel switches

Definitions

  • mapping method and device mapping method and device
  • This relates to optical communication teaching, and in particular to methods and devices for teaching God mapping, and methods and devices for teaching mapping. Background
  • the ⁇ channel (C, be Cha e ) is in the middle, because the rate of C is lower than (OT, Op ca Ta po ewo ), so C cannot be directly mapped to OT, especially 1 C, because the rate of 1 C is only 1.0625Gbp, which is 42% of the channel teaching sheep (O, Op ca Cha e aa), if the direct rate is matched, the wave is 58%.
  • G P Ge e c a g oced
  • encapsulation rate is assigned to the channel sheep (OP, Op ca Cha e ayoad) rate, and then P, then the channel sheep (T, Op ca Cha e Ta po), the most OT. Not only the way, but also the G FC of the sheep.
  • users can eliminate the need to assign the 1 C G package rate to the rate of P, and can directly package the 1 C G package.
  • T the most OT.
  • the main purpose of this is to provide a method and device for mapping the gods, and a method and device for teaching the mapping, and to reduce the C system.
  • the wood plan is
  • the 9 C's C contains 8 valid, where, the valid 0 in the 0, 0 and the remaining 7 are in turn C, and the rest are in turn C is valid.
  • the method steps include all the teachings of Zhanyi are less than, all, will be the full effort OP
  • the method step includes
  • the method step includes 0 0 of the received OP in the OP, and the OP is full and full.
  • the method step includes parsing O in the OT, and then
  • mapping gods including compilation, rate, and control.
  • the rate, the received, and all the teachings are not less than the effective force of the OP
  • Step the device includes O and OT, where O, the OP at the receiving rate, will receive the OP O OT, in receiving O out of O, will receive O OT.
  • the mapping device includes a solution, where, in the receiving OP, 0 0 of the received OP, OP full, all received 0 0 of the OP, the OP force is valid, valid
  • Step the device includes OT and O, where OT, parses O O in the received OT, and receives OP from the received OT, and parses the OP in the received O.
  • the method of mapping this C OP mapping includes the following steps
  • step 101 C is taught, and 0 of 0 is.
  • C 9 (b ) which contains 8 valid and .
  • C is arranged in order, as shown in the upper part of 2, representing 1 , each C, the bottom of each is C, and the top is C. , move C to the front to form 0, because only, so 0 at 0 fixed padding, it forms 0.
  • C's 8 is effective, and the media contains 8 valid, 2 times The half is shown.
  • Step 102 104 receiving valid, all valid If it is less than, if it is, then it is full, and then proceed to step 105. Otherwise, it is valid.
  • Steps 102 to 104 rate, rate out of valid or full OP.
  • the size of the value is determined by the rate of C and OP, and the value of the value is the size of the rate in which the value can be required.
  • Steps 101 to 104 show the mapping from C to OP.
  • Step 105 106 receiving the OP, receiving the received OP O receiving O, and receiving the O OT
  • OP can be effective or capable.
  • OP O what will be OP O, and what will be O
  • steps 101 to 106 are operations that can be performed on C on OT.
  • Step 107 Receive OT, and parse out the received OT.
  • Step 108 Receive the parsed O, and parse the OP in the received O.
  • Step 109 receiving the parsed OP, 0 of 0 in the OP is 0, if yes, then the OP is complete, and then performing step 110 otherwise, the OP force is valid, and then step 5 is performed, step 110, all, before Process.
  • Step 11 the output is valid, and the execution step 112 is performed.
  • Step 112 receiving valid, will receive a valid C, which will contain 8 of the seven.
  • step 09 to step 2 have the demapping of OP to C.
  • Steps 107 to 112 are operations required to receive the OT exchange for C when the OT is received.
  • each includes C, , rate, control, O and OT receiving, each including OT, O, and C. Where, rate, and control C are mapped to the OP device
  • the mediation C contains 8 valid and .
  • the main function of the rate is to assign the rate of C to P.
  • Control as received, to rate control.
  • the main function of the control is to monitor the rate of the rate, the effective amount of the rate is less than, and the control control rate stops to be valid, but all.
  • the rate is valid or full OP.
  • the O is parsed in the received OT.
  • the OP in receiving O from the OT, The OP is parsed out in O.
  • the 0 of the received OP 0 is 0, if it is, then OP full, all Otherwise, the OP force is valid, valid. Will only be valid. Hugh, first set, but 0 in the 0 is no or not, if yes, then all else, will.
  • C 9 contains 8 valid and .
  • Step 202 Receive 2 C of each C, and receive the received C of the C into a valid of 8 and then valid, 0 1 of the valid 0. 2 and 4 are shown.
  • step 203 the rate is valid, and the rate of 2 C is assigned to P. Since the rate of P is faster than the rate of 2G C, there may be a case where the effective number is empty, so the rate needs to fill the same rate of 2G C P with the full packet, the rest force
  • Step 204 O receives the rate of P, and will receive the P, of course.
  • Step 205 the OT receives the O, and receives the received T, and then receives each T.
  • Step 206 the OT receives the T of the OT, and parses the received T, and then.
  • Step 207 O receives the OT out, and parses the received P, and then P.
  • Step 208 receiving O out of P, 0 of 0 in the received P is 0, if yes, then OP is full, all else, otherwise P force is effective and effective.
  • Step 209 the valid is received, and the valid C will be received, and Hugh will contain 8 of C 9 C.
  • Step 210 C each receives C.
  • each 2 C maps to P, then T, then T, receives each T
  • 1 C P has the same mapping mapping
  • 8G C P 2 has the same mapping mapping
  • 2G C P has the same mapping mapping, and is no longer here.

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  • Engineering & Computer Science (AREA)
  • Computer Networks & Wireless Communication (AREA)
  • Signal Processing (AREA)
  • Data Exchanges In Wide-Area Networks (AREA)
  • Communication Control (AREA)

Abstract

The present invention discloses a method and device for data mapping. The method includes: encoding Fiber Channel (FC) data into valid data packets, and outputting the encoded valid data packets, wherein the 0th bit of the 0th byte of each valid data packet is 1; receiving and buffering the valid data packets, and outputting the buffered valid data packets as Optical channel Payload Unit-k (OPUk) data when the number of all the buffered valid data packets is not less than a predetermined threshold. The present invention also discloses a method and device for data demapping. Application of the present invention can avoid a Generic Framing Procedure (GFP) encapsulation process, thereby reducing complexity of the implementation of an FC system.

Description

映射的方法及裝置、 以及 映射的方法及裝置 木領域  Mapping method and device, and mapping method and device
本 涉及光通信教 木, 尤其涉及 神教 映射的方法及裝 置、 以及教 映射的方法及裝置。 背景 木  This relates to optical communication teaching, and in particular to methods and devices for teaching God mapping, and methods and devices for teaching mapping. Background
目前, 仟通道 ( C, be Cha e ) 在 中, 由于 C的 速率比 (OT , Op ca Ta po ewo ) 低, 因此 C不能 直接映射到 OT 的 中 特別是 1 C, 由于 1 C 的速率只有 1.0625Gbp , 是 通道教 羊 (O , Op ca Cha e aa ) 的42%, 果將 直接速率 配到 , 則 浪 58%的 。  At present, the 仟 channel (C, be Cha e ) is in the middle, because the rate of C is lower than (OT, Op ca Ta po ewo ), so C cannot be directly mapped to OT, especially 1 C, because the rate of 1 C is only 1.0625Gbp, which is 42% of the channel teaching sheep (O, Op ca Cha e aa), if the direct rate is matched, the wave is 58%.
現有 木中, 通常的做法是 在 的 , 將 G C  Existing wood, the usual practice is in , will G C
通用 規程 (G P, Ge e c a g oced )封裝 速率 配到 通道 羊 (OP ,Op ca Cha e ayoad )的速率,再將 P ,然 將 通道 羊 ( T ,Op ca Cha e Ta po ) , 最 OT 。 理方式 不但 , 而且不能 羊 的 G FC 。 但是, 在大容量交叉 木和 通信 組 ( T ) 于G7 9 中 的 出現 , 用戶可以不必再將 1 C G 封裝 速 率 配到 P 的速率, 而可以將 路1 C G 直接封裝 The general procedure (G P, Ge e c a g oced ) encapsulation rate is assigned to the channel sheep (OP, Op ca Cha e ayoad) rate, and then P, then the channel sheep (T, Op ca Cha e Ta po), the most OT. Not only the way, but also the G FC of the sheep. However, in the presence of large-capacity cross-over wood and communication group (T) in G7 9, users can eliminate the need to assign the 1 C G package rate to the rate of P, and can directly package the 1 C G package.
配到 P , 再將 P , 然 將 Match P, then P, then
T , 最 OT 。  T, the most OT.
上 理方式 然 了 ,但是 的東說,在 C , 目前近需先將 C G P封裝 速率 配到 OP ,再將OP O , 然 將O OT , 最 T , 但是, G P封裝 , C 統的 也很 , 而浪 多的資源。 內容 The way to go, but East said, in C, it is now necessary to first set the CGP encapsulation rate to OP, then OP O, then O OT, the most T, but, GP package, C system is also very, and more resources. content
5 有 于此, 本 的主要目的在于提供 神教 映射的方法及裝置、 以及教 映射的方法及裝置, 兔 封裝 , 而降低 C 統 的 。 5 In this case, the main purpose of this is to provide a method and device for mapping the gods, and a method and device for teaching the mapping, and to reduce the C system.
到上 目的, 本 的 木方案是 的  For the purpose of the purpose, the wood plan is
神教 映射的方法, 包括 God's mapping method, including
C 有效 , 的有效 , 有效 中 0 的 0  C is valid, effective, effective 0 of 0
接收 有效 , 的所有有效 不小于 的 , 將 的有效 作力 OP 。  Receive valid, all valid not less than, will be the effective force OP.
其中, 將 C 有效 休 Among them, C will be effective
5 將 9 的 C 成 介包含8 的有效 , 其中, 有效 中 0 的 0 1, 0 的 余7 依次 各 C 的 , 其余 依次 各 C 的有效 。 5 The 9 C's C contains 8 valid, where, the valid 0 in the 0, 0 and the remaining 7 are in turn C, and the rest are in turn C is valid.
其中, 方法 步包括 占鍰 的所有教 小于 , 全 的 , 將 全 的 作力 OP Among them, the method steps include all the teachings of Zhanyi are less than, all, will be the full effort OP
0 。 0.
其中, 在 OP , 方法 步包括  Wherein, in the OP, the method step includes
接收OP ,將收到的OP O ,再將O Receive OP, will receive OP O, then O
OT 。  OT.
神教 映射的方法, 包括 God's mapping method, including
5 OP 中 0 的 0 1 , OP 力有效 , 出有效 5 0 in 0 0 of OP, OP force is valid Effective
接收有效 , 將收到的有效 C 。  Receive valid, will receive a valid C.
其中, 將收到的有效 C 休  Among them, will receive a valid C
將 包含8 的 成七 9 的 C 。  Will contain 8 of C 7 to 9 .
其中, 方法 步包括 占收到的 OP 中 0 的 0 0 , OP 全 的 , 全 的 。  The method step includes 0 0 of the received OP in the OP, and the OP is full and full.
其中, 在 OP 力有效 或全 的 前, 方法 步包括 OT 中解析出 O , 再 O  Wherein, before the OP force is valid or full, the method step includes parsing O in the OT, and then
中解析出 OP 。 Parsing OP out.
神教 映射的裝置, 包括 編碉 、 速率 及控制 其中,  a device for mapping gods, including compilation, rate, and control.
, 于將 C 有效 , 的有效 , 有效 中 0 的 0  , valid for C, valid, effective 0
速率 , 于接收 出的 , 的所 有教 不小于 的 , 將 的有效 作力 OP  The rate, the received, and all the teachings are not less than the effective force of the OP
的所有教 小于 的 , 向控制 控制 , 于在收到 , 向 速率 控制 速率 , 近 于根 收到的控制 , 全 的 , 將 全 的 作力 OP 。  All of the teachings are less than, to the control control, at the rate of control, to the rate control rate, near the root received control, all, will be the full force OP.
步 , 裝置近包括O 和OT 其中, O , 于接收速率 出的 OP , 將收到 的OP O OT , 于接收O 出的 O , 將收 到的O OT 。 Step, the device includes O and OT, where O, the OP at the receiving rate, will receive the OP O OT, in receiving O out of O, will receive O OT.
映射的裝置, 包括 和解 其中, , 于接收OP , 收到的 OP 中 0 的 0 0 , OP 全 的 , 全 的 收到的OP 中 0 的 0 1 , OP 力有效 , 出有效  The mapping device includes a solution, where, in the receiving OP, 0 0 of the received OP, OP full, all received 0 0 of the OP, the OP force is valid, valid
, 于接收 出的有效 , 將收到的 有效 C 。  , valid after receiving, will receive a valid C.
步 , 裝置近包括OT 和O 其中, OT , 于 收到的 OT 中解析出 O O , 于接收OT 出的 O , 收 到的O 中解析出 OP 。  Step, the device includes OT and O, where OT, parses O O in the received OT, and receives OP from the received OT, and parses the OP in the received O.
現有 木使用 G P 封裝以及速率 , 比較 , 資源 比較大,而本 只需執行 以及速率 步驟就可以 C 到 OP 的映射, 只需執行 和解 步驟就可以 OP 到 C 的解映射, 比較 羊, 而降低了 C 統 的 , 了更多的資源。 1力本 C OP 映射 映射的方法的 程示意 2力本 C 的 規則示意 Existing wood uses GP encapsulation and rate, comparison, and resource is relatively large, but this only needs to perform and rate step can be C to OP mapping, just perform the resolving step to OP to C demapping, compare sheep, and reduce C, more resources. 1 force this C OP mapping mapping method of the schematic diagram 2 force this C rules
3力本 C 各 OT 的示意 4力本 C 的 羊示意 。 休 方式 It is indicated by the sheep of the C of the C. Hugh way
下面結合 本 的 木方案 細說明。 The following is a detailed description of the wood scheme.
1所示,本 C OP 映射 映射的方法包括以 下步驟  As shown in Figure 1, the method of mapping this C OP mapping includes the following steps
步驟101, 將 C 成教 , 的 , 中 0 的 0 1。  In step 101, C is taught, and 0 of 0 is.
休 將 9 的 C 成 介包含8 的 , 其中, 中 0 的 0 1, 0 的 余7 依次 各 C 的 , 其余 依次 各 C 的有效 。  Hugh will divide 9 C into 8 , where 0 in 0, 0 in the remaining 7 in order of C, and the rest in order for each C to be valid.
本文中, 了 此 的 下文提到的全 的 , 將 C 形成的 力有效 。  In this paper, the full force mentioned below is effective for the force formed by C.
下面結合 2 細說明 C 的 規則。 C 9 (b ), 其中包含8 有效 和 。 C 依次排列, 2上 半部分所示, 代表1 , 每 介 C , 每 的最下面 是 C 的 , 最上面的 代表 C 的 。 , 將 C 的 依次移到最前面用以形成 0 , 由于 只有 , 因此在 0 的 0 固定填充1, 就形成了 0 另外, C 的 8 有效 成 介 , 就形成了 介包含8 的有 效 , 2下半部分所示。  The rules of C are described in detail below in conjunction with 2. C 9 (b ), which contains 8 valid and . C is arranged in order, as shown in the upper part of 2, representing 1 , each C, the bottom of each is C, and the top is C. , move C to the front to form 0, because only, so 0 at 0 fixed padding, it forms 0. In addition, C's 8 is effective, and the media contains 8 valid, 2 times The half is shown.
另外, 規定 中 0 的 0 0 , 的 全 的 , 此 都是全 。 全 的  In addition, the total 0 0 of 0 is specified, and this is all. complete
填充 C和OP 同的速率 , 其也 力空 。 其中, 有不同 的 , 可以 0、 1或2, 且, 不同的 代表不同的 P 速率等 , 此 現有 木 在此不再贅述。  Fill the same rate as C and OP, which is also empty. Among them, there are different, can be 0, 1 or 2, and different represent different P rates, etc., and the existing wood will not be described here.
步驟102 104,接收 有效 , 的所有有效 是否小于 的 , 果是, 則 全 的 , 接 看執行步驟105 否則, 的有效 , 接看執行步驟1 5 Step 102 104, receiving valid, all valid If it is less than, if it is, then it is full, and then proceed to step 105. Otherwise, it is valid.
步驟 102至 104 速率 , 速率 出的有效 或全 的 OP 。  Steps 102 to 104 rate, rate out of valid or full OP.
另外, 由于 OP 的速率比較快, 可能出現 的有效 空 的情況, 因此預先 , 的有效 小于 , 停 止 出有效 , 而 全 的 , 以 OP  In addition, because the rate of OP is relatively fast, there may be a valid empty condition, so the pre-, effective is less than, the stop is effective, and the whole, to OP
。 其中, 值的大小由 C和OP 的速率 , 值的 倍則 速率 的 的大小 在 中可以 需要 值的大 小。  . Among them, the size of the value is determined by the rate of C and OP, and the value of the value is the size of the rate in which the value can be required.
步驟101至步驟104 了 C 到 OP 的映射。  Steps 101 to 104 show the mapping from C to OP.
步驟 105 106, 接收OP , 將收到的 OP O 接收O , 將收到的O OT  Step 105 106, receiving the OP, receiving the received OP O receiving O, and receiving the O OT
 .
其中, OP 可能力有效 , 也可能力全 的 。 其中, 何將OP O 、 以及 何將O Among them, OP can be effective or capable. Among them, what will be OP O, and what will be O
OT 現有 木, 在此不再 。  OT existing wood, no longer here.
另外, 步驟101至步驟 106是 了能在OT 上 C , 所需執行的操作。  In addition, steps 101 to 106 are operations that can be performed on C on OT.
步驟 107, 接收OT , 收到的 OT 中解析出 O  Step 107: Receive OT, and parse out the received OT.
 .
步驟 108, 接收解析出的 O , 收到的 O 中解析出 OP 。  Step 108: Receive the parsed O, and parse the OP in the received O.
何 OT 中解析出 O 、 以及 何 O 中 析出 OP 現有 木, 在此不再 。 步驟109,接收解析出的OP , OP 中 0 的 0 是否 0, 果是, 則可 OP 全 的 , 接 看執行步驟 110 否則, 可 OP 力有效 , 接看執行步驟 5 步驟110, 全 的 , 前流程。 In the OT, the O is analyzed, and the OP is extracted from the existing wood. Step 109, receiving the parsed OP, 0 of 0 in the OP is 0, if yes, then the OP is complete, and then performing step 110 otherwise, the OP force is valid, and then step 5 is performed, step 110, all, before Process.
步驟 11, 出有效 , 接看執行步驟112  Step 11, the output is valid, and the execution step 112 is performed.
步驟112, 接收有效 , 將收到的有效 C , 即將 包含8 的 成七 9 的 。  Step 112, receiving valid, will receive a valid C, which will contain 8 of the seven.
程力步驟 101 中 的 , 結合 2本領域 木0 很容易理解 何 行解 , 在此不再 。  Chengli in step 101, combined with 2 fields of wood 0 is easy to understand what the solution, no longer here.
其中, 步驟 09至步驟 2 了 OP 到 C 的解映射。 步驟 107至步驟 112是接收 在收到 OT , 將OT 特 換成 C 所需執行的操作。 Among them, step 09 to step 2 have the demapping of OP to C. Steps 107 to 112 are operations required to receive the OT exchange for C when the OT is received.
3力本 C 各 OT 的示意 。5 其中, 各包括 C 各、 、 速率 、 控制 、 O 及 OT 接收 各包括 OT 、 O 、 、 及 C 各。其中, 、 速率 及控制 C 映射到 OP 的裝置  3 force this C OT's indication. 5 Among them, each includes C, , rate, control, O and OT receiving, each including OT, O, and C. Where, rate, and control C are mapped to the OP device
及解 OP 映射到 C 的裝置。 Resolve the device that the OP maps to C.
C 各 C 的 東說明 3所示 的裝置 何 C OP 同的映射。 C The east of each C indicates the same mapping of the device shown in Figure 3.
C 各, 于 C 。 其中, 介 C 包含8 有效 和 。  C each, at C. Among them, the mediation C contains 8 valid and .
, 于接收 C 各 出的 C , 將 C 有5 ,有效 中 0 的 0 然 的有效 。 , C for each C, C with 5, valid 0 for 0 is valid .
速率 , 于接收 出的有效 , 的所有有效 不小于 的 , 的有效  Rate, valid for receipt, valid for all valid not less than
的所有有效 小于 的 , 向控制 , 控制 的控制 , 全 的 。  All of the effective less than, control, control, and full.
其中, 是速率 的 神狀 , 在說明速率  Which is the god of the rate, indicating the rate
的所有有效 已小于 的 。  All valid is already less than .
速率 的主要功能是將 C的速率 配到 P  The main function of the rate is to assign the rate of C to P.
控制 , 于在收到的 , 向速率 控制 。 控制 的主要功能是監控速率 的速率 , 速率 的有效 數量小于 的 , 控制 控制速率 停止 出有效 , 而 全 的 。  Control, as received, to rate control. The main function of the control is to monitor the rate of the rate, the effective amount of the rate is less than, and the control control rate stops to be valid, but all.
速率 出的有效 或全 的 OP 。  The rate is valid or full OP.
上 了 C 到 OP 的映射。  The mapping from C to OP is performed.
O , 于接收速率 出的 OP , 將收到 的OP O 。  O, at the receiving rate out of the OP, will receive the OP O.
OT , 于接收O 出的 O , 將收 到的O OT 。  OT, in receiving O out of O, will receive O OT.
至此, 各完成了 C教 到 OT 的特 , , 就可以 在OT 上 C 。  At this point, each completed the C to OT special, you can on the OT C.
下面 接收 各 何將收到的OT 特換成 C 。  The following will receive the OT special for each received.
OT , 于接收OT 出的 OT , 收 到的OT 中解析出 O 。 OT, in the OT received by the OT, the O is parsed in the received OT.
O , 于接收OT 出的 O , 到的O 中解析出 OP 。 O, in receiving O from the OT, The OP is parsed out in O.
, 于接收O 出的 OP , 收到的OP 中 0 的 0 是否 0, 果是,則得 OP 全 的 , 全 的 否則,得 OP 力有效 , 出有效 。 只將有效 。 休 , 先定出 , 然 中 0 的 0 是否 0 是 否 全 的 , 果是, 則 全 的 否則, 將 。  , in the OP received O, the 0 of the received OP 0 is 0, if it is, then OP full, all Otherwise, the OP force is valid, valid. Will only be valid. Hugh, first set, but 0 in the 0 is no or not, if yes, then all else, will.
, 于接收 出的有效 , 將收到的 有效 C 。  , valid after receiving, will receive a valid C.
和解 了 OP 到 C 的解映射。 C 各, 接收 出的 C 。  The demapping of OP to C is reconciled. C each, received C.
至此, 完成了 C敖 在 C 各 同的 。  At this point, C敖 is completed in C.
由以上 木方案 現有 木相比可以看出 現有 木中, 由于G P 本身比較 , 需要執行64b/65b , 包含有 管理和制, 而且近包 含了多通道的 , 因此 G P C OP 同的 映射 映射 , 占用資源比較大 而本 使用 和速率  Compared with the existing wood schemes of the above wood schemes, it can be seen that the existing woods, due to the comparison of the GP itself, need to execute 64b/65b, including management and system, and nearly contain multiple channels, so the GPC OP has the same mapping mapping, occupying resources. Larger and used and rate
就可以 C 到 OP 的映射,使用 和解 就可以 OP 到 C 的解映射, 比較 羊, 因此可 以 占用的資源。  It is possible to map from C to OP, and use the solution to demap the OP to C and compare the sheep, so the resources can be occupied.
下面 介 休 , 何 2G C P 同 的映射, 以使本領域 木 步了解本 。 Let's take a look at the mapping of the 2G C P in the following, so that the field can understand this.
2G C P 同的映射 下 步驟201, C 各 2 C 。 2G CP under the same mapping Step 201, C each 2 C .
其中, C 9 , 包含8 有效 和 。  Among them, C 9 , contains 8 valid and .
步驟202, 接收 C 各 出的2 C , 將收到的 9 的 C 成 介包含8 的有效 ,然 出有效 其中, 有效 中 0 的 0 1。 2及 4所示。  Step 202: Receive 2 C of each C, and receive the received C of the C into a valid of 8 and then valid, 0 1 of the valid 0. 2 and 4 are shown.
步驟203,速率 接收 出的有效 , 責將2 C的速率 配到 P 。 由于 P 的速率比2G C的速率快, 因此可能出現 的有效數 空的情況, 所以速率 需要用 全 的 包東填充2G C P 同的速率 , 休 程力  In step 203, the rate is valid, and the rate of 2 C is assigned to P. Since the rate of P is faster than the rate of 2G C, there may be a case where the effective number is empty, so the rate needs to fill the same rate of 2G C P with the full packet, the rest force
的所有有效 數量是否小于 的 , 果是, 則向控制 , 控制 在收到 , 返 控制 , 速率 控制 全 的 , 以 到速率 的目 的 否則, 的有效 。  Whether all the effective quantities are less than, if it is, then to control, control in the receipt, return control, rate control, and to the rate of the target, otherwise effective.
步驟204, O 接收速率 出的 P , 將 收到的 P , 然 。  Step 204, O receives the rate of P, and will receive the P, of course.
步驟205, OT 接收O 出的 , 將 收到的 T ,然 向接收 各 T 。  Step 205, the OT receives the O, and receives the received T, and then receives each T.
步驟206, OT 接收OT 的 T , 收到的 T 中解析出 , 然 。  Step 206, the OT receives the T of the OT, and parses the received T, and then.
步驟207, O 接收OT 出的 , 收到的 中解析出 P , 然 P 。  Step 207, O receives the OT out, and parses the received P, and then P.
步驟208, 接收O 出的 P , 收到的 P 中 0 的 0 是否 0, 果是, 則得OP 全 的 , 全 的 否則, 得 P 力有效 , 出有效 。 Step 208, receiving O out of P, 0 of 0 in the received P is 0, if yes, then OP is full, all else, otherwise P force is effective and effective.
步驟209, 接收 出的有效 , 將收到 的有效 C , 休是 將 包含8 的 成 七 9 的 C 然 C 。  Step 209, the valid is received, and the valid C will be received, and Hugh will contain 8 of C 9 C.
步驟210, C 各接收 出的 C 。  Step 210, C each receives C.
上 中, 各 了 2 C 映射到 P , 再 T , 然 T , 接收 各 了 T In the above, each 2 C maps to P, then T, then T, receives each T
P , 然 P 映射出 2 C 。  P , then P maps 2 C .
另外, 1 C P 同的映射 映射, 以及8G C P 2 同的映射 映射, 2G C P 同的 映射 映射 似, 在此不再 。  In addition, 1 C P has the same mapping mapping, and 8G C P 2 has the same mapping mapping, and 2G C P has the same mapping mapping, and is no longer here.
以上 , 力本 的較佳 而已, 非 于限定本 的保  Above, the best of the power, not the limit

Claims

要求 Claim
1、 神教 映射的方法, 其特 在于, 方法包括  1. The method of mapping of gods, the method of which includes
將光仟通道 C 有效 , 的有效 , 有效 中 0 的 0  Effective for the pupil channel C, valid, effective 0
接收 有效 , 的所有有效 不小于 的 , 將 的有效 作力 通道 羊 OP 。  Receive valid, all valid not less than, will be the effective force channel sheep OP.
2、 要求 1 的 映射的方法, 其特 在于, 將 C 有效 休  2, the method of mapping 1 is required, and it is effective to take C
將 9 的 C 成 介包含8 的有效 , 其中, 有效 中 0 的 0 1, 0 的 余7 依次 各 C 的 , 其余 依次 各 C 的有效 。  The C of 9 is composed of 8 valid, where the valid 0 of 0, 0 of the remaining 7 are successively C, and the rest are successively valid for each C.
3、 要求 1或2 的 映射的方法, 其特 在于, 方 法 步包括  3. A method of mapping 1 or 2, the method of which includes
占鍰 的所有教 小于 , 全 的 , 將所述全 的 作力 OP 。  All the teachings of Zhanyi are less than, and all of them will be OP.
4、 要求3 的 映射的方法,其特 在于,在 OP , 方法 步包括  4. The method of mapping 3 is required, in particular, in the OP, the method step includes
接收OP , 將收到的 OP 通道教 羊 O , 再將O 通道 羊 OT 。 Receiving the OP, will receive the OP channel to teach the sheep O, and then the O channel sheep OT.
5、 神教 映射的方法, 其特 在于, 方法包括  5. The method of mapping the gods, the method of which includes
占 OP 中 0 的 0 1 , OP 力有效 , 出有效  0 1 of 0 in OP, OP force is valid, valid
接收有效 , 將收到的有效 C 。  Receive valid, will receive a valid C.
6、 要求5 的 映射的方法, 其特 在于, 將 到的有效 C 休 將 包含8 的 成七 9 的 C 。 6, the method of requesting 5 mapping, its special feature, will be effective C Will contain 8 of C 9 to 9.
7、 要求5或6 的 映射的方法, 其特 在于, 方法 步包括  7. A method of mapping 5 or 6 is required, the method step comprising
占收到的 OP 中 0 的 0 0 , OP  Occupies 0 of 0 in the received OP, OP
全 的 , 全 的 。 All, all.
8、 要求 7 的 映射的方法, 其特 在于, 在 OP 力有效 或全 的 前, 方法 步包括 8. The method of mapping 7 is characterized in that, before the OP force is valid or full, the method step includes
OT 中解析出 O , 再 O 中解析出 OP 。 O is parsed in OT, and OP is parsed in O.
9、 神教 映射的裝置, 其特 在于, 裝置包括 編碉 、 速 率 及控制 其中,  9. A device for mapping a god, characterized in that the device includes a compilation, a rate, and a control thereof.
, 于將 C 有效 , 的有效 , 有效 中 0 的 0  , valid for C, valid, effective 0
速率 , 于接收 出的 , 的 有教 不小于 的 , 將 的有效 作力 OP  The rate, the received, the less than, the effective force of the OP
的所有教 小于 的 , 向控制 控制 , 于在收到 , 向 速率 控制 速率 , 近 于根 收到的控制 , 全 的 , 將 全 的 作力 OP 。  All of the teachings are less than, to the control control, at the rate of control, to the rate control rate, near the root received control, all, will be the full force OP.
10、 要求 9 的 映射的裝置, 其特 在于, 裝置 近包括O 和OT 其中,  10. A device for mapping 9 that is characterized in that the device comprises O and OT,
O , 于接收速率 出的 OP , 將收到 的OP O T , 于接收O 出的 O , 將收 到的O OT 。 O, at the receiving rate out of the OP, will receive the OP O T, in the O received O, will receive O OT.
11、 神教 映射的裝置, 其特 在于, 裝置包括  11. A device for mapping a god, characterized in that the device comprises
和解 其中,  Reconciliation
, 于接收OP , 收到的 OP 中 0 的 0 0 , OP 全 的 , 全 的 收到的OP 中 0 的 0 1 , OP 力有效 , 出有效  , in the receiving OP, 0 0 of the received OP, OP full, all received 0 0 of the OP, OP force is valid, valid
, 于接收 出的有效 , 將收到的 有效 C 。  , valid after receiving, will receive a valid C.
12 要求 11 的數 映射的裝置, 其特 在于, 裝 置近 OT 和O 其中,  12 means for number 11 mapping, which is characterized by a device near OT and O,
OT , 于 收到的 OT 中解析出 O O , 于接收OT 出的 O , 收 到的O 中解析出 OP 。  OT, parses O O in the received OT, and parses the OP in the O received by the OT.
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