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common H-RNTI 如何產(chǎn)生?
elvis130
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發(fā)表于 2011-12-09 16:49:12  只看樓主 
請教一下各位高手:
common H-RNTI是如何產(chǎn)生的呢?
根據(jù)3GPP的定義:
The detection of HS-SCCH is based on Common H-RNTI listed in System Information Block type 5 (SIB 5) or System Information Block type 5bis (SIB 5bis). The UE shall select the Common H-RNTI to be used for reception of the RRC CONNECTION SETUP message according to the following rule:
-
the UE shall select a Common H-RNTI from the Common H-RNTIs listed in System Information Block type 5 (SIB 5) or System Information Block type 5bis (SIB 5bis) based on "Initial UE Identity" as follows:


"Index of selected Common H-RNTI" = "Initial UE Identity" mod K,

-
where K is equal to the number of listed Common H-RNTI. These Common H-RNTIs shall be indexed from 0 to K-1 in the order of their occurrence in SIB 5 or SIB 5bis, and "Index of selected Common H-RNTI" identifies the selected Common H-RNTI. "Initial UE Identity" refers to the Information Element included by the UE into the RRC CONNECTION REQUEST message. In the above formula, the parameter "Initial UE Identity" shall be interpreted as follows, depending on the choice of UE-Id type of the respective IE:


For UE-Id type "IMSI (GSM-MAP)", i.e. the IE is given as sequence of digits of type Integer(0..9), "Initial UE Identity" shall be interpreted as an integer number, where the first digit given in the sequence represents the highest order digit.


For UE-Id types "TMSI and LAI (GSM-MAP)" or "P-TMSI and RAI (GSM-MAP)", only the TMSI or P-TMSI parts given as Bitstring(32) shall be used, and "Initial UE Identity" shall be interpreted as a binary representation of an integer number, where the first bit in the Bitstring represents the highest order bit.

For UE-Id type "IMEI", i.e. the IE is given as sequence of hexadecimal digits of type Integer(0..15), "Initial UE Identity" shall be interpreted as a hexadecimal representation of an integer number, where the first digit given in the sequence represents the highest order digit.

For UE-Id type "IMSI (DS-41)", i.e. the IE is given as octet string, "Initial UE Identity" shall correspond to the decoded decimal representation of the IMSI-S part included in the octet string (see TIA/EIA/IS-2000-5).


For UE-Id types "ESN (DS-41)" or "TMSI (DS-41)", i.e. the IE is given as Bitstring(32), "Initial UE Identity" shall be interpreted as a binary representation of an integer number, where the first bit in the Bitstring represents the highest order bit.


For UE-Id type "IMSI and ESN (DS-41)" only the ESN part shall be used as "Initial UE Identity", as defined above.

After detecting the HS-SCCH with Common H-RNTI, the UE start reception of the corresponding HS-PDSCH(s) TTI.

那么這個 "Index of selected Common H-RNTI" = "Initial UE Identity" mod K 的這個K是2么?指的是common H-RNTI的數(shù)量么?通常在SIB5里面的common H-RNTI都是固定的那兩個(9B CF 和 37 9E)。

另外這個Initial UE Identity它選用的是哪一個?有沒有什么規(guī)則?
如果上面的那個"K”就是2的話,那么這個"Index of selected Common H-RNTI"如何算數(shù)來呢?

請高手們出來幫忙解答解答啊!
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    elvis130
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    發(fā)表于 2011-12-12 10:15:15  只看樓主 
    有沒有高手出來說說呀,那么多天了,竟然一個回復(fù)的人都沒有!

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    發(fā)表于 2011-12-12 12:59:30 
    1) common H-RNTI作用
    為了能從HS-DSCH接收數(shù)據(jù),UE需要一個標(biāo)識符,用它來對應(yīng)在共享信道HS-DSCH上為UE指定的資源,在CELL_DCH狀態(tài)這個標(biāo)識符為dedicate H-RNTI,在CELL_FACH下則是common H-RNTI。
    2) common H-RNTI由何而來?
    系統(tǒng)消息HS-DSCH Common system information中會包含一個common H-RNTI列表,UE會從中選擇一個使用
    the UE shall select a Common H-RNTI from the Common H-RNTIs listed in System Information Block type 5
    3) index of comman H-RNTI(index選擇規(guī)則)
    a>UE是連接模式(Connected mode)
    UE通過U-RNTI獲。"Index of selected Common H-RNTI" = U-RNTI [mod K]
    b>非連接模式
    UE通過RRC Conection Request 消息獲得H-RNTI list和一個Initial UE Identity,相似的方式獲得其index:"Index of selected Common H-RNTI" = "Initial UE Identity" [mod K]
    而這條消息會告知Initial UE Identity類別,在樓主提供的信息中也有詳細(xì)介紹:For UE-Id type "IMSI (GSM-MAP)", "TMSI and LAI (GSM-MAP)" , "IMEI"......

    綜上,
    1)Initial UE Identity根據(jù)UE-Id由RRC Con. setup告知,算法其實樓主給的文檔里面也已經(jīng)指明。
    2)K為H-RNTI list中H-RNTI的個數(shù),最大取值應(yīng)該是4
    3)核心公式"Index of selected Common H-RNTI" = "Initial UE Identity" mod K
    舉例Initial UE Identity=25,K=4,則Index of selected Common H-RNTI=25 mod 4,取模計算法則樓主如果不懂可以去翻翻數(shù)學(xué)書籍,很簡單的

    [ 本帖最后由 魚兒 于 2011-12-12 13:02 編輯 ]

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    發(fā)表于 2011-12-21 18:09:02  只看樓主 
    首先先非常感謝 魚兒 的回帖。
    還有些疑問想跟你請教喔:上面的那個K的確為H-RNTI list中H-RNTI的個數(shù),
    但通常在SIB5消息里面,只能找到其中兩個,那么這種情況那個K值應(yīng)該取2還是4呢?
    我個人覺得應(yīng)該是2.
    在一些資料里面,找到下面的這個算法,
    H-RNTI shall be allocated by using following algorithm:
    p(0) = 0
    p(n) = mod(p(n-1)+39887, 65536)
    , where p(n) is H-RNTI. n shall be selected from range 0 - 3 for common H-RNTIs so that:
    - value 0 is used for BCCH Specific HS-DSCH-RNTI
    - value 1 is used for CCCH common H-RNTI
    - value 2 is used for DCCH SRB1 common H-RNTI
    - (value 3 spare for future use)
    那么按照這個算法的話,好像就是說只有value1 和value2這兩種情況產(chǎn)生的H-RNTI才是common H-RNTI,是吧?

    另外,“Initial UE Identity” 可以是 "IMSI (GSM-MAP)", "TMSI and LAI (GSM-MAP)" , "IMEI"...... 那么如果這個值是"TMSI and LAI (GSM-MAP)" =22 01 77 C3 26 2F 03 00 66 , K=2或K=4的話,那么"Index of selected Common H-RNTI" =??

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