5.1a.6 Packet data block type 17 (UAS-10)

3GPP45.003GSM/EDGE Channel codingRelease 17TS

5.1a.6.1 Block constitution

If the message delivered to the encoder does not include a PAN, it has a fixed size of 1402 information bits {d(0),d(1),…,d(1401)}. If the message delivered to the encoder includes a PAN, it has a fixed size of 1427 information bits {d(0),d(1),…,d(1426). If the message delivered to the encoder includes a PAN and an eTFI, it has a fixed size of 1430 information bits {d(0),d(1),…,d(1429).

NOTE: The presence of the PAN is indicated by the PANI field in the header (see 3GPP TS 44.060).

The message is separated into the following parts:

h(k) = d(k) for k = 0,…,51

i1(k-52) = d(k) for k = 52,…,501

i2(k-502) = d(k) for k = 502,…,951

i3(k-952) = d(k) for k = 952,…,1401

And if a PAN is included:

pn(k-1402) = d(k) for k = 1402,…,1426

And if a PAN and an eTFI are included:

et(k-1427) = d(k) for k = 1427,…,1429

5.1a.6.2 Header coding

The header {h(0),…,h(51)} is coded as defined in subclause 5.1a.1.1, with N=52, resulting in a block of 180 bits, {C(0),…,C(179)}.

The code is punctured in such a way that the following coded bits:

{ C(15*k+12) for k=0,…,11} are not transmitted

This results in a block of 168 bits, {hc(0),…,hc(167)}.

5.1a.6.3 Data coding

Each data part, {i1(0),…,i1(449)}, {i2(0),…,i2(449)} and {i3(0),…,i3(449)}, is coded as defined in subclause 5.1a.1.2, with N=450, resulting in three coded blocks of 1404 bits, {C1(0),…,C1(1403)}, {C2(0),…,C2(1403)} and {C3(0),…,C3(1403)}.

Each coded block is punctured depending on the value of the CPS field as defined in 3GPP TS 44.060. Three puncturing schemes named P1, P2 or P3 are applied in such a way that the following coded bits are punctured:

Always punctured

Punctured only if a PAN is included

P1

C(18*k+j) for k=0,…,77, j=2, 4, 5, 7, 8, 9, 11, 12, 13, 15 and 16,
except C(18*k+7) for k=2, 8, 13, 19, 24, 30, 35, 41, 46, 52, 57, 63, 68 and 74 that are not punctured.

C(18*k+7) for k=2, 8, 13, 19, 24, 30, 35, 41, 46, 52, 57, 63, 68 and 74; and

C(18*k) for k=0, 6, 13, 19, 26, 32, 39, 45, 52, 58, 65 and 71

P2

C(18*k+j) for k=0,…,77, j=0, 1, 3, 4, 6, 8, 10, 11, 13, 14 and 17,
except C(18*k+3) for k=4, 10, 15, 21, 26, 32, 37, 43, 48, 54, 59, 65, 70 and 76 that are not punctured.

C(18*k+3) for k=4, 10, 15, 21, 26, 32, 37, 43, 48, 54, 59, 65, 70 and 76; and

C(18*k+15) for k=2, 8, 15, 21, 28, 34, 41, 47, 54, 60, 67 and 73

P3

C(18*k+j) for k=0,…,77, j=1, 2, 3, 5, 6, 7, 9, 10, 14, 16 and 17,
except C(18*k+1) for k=0, 6, 11, 17, 22, 28, 33, 39, 44, 50, 55, 61, 66 and 72 that are not punctured.

C(18*k+1) for k=0, 6, 11, 17, 22, 28, 33, 39, 44, 50, 55, 61, 66 and 72; and

C(18*k+12) for k=4, 10, 17, 23, 30, 36, 43, 49, 56, 62, 69 and 75

If a PAN is not included, the result is three blocks of 560 bits, {c1(0),…,c1(559)}, {c2(0),…,c2(559)} and {c3(0),…,c3(559)}, where c1 corresponds to i1, c2 to i2 and c3 to i3.

If a PAN is included, the result is three blocks of 534 bits, {c1(0),…,c1(533)}, {c2(0),…,c2(533)} and {c3(0),…,c3(533)}, where c1 corresponds to i1, c2 to i2 and c3 to i3.

NOTE: C1 and c1 correspond to i1, C2 and c2 to i2, and C3 and c3 to i3.

5.1a.6.4 PAN coding

The PAN coding is the same as for UAS-7 as specified in subclause 5.1a.3.4.

5.1a.6.5 Interleaving

a) Header

The header, {hc(0),…,hc(167)}, is interleaved as defined in subclause 5.1a.2.1, with NC=168 and a=25, resulting in a block of 168 bits, {hi(0),…,hi(167)}.

b) Data and PAN

If a PAN is not included, data are put together as one entity as described by the following rule:

dc(k) = c1(k) for k = 0,…,559

dc(k) = c2(k-560) for k = 560,…,1119

dc(k) = c3(k-1120) for k = 1120,…,1679

If a PAN is included, data and PAN are put together as one entity as described by the following rule:

dc(k) = ac(k) for k = 0,…,77

dc(k) = c1(k-78) for k = 78,…,611

dc(k) = c2(k-612) for k = 612,…,1145

dc(k) = c3(k-1146) for k = 1146,…,.1679

The block {dc(0),…,dc(1679)} is interleaved as defined in subclause 5.1a.2.1, with NC=1680 and a=173, resulting in a block of 1680 bits, {di(0),…,di(1679)}.

5.1a.6.6 Mapping on a burst

a) Straightforward Mapping

The mapping is given by the rule:

For B=0,1,2,3, let

e(B,j) = di(420B+j) for j = 0,…,211

e(B,j) = hi(42B+j-212) for j = 212,…,231

e(B,j) = q(2B+j-232) for j = 232,233

e(B,j) = hi(42B+j-214) for j = 234,…,255

e(B,j) = di(420B+j-44) for j = 256,…,463

where

q(0),q(1),…,q(7) = 1,1,1,1,1,1,1,1 identifies the coding scheme UAS-10 or UAS-11.

b) Bit swapping

After this mapping the following bits are swapped:

For B = 0,1,2,3,

Swap e(B,192+k) with e(B,214+k) for k=0, 1, 4, 5, 8, 9, 12, 13, 16, 17, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62 and 63.

c) PAN bit swapping

In case a PAN is included in the radio block, the following additional bits are swapped after the bit swapping in b):

For B = 0

Swap e(B,50) with e(B,120)

Swap e(B,75) with e(B,81)

Swap e(B,99) with e(B,177)

Swap e(B,174) with e(B,40)

Swap e(B,198) with e(B,160)

Swap e(B,267) with e(B,368)

Swap e(B,366) with e(B,305)

Swap e(B,390) with e(B,444)

Swap e(B,391) with e(B,280)

Swap e(B,415) with e(B,405)

For B = 1

Swap e(B,7) with e(B,0)

Swap e(B,31) with e(B,177)

Swap e(B,106) with e(B,21)

Swap e(B,130) with e(B,120)

Swap e(B,155) with e(B,40)

Swap e(B,226) with e(B,160)

Swap e(B,251) with e(B,368)

Swap e(B,298) with e(B,280)

Swap e(B,322) with e(B,444)

Swap e(B,323) with e(B,321)

Swap e(B,347) with e(B,405)

Swap e(B,446) with e(B,305)

For B = 2

Swap e(B,38) with e(B,40)

Swap e(B,62) with e(B,160)

Swap e(B,87) with e(B,120)

Swap e(B,186) with e(B,81)

Swap e(B,210) with e(B,177)

Swap e(B,211) with e(B,0)

Swap e(B,279) with e(B,405)

Swap e(B,378) with e(B,368)

Swap e(B,403) with e(B,305)

Swap e(B,427) with e(B,444)

For B = 3

Swap e(B,19) with e(B,160)

Swap e(B,118) with e(B,81)

Swap e(B,143) with e(B,40)

Swap e(B,167) with e(B,177)

Swap e(B,214) with e(B,120)

Swap e(B,239) with e(B,280)

Swap e(B,310) with e(B,368)

Swap e(B,335) with e(B,305)

Swap e(B,359) with e(B,444)

Swap e(B,434) with e(B,321)

Swap e(B,458) with e(B,405)