Wireless Communications

(1)

ةبيط ةعماج

Wireless Communications

Lecture 3: Global Services Mobile (GSM) Transmitter Encoding

Omar Siddiqui

Department of Electrical Engineering College of Engineering

Taibah University Madinah

Email:[email protected]

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The Air Interface-Mobile-BTS Link

College of Engineering, Taibah University Wireless Communications

TB-3

-The Mobile and BTS communicate on the interface with digitally encoded TDMA frames

- Each user is allotted a 0.577 ms slot on the uplink and downlink “TDMA frames” in which the data or control “bursts” are sent to exchange information - The GSM burst is formed by a complicated process of speech and channel coding

The uplink frame transmission is delayed by 3 slots

3 57 1 26 1 57 3 8.25

0.577 ms data burst occupies 1 slot

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6

0.577 ms

TDMA frames

4.616 ms TDMA

frame

slot

(3)

The GSM Burst (Frame) Generation

• The GSM frames are generated in several steps of coding and signal processing to accomplish the following:

 Increase compression to save bandwidth

 Improve quality by error reduction and equalization etc

 Improve security by encryption etc

 Reduce cost and operation and maintenance issues

Speech Coding

Channel Coding

Interleaving

Burst Formatting

RF Modulation

RF

Signaling

Voice

Transmitter

Voice

A/D, PCM

2080 bits

20ms

260 bits

456 bits

PCM Speech Coding

Channel Coding

Interleaving

TB 3

Data Bits 57

TS 26 RB

1

DB 57

TB 3

GB 8.25 RB

1

Burst Formatting

57 57

0.577ms burst

Frame View of the Transmitte

(4)

The GSM Burst (Frame) Reception

College of Engineering, Taibah University Wireless Communications

• The reverse procedure is followed for reception

Speech Coding

Channel Coding

Interleaving

Burst Formatting

RF Modulation

RF

Signaling

Voice

Speech deCoding

Channel deCoding

De-Interleaving

Burst De-Formatting

RF DeModulation

RF

Signaling

Voice

Transmitter Receiver

Frame Structuring

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The GSM Speech Coding

College of Engineering, Taibah University Wireless Communications

• The first step is the Speech Encoding which is done in two steps:

Voice is first digitized and encoded using 13 bit 104 kB/s Pulse code Modulation (PCM)

 The 104 Kbps PCM bit stream is then compressed and coded using the RPE-LPC (Regular Pulse Excited – Long Term Predictor) voice encoding

Speech Coding Channel

Coding

Interleaving

Burst Formatting

RF Modulation RF

Signaling Voice

Low-Pass Filter

Sampling and

quantization PCM RPE-LTP

Analog speech

104 kbps

13 kbps

T

_s

0 00 1

0 1 1

0 0 1 1

0 1

1 11 11

1 11 1

To channel Coding

8000 samples/s

(6)

The GSM Speech Coding: PCM Encoder

College of Engineering, Taibah University Wireless Communications

Coding Interleaving

Burst Formatting RF ModulationRF

Signaling Voice

 The Voice signal is sampled at 8000 samples per s

 which means one slot of 0.125 ms

Pulse Code Modulation using 13 bit or 2

¹³

= 8192 levels

0 1 8190 8191

0.125ms

Sampling/Quantization

S (Sampling)

Q (Quantization)

3 2 4095 4096

13 bit PCM

0.125ms

Output Data Rate of Speech coder

Data rate = 13 / 0.125 x 10

^-3

= 104kbps Voice

Sampling and quantization

PCM

To RPE-LTP

(7)

The GSM Speech Coding: RTE-LTP Encoder

College of Engineering, Taibah University Wireless Communications

Coding Interleaving

Burst Formatting

RF Modulation RF

Signaling Voice

 The RPE-LTP employs three types of coding schemes in parallel

 The Linear Prediction Coder (LPC) produces 36 bits per 20 ms

 The RPE coder produces 188 bits per 20 ms

 The LTP coder produces a stream of 36 bits per 20ms

 The three outputs are multiplexed into a single stream of 13kbps

104kbps

2080 bits / 20ms

36 LPC

bits per 20 ms

188 RPE bits per 20 ms

36 LTP

bits per 20 ms

MUX

188+36+36 bits per 20ms 260 bits per 20ms

13kbps

Input = 104kbps Output = 13kbps

To Channel Coding

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The GSM Channel Coding

College of Engineering, Taibah University Wireless Communications

 The Channel is done to reduce the errors when the bit stream is transmitted over the channel

 Two types of error correction schemes are used

 The 260 bits in the 20ms speech coded frame is divided into three streams

 the 50 class 1a bits are CRC encoded using a 4-bit generator polynomial that generates a 53 bit stream

 The 132 class 1b bits and the 53 class 1a bits are coded using (2,1,5) convolution coder and then passed to

interleaver stage

 The remaining 78 class 2 bits are directly passed to Bit interleaver stage without any coding

104kbps 2080 bits

/ 20ms

Channel Coding Block Diagram

RPE- LTP Speech

Coding

260 bits / 20ms

13kbps

Demux

50 class 1a bits

3-bit CRC 132 class

1b bits

Convolution Coder (2,1,4)

456 Bits /20ms

78 class 2 bits

Interleaving

456Bits/ms22.8 kbps

Coding Interleaving

Burst Formatting

RF Modulation RF

Signaling Voice

Bit Interleaver

Class 1a 50 bits

Class 1b 132 bits

Class 2 78 bits

50 3 132 4

378 78

456 bits per 20ms frame Convolution Coding

4 tail bits 3 parity bits

189 bit frame to conv. coder CRC

GSM Frame Hierarchy

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The GSM Channel Coding-The 3 bit CRC Generator

College of Engineering, Taibah University Wireless Communications

 The first 50 Class 1a bits are protected by 3 parity bits that are added by a Cyclic Redundancy Check (CRC) encoder with the generator sequence g(D) = D

³

+ D + 1 , where r = 3

The parity bits are the remainder when the D

^r

X(D) polynomial is divided by the generator polynomial

The output sequence is given by :

Class 1a 50 bits

Class 1b 132 bits

Class 2 78 bits

50 3 132 4

378 78

456 bits per 20ms frame Convolution Coding

   

) (D g

D X rem D

D p



r

CRC

1001 1001101

  ^D

X

   

) (D g

D rem X

D p 

  ^D

g 1011

 

  ^D ^p ^(D ⁾

X D

D Y

r



  ^D ^D ^X   ^D ^p ^(D ⁾ 

Y 

^r



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The GSM Channel Coding-The 3 bit CRC Generator Example

College of Engineering, Taibah University Wireless Communications

Q. Calculate the parity bits when the message polynomial is given by M(D) = D³+1 and the generator polynomial is g(D) = D³+D+1. What is the output bit sequence

Solution:

1. Write the polynomials into bit sequences M(x) 1001

G(D)  1011

2. Append ‘r’ zeroes to the input sequence, where ‘r’ represents the order of the generator polynomial. In this case r= 3, i.e.

3. Perform the division operation and find the remainder D³M(x) 1001000

1001000 1011

1  1011 10 XOR

Don’t write

zeroes at front

0 0

0 1

1011 110 3-bit

CRC

1001 1001110

1011

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The GSM Channel Coding-The 3 bit CRC Generator Example-Receiver side

College of Engineering, Taibah University Wireless Communications

On the receiver side the CRC is performed again. If the remainder is zero, the frame has no errors. If the remainder is not zero, the frame is discarded

1011

1  1011 101 0

1 1

1011 0 1001110

Sequence with Error Correct Sequence

1011

 1011 1 1 1011

0 1000110 1011

1 1 100 1011

11 Zero Remainder

Non-Zero Remainder

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The GSM Channel Coding-The Convolution Encoder

 The Convolution encoder is a logic error correcting circuit characterized by three parameters written as (n,k,m), where:

k: the number of bits present at the input at a particular instant of time (k) .

n: The number of output bits obtained by Modulo-2 addition

m: The number of memory or flip-flop stages i.e. m.

 Hence the GSM Conv Encoder is characterized by (2,1,4)

 At any instant of time, there are k-input bits and n output bits which fixes the rate of the encoder, given by k/n

 The GSM encoder rate is ½

 The output of is a convolution between the input X(D) and the generator polynomial G(D) i.e.

FF1 FF2 FF3 FF4

X(D)

1 Modulo-2 Adder

Y(D) GSM Convolution Encoder

2   ^D ^X   ^D ^G ^(D ⁾

Y  

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The Convolution Encoder –Generator Sequence

 For each output there is a generator sequence

 The generator sequence is the impulse response of the coder and can be determined by:

Either applying the sequence 1000… at the input (number of zeroes = m)

Or by simply specifying the ‘connection’ from the shift register to the adder by ‘1’ and ‘no- connection’ by ‘0’

FF1 FF2 FF3 FF4

X(D)

1 Modulo-2 Adder

Y(D) GSM Convolution Encoder

2 GSM Generator Sequences

 ¹¹⁰¹¹ 

 g

o

 10011 

) 1

(



g

Polynomial form of the Generator Sequences

1 )

( D  D

⁴

 D

³

 D  g

^o

1 )

(

⁴

1

D  D  D 

g

Matrix Form

] 1 1

[ )

( D  D

⁴

 D

³

 D  D

⁴

 D 

G

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Examples of Convolution Encoders

College of Engineering, Taibah University Wireless Communications

Determine the encoder type (n,k,m) , the rate k/n of the following encoders and the generator polynomials

(2,1,3) (2,1,3) Rate = 1/2

Rate = 1/2

 1011 

) 1

(



g

 1101 

) 2

(



g

Transfer Matrix Function

 1011 

) 1

(



g

 1111 

) 2

(



g

] 1 1

[ )

( D  D

³

 D  D

⁴

 D

³

 D

²

 G

Transfer Matrix Function

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

k: Input bits at any instant . n: Output bits at any instant m: Number of flip-flop stages

(15)

Examples of Convolution Encoders

College of Engineering, Taibah University Wireless Communications

(3,2,1) Rate = 2/3 (4,3,2) Rate = 3/4

  ¹¹

) 1 (

1



g ^g

₁⁽²⁾

^   ⁰¹ ^g

₁⁽³⁾

^   ¹¹

  ⁰¹

) 1 (

2



g g

₂⁽²⁾

   10 ^g

₂⁽³⁾

^   ¹⁰

 

 



  

 D D

D D D

G 1

1 1

) 1 (

Transfer Matrix Function

(16)

Examples of Convolution Encoders

College of Engineering, Taibah University Wireless Communications (2,1,3) Rate = 1/2

(3,2,2) Rate = 2/3

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Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

Let us calculate the output when a bit stream of 10110 enters a (2,1,3) convolution encoder

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

1 0 1 1 0

Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0 0 0

1 1 0

0 0 0

1

(18)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

1 0

1 1 0

Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0 0

0 0 0

1 0 0

0 1 1

1

(19)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

0 1 1

1 0

Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0

1 1 1

0 0 0

0 1 1

1 0

1

(20)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

1 1 0

1 0

Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

1 1 0

1 1 1

0 1 1

1 0

1

(21)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

1 0 1

0 Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0 0 1

1 0 0

0 1 1

1 0

1 1

1

(22)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

1 1 0

Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0 0 1

0 1 1

1 0

1 1

1 0

1

(23)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

1 0

Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0 0 0

0 1 1

1 0

1 1

1 0

1

(24)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input:

0 Let us initialize the registers to zero and reverse the stream so that the earliest bit enters first

Rate = 1/2 (2,1,3) 1 0 1 1 0

0 0 0

0 1 1

1 0

1 1

1 0

1 1 0

0

0 Input: 1 0 1 1 0 Output: 11 01 01 01 11 01 11 00

(25)

Calculation of a Convolution Encoder’s Output

College of Engineering, Taibah University Wireless Communications

] 1 1

[ )

( D  D

³

 D  D

³

 D

²

 G

Input: 1 0 1 1 0 Rate = 1/2 (2,1,3)

Input: 1 0 1 1 0 Output: 11 01 01 01 11 01 11 00

The length of each output bit stream is given by: L+m (L=message length) The total length of the output = n(L+m)

In this case the total output length is 2(5+3)=16

Y

₁

Y

₂

The output bit stream appearing at each output is:

Y

₁

: 1 0 0 0 1 0 1 0 Y

₂

: 1 1 1 1 1 1 1 0

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Calculation of a Convolution Encoder’s Output- Using Convolution

College of Engineering, Taibah University Wireless Communications

Calculate the output when a bit stream of 10110 enters a (2,1,3) convolution encoder using the convolution

Input:

Rate = 1/2 (2,1,3)

1 0 1 1 0 Y

₁

Y

₂

16 5

 ¹⁰¹¹ 

) 1

(



g ^g

⁽²⁾

^  ¹¹⁰¹ 

2. Calculate the output length which is the number of columns of the generator matrix : n(L+m) =2(5+3)=16

3. Construct the generator matrix which has an order of Lx n(L+m) or 5x16:

• the first row consists of combining the elements of the generator sequences

• the first two elements consist of the first element of each sequence

• Fill the remaining columns by zeroes

• the second row will start from the last two elements of the first row. Then copy the rest in the same order

1. Write the generator sequences :

 ] [G

11 10 01 11

00 11 01 10 11 00 00 00 00 00 11 01 10 11 00 00 00 00 00 11 01 10 11 ⁰⁰

00 00 00 00 11 01 10 11

00 00 00 00

(27)

Calculation of a Convolution Encoder’s Output- Using Convolution

College of Engineering, Taibah University Wireless Communications Rate = 1/2 (2,1,3)

Y

₁

Y

₂













11 10 01 11 00 00 00 00

00 11 10 01 11 00 00 00

00 00 11 10 01 11 00 00

00 00 00 11 10 01 11 00

00 00 00 00 11 10 01 11

 ] [Y

] 0 1 1 0 1 [



 ] [Y  11 01 01 ⁰¹ 11 01 11 ⁰⁰ 

4. Find the output by doing a matrix multiplication of the input bit array and the generator matrix:

5. Since the convolution coder is rate ½ coder, the first 10 bits form the coded output.



 ] [C  11 01 01 ⁰¹ ¹¹ 

(28)

Back to the GSM Convolution Coder

College of Engineering, Taibah University Wireless Communications

GSM Frame Hierarchy

Class 1a 50 bits

50 3 132 4

378 78

456 bits per 20ms frame Convolution Coding 3 parity bits

Class 1b 132 bits

Class 2 78 bits

4 tail bits

0 1     187 188

Rate = 1/2 (2,1,4)

FF1 FF2 FF3 FF4

X(D)

1

Y(D)

2 g^o 



11011





¹⁰⁰¹¹



) 1

( 

g

Message length = L=189 bits Output length = 2(L+M)

=2(189+4)

= 386 bits

L=189 bits

382-8=378 bits

(29)

Back to the GSM Convolution Coder- The Generator Matrix

Rate = 1/2 (2,1,4)

FF1 FF2 FF3 FF4

X(D)

1

Y(D)

2 ^g^o ^



¹¹⁰¹¹





¹⁰⁰¹¹



) 1

( 

g

 ] [G

00 00 00



00 00 00

Generator Matrix

11 10 00 11 11

10 bits Message length

= L=189 bits

Output length = 386 bits

372 bits



  00



 00

00 00 00 00

11 10 00 11 11   ⁰⁰ ⁰⁰ ⁰⁰ ⁰⁰

11 ₁₀ 00 11 11 00 00   

372 bits 10 bits

189 rows

386 189 



  



 

  



(30)

Output of the GSM Conv Coder

 ]

[Y

0 1     187188

189 1 

 ]

[Y

0 1     187188

386 1 

189 190   376377

The final code consist of Y – 2m = 378bits

] 

[C

0 1     187188

378 1 

189 190   376377



378 379

  378bits

385

386 189 

(31)

Back to the GSM Frame Hierarchy where we had left

]  [C

378 1 

189 190   376377

GSM Frame Hierarchy

Interleaving Stage

Output frame of the convolution coder stage

Class 1a 50 bits

Class 1b 132 bits

Class 2 78 bits

50 3 132 4

378 78

Convolution Coding

78 Class 2 bits are added and the 456 bit frame is passed to the Interleaving stage

GSM Transmitter

0 1     187188

189 190   376377

0 1     187188 378379   454455

78bits

20ms

(32)

The Bit Interleaving stage

-The bit interleaving is the dispersion of the convoluted code in various locations of the GSM bursts to minimize consecutive data or speech loss

- Without interleaving, if some consecutive bits are lost due to noise or fading, the speech will become intangible

189 190   376377

0 1     187188 378379   454455

20ms

College of Engineering, Taibah University Wireless Communications Bits lost due to fading or

similar mechanisms This part of speech

will be lost

Two Levels of Interleaving

• First Level: The 456 encoded frame is divided into 8 blocks of 57 bits

• Second Level: Half of the 57-bit blocks are grouped together with blocks of previous 456-bit frame and half with the next 456-bit frame to form the TDMA burst

3 57 1 26 1 57 3 8.25

57 bits from Frame A and B 57 bits from Frame A and B

Block 1 Block 2

57 bits 57 bits

Block 8

57 bits

    First Level

Second

Level

(33)

First Level of Bit Interleaving

189190   376377

0 1     187188 378379   454455

20ms

-First Block: Start from 0 and add 64 to find the next bit location until 57 bits are reached. If the sum is more than 455, subtract 456

-Second and up blocks: Add 57 to each of the parallel entry. If the sum is more than 455, subtract 456

0 64 128 192 256 320 384 448 56

8 72 136 200 264 328 392



57 121 185 249 313 377 441 49 113



65

129 193 257 321 385 449

122 186 250 314 378 442 50



114 178 242 306 370 434 42 106 170

171 235 299 363 427 35 99 163 227

228 292 356 420 28 92 156 220 284

285 349 413 21 85 149 213 277 341

342 406 470 78 142 206 270 334 398

399 7 71 135 199 263 327 391

 455







407 471 79 143 207 271 335 350

414 22 86 150 214 278 293

357 421 29 93 157 221 236

300 364 428 36 100 164 179

243 307 371 435 43 107

57 bits

Block B_o Block B₁ Block B₂ Block B₃ Block B₄ Block B₅ Block B₆ Block B₇ 448+64

-456

(34)

Second Level of Bit Interleaving or Burst Formatting

-Half of the present 57-bit blocks (B) are grouped together with blocks of previous 456-bit frame (A) and half with the next 456-bit frame (C) to form the TDMA burst

- In each burst, bits of two successive blocks are further interleaved in an even/odd order, as shown below

20 ms 456 bit Frame A 20 ms 456 bit Frame B 20 ms 456 bit Frame C

0 57 114 171 228 285 342 399 64 121 178 235 292 349 406 7 128 185 242 299 356 413 470 71 192 249 306 363 420 21 78 135 256 313 370 427 28 85 142 199 320 377 434 35 92 149 206 263 384 441 42 99 156 213 270 327 0 57 114 171 228 285 342 399

64 121 178 235 292 349 406 7 128 185 242 299 356 413 470 71 192 249 306 363 420 21 78 135 256 313 370 427 28 85 142 199 320 377 434 35 92 149 206 263 384 441 42 99 156 213 270 327

B_o B₁ B₂ B₃ B₄ B₅ B₆ B₇

A_o A₁ A₂ A₃ A₄ A₅ A₆ A₇ C_o C₁ C₂ C₃ C₄ C₅ C₆ C₇

0 57 114 171 228 285 342 399 64 121 178 235 292 349 406 7 128 185 242 299 356 413 470 71 192 249 306 363 420 21 78 135 256 313 370 427 28 85 142 199 320 377 434 35 92 149 206 263 384 441 42 99 156 213 270 327

       

72 129 186 243 300 357 414 471 136 193 250 307 364 421 22 79 200 257 314 371 428 29 86 143 264 321 378 435 36 93 150 207 328 385 442 43 100 157 214 271 392 449 50 107 164 221 278 335

3 BAB……AB 28 ABA……BA11.25 3 CBC……BC 28 BCB……CB11.25

BAB ABA BAB ABA BAB ABA CBC BCB

A₄,B_o A₅,B₁ A₆,B₂ A₇,B₃ B₄,C₀ B₅,C₁

CBC BCB

B₆,C₂ B₇,C₃

CBC BCB

User 1

Traffic Bursts

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

TDMA Frame 1 TDMA Frame 2 TDMA Frame 3 TDMA Frame 4 TDMA Frame 5 TDMA Frame 6 TDMA Frame 7 TDMA Frame 8

(35)

Summary of the Burst Formation Steps

A₄,B_o A₅,B₁ A₆,B₂ A₇,B₃ B₄,C₀ B₅,C₁ B₆,C₂ B₇,C₃

0 1 2 3        2077 2078 2079 0 1 2 3        2077 2078 2079 0 1 2 3       

0 1   ²⁵⁸ 259

104 kbps

13 kbps

2077 2078 2079

0 1   ²⁵⁸ 259 0 1   ²⁵⁸ 259

0 1 2   453 ⁴⁵⁴ ⁴⁵⁵ 0 1 2   453 ⁴⁵⁴ ⁴⁵⁵ 0 1 2   453 ⁴⁵⁴ ⁴⁵⁵

22.8 kbps

0 12 3 4 56 70 1 2 3 4 56 7 0 1 2 3 4 56 7 0 1 2 3 4 56 7 0 1 2 3 4 567 0 1 2 3 4 56 7 0 1 2 3 4 56 71 0 1 2 3 4 56 7 0 1 2 3 4 56 7 0 1 2 3 4 56 7 0 1 2 3 4 56 7 0 1 2 3 4 56 7 0 1 2 3 4 56 7 B_o B₁ B₂ B₃ B₄ B₅ B₆ B₇ C_o C₁ C₂ C₃ C₄ C₅ C₆ C₇

22.8 kbps

20ms

270 kbps ¹ ^{3 45 7}

SECOND INTERLEAVING FIRST INTERLEAVING

CONVOLUTION CODING SPEECH CODING

PCM CODING

1 slot

= 0.577ms

1 TDMA Frame = 4.616ms 0

64

392



 57 121

449



 114 178

50



 171 235

107



285 349

221



 342 406

278



 399

7

335



 228

292

164



57 it blocks

A_o A₁ A₂ A₃ A₄ A₅ A₆ A₇ 0 64

392



 57 121

449



 114 178

50



 171 235

107



285 349

221



 342 406

278



 399

7

335



 228

292

164



0 64

392



 57 121

449



 114 178

50



 171 235

107



285 349

221



 342 406

278



 399

7

335



 228

292

164



20ms 33.8 kbps

B A B A B A

3 28

11.25

C B C B C B 3

11.25

28

4.03ms 4.03ms 4.03ms

B A B A B A

3 28

11.25

B A B A B A

3 28

11.25

C B C B C B 3

11.25

B A B A B A 28

3 28

11.25

C B C B C B 3

11.25

28 3C B C B C B

11.25

28

(36)

Radio Frequency Stage

Speech Coding

Channel Coding

Interleaving

Burst Formatting

RF Modulation RF

Signaling

Voice

(37)