BPSK with Block Coding Code rate is { } m b - - PowerPoint PPT Presentation

BPSK with Block Coding

• cos(2

)

c

V f t π ±

{ }

ˆ

k

b ( ) t r ( ) t w AWGN, 2-sided PSD of 2 N Coded bit error probability is ε Channel coding by mapping bits to bits m n

{ }

(1) ( )

, ,

m k k

b b ⋯ ⋯ ⋯

{ }

(1) ( )

, ,

n k k

c c ⋯ ⋯ ⋯

• Code rate is

<1 m R n =

Block coding adds redundancy by mapping each block of m information bits into a block of n coded bits, where n > m. If properly designed, can help reduce energy per bit (Eb), while providing the same information rate (rb) with the same bit error probability (P[error]), but by increasing the transmission bandwidth. Equivalently: For a fixed transmission bandwidth, can reduce Eb to achieve the same P[error] by slowing down the information rate rb. Note that ǫ = Q

• 2Eb·R

N0

• , where R = m

n is the code rate.

EE456 – Digital Communications Ha H. Nguyen

BPSK with Repetition Coding

• { }

k

b cos(2 )

c

V f t π ±

{ }

ˆ

k

b ( ) t r

{ }

k k

b b ⋯ ( ) t w AWGN, 2-sided PSD of 2 N Coded bit error probability is ε

This is a special form of block coding by simply repeating one information bit (m = 1) n times. Let k represent the number of coded bits equal to 1, then the majority voting decision is: k

ˆ bk=1

• ˆ

bk=0 n 2 .

P[error] =

n

• k= n+1

2

n k

• ǫk(1 − ǫ)n−k,

ǫ = Q

• 2Eb

nN0

• .

EE456 – Digital Communications Ha H. Nguyen

Performance of BPSK with Repetition Coding

2 4 6 8 10 10

−6

10

−5

10

−4

10

−3

10

−2

10

−1

10 Eb/N0 (dB) P[error] Uncoded BPSK BPSK with repetition code (7,4) Hamming code n=[3,9,13,17,21]

Repetition code turns out to be not useful at all. It is too simple. Upon reflection one realizes that the transmitted signals after repetition coding are still a BPSK signal set over a duration of Tb sec (the transmission bandwidth is actually unchanged!). As such optimum BPSK detection surely

• utperforms the detection based on majority voting.

EE456 – Digital Communications Ha H. Nguyen

BPSK with (7,4) Hamming Code

• cos(2

)

c

V f t π ±

{ }

ˆ

k

b ( ) t r ( ) t w AWGN, 2-sided PSD of 2 N Coded bit error probability is ε Channel coding by mapping bits to bits m n

{ }

(1) ( )

, ,

m k k

b b ⋯ ⋯ ⋯

{ }

(1) ( )

, ,

n k k

c c ⋯ ⋯ ⋯

• Code rate is

<1 m R n =

b(1)

k

b(2)

k

b(3)

k

b(4)

k

c(1)

k

c(2)

k

· · · c(7)

k

(0000) (0000000) (1000) (1101000) (0100) (0110100) (1100) (1011100) (0010) (1110010) (1010) (0011010) (0110) (1000110) (1110) (0101110) (0001) (1010001) (1001) (0111001) (0101) (1100101) (1101) (0001101) (0011) (0100011) (1011) (1001011) (0111) (0010111) (1111) (1111111)

ck =

• c(1)

k c(2) k

· · · c(7)

k

• =
• b(1)

k , b(2) k , b(3) k , b(4) k

• ·

    1 1 1 1 1 1 1 1 1 1 1 1 1    

• Generator matrix G

Note: Multiplication and addition are modulo-2 operations.

EE456 – Digital Communications Ha H. Nguyen

Decoding (7,4) Hamming Code I

For a (n, m) block code, there is (n − m) × n parity-check matrix H such that every valid codeword c =

• c(1), c(2), · · · , c(n)

satisfies c · H⊤ = 0. This parity-check matrix H can be used for decoding. The parity-check matrix of the (7,4) Hamming code is H =   1 1 1 1 1 1 1 1 1 1 1 1  

EE456 – Digital Communications Ha H. Nguyen

Decoding (7,4) Hamming Code II

Let y be the length-n vector of the decoded bits after BPSK

• demodulator. The decoding consists of three steps:

1

Compute the syndrome of y, namely y · H⊤.

2

Locate the coset leader el whose syndrome is equal to y · H⊤. Then el is taken to be the error pattern caused by the channel.

3

Decode the vector y into the codeword c∗ = y + el. From c∗ find the corresponding k information bits by inverse mapping.

Syndrome Coset leaders (100) (1000000) (010) (0100000) (001) (0010000) (110) (0001000) (011) (0000100) (111) (0000010) (101) (0000001)

EE456 – Digital Communications Ha H. Nguyen

Performance of BPSK with (7,4) Hamming Code

2 4 6 8 10 10

−6

10

−5

10

−4

10

−3

10

−2

10

−1

10 Eb/N0 (dB) P[error] Uncoded BPSK BPSK with repetition code (7,4) Hamming code n=[3,9,13,17,21]

P [error] = 9 · ǫ2 · (1 − ǫ)5 + 19 · ǫ3 · (1 − ǫ)4 + 16 · ǫ4 · (1 − ǫ)3 + 12 · ǫ5 · (1 − ǫ)2 + 7 · ǫ6 · (1 − ǫ) + ǫ7, ǫ = Q  

• 2Eb

4

7

• N0

  .

EE456 – Digital Communications Ha H. Nguyen

A 4-State Rate-1/3 Convolutional Encoder and Its State Diagram

1 k

b −

2 k

b −

k

b

(3) k

c

(2) k

c

(1) k

c

• 1

2 k k

b b

− − k

b = 1

k

b = S

1

S

2

S

3

S 000 111 101 010 011 100 110 001

(1) (2) (3) k k k

c c c

EE456 – Digital Communications Ha H. Nguyen

Trellis Diagram and Viterbi Decoding

00 10 01 11

• 1

2 k k

b b

− − (1) (2) (3) k k k

c c c

• S

2

S

1

S

3

S

• ⋯
• S

1

S

2

S

3

S

000 111 011 100 101 110 001 010

Let {r(1)

k , r(2) k , r(3) k } be the received signal samples (at the output of

the matched filter) corresponding to the coded bits {c(1)

k , c(2) k , c(3) k }.

Then it can be shown that the branch metric is 3

j=1 r(j) k [2c(j) k − 1].

With the above branch metric computation, the Viterbi algorithm finds that path through the trellis that maximizes the path metric.

EE456 – Digital Communications Ha H. Nguyen

Performance Comparison with Uncoded BPSK and (7,4) Hamming-Coded BPSK

2 4 6 8 10 10

−7

10

−6

10

−5

10

−4

10

−3

10

−2

10

−1

10 Eb/N0 (dB) P[error] Rate−1/3 4−state convolutional code Uncoded BPSK (7,4) Hamming code

EE456 – Digital Communications Ha H. Nguyen

Convolutional Code used in CDMA2000

EE456 – Digital Communications Ha H. Nguyen

Convolutional Code used in WCDMA

• ❍

EE456 – Digital Communications Ha H. Nguyen

Convolutional Code used in WiMAX

EE456 – Digital Communications Ha H. Nguyen