數位通信練習題解答

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數位通信簡介 ‐‐ 練習題參考解答

1. Choose (c) 2 ) ( 4 1 2 2 ; 2 2 0 ; 2 0 2 ; 0 (b) ) 2 , 2 ( ); ( 2 ) ( 2 ) ( ) 2 , 0 ( ); ( 2 ) ( 0 ) ( ) 0 , 2 ( ); ( 0 ) ( 2 ) ( ) 0 , 0 ( ); ( 0 ) ( 0 ) ( 2 1 1 ) ( ) ( : satisfy to needs (a) 4 3 2 1 2 2 4 2 2 3 2 2 2 1 4 2 1 4 3 2 1 3 2 2 1 2 1 2 1 1 2 / 0 2 0 2 2 0 2 1 T E E E E E T T T E T T E T T E E T T s t T t T t s T s t T t t s T s t t T t s s t t t s T A dt A dt t dt t A s T T T                                                           



         

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2. Choose

 

(c) 4 3 ) ( 4 1 ; 2 2 0 ; 0 (b) ) 2 , 0 ( ); ( 2 ) ( 0 ) ( ) 0 , ( ); ( 0 ) ( ) ( ) 2 , 0 ( ); ( 2 ) ( 0 ) ( ) 0 , ( ); ( 0 ) ( ) ( 2 1 1 1 1 ) ( ) ( : satisfy to need (a) 4 3 2 1 2 2 4 2 2 2 3 1 4 2 1 4 3 2 1 3 2 2 1 2 1 2 1 1 4 / 0 3 /4 2 2 0 2 0 2 2 0 2 1 T E E E E E T T E E T T E E T s t T t t s T s t t T t s T s t T t t s T s t t T t s T B dt B dt B T A dt A dt t dt t s T T T T T T                                               



         

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3. (a)

s

1

s

2

s

3

s

4

11

01

10

00 Z

1

Z

2

Z

3

Z

4 (b)

   





1 2 2 ₂ 2 4 2 2 2 3 2 1 2 3 4 0 2 2 4 ; 0 2 2 8 ; 1 ( ) 4 4 s E E E E E E E E E                     (c) 12 2 1 2 0 0 0 2 2 2 0 0 1 1 2 1

Let ( , ) erfc erfc erfc

2 2 2 2 2 1 1 (1 ) 2 erfc erfc 4 e d p P s s N N N P p p p N N              _ _ _ _ _ _                    _ _ _ _     2 1 2 2 1 4 0

another solution: by union bound

P_e P s s( , ) P s s( , ) erfc N       _ _   (d) 格雷編碼示於圖中 (編碼方式不唯一) 2 1 2 0 1 ( , ) erfc 2 b P P s s N      _ _   (e) X 0 T dt





1

(t)

+

decision

b

0

= 1, if y

1

> 

1

b

0

= 0, if y

1

< 

1

y

1

^

1 2



 



X 0 T dt





2

(t)

+

decision

b

1

= 1, if y

2

> 

2

b

1

= 0, if y

2

< 

2

y

2

^

2

2 





x(t)

_

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4. (a) 2 2 1 ₀ 1 2 2 1 2 1 2 12 12 ,BASK 2 1 2 0 0 0 2 ( ) cos(2 ), 0 1 ( ) 2 2 0 4 2 4 4 ( ) cos(2 ) 2 ( ) ( ) 0 2 2 1 1 1

( , ) erfc erfc erfc

2 2 2 2 2 2 b c b b T b b b b b b c b b b b b b t f t t T T A T E s t dt E A T E E E E A T E s t f t E t T s t d E E E d P P s s N N N                          _ _ _ _    



        (b) 1 2 1 2 12 12 ,BPSK 2 1 2 0 0 0 ,BPSK 2 ( ) cos(2 ), 0 2 ( ) cos(2 ) ( ) 2 ( ) cos(2 ) ( ) 2 2 1 1 1

( , ) erfc erfc erfc

2 2 2 2 2 compare c b b b c b b b c b b b b b b b b t f t t T T E s t f t E t T E s t f t E t T E E E d E E E d P P s s N N N P                           _ _ _ _ _ _       ,BASK

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8. (a) 2 / 4 / 3 11 4 / 3 6 4 / 3 4 / 3 01 2 / 5 2 / 4 / 00 4 / 4 4 / 4 / 10 2 / 3 2 / 4 / 10 4 / 3 2 4 / 3 4 / 3 01 0 1 dibit input 4 / 10 4 / 3 11 4 / 3 01 4 / 00 dibit : define 1                                     k k k k (b) 9. (a)

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EQV

D

{r

k

}

{b

^

k

}

Decoder:

(b) 0 0 0 1 1 1 0 0 1 } { 1 1 0 1 1 0 1 0 } { k k d b (c) 1 1 0 1 0 1 1 0 } { 0 0 0 1 1 0 0 0 1 } { k k b r  (d) 參閱課本 pp. 9-15~9-16. 10. (a)





0 but 4 sin 2 similarly, cos 2 ) 2 cos( ) ( ) 4 cos( cos 2 ) 2 cos( ) 2 cos( ) 2 cos( ) ( ly. respective , ) 2 sin( and ) 2 cos( are outputs LO ) 2 cos( ) ( . negligible is noise additive and sent, was ) ( assume 2 2 2 2 2 1 0 1 1 1 1 1 1 1 1 1                    



y T A y y y T A y T A dt t f t x y t f A t f t f A t f t x t f t f t f A t x t s c Q I c Q c T I c c c                  (b) 參閱課本 pp. 9-6~9-9, 9-14, 9-22~9-23.

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12. (a) 暫存器的狀態變化 (1000)(1100)(1110)(1111) (0111)(1011)(0101)(1010) (1101)(0110)(0011)(1001) (0100)(0010)(0001) (1000)重複 PN 序列為 (0001 1110 1011 001) m = 4 N = 15 = 2m_{− 1} 所以是 m-sequence (b) 自相關函數

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14. (a)





l bits/symbo 171 . 2 1 . 0 log 1 . 0 2 2 . 0 log 2 . 0 3 . 0 log 3 . 0 2 ) ( ₂ ₂ ₂           S H (b) symbol s1 s2 s3 s4 s5 pk 0.3 0.3 0.2 0.1 0.1 ck 00 01 10 110 111 0.2 0.4 0.6 1 1 1 1 1 0 0 0 0 (c) % 68 . 98 2 . 2 171 . 2 ) ( l bits/symbo 2 . 2 3 2 . 0 2 8 . 0         L S H L  (d) 是 PF code, 也是 UD code.

(e) ( 2)_2 ( )_2_2.171_4.342 bits/(2symbols)

S H S

H

15. (a) Code 1 and Code 4 are PF codes, Code 2 and Code 3 are not PF codes. (b)

For Codes 1, 2, and 4,



2lk 1,

For Code 3, 1 8 9 2  



lk . Code 3 不可能為 UD code. 反例： (0110) 可解為 (0,110) 或 (011,0) (c) Code 2 為即時碼 (instantaneous code)

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16. (a)

For data stream: {11101001011010110001011…} build code book 0 1 11 10 100 101 1010 110 00 1011 position 0 1 2 3 4 5 6 7 8 9 position +code 11 10 30 31 50 20 00 51 codeword 0011 0010 0110 0111 1010 0100 0000 1011 (b)

1. LZ code does not need the knowledge of source probability distribution; 2. LZ achieves higher compression efficiency in general;

3. LZ is fixed-length code, but Huffman code is variable-length.

4. but LZ needs to build a code book (more complex than Huffman code).

17. (a)            1 1 0 1 1 0 0 1 0 1 1 0 1 0 1 1 1 0 0 0 1 H (b) m1 = (1011), ∴ c1 = m1 G = (0011011) m2 = (0101) , ∴ c2 = m2 G = (0010101) (c) m c m c 0000 0000000 1000 0111000 0001 1110001 1001 1001001 0010 1010010 1010 1101010 0011 0100011 1011 0011011 0100 1100100 1100 1011100 0101 0010101 1101 0101101 0110 0110110 1110 0001110 0111 1000111 1111 1111111 dmin = wmin = 3 t = 1 (d) ) 1010 ( ˆ ) 1101010 ( ˆ ˆ ) 0100000 ( ˆ ) 010 ( ) 1001010 (          m e r c e rH s r T

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18. (a)





l bits/symbo 9219 . 0 1 . 0 log 1 . 0 1 . 0 log 1 . 0 8 . 0 log 8 . 0 ) ( ₂ ₂ ₂         S H (b)

symbol

s

1

s

2

s

3

p

k

0.8

0.1

0.1 c

k

0

10

11

0.2

1

0

(c) % 8 . 76 2 . 1 9219 . 0 ) ( l bits/symbo 2 . 1 2 2 . 0 1 8 . 0         L S H L  (d) symbol s1 s2 s3 s4 s5 s2 s3 s4 s5 pk 0.64 0.08 0.08 0.08 0.08 0.01 0.01 0.01 0.01 ck 0 110 100 101 1110 111100 111101 111110 111111 0.36 1 1 1 1 1 0 0 0 0 0.02 1 1 0 0.02 0 1 0 0.04 0.12 0.16 0.20 1 0 % 03 . 96 92 . 1 9219 . 0 2 ) ( 2 ) ( symbols) bits/(2 92 . 1 6 04 . 0 4 08 . 0 3 24 . 0 1 64 . 0 2               L S H L S H L 

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19. (a) m1 = (0), c1 = (00000)

m2 = (1) , c2 = (11111)

dmin = wmin = 5

t = 2

(b) If “number of 1’s” in r is more than “number of 0’s”, then choose cˆ(11111) and mˆ (1),

otherwise, choose cˆ(00000) and mˆ (0).

m1 = (1011), ∴ c1 = m1 G = (0011011) m2 = (0101) , ∴ c2 = m2 G = (0010101) (c) r(10011)  cˆ(11111), mˆ (1) (d) 3 2 3 4 5 3 5 3 4 5 4 5 10 then , 1 if ) 1 ( 10 ) 1 ( 5 ) sent was 1 | ( symmetry, to due ) 1 ( 3 5 ) 1 ( 4 5 ) sent was 1 | ( sent was (11111) assume p P p p p p p p e P P p p p p p e P e e                            20. (a)        1 0 1 1 0 0 1 0 1 1 G (b) m c 00 00000 01 01101 10 11010 11 10111 ) 11 ( ˆ ), 10111 ( ˆ or ) 00 ( ˆ ), 00000 ( ˆ ) 00011 ( ) 11 ( ˆ ), 10111 ( ˆ or ) 00 ( ˆ ), 00000 ( ˆ ) 10001 ( ) 11 ( ˆ ), 10111 ( ˆ ) 11111 ( ) 11 ( ˆ ), 10111 ( ˆ ) 10101 (                     m c m c r m c m c r m c r m c r

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21.                       1 0 1 1 1 1 1 1 0 0 1 1 ) 1000110 ( ) 1000 ( : left shift (0100011) (0100) : ) 3 ( ) 1 ( (3) ) 0101110 ( ) 0101 ( : left shift (0010111) (0010) : ) 2 ( ) 1 ( (2) ) 0011010 ( ) 0011 ( : left shift (1) ) 0001101 ( ) 0001 ( P 22. (a)                1 0 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 0 1 1 (0011010) (0110100), (1101000), (1010001), (0100011), (1000110), (0001101), (3) (1111111) (0000000), (2) (1100101), (1001011), (0010111), (0101110), (1011100), (0111001), (1110010), (1) are codewords : is that shifts cyclic its and (1111111), (1110010) (3) (1111111) and (0000000) (2) shifts cyclic its and (1110010) (1) : include codewords G (b) dmin = wmin = 3 t = 1 (c) ( 0001101, 1110010, 0001101 ) 23. 編碼器方塊圖 output: (11 00 10 10 00 11 01 11 00) state: (000)(100)(110)(011) (001)(100)(010)(001)(000)

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