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Instructor’s Guide to Accompany Digital Signal Processing: Fundamentals and Applications

Li Tan Jean Jiang

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

Chapter 2 2.1

 e j 2 1500t  e j 2 1000t  j 2 1500t     2.5 e j 2 1500t 5cos(2 1000 t) 5    2.5 e 2   c1  2.5 a.

X(f ) 2.5

f kHz 1.5

1.5

b.

Xs ( f ) 2.5/T

f 1.5 1.5

9.5 8 6.5

6.5 8 9.5

kHz

14.5 16 17.5

2.2 2  3200 t

x (t )  e  j

 2.5e  j2 2500 t  2.5e

j2  2500 t

e

j2  3200 t

a.

Xs( f ) 2.5 / T

f kHz 11.2 10.5

5.5 4.8 3.2 2.5

2.5

3.2 4.8

5.5

8

10.5

11.212.8 13.5 16 18.5 19.2

b.

Y( f )

f 3.2 2.5

2.5 3.2

kHz

c1  2.5 and

2

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

2.3 a.

x (t )  e  j 2  2200t  1.5e  j 2 1500t  1.5e j 2  1500t  e

j 2  2200 t

Xs ( f ) 1.5 / T

10.29.5 8 7.5 5.8

2.21.5

1.5 2.2

5.8 7.5 8

9.5 10.2

13.8 14.5 16 17.518.2

f kHz

b. Y( f )

f kHz

2.21.5

2.4 a.

1.5 2.2

x (t )  e  j 2  4200t  1.5e  j 2 1500t  1.5e j 2  1500t  e Xs ( f )

j 2  4200 t

Aliasing noise

1.5 / T

11.8 9.5 8 7.5 4.2 3.8 1.5

1.5

3.8 4.2

7.5 8

9.5 11.8 12.2 14.5 16 17.5 19.820.2

b. Y( f )

3.8 1.5

Aliasing noise

f kHz 1.5

3.8

f kHz

3

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

2.5 x (t )  e  j 2  4500t  2.5e  j 2  2500t  2.5e j 2  2500 t  e

j2  4500 t

a.

X s(f )

Aliasing noise

2.5 / T

f 11.5 10.5

5.5 4.5 3.5 2.5

2.5

3.5 4.5

5.5 8

kHz

10.5 11.5 12.5 13.5 16 18.519.5

b.

Y( f ) Aliasing noise

f 3.5 2.5

2.5

kHz

3.5

c. The aliasing frequency = 3.5 kHz 2.6 x (t ) 

2.5 2.5  j 2  4500t  5e  j 2  5500t  5e j 2  5500t  e e j j

j 2  7500t

a.

X s(f ) 5/T

f kHz 10.5 8.5 7.5 5.5 2.5 0.5 0.5

2.5

5.5

7.5 8 8.5 10.5

13.5 15.5 16 16.518.5

b.

Y( f )

2.5 0.5

0.5

2.5

c. The aliasing frequencies: 0.5 kHz and 2.5 kHz.

f kHz

4

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

2.7

x (t ) 

2.5 2.5  j 2  7000t  4e  j 2  5000t  4e j 2  5000t  e e j j

j 2  7000t

a. Aliasing noise

Xs ( f )

4 /T

11

9 8 7

5

3

1

1

3

5

7

8

9

11

13

15 16 17

19

f kHz

b. Aliasing noise

Y( f )

3

1

f kHz 1

3

c. The aliasing frequencies: 1 kHz and 3 kHz. 2.8

x (t ) 

2.5  j 2  7500t 2.5 e e  5e  j 2  5000t  5e j 2  5000t  j j

j 2  7500t

a. Aliasing noise

Xs ( f )

5/T

11

b.

8.5 8 7.5 5

3

0.5 0.5

3

5

7.5 8 8.5

11

13

15.51616.5

19

f kHz

5

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

Aliasing noise

Y( f )

3

f kHz

0.5 0.5

3

c. The aliasing frequencies: 0.5 kHz and 3 kHz. 2.9 Choose C2  0.1 F 1.4142 1.4142 R1  R2    2251  C 2  2  f c  0.1 10 6  2  1000 1 1 C1    0.05  F 2 2 6 R1R 2 2 f c  2251 2251 0.1 10  2  1000  0.1  F Vin 2.25 k

2.25 k

Vo

  0.05 F

2.10

% aliasing level 

 f  1  a   fc 

2n

2n

 f  fa  1  s   fc 



 500  1   1000 

4

4

 4000  500  1    1000 

 8.39%

6

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

2.11 Choose C2  0.1 F 1.4142 1.4142   2813  R1  R2  C 2  2  f c  0.1 10 6  2   800 1 1   0.05 F C1  2 2 R1R 2 2 f c  2813  2813  0.110 6  2   800  0.1  F Vin 2.813 k 2.813 k

Vo

  0.05 F

2.12  f  1  a   fc 

% aliasing level 

2n

 400  1   800 



2n

4

 4000  400  1   800  

 f  fa  1  s   fc 

4

 6.43%

2.13

a. % aliasing level 

 f  1  a   fc 

2n

2n



 f  fa  1  s   fc 

b. % aliasing level 

 f  1  a   fc 

2n

4

4

 57.44%

 8000  3200  1    3200 

2n

 f  fa  1  s   fc 

 3200  1    3200 



1000  1     3200 

4

4

 8000  1000  1    3200 

 20.55%

7

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

2.14 When n  7

% aliasing level 

 f  1  a   fc 

2n

 3200  1    3200 

27

  8.26% 2n 2 7  fs  fa   8000  3200  1  1    3200    fc  The order of the anti-aliasing filter should be seven (7).

2.15

a. % aliasing level 

 f  1  a   fc 

2n

2n



b. % aliasing level 

4

2n

2n

4

 8000  3100  1    3100 

 f  fa  1  s   fc 

 f  1  a   fc 

 3100  1    3100 

 900  1    3100 



 f  fa  1  s   fc 

 52.55%

4

4

 18.79%

 8000  900  1    3100 

2.16 When n  6

% aliasing level 

 f  1  a   fc 

2n

 3100  1    3100 

26

  9.05% 2n 2 6  fs  fa   8000  3100  1  1    3100    fc  The order of the anti-aliasing filter should be six (6).

2.17 a. fT  3200 /8000  0.4  sin( fT  )  % distortion   1   100%   fT   b. fT  1500 /8000  0.1875  sin( fT  )   100%  % distortion   1  fT  

  sin(0.4 )  100%  24.32% 1 0.4      sin(0.1875 )  100%  5.68% 1 0.1875   

8

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition

9

2.18

sin(0.2 )  0.9355 ( 0.58 dB) 0.2 sin(0.8 )  0.2339 ( 12.62 dB) f  16000 Hz, fT  16000 1/ 20000  0.8 , and gain  0.8 Maximum allowable gain variation from 0 to 4000 Hz = 2-0.58=1.42dB 40 -12.62 = 27.38 dB rejection at frequency 16000 Hz. f  4000 Hz, fT  4000 1/ 20000  0.2 , and gain 

  20log 1  16000 / f  

20log 1   4000 / f c 

2n 1/ 2

 1.42

2 n 1/ 2

c

 27.38

Then 1 n  log (10 2.738  1) /(10 0.142  1)  / log 16000 / 4000   2.6158  3 2 4000 Hz fc  1/ 6  4686 0.142  10 1  

2.19 a. fT  3000 /8000  0.375

b.

 sin( fT )  % distortion   1   100%   fT   fT  1600 /8000  0.2

 sin(0.375 )  1  100%  21.58% 0.375   

 sin( fT )  % distortion   1   100%   fT  

 sin(0.2 )  1  100%  6.45% 0.2   

2.20 f  4000 Hz, fT  4000 1/ 22000  0.18182 , and gain 

sin(0.18182 )  0.9465 ( 0.48 dB) 0.18182

f  16000 Hz, fT  18000 1/ 22000  0.818182 , and sin(0.81812  ) gain   0.2104 (13.54 dB) 0.81812 Maximum allowable gain variation from 0 to 4000 Hz = 2-0.48=1.52dB 40 -13.54 = 26.46 dB rejection at frequency 18000 Hz.

  20log 1  16000 / f  

20log 1   4000 / f c 

1/ 2 2n

2n

c

 1.52

1/ 2

 26.46

Then 1 n  log (10 2.646 1) /(10 0.152  1)  / log 18000 / 4000   2.3138  3 2 4000 fc   4624 Hz 1/ 6 0.152 10  1

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 10

2.21 b1b0=01 2.22 1  1  1 1 V 0 VR  b1  b0   5    0   1  1.25 Volts 4  4  2 2 2.23 b1b0=10

2.24 1  1  1 1 For b1b0=11, V 0 V R  b1  b0   5    1  1  3.75 Volts 4  4  2 2 1  1 1 1  For b1b0=10, V 0 V R  b1  b0   5    1  0   2.5 Volts 4  4 2 2 

2.25 a. L  2 4  16 levels x x 5 b.   max min   0.3125 16 L c. x q  0 10 0.3125 3.125 3.2  x  xmin   10.24  , i  round  0.3125   code =1010 e. eq  0.075

d.

  10.24   0    round    round 10.24  10 binary    

2.26 5 a. L  2  32 levels 4 x x  0.125 b.   max min  L 32 c. x q  0 10 0.125 1.25 1.2  x  xmin   9.6  , i  round  0.125   =1010 e. eq  0.05

d.

  9.6   0    round    round  9.6  10 binary code    

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 11

2.27 a. L  23  8 levels x  xmin 5   0.625 b.   max L 8 c. x q  2.5  2 0.625  1.25  1.2    1.92 , d. x  0.625  x  xmin   1.92  (4)  i  round    round    round  2.08  2       binary code =010 e. eq  0.05 2.28 a. L  28  256 levels 5 x x b.   max min   0.01953125 L 256 c. x q  2.5  205 0.01953125  1.5039 1.5   76.8 , 0.01953125  x  xmin   76.8  (128 )  i  round    round    round  204.8  205       binary code =11001101 e. eq  0.0039

d. x 

2.29 xmax  xmin 20   0.3125 L 64 c. SNRdB  1.76 6.02 6  37.88 dB

a. L  26  64 levels b.  

2.30 5 xmax  xmin   0.078125 L 64 SNRdB  4.77  20  log xrms / x max  6.02  6 c. dB  4.77  20  log 0.25   36.12  28.85

a. L  26  64 levels b.  





Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 12

2.31 a. See Program 2.1 b. 5 4 3

Quantized x(n)

2 1 0 -1 -2 -3 -4 -5

0

0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 Time (sec.)

c. SNR = 37 dB 2.32 a. Use Program 2.1 b. 4 3 2

Quantized x(n)

1 0 -1 -2 -3 -4 -5

0

c. SNR = 35 dB

0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 Time (sec.)

0.02

0.02

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 13

2.33 a. See Program 2.2 b.

Quantized error

Quantized speech

Original speech

we.dat: "we" 5

0

-5

0

0.05

0.1

0.15

0.2

0.25

0

0.05

0.1

0.15

0.2

0.25

0

0.05

0.1

0.15 Time (sec.)

0.2

0.25

5

0

-5 1

0

-1

c. SNR = 26.9 dB

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 14

Chapter 3 3.1 0

6

-0.5

(a)

(b)

4

2

-1 -1.5

0 -5

0 n

-2 -5

5

0

5

10

5

10

n 6

-2

4

(c)

(d)

0

-4

2

-6 -5

0

5

10

0 -5

0

n

n

3.2 a. n x( n)

0 1.000

b. n x( n)

0 0.0000

1 2.9389

2 4.7553

3 4.7553

4 2.9389

5 0.0000

6 -2.9389

7 -4.7553

c. n x( n)

0 4.3301

1 3.3457

2 2.0377

3 0.5226

4 -1.0396

5 -2.5000

6 -3.7157

7 -4.5677

d. n x( n)

0 0.0000

1 2 0.5000 0.2500

3 4 0.1250 0.0625

1 2 3 4 1.1588 1.6531 1.7065 1.5064

5 6 7 0.0313 0.0156 0.0078

5 6 7 1.1865 0.8463 0.5400

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 15

5

(b)

(a)

1

0.5

0

0

2

4 n

6

0

-5

8

5

0

2

4 n

6

8

0

2

4 n

6

8

2

(d)

(c)

1.5 0

1 0.5

-5

0

2

4 n

6

8

0

8

0

6

-1

(b)

(a)

3.3

4 2 0 -5

-2 -3

0

5

-4 -5

10

0

n

5

10

5

10

n 0

4

-2

(c)

(d)

6

2

0 -5

-4

0

5 n

10

-6 -5

0 n

3.4 xa =[ 1.0000 0.2500 0.0625 0.0156 0.0039 0.0010 0.0002 0.0001] xb =[ 0 2.8532 1.7634 -1.7634 -2.8532 -0.0000 2.8532 1.7634] xc =[ 5.1962 2.4404 -1.2475 -4.4589 -5.9671 -5.1962 -2.4404 1.2475] xd =[ 0 0.6180 0.5878 0.4045 0.2378 0.1250 0.0594 0.0253]

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 16

1

4

(b)

(a)

2 0.5

0 -2

0

2

4 n

6

-4

8

0

10

0.8

5

0.6 (d)

(c)

0

0 -5 -10

0

2

4 n

6

8

2

4 n

6

8

0.4 0.2

2

4 n

6

8

0

0

3.5 10

(a)

5

0

-5 -8

-6

-4

-2

0 n

2

4

6

8

-6

-4

-2

0 n

2

4

6

8

6

(b)

4 2 0 -2 -8

3.6 a. x( n)  3 ( n)   ( n 1)  2 ( n 2)   ( n 3)   (n 5) b. x( n)   ( n 1)   ( n  2)   ( n 4)   ( n 5)

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 17

3.7 2

(a)

0 -2 -4 -6 -5

0

5

10

5

10

n 6

(b)

4 2 0 -2 -5

0 n

3.8 a. x( n)   ( n 1)  0.5 ( n 1)  3 ( n)  2.5 ( n 2)   ( n 3)  0.5 ( n 4)  ( n 5) b. x( n)  0.5 ( n 1)  0.5( n)  ( n  1)  0.5 ( n  2)   ( n 4)   ( n 5) 3.9 a. x (n)  e 0.5nu (n )  (0.6065)n u (n ) b. x( n)  5sin(0.2 n)u( n) c. x(n)  10cos(0.4 n   / 6)u( n)  d. x (n)  10e n sin(0.15n )u (n )  10(0.3679) n sin(0.15n )u (n ) 3.10 a. Let y1 ( n)  5 x1( n)  2 x12 (n), y2 ( n)  5 x2 ( n)  2 x22 ( n) y1( n)  y2 ( n)  5 x1 ( n)  2 x21 ( n)  5 x2 ( n)  2 x22 (n)

For x( n)  x1 ( n)  x2 (n) 2 y( n)  5 x( n)  2 x ( n)  5  x1 ( n)  x2 ( n)  2  x1 ( n)  x2 ( n)

 5x1 (n )  5x2 (n)  2x12 (n ) 2x22 (n ) 4x1 (n )x2 (n ) Since y1 ( n)  y2 ( n)  y( n) , the system is a nonlinear system. b. Let y1 ( n)  x1 ( n 1)  4 x1 (n) , y2 ( n)  x2 ( n 1)  4 x2 (n) y1( n)  y2 ( n)  x1 ( n 1)  x2 ( n  1)  4 x1 ( n)  4 x2 (n) For x( n)  x1 ( n)  x2 (n)

2

Instructor’s Guide to Accompany to Digital Signal Processing, Fundamentals and Applications, Second Edition 18

y( n)  x( n  1)  4 x( n)   x1( n  1)  x2 ( n  1)  4  x1( n)  x2 ( n)   x1 (n  1)  x2 (n  1)  4x1 (n ) 4x2 (n) Since y1 ( n)  y2 ( n)  y(n) , the system is a linear system. c. Let y1 ( n)  4 x13 ( n)  2 x1 (n) , y2 ( n)  4 x23( n)  2 x2 (n) y1( n)  y2 ( n)  4 x13 ( n)  2 x1 ( n)  4 x32 ( n)  2 x2 (n)

For x( n)  x1 ( n)  x2 (n)

y( n)  4 x3( n)  2 x( n)  4  x1( n)  x 2( n)   2  x1( n)  x 2 (n)  3

 4x13 (n )  8x12 (n )x2 (n )  8x1 (n )x22 (n ) 4x23 (n ) 2x1 (n ) 2x2 (n ) Since y1 ( n)  y2 ( n)  y( n) , the system is a nonlinear system. 3.11 0.5 a. x (n)  e  nu (n )  (0.6065)n u (n ) b. x( n)  4sin(0.3 n)u(n) c. x(n)  7.5cos(0.1 n   / 3)u( n) d. x (n )  20e n sin(0.3 n )u (n )  20(0.3679)n sin(0.3 n )u (n ) 3.12 3 3 a. Let y1 ( n)  4 x1 ( n)  8 x1 (n), y2 ( n)  4 x2 ( n)  8 x2 ( n) y1( n)  y2 ( n)  4 x1 ( n)  8 x31 ( n)  4 x2 ( n)  8x32 (n)

For x( n)  x1 ( n)  x2 (n) y( n)  4 x( n)  8 x 3( n)  4  x1 ( n)  x2 ( n)  8 x1 ( n)  x2 (n)

3

Since y1 ( n)  y2 ( n)  y( n) , the system is a nonlinear system. b. Let y1 ( n)  x1 ( n  3)  3x1 (n) , y2 ( n)  x2 ...