Digital signal processing solution mannual chapter Bonus Oppenheim

7/29/2019 Digital signal processing solution mannual chapter Bonus Oppenheim

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EE141 Digital Signal Processing (Fall 2003)Solution Manual

Prepared by Xiaojun Tang and Zhenzhen Ye

Text: A.V. Oppenheim, R.W. Schafer, and J. R. Buck,Discrete-Time SignalProcessing, 2nd edition, Prentice-Hall, 1999.

HW#1: P2.1 (a), (c), (e), (g); P2.4; P2.24

HW#2: P2.5; P2.18; P2.29 (a), (c), (e)HW#3: P2.40; P2.41; P3.27 (a), (c)

HW#4: P3.6 (d), (e); P3.20; P4.1; P4.3

HW#5: P4.5; P4.7; P5.2; P5.3HW#6: P5.10; P5.12; P5.15

HW#7: P6.7; P6.8; P6.11; P6.25; P7.15

HW#1: P2.1 (a), (c), (e), (g); P2.4; P2.24

P2.1

(a) T(x[n]) = g[n]x[n];

Stable if g[n] is bounded; Causal output is not decided by future input; Linear T(ax[n] + by[n]) = ag[n]x[n] + bg[n]y[n] = aT(x[n]) + bT(y[n]); Time variant T(x[n-m]) = g[n]x[n-m]; Memoryless output only depends on x[n] with same n;

(c)

+

==

0

0][])[(

nn

nnk kxnxT ;

Stable; Causal only if n0 = 0, else non-causal; Linear T(ax[n] + by[n]) = aT(x[n]) + bT(y[n]); Time Invariant +

=

+

=

==0

0

0

0 '

]'[][])[(nn

nnk

nmn

nmnk

mkxkxmnxT ;

Memoryless only if n0 = 0;(e) ][])[( nxenxT = ;

Stable; Causal; Nonlinear T(ax[n] + by[n]) aT(x[n]) + bT(y[n]); Time Invariant ][])[( mnxemnxT = ; Memoryless output only depends on x[n] with same n;

(g) T(x[n]) = x[-n];


2/14

Stable; Non-Causal output depends on future input; Linear T(ax[n] + by[n]) = aT(x[n]) + bT(y[n]); Time Variant; Not Memoryless;

P2.4

As ]1[2]2[8

1]1[

4

3][ =+ nxnynyny , the Fourier transform is

jwjwjwjwjwjwjw eeXeeYeeYeY =+ )(2)(8

1)(

4

3)(

2

When 1)(][][ == jweXnnx , thus

=

+=

jwjwjwjw

jwjw

eeee

eeY

)4/1(1

1

)2/1(1

18

)8/1()4/3(1

2)(

2

][41

218][ nuny

nn

=

P2.24

As h[n] = [1 1 1 1 2 2] for n from 0 to 5 and x[n] = u[n-4], the system response is:

y[n] = x[n] * h[n] = [0 0 0 0 1 2 3 4 2 0 ..]; The sketch is shown as follows:

HW#2: P2.5; P2.18; P2.29 (a), (c), (e)

P2.5

(a) The roots for polynomial 0651 21 =+ zz are 2 and 3, so the homogeneousresponse for the system is:

nn AAny 32][ 21 +=


3/14

(b) As ]1[2]2[6]1[5][ =+ nxnynyny and ][][ nnx = , the impulse response of

the system is:

][)23(2][

21

1

31

12

651

2)(

2

nunh

eeee

eeH

nn

jwjwjwjw

jwjw

=

=

+=

(c) As ]1[2]2[6]1[5][ =+ nxnynyny and ][][ nunx = , the step response of the

system is:

( )( )][)123(][

3,31

3

21

4

1

1

1651

2)()()(

21

111121

1

nuny

zzzzzzz

zzXzHzY

nn +=

>

+

=+

==

++

P2.18

(a) ][)2/1(][ nunh n=

Causal, the output of the system does not depend on future input;

(b) ]1[)2/1(][ = nunh n

Causal, the output of the system does not depend on future input;

(c)n

nh )2/1(][ =

Non-Causal, the output of the system depends on future input;

(d) ]2[]2[][ += nununh


(e) ]1[3][)3/1(][ += nununh nn


P2.29

As x[n] = [1 1 1 1 1 1/2] for n from 1 to 4,

(a) x[n-2] = [1 1 1 1 1 1/2] for n from 1 to 6; The sketch is:


4/14

(c) x[2n] = [1 1 1/2] for n from 0 to 2; The sketch is:

(e) x[n-1][n-3] = x[2]; The sketch is:

HW#3: P2.40, P2.41, P3.27 (a), (c)

P2.40


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][)1(][)cos(][ nununnx n==

][)2

(][ nuj

nh n=

)2/(1

)2/(1)1(

)2

()1(][)1]([)2

(][][][*][][

10

j

j

jknuku

jknxkhnxnhny

nn

k

n

k

knknk

k

=

====

+

= =

=

Since2/1

1

)2/(1

)2/(1lim

1

jj

j n

n +=

+

The steady state response to the excitation ][)1(][ nunx n= is

2/1

)cos(

2/1

1)1(

j

n

j

n

+=

+

P2.41

Given a periodic impulse train

=

+=k

kNnnx ][][ , we can write its Fourier transform as

=

+=k

j NkN

eX )/2(2

)(

(1)

(Refer to Signal and Systems, 2nd edition by A.V. Oppenheim and A.S. Willsky, Page 371 for its proof)

In problem, 2.41,N=16, so its Fourier transform is

=

+=k

j keX )16/2(16

2)(

(2)

Let )(

j

eY denotes the output of the system, then)()()(

jjj eHeXeY = (3)

If 8/3|| < , 3)( jj eeH =

)]8/()8/()([16

2)16/2(

16

2 8/38/33

+++=+=

=

jjk

j eeke (4)

If 8/3|| , 0)( =jeH , thus 0)( =jeY , (5)

So )]8/()8/()([16

2)( 8/38/3

+++= jjj eeeY (6)

Take the inverse Fourier transform, we can get

))3(8

cos(8

1

16

1))3(

8cos21(

16

1

)1(16

1)1(16

1][

8/)3(8/)3(8/8/38/8/3

+=+=

++=++=

nn

eeeeeenynnnjnj

(7)

Note: Take a look at (3), )(jeH is band limited, )(

jeX is infinite pulse train. If we

multiply them together, we can only consider those pulses falling into the band

( 8/3 , 8/3 ). The rest pulses are cancelled due to the multiplication with 0. There are


6/14

three pulses falling into the band )(16

2

, )8/(

16

2

+ , )8/(

16

2

, so we get

(6)

P3.27

(a)

111211121 3121

2

11)

2

11()31)(21()

2

11(

1)(

+

+

+

+

+

=

+

=z

D

z

C

z

B

z

A

zzz

zX

X (z)s poles are z=-1/2, 2, 3, if it is stable, the ROC is )2,2/1(|| z

35

1

)31)(21(

1)

2

11)(( 2/1112/1

21 =

=+= ==

zzzz

zzXA

1225

1568

)31()2

11(

1)21)(( 2

1212

1 =

+

== =

=

zz

zz

zzXC

1225

2700

)21()2

11(

1)31)(( 3121

3

1 =

+

== =

=

zz

zz

zzXD

Also, Let 01 =z at both sides,

DCBAzXz

+++===

1)(01

Thus,1225

581 == DCAB

111211121 31

1225/2700

21

1225/1568

2

11

1225/58

)

2

11(

35/1

)31)(21()

2

11(

1)(

+

+

+

+

=

+

=zz

zzzzz

zX

Since the ROC is )2,2/1(|| z ,

]1[31225

2700]1[2

1225

1568][)

2

1(

1225

58]1[)

2

1)(1(

35

1][ ++++= nununununnx nnnn

Note: To get the inverse Z-Transform of second-order term or multiple order term, we

can use the differentiation propertydz

zdXznnx

)(][ (Refer to page122 of textbook for

its proof)

E.g. right side sequence1

1

1][][

=

az

nuanx n (ROC: |||| az > )

21

11

)1(]

)1

1(

[][

=

az

azdz

azd

znna n , So21

11

)1(][

az

znna n

(c)

1

23

21

22

2

2)(][

++=

=

zzz

z

zzzXnx


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X(z) has its only pole at z=2. If x[n] is a left-sided sequence, the ROC is 2||

=

zz

z

zX

Partial Fraction Expansion:

][2

1][

)2/1(1

1

))2/1(1)()2/1(1(

)2/1(1

)4/1(1

)2/1(1)(

111

1

2

1

nunx

zzz

z

z

zzX

n

=

+=

+

=

=

Power Series Expansion:

][2

1][

)16/1()8/1()4/1()2/1(1)4/1(1

)2/1(1

)(

4321

2

1

nunx

zzzzz

z

zXn

=

+++=

=

L

Fourier Transform exists as the ROC including unit circle.

(e) azaz

azzX /1

1)(

1

1

>

=

Partial Fraction Expansion:

]1[1

][1

][1

][][1

][

)/1(1)/1(1

/1111)(

1111

111

2

11

1

11

1

+

=

+

=

+

=

=

=

=

++

nua

nua

nua

nanua

nx

za

aa

za

a

az

aa

azaz

az

azaz

azzX

nnnn

Power Series Expansion:


8/14

]1[1

][1

][2][1][

]1[1][2111

][1

][11

][1111

111

11)(

11

14

3

3

2

21

1

1

1

3

3

2

2

1

1

1

1

11

1

+

=+=

=++++=

=

=

++++

=

=

=

+

+

nua

nua

nxnxnx

nua

nxza

za

za

zaz

az

nua

nuaa

nxza

za

zaaaz

az

az

azaz

azzX

nn

n

nn

L

L

Fourier Transform exists when the ROC including unit circle, which means1 3/4, and the ROC ofY(z) is |z| > 2/3, the ROC ofH(z)should be |z| > 2/3 ;

(b)As the ROC ofX(z) is |z| < 1/3, and the ROC ofY(z) is 1/6 < |z| < 1/3, the ROC ofH(z) should be |z| > 1/6 ;

P4.1

As [ ])100(2sin)( ttxc = and T = 1/400 sec,

[ ]

===

2sin)100(2sin)(][ nnTnTxnx c

P4.3

As [ ]ttxc 4000cos)( = and

=

3cos][

nnx ,

(a)Let000,12

1)(][ == TnTxnx c

(b)T is not unique, for example,000,12

5=T

HW#5: P4.5; P4.7; P5.2; P5.3

P4.5

(a) From Nyquist Sampling theorem, to avoid aliasing in the C/D converter, the sampling

frequency HzHzT

m

s

s

4105000*221

=== , so sTs410


9/14

(b) Hzf

s

cutoff

cutoff 625102

8/

2

4 ===

(c) kHzHzf

s

cutoff

cutoff 25.112501022

8/

2

4 ====

Note: The relation between digital frequency f and analog frequency is 2

f

s=

,

where s is the sampling frequency, f is in radians.

P4.7

(a)

)()()( dccc tststx +=

)1)(()( dj

cc ejSjX

+=

Consider sampling, )(][ nTxnx c= , in frequency domain (refer to Eq4.19 in textbook, P147),

( ) ))2((1

=

=k

c

Tj

T

kjX

TeX

( ) ( )

))2

(()1(1

))2

(()1(1

))2

((1

/

=

=

=

+=

+===

k

c

Tj

k

c

j

k

c

Tjj

T

k

TjSe

T

T

kjSe

TT

kjX

TeXeX

d

d

(b)Tjj deeH

/1)(

+=

(c)

)/(

])/sin[()sin(

2

1

2

1)1(2

1)(2

1][)/(/

Tn

Tn

n

n

dededeedeeHnh

d

d

TnjjnjnTjjnj dd

+=

+=+==

if Td = , ]1[][)1(

])1sin[()sin(][ +=

+= nn

n

n

n

nnh

if 2/Td = ,

)1(

])2/1sin[(][

)1(

])2/1sin[()sin(][

+=

+=

n

nn

n

n

n

nnh

P5.2

][]1[][3

10]1[ nxnynyny =++

)()()(3

10)(

1zXzzYzYzYz =+


10/14

1111

1

1 31

8/3

)3/1(1

8/3

)31)()3/1(1(

3

10

1

)(

)()(

+

=

=

+

==zzzz

z

zzzX

zYzH

(a) H (z) has two zeroes: 0, ; two poles: 1/3, 3

(b) The system is stable, so the ROC includes the unit circle. The ROC is 1/3


11/14

P5.12

(a)

As the poles of H(z) are 0.9, -0.9, ROC includes the unit circle, the system is stable.

(b)

)()(

))3/1(1)()3/1(1(

)3/1)(3/1(

)9.01)(9.01(

))3/1(1)()3/1(1)(2.01(9

)9.01)(9.01(

)31)(31)(2.01()(

1

)(

11

11

)(

11

111

11

111

1

zHzH

zz

zz

zjzj

zzz

zjzj

zzzzH

ap

zHzH ap

=

+

+

+

++=

+

++=

4444 34444 214444444 34444444 21

P5.15

Generalized Linear Phase GLP;

Linear Phase LP;

(a) As jwewjwHnnnnh+=++= )cos41()(]2[2]1[][2][ ,

wjwA cos41)( += and 0,1 == , it is a GLP, but not LP as )( jwA is not always

nonnegative for all w;

(b) It is not a GLP or LP as it is not a symmetric filter;

(c) As jwewjwHnnnnh+=++= )cos23()(]2[]1[3][][ ,

wjwA cos23)( += and 0,1 == , it is a GLP and also LP as )( jwA is always


(d) As )2/()2/cos(2)(]1[][][ wjewjwHnnnh =+= ,


12/14

)2/cos(2)( wjwA = and 0,2/1 == , it is a GLP, but not LP as )( jwA is not always


(e) As )2/(sin2)(]2[][][ == wjwejwHnnnh ,

wjwA sin2)( = and 2/,1 == , it is a GLP, but not LP as )( jwA is not always


HW#7: P6.7; P6.8; P6.11; P6.25; P7.15

P6.7

The difference equation is: ][4

1]2[]2[

4

1][ nxnxnyny =

Z-Transform: )(4

1)()(

4

1)(

22zXzzXzzYzY =

Transfer function:2

2

411

4

1

)(

)()(

+==

z

z

zX

zYzH

P6.8]2[]1[3]2[2][ += nxnxnyny

P6.11

1

321

1

111

2

11

86

2

11

)41)(21()(

+=

=

z

zzz

z

zzzzH

(a)

-1/4

1z

1z

][nx ][ny

1/4

1z

1z

][ny

1/2

-6

1z8


13/14

(b)

P6.25

(a) )1

1)](21(

8

7

8

31

2

11

[)()]()([)(1

21

21

1

321

+++

+

=+=

zzz

zz

z

zHzHzHzH

The Z-Transfer function:

321

4321

8

7

4

5

8

111

87

811

89

892

)(

+

++++=

zzz

zzzz

zH

(b)

The difference equation:

]4[8

7]3[

8

11]2[

8

9]1[

8

9][2]3[

8

7]2[

4

5]1[

8

11][ ++++=+ nxnxnxnxnxnynynyny

(c)

P7.15Specifications:

(a) Pass band ripple: 05.0=p , dBA pp 02.26log20 ==

Stop band ripple: 1.0=s , dBA ss 20log20 ==

Pass band edge: 25.0=p

Stop band edge: 35.0=s

][nx

1z

1

z

][ny

-6

1/2

1z8

2

1z

1z

][nx ][ny

-5/4

11/8 9/8

9/8

1z

1z

7/8 11/8

7/8


14/14

Cutoff: 3.0=cThe peak approximate error dB02.26log20 10

Digital signal processing solution mannual chapter Bonus Oppenheim

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