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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];
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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 +=
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(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
Non-Causal, the output of the system depends on future input;
(e) ]1[3][)3/1(][ += nununh nn
Non-Causal, the output of the system depends on future input;
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:
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(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
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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:
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]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
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(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 =+
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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
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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
nonnegative for all w;
(d) As )2/()2/cos(2)(]1[][][ wjewjwHnnnh =+= ,
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)2/cos(2)( wjwA = and 0,2/1 == , it is a GLP, but not LP as )( jwA is not always
nonnegative for all w;
(e) As )2/(sin2)(]2[][][ == wjwejwHnnnh ,
wjwA sin2)( = and 2/,1 == , it is a GLP, but not LP as )( jwA is not always
nonnegative for all w;
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
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(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
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Cutoff: 3.0=cThe peak approximate error dB02.26log20 10