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13.3
z
_1
2 2
2 2
z
_1
X ( z)
Y ( z)
V ( z)
W ( z)
V ( z)u
W ( z)u
Analysis yields ),()()( 2 / 1
2
12 / 1
2
1 z X z X zV
),()()(),()()(2
1
2
12 / 1
2
2 / 1
2
2 / 12 / 1 z X z X zV z X z X zW u
z z
).()()(22
11 z X z X zW
z zu Hence, ),()()()( 11 z X z zW zV z zY uu
= or in other
words ].1[][ n x n y
13.4 .1
111
][
1
0⎟⎠
⎞
⎜⎝
⎛
=∑=n M
nM M
M
k
kn M W
W
M W M nc Hence, if .01
111
][, =⎟⎠
⎞
⎜⎝
⎛
= n M W M ncrM n On
the other hand, if ,rM n = then .11111
][
1
0
1
0
1
0
=∑
=
=
= M
M
M W
M W
M nc
M
k
M
k
krM M
M
k
kn M
13.5 (a) For and6 M ,5 L
},,,,,{}{}{ 56
46
36
26
16
066
= W W W W W W W W k k M , and
},,,,,{}{}{ 256
206
156
106
56
06
56
= W W W W W W W W k Lk M
}.{},,,,,{6
1
6
2
6
3
6
4
6
5
6
0
6
k W W W W W W W =
(b) For to have same set of values for}{ k M W ,10 M k as , each should
have unique values for each k . Therefore
}{ kL M W
nL M
kL M W W ≠ for all or
for any positive integer r , which implies that L and M should be
relatively prime.
],1,0[, M nk
rM Lnk ≠)(
13.6
M M H ( z) H ( z ) M X ( z) X ( z)V ( z)1
Y ( z)1
V ( z)2Y ( z)2
For the figure on the left-hand side we have ),(1
)(1
0
/ 11 ∑
=
= M
k
k M
M W z X M
zV and
).()(1
)(
1
0
/ 11 ∑
=
= M
k
k M
M W z X z H M
zY For the figure on the right-hand side we have
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)()(1
)(),()()( / 11
0
22k M
M M
k
kM M
M W z X zW H M
zY z X z H zV ∑=
= ).()(1 / 1
1
0
k M
M M
k
W z X z H M ∑
=
=
Hence, ).()( 21 zY zY =
L L H ( z) H ( z ) L X ( z) X ( z)V ( z)1 Y ( z)1V ( z)2 Y ( z)2
For the figure on the left-hand side we have For the).()()(),()( 11 L L L z X z H zY z X zV =
figure on the right-hand side we have Hence,).()()(),()()( 22 L L z X z H zY z X z H zV =
).()( 21 zY zY =
13.7 (a) The system of Figure P13.1 with internal variables labeled is shown below:
L LG( z) X ( z)V ( z)
Y ( z)U ( z)
Analysis yields ),()()(),()( zV zG zU z X zV L = and ).(1
)(
1
0
/ 1∑=
= L
k
k L
LW zU L
zY
Substituting the first equation in the second equation we get
Substituting this equation in the expression for in the above we get
).()()( L z X zG zU =
)( zY
),()(1
)()(1
)(1
0
/ 11
0
/ 1 z X W zG L
W z X W zG L
zY
L
k
k L
LkL L
L
k
k L
L ∑
=
=
⋅ since .1kL LW
Therefore, ).(1
)(
)()(
1
0
/ 1
∑= =
L
k
k
L
LW zG L z X
zY z H Hence, Figure P13.1 is a LTI system.
(b) It follows from the last equation given above, if ,1)(1
1
0
/ 1 ==
L
k
k L
LW zG L
then
i.e., or,1)( = z H ),()( z X zY = ].[][ n x n y = Or, in other words, the system of Figure
P13.1 is an identity system for .1)(1
1
0
/ 1 ==
L
k
k L
LW zG L
13.8 Consider the multirate structure shown below. Analysis yields
L L G( z)u[n]
x [n] H ( z) y[n]
and),()()( L z X z H z U = ).()()( // k L L
L
k
k L L W z U W z G
L z Y 1
1
0
11∑−
=
= Substituting the first
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equation into the second we get ).()()()( // z X W z H W z G L
z Y k L L
L
k
k L L
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡= ∑
−
=
11
0
11 Hence,
if ,)()( //1
1 11
0
1 =
∑
−
=
k L L
L
k
k L L W z H W z G
L
we have ),()( z X z Y = or in other words, the
above multirate structure is an identity system. We break the system as shown below
into two parts with the transfer functions and satisfying the relation
L L G( z)u[n] x [n] H ( z) x [n]u[n]
.)()( //1
1 11
0
1=∑
−
=
k L
L
L
k
k L
LW z H W z G
L If we now place the second system in front of the
first we arrive at the system shown in Figure P13.2 which is an identity system
provided .)()( //1
1 11
0
1=∑
−
=
k L
L L
k
k L
LW z H W z G
L
13.9 Making use of the multirate identities we simplify the structure of Figure P13.3 asindicated below:
53 x [n] y[n]15
53 x [n] y[n]53
5 x [n] y[n]5v[n]
Analysis of the last structure yields, ],5[][ n x nv and
or,⎩
±
otherwise,,0
,,10,5,0],5 / [][
Knnvn y
⎩
±
otherwise.,0
,,10,5,0],[][
Knn x n y
13.10
L L x [n] y[n]v[n]
Analysis yields ],[][ Ln x nv = and or,⎩
±=
otherwise,,0
,,2,,0], / [][
K L Ln Lnvn y
⎩
±=
otherwise.,0
,,2,,0],[][
K L Lnn x n y
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13.11 Since 3 and 4 are relatively prime, we can interchange the positions of the factor-of-3
down-sampler and the factor-of-4 up-sampler as indicated below:
x [n] y[n] z_ 6 3 42
which simplifies to the structure shown below:
x [n] y[n] z_ 6 46
Using the Noble identity of Figure 13.14(a) we redraw the above structure as indicated
below:
x [n] y[n]46 z_1
u[n] v[n]
Analysis yields ],16[]1[][],6[][ n x nunvn x nu and
⎩
±
⎩
±
otherwise.otherwise, ,0
,,8,4,0],1)2 / 3[(
,0
,,8,4,0],4 / [][
KK nn x nnvn y
13.12 As outlined in Section 8.2, the transpose of a digital filter structure is obtained byreversing all paths, replacing the pick-off node with an adder and vice-versa, and
interchanging the input and the output nodes. Moreover, in a multirate structure, the
transpose of a factor-of- M down-sampler is a factor-of- M up-sampler and vice-versa.Applying these operations to the factor-of- M decimator shown below on the left-hand
side, we arrive at a factor-of- M up-sampler shown below on the right-hand side.
x [n] y[n] H ( z) M x [n] y[n] H ( z) M
13.13 (a) To prove Eq. (13.20), consider the fractional-rate sampling rate converter of
Figure 13.16(b) with internal variables labeled as shown below:
L M x [n] y[n] H ( z)v[n] x [n]u
Analysis yields K,2,1,0],[][ ±nn x Ln x u and ].[][][ ll
l
∑ ∞
u x nhnv
Substituting the first equation in the second we get Finally,].[][][ m x Lmnhnv
m
∑ ∞
∑ ∞
m
m x Lm Mnh Mnvn y ].[][][][
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(b) Next, to prove Eq. (13.21), we make use of the z-domain relations of the down-
sampler and the up-sampler. From Eq. (13.17) we have
).()(1
)( / 11
0
/ 1 k M
M u
k M
M
k
M W z X W z H M
zY
=∑ But Hence,).()( L
u z X z X =
).()(1)( / 1
0
/ 1 Lk M
M Lk M
M
k
M W z X W z H M
zY
=∑
13.14
2
2
3 E ( z)00
E ( z)01
E ( z)02
E ( z)10
E ( z)11
E ( z)12
z_1
z_1
3
3
3
3
3
+
z_1
+
z_1
+
z_1
+
z_1
+
x [n]
y[n]
13.15
x [n] y[n] H ( z)
400 Hz
8 15
3200 Hz 3200 Hz 213.3333 Hz
v[n] x u n[ ]
(a) 400T F Hz, L = 8, M = 15. Now, the sampling rate of and is
kHz. Hence, the sampling rate of is
Hz.
][n x u ][nv
2.38400400 = L ][n y 15 / 3200 / 3200 = M
33.213
(b) The normalized stopband edge angular frequency of (for no aliasing))( z H
.15
,min π π π π
ω =⎠
⎞⎜⎝
⎛=
M M Ls Hence, the stopband edge frequency is
30
1=
sF Hz.
13.16
x [n] y[n] H ( z)8 15v[n] x u n[ ]
650 Hz 3.25 kHz 3.25 kHz 361.1 Hz
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(a) 650T F Hz, L = 5, M = 9. Now, the sampling rate of and is][n x u ][nv
kHz. Hence, the sampling rate of is
Hz.
25.35650650 = L ][n y 9 / 3250 / 3250 = M
1.361
(b) The normalized stopband edge angular frequency of (for no aliasing))( z H
.9,min π π π π
ω =⎠
⎞⎜⎝
⎛= M M L
s Hence, the stopband edge frequency is Hz.
13.17 Applying the transpose operation to the M-channel analysis filter bank shown belowon the left-hand side, we arrive at the M-channel synthesis filter bank shown below on
the right-hand side.
x [n] v [n]0
v [n]1
v [n] M 1_
M
M
M
H ( z)0
H ( z)1
H ( z) M 1_
y[n] v [n]0
v [n]1
v [n] M 1_
M
M
M
H ( z)0
H ( z)1
H ( z) M 1_
+
+
y[n]v [n]0
v [n]1
v [n] M 1_
M H ( z)0
H ( z)1
H ( z) M 1_
+
+
M
M
13.18 Specifications for H ( z ) are: 180 pF Hz, 200
sF Hz, .001.0,002.0 =
s p δδ
H ( z) 30
12 kHz 12 kHz 400 Hz
)()()( 5 zF z I z H
30 I ( z) F ( z )5
I ( z) F ( z )5 5 6
I ( z) 5 6F ( z)
12 kHz 12 kHz 2.4 kHz 2.4 kHz 400 Hz
Specifications for F ( z ) are: 900 pF Hz, 1000sF Hz, .001.0,001.0 s p δδ
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Here .12000
100= f ∆ Using Eq. (10.3) we arrive at the order of F ( z ) given by
3873.386
12000
1006.14
13001.0001.0log20 10 =
⎟
⎠
⎞⎜
⎝
⎛
= F F N N
Specifications for I ( z ) are: 180 pF Hz, 2200sF Hz, .001.0,001.0 =s p δδ
Here .12000
2020= f ∆ Using Eq. (10.3) we arrive at the order of I ( z ) given by
20712.19
12000
20206.14
13001.0001.0log20 10 =
⎟⎠
⎞⎜⎝
⎛
= F I N N
Hence, 200,1556
2400)1387(, =F M R mps, 400,50
5
12000)120(, = I M R mps.
The above realization requires a total of 205,600 mps. As a result, the computational
complexity is slightly higher than in Example 13.10.
13.19 Specifications for H ( z ) are: 180 pF Hz, 200sF Hz, .001.0,002.0 =s p δδ
)()()( 3 zF z I z H =
30 I ( z) F ( z )3
I ( z) F ( z )3 3 10
I ( z) 3 10F ( z)
12 kHz 12 kHz 4 kHz 4 kHz 400 Hz
Specifications for F ( z ) are: 540 pF Hz, 600sF Hz, .001.0,001.0 =s p δδ
Here .12000
60= f ∆ Using Eq. (10.3) we arrive at the order of F ( z ) given by
6448.643
12000
606.14
13001.0001.0log20 10 =
⎟⎠
⎞⎜⎝
⎛
= F F N N
Specifications for I ( z ) are: 180 pF Hz, 3800sF Hz, .001.0,001.0 =s p δδ
Here .12000
3260= f ∆ Using Eq. (10.3) we arrive at the order of I ( z ) given by
1285.11
12000
32606.14
13001.0001.0log20 10 =
⎟⎠
⎞⎜⎝
⎛
= F I N N
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Hence, 000,26210
4000)1644(, =F M R mps, 000,52
3
12000)112(, = I M R mps.
The above realization requires a total of 314,000 mps. As a result, the computational
complexity is higher than in Example 13.10 and that in Problem 13.18.
13.20 (a) The desired down-sampling factor is .20000,2
40000 = M The general structure of
the desired decimator is thus as shown below:
20 H ( z)
40 kHz 40 kHz 2 kHz
Now, the normalized stopband edge angular frequency of a factor-of-20 decimator is
.20
π ω =s Hence, the desired stopband edge frequency in this case is
100020
20000 =sF Hz. The specifications of the decimation filter H ( z ) is thus as
follows: Hz,800 pF 1000sF Hz, .002.0,002.0 =s p δδ Here .40000
200= f ∆
Using Eq. (10.3) we arrive at the order of H ( z ) given by
.5624.561
40000
2006.14
13002.0002.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
= H N Therefore, the computational
complexity is given by 000,112620
40000)1562(, = H M R mps.
(b) For a two-stage realization of the decimator, there are 4 possible realizations of thedecimation filter:
Realization #1 – )()()( z I z F z H 2=
2 I ( z) F ( z )2 10
40 kHz 40 kHz 40 kHz 10 kHz 2 kHz
2 I ( z) 10
40 kHz 20 kHz 2 kHz
F ( z)
20 kHz40 kHz
Specifications for and are as follows:)( zF )( z I
1600:)( = pF zF Hz, Hz,2000sF ,002.0,001.0 =s p δδ and thus, .40000
400= f ∆
800:)( = pF z I Hz, Hz,000,19sF ,002.0,001.0 =s p δδ and thus, .40000
18200= f ∆
Orders of and are given by)( zF )( z I
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30229.301
40000
4006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
=F N
762.6
40000182006.14
13002.0001.0log20 10 ⇒
⎟⎠⎞⎜
⎝⎛
= I N
Computational complexities of the two sections are:
000,60610
20000)1302(, =F M R mps and 000,160
2
40000)17(, = I M R mps.
Hence, the total computational complexity of the two-stage realization is
mps.000,766,, = I M F M M R R R
Realization #2 – )()()( z I z F z H 4=
4 I ( z) F ( z )4 5
40 kHz 40 kHz 40 kHz 10 kHz 2 kHz
4 I ( z) 5
40 kHz 10 kHz 2 kHz
F ( z)
10 kHz40 kHz
Specifications for and are as follows:)( zF )( z I
3200:)( = pF zF Hz, 4000sF Hz, ,002.0,001.0 =s p δδ and thus, .40000
800= f ∆
800:)( = pF z I Hz, Hz,000,9sF ,002.0,001.0 =s p δδ and thus, .400008200= f ∆
Orders of and are given by)( zF )( z I
15165.150
40000
80006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
=F N
1569.14
40000
82006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
= I N
Computational complexities of the two sections are:
000,3045
10000)1151(, =F M R mps and 000,160
4
40000)115(, = I M R mps.
Hence, the total computational complexity of the two-stage realization is
mps.000,464,, = I M F M M R R R
Realization #3 – )()()( z I z F z H 5
=
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5 I ( z) F ( z )5 4
40 kHz 40 kHz 40 kHz 4 kHz 2 kHz
5 I ( z) 4
40 kHz 8 kHz 2 kHz
F ( z)
8 kHz40 kHz
Specifications for and are as follows:)( zF )( z I
4000:)( = pF zF Hz, 5000sF Hz, ,002.0,001.0 =s p δδ and thus,
.40000
1000= f ∆
800:)( = pF z I Hz, Hz,000,7sF ,002.0,001.0 =s p δδ and thus, .40000
6200= f ∆
Orders of and are given by)( zF )( z I
12152.120
40000
10006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
=F N
2044.19
40000
62006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
= I N
Computational complexities of the two sections are:
000,2444
8000)1121(, =F M R mps and 000,168
5
40000)120(, = I M R mps.
Hence, the total computational complexity of the two-stage realization ismps.000,412,, = I M F M M R R R
Realization #4 – )()()( z I z F z H 10
=
10 I ( z) F ( z )10 2
40 kHz 40 kHz 40 kHz 4 kHz 2 kHz
10 I ( z) 2
40 kHz 4 kHz 2 kHz
F ( z)
4 kHz40 kHz
Specifications for and are as follows:)( zF )( z I
8000:)( = pF zF Hz, Hz,000,10sF ,002.0,001.0 =s p δδ and thus, .40000
2000= f ∆
800:)( = pF z I Hz, Hz,000,3sF ,002.0,001.0 =s p δδ and thus, .40000
2200= f ∆
Orders of and are given by)( zF )( z I
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6126.60
40000
20006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
=F N
5578.54
4000022006.14
13002.0001.0log20 10 ⇒
⎟⎠⎞⎜
⎝⎛
= I N
Computational complexities of the two sections are:
000,1242
4000)161(, =F M R mps and 000,224
10
40000)155(, = I M R mps.
Hence, the total computational complexity of the two-stage realization is
mps.000,348,, = I M F M M R R R
Hence, the optimum two-stage design with the lowest computational complexity is theRealization #4.
13.21 (a) The desired up-sampling factor is .50480
24000= L The general structure of the
desired interpolator is thus as shown below:
H ( z)50
480 Hz 24 kHz 24 kHz
Now, the normalized stopband edge angular frequency of a factor-of-50 interpolator is
.50
π ω =s
Hence, the desired stopband edge frequency in this case is
24050
12000=sF Hz. The specifications of the decimation filter H ( z ) is thus as
follows: Hz,190 pF 240sF Hz, .002.0,002.0 =s p δδ Here .24000
50= f ∆
Using Eq. (10.3) we arrive at the order of H ( z ) given by
.134827.1347
24000
506.14
13002.0002.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
= H N
Computational complexity is thus 520,64750
24000)11348(, = H M R mps.
(b) ).()()( 10 z I zF z H =
Specifications for and are as follows:)( zF )( z I
1900:)( = pF zF Hz, Hz,2400sF ,002.0,001.0 =s p δδ and thus, .24000
500= f ∆
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10
I ( z)50 F ( z )10
I ( z)5 F ( z)
480 Hz 24 kHz 24 kHz 24 kHz
480 Hz 2400 Hz 2400 Hz 24 kHz 24 kHz
190:)( = pF z I Hz, Hz,2160sF ,002.0,001.0 =s p δδ and thus, .24000
1970= f ∆
Orders of and are given by)( zF )( z I
14562.144
24000
5006.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
=F N
3771.36
24000
19706.14
13002.0001.0log20 10 ⇒
⎟⎠
⎞⎜⎝
⎛
= I N
Computational complexities of the two sections are:
080,705
2400)1145(, =F M R mps and 200,91
10
24000)137(, = I M R mps.
Hence, the total computational complexity of the two-stage realization is
mps.280,161,, = I M F M M R R R
Therefore, the complexity in a two-stage design with is
approximately 25% of that of the single-stage design.
).()()( 10 z I zF z H =
13.22 A computationally efficient realization of a factor-of-3 interpolator as shown below
3 H ( z)
is obtained from the 3-band polyphase decomposition of H ( z ) given by
).()()()( 32
231
130 z E z z E z z E z H The general form of the polyphase
representation of the interpolator is as shown below:
E ( z) 30
E ( z) 31
E ( z) 32
1_ z
+
+
1_ z
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Since H ( z ) is a length-15 linear-phase transfer function, 1]1[]0[)( zhh z H
98765432 ]5[]6[]7[]6[]5[]4[]3[]2[ zh zh zh zh zh zh zh zh
,]0[]1[]2[]3[]4[ 1413121110 zh zh zh zh zh the transfer functions of the
sub-filters are as follows:
,]2[]5[]6[]3[]0[)( 4321
0
zh zh zh zhh z E ,]1[]4[]7[]4[]1[)( 4321
1 zh zh zh zhh z E
.]0[]3[]6[]5[]2[)( 43212
zh zh zh zhh z E
A computationally efficient realization of the factor-of-3 interpolator is obtained bysharing common multipliers as shown below:
1
_
z
+
+
1_ z
31_ z + 1_
z + 1_ z + 1_
z +
31_ z + 1_
z + 1_ z + 1_
z +
31_ z + 1_
z + 1_ z + 1_
z +
h[0]
h[3]
h[6]
h[5]
h[2]
h[4]
h[1]
h[7]
x [n]
y[n]
13.23 A computationally efficient realization of a factor-of-4 decimator as shown below
4 H ( z)
is obtained from the 4-band polyphase decomposition of H ( z ) given by
).()()()()( 33
332
231
130 z E z z E z z E z z E z H The general form of the
polyphase representation of the interpolator is as shown below:
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4 E ( z)0 +
4 E ( z)1 +
z_
1
4 E ( z)2 +
z_
1
4 E ( z)3
z_
1
Since H ( z ) is a length-16 linear-phase transfer function, 1]1[]0[)( zhh z H
98765432 ]6[]7[]7[]6[]5[]4[]3[]2[ zh zh zh zh zh zh zh zh
,]0[]1[]2[]3[]4[]5[ 151413121110 zh zh zh zh zh zh the transfer
functions of the sub-filters are as follows:
,]3[]7[]4[]0[)( 3210 zh zh zhh z E
,]2[]6[]5[]1[)( 3211
zh zh zhh z E
.]1[]5[]6[]2[)( 3212
zh zh zhh z E
.]0[]4[]7[]3[)( 3213
zh zh zhh z E
A computationally efficient realization of the factor-of-4 decimator is obtained by
sharing common multipliers as shown below
z
_1
4
4
+
+
+
+
4
z
_1
z
_1
z
_1
z
_1
z
_1
z
_1
z
_1
+
+
+
z
_1
z
_1
z
_1
z
_1 z
_1 z
_1
z
_1
4 +
+
+
+
h[0]
h[1]
h[2]
h[3]
h[4]
h[5]
+
+
+
h[7]
h[6]
13.24 )()()1()( )1()2(3211
0
=
∑ N N N
i
i z z z z z z z H L
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),()1()1)(1( 21)2(221 zG z z z z z N L where
Using a similar technique we can show that Therefore we
can write where
Continuing this decomposition process further we arrive at
where
.)(
1)2 / (
0
∑
=
= N
i
i z zG
.)1()(
1)4 / (
0
21⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ∑
=
N
i
i z z zG
),()1)(1()1)(1()( 4211)4 / (
0
421 zF z z z z z z H
N
i
i
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ∑
.)(
1)4 / (
0
∑
=
= N
i
i z zF
),1()1)(1()( 1221 K
z z z z H L .2K N =
The transfer function of a box-car decimation filter of length-16 can be expressed as:
.1
1)( 1
1615
0
=
=∑ z
z z z H
i
i
As a result, a computationally efficient realization of a factor-of-16 decimator using a
length-16 boxcar decimation filter is as shown below:
z 1_
++
z1_
16
13.25 Let denote the output of the factor-of- L interpolator. Then][nu
,
][
])1[][(
2
2
∑
∑∞
∞
∞
∞
=
n
n
nu
nunu
E (13-1)
and
.
][
]1[][
2 nu
nunu
C
n
n
∑
∑
∞
∞
∞
∞
= (13-2)
Substituting Eq. (13-2) in Eq. (13-1) we get ).1(2 C E Hence, as i.e., as the
signal becomes highly correlated,
,1C
][nu .0E
Now, by Parsevals’relation, ,)(*)(2
1][][ ω
π
ω
π
π
ω d eV eU nvnu j j
n∫
∞
∞
= where )( ω jeU
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and are the DTFTs of and , respectively. If we let)( ω jeV ][nu ][nv ]1[][ nunv in
the numerator of Eq. (13-1), and ][][ nunv = in the denominator of Eq. (13-1), then we
can write
,
)(
)cos()(
)(2
1
)(
2
1
0
2
0
2
2
2
∫∫
∫∫ =
π ω
π ω
π
π
ω
π
π
ωω
ω
ωω
ωπ
ω
π
d eU
d eU
d eU
d eeU
E
j
j
j
j j
assuming to be a real sequence. If][nu ][n x is assumed to be a broadband signal with
a flat magnitude spectrum, i.e., 1)( =ω je X for ,0 π ω ≤ then the magnitude
spectrum of is bandlimited to the range , / 0 Lπ ω ≤ i.e.,
⎩<=
otherwise.,0, / 0,1)( LeU j π ωω Therefore, .
) / () / sin(
)cos(
/
0
/
0
L L
d
d
C L
L
π π
ω
ωω
π
π
=
∫
∫ Hence, as
.1, →C L
13.26
z L H ( z) M M – X (e ) jω
W (e ) jω R(e ) jω
Y (e ) jω S (e ) jω
Analysis of the above structure yields ),()( M j j e X eW ωω =),()()( M j j j e X e H e R
ωωω = ),()()()( M j j L j j L j j e X e H ee ReeS ωωω−ωω−ω ==
.)()( //∑−
=
ωω−ω =1
0
1 M
k
M j M k j j eeS M
eY If the filter is assumed to be close to
an ideal lowpass filter with a cutoff at
)( ω je H
,/ M ω we can assume that all images of
are suppressed leaving only the term in the expression for Hence, we
can write
)( ω je X ).( ω jeY
).()()()( /// ωω−ωωω == j M L j M j M j j e X ee H eS eY 11
Since is a
Type 1 FIR filter with exact linear phase and a delay of
)( z H
KM N =− 21 /)( samples and amagnitude response equal to M in the passband, we have
Thus, the structure of Figure P13.7 Is
approximately an allpass filter with a fixed delay of K samples and a variablenoninteger delay of L/ M samples.
).()( / ωω−ω−ω = j M L j K j j e X eeeY
13.27 An ideal M -th band lowpass filter is characterized by a frequency response)( z H
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⎩⎨⎧ π≤ω≤π−=ω
otherwise.,,//,
)(01 M M
e H j The transfer function can be expressed in
an M -branch polyphase form as From the above we observe
Therefore,
)( z H
).()( M k
M
k
k z H z z H ∑−
=
−=1
0
).()( M M
r
r M z H M zW H 0
1
0
=∑−
=
.)()( /(
M e H
M e H
M
r
M r j M j 111
0
20 == ∑
−
=
π−ωω
Or in other words, is an allpass function.)( M z H 0
13.28 An equivalent realization of the structure of Figure P13.38 obtained by realizing the
filter in a Type 1 polyphase form is shown below on the left. By moving the down-
sampler through the system and invoking the cascade equivalence #1 of Figure 13.14 wearrive at the structure shown below on the right.
z
_1
+
+
+
z
_1
z
_1
0 E ( z ) L
1 E ( z ) L
2 E ( z ) L
L 1_ E ( z ) L
L L
z
_1
+
+
+
z
_1
z
_1
0 E ( z)
1 E ( z)
2 E ( z)
L 1_ E ( z)
L L
L
L
L
The structure on the right hand side reduces to the one shown below on the left from
which we arrive at the simplified equivalent structure shown below on the right.
L L E ( z)0 E ( z)0
13.29 (a) Let and
Then
(b)
.][)(
1
0
∑=
= N
n
n znh z H ∑
=
1)2 / (
0
20 ])12[]2[()(
N
i
i zihih z H
.])12[]2[()(
1)2 / (
0
21 ∑
=
N
i
i zihih z H
).(1]12[]2[)()1()()1(
1)2 / (
0
21)2 / (
0
221
120
1 z H zih zih z H z z H z
N
i
i N
i
i =∑
=
=
)()1()()1()( 21
120
1 z H z z H z z H
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).()()()()()( 21
120
21
20
121
20 z E z z E z H z H z z H z H Therefore,
and)()()( 100 z H z H z E ).()()( 101 z H z H z E
(c) Now, [ ] [ ] ⎥⎦
⎤⎢⎣
⎡
=
⎦
⎤⎢⎣
⎡
⎥
⎤
⎢
⎡
)()(
)()(1
)(
)(
11
111)(
2120
21
201
21
201
z H z H
z H z H z
z H
z H z z H
).()1()()1()()()()( 21
120
121
20
121
20 z H z z H z z H z H z z H z H
(d) If i.e., then we can express,2 L ,22= N
)()1()()1()( 401
2400
220 z H z z H z z H and
).()1()()1()( 411
2410
221 z H z z H z z H Substituting these expressions in
)()1()()1()( 21
120
1 z H z z H z z H we get
[ ])()1()()1()1()( 401
2400
21 z H z z H z z z H
[ ])()1()()1()1( 411241021 z H z z H z z
)()1)(1()()1)(1( 401
21400
21 z H z z z H z z
)()1)(1()()1)(1( 411
21410
21 z H z z z H z z
[ ] [ ] .
)(ˆ)(ˆ)(ˆ)(ˆ
1
)(
)(
)(
)(
111111111111
1111
1
43
42
41
40
4321
411
410
401
400
321
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
z H
z H
z H
z H
z z z
z H
z H
z H
z H
z z z R
Continuing this process it is easy to establish that for we have
[ ]⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡=
)(ˆ
)(ˆ)(ˆ
1)(
1
1
0
)1(1
L L
L L
L L
z H
z H z H
z z z H M
L R .
13.30 Now
Therefore,
[ ] [ ] .
)(
)(
)(
)(
1
)(ˆ)(ˆ)(ˆ)(ˆ
1)(
43
42
41
40
321
43
42
41
40
4321
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
=
z E
z E
z E
z E
z z z
z H
z H
z H
z H
z z z z H R
.
)(
)(
)()(
111111111111
1111
4
1
)(
)(
)()(
4
1
)(
)(
)()(
)(ˆ)(ˆ)(ˆ)(ˆ
3
2
1
0
3
2
1
0
4
3
2
1
0
14
3
2
1
0
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
z E
z E
z E z E
z E
z E
z E z E
z E
z E
z E z E
z H
z H
z H z H
RR
A length-16 Type 1 linear-phase FIR transfer function is of the form9
68
77
76
65
54
43
32
21
10)( zh zh zh zh zh zh zh zh zhh z H
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.150
141
132
123
114
105
zh zh zh zh zh zh
Hence, ,)(,)( 32
26
1511
33
27
1400
zh zh zhh z E zh zh zhh z E
.)(,)( 30
24
1733
31
25
1622
zh zh zhh z E zh zh zhh z E Thus,
,)(ˆ,)(ˆ 32
23
1321
30
21
1100
zg zg zgg z H zg zg zgg z H
,)(ˆ,)(ˆ 36
27
1763
34
25
1542
zg zg zgg z H zg zg zgg z H where
),(4
1),(
4
1),(
4
1321027654132100 hhhhghhhhghhhhg
),(4
1),(
4
1),(
4
1765453210476543 hhhhghhhhghhhhg
).(4
1),(
4
17654732106 hhhhghhhhg Note that and
are Type 1 linear-phase FIR transfer functions, whereas, and are Type 2
linear-phase FIR transfer functions. A computationally efficient realization of a factor-
of-4 decimator using a four-band structural subband decomposition of the decimationfilter is shown below:
)(ˆ0 z H )(ˆ
3 z H
)(ˆ1 z H )(ˆ
2 z H
)( z H
z
_1
z
_1
z
_1
+
+
+
M
M
M
M
R4
0 H ( z)^
1 H ( z)^
2 H ( z)^
H ( z)3
^
Because of the symmetry or anti-symmetry in the impulse responses of the subbandfilters, each subband filter can be realized using only 2 multipliers. Hence, the final
realization uses only 8 multipliers. Note also that by delay-sharing, the total number of
delays in implementing the four subband filters can be reduced to 3.
13.31 ]2[)(]1[)(][)(]1[)(]2[)(][ 21012 n x Pn x Pn x Pn x Pn x Pn y ααααα
where ),22()22)(12)(2)(12(
)2)(1()1(
)(
234
24
1
2 αααα
αααα
α =
P
),44()21)(11)(1)(21(
)2)(1()2()( 234
6
11 αααα
ααααα =P
),45()20)(10)(10)(20(
)2)(1)(1)(2()( 24
4
10 = αα
αααααP
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),44()21)(01)(11)(21(
)2()1)(2()( 234
6
11 αααα
ααααα =P
).22()12)(02)(12)(22(
)1()1)(2()( 234
24
12 αααα
ααααα =P
We consider the computation of ]3[],2[],1[],[ n yn yn yn y using 5 input samples:
through]2[ n x ].2[ n x
For ,0)(,1)(,0)(,0)(,0 010001020 = ααααα PPPP and For.0)( 02 =αP
,3428.0)(,127.0)(,022.0)(,4 / 5 1011121 αααα PPP
and
,1426.1)( 11 =αP
.0952.0)( 12 =αP
For ,9531.2)(,4062.1)(,2734.0)(,4 / 10 2021222 = αααα PPP
and,2812.3)( 21 αP .4609.2)( 22 =αP
For ,86.32)(,2949.17)(,5718.3)(,4 / 15 3031323 = αααα PPP
and,873.29)(31 αP .7358.11)(
32 =αP
The block filter implementation is thus given by
Another
implementation is given by
.
]2[]1[
][]1[]2[
7358.11873.2986.322949.175718.34609.22812.39531.24062.12734.00952.01426.13428.0127.0022.0
00100
]3[
]2[
]1[
][
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
n x n x n x
n x n x
n y
n y
n y
n y
⎟⎠⎞⎜
⎝⎛ ]2[]1[][]1[]2[][
24
1
6
1
4
1
6
1
24
14 n x n x n x n x n x n y α
⎟⎠⎞⎜
⎝⎛ ]2[]1[]1[]2[
12
1
6
1
6
1
12
13 n x n x n x n x α
⎟⎠⎞⎜
⎝⎛ ]2[]1[][]1[]2[
24
1
6
4
4
5
6
4
24
12 n x n x n x n x n x α
].[]2[]1[]1[]2[12
1
6
4
6
4
12
1n x n x n x n x n x
⎠⎞⎜
⎝⎛ α
The Farrow structure implementation of the interpolator is shown below:
y[n]
x [n]
H 3( z)
H 1( z) H 2( z)
α α α
H 0( z)
α
where ,)(,)( 2
24
1
6
11
6
12
12
11
2
24
1
6
1
4
11
6
12
24
10 z z z z z H z z z z z H
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,)( 2
24
1
6
4
4
51
6
42
24
12 z z z z z H and .)( 2
12
1
6
41
6
42
12
13 z z z z z H
13.32 From Eq. (13.75), we have .1111
1111
33
32
31
30
23
22
21
20
3210
33323134
23
22
21
24
3214
40
t t t t
t t t t
t t t t
t t t t
t t t t
t t t t
aa Both the numerator and the
denominator are determinants of Vandermonde matrices and have a nonzero value if
From the solution of Problem 6.28, we get., jit t ji ≠≠
))()()()()((
))()()()()((
231312030201
23131243424140
t t t t t t t t t t t t
t t t t t t t t t t t t aa
−−−−−−
−−−−−−= ,
))()((
))()((
030201
4342414
t t t t t t
t t t t t t a
−−−
−−−= or
.))()((
))()((
302010
34241440
t t t t t t
t t t t t t aa
13.33 From Eq. (13.74) we get))()()()()((
))()()()()((
231312030201
23434203020441
t t t t t t t t t t t t
t t t t t t t t t t t t aa
−−−−−−
−−−−−−−=
.))()((
))()((
131201
4342044
t t t t t t
t t t t t t a
−−−
−−−−= Substituting the value of given by Eq. (13.77) in
the above we arrive at
4a
.))()()(( 41312101
11
t t t t t t t t a
−−−−= Likewise, from Eq. (13.74)
we get))()()()()((
))()()()()((
231312030201
43131403040142
t t t t t t t t t t t t
t t t t t t t t t t t t aa
−−−−−−−−−−−−
=
.))()((
))()((
231202
4314044
t t t t t t
t t t t t t a
−−−
−−−= Substituting the value of given by Eq. (13.77) in the
above we arrive at
4a
.))()()(( 42321202
01
t t t t t t t t a
−−−−= Finally, from Eq. (13.74) we
get))()()()()((
))()()()()((
231312030201
24141204020143
t t t t t t t t t t t t
t t t t t t t t t t t t aa
−−−−−−
−−−−−−−=
.))()(())()((
2313032414044t t t t t t t t t t t t a −−− −−−−= Substituting the value of given by Eq. (13.77) in the
above we arrive at
4a
.))()()(( 43231303
31
t t t t t t t t a
−−−−=
13.34 ., 4+≤≤= mimit i
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,))()()(( 24
1
4321
1=
−−−−−−−−=
mmmmmmmmam
,))()()(( 6
1
4131211
11 −=
−−+−−+−−+−+=+
mmmmmmmmam
,))()()(( 4
1
4232122
12 =−−+−−+−−+−+=+ mmmmmmmmam
,))()()(( 6
1
4323133
13 −=
−−+−−+−−+−+=+
mmmmmmmmam
.))()()(( 24
1
3424144
14 =
−−+−−+−−+−+=+
mmmmmmmmam
From Eq. (13.69), we have the expressions for given by)()( t Bm3
⎪⎪⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
+≥
+<≤+−−−−−+−−−−
+<≤+−−+−−−−
+<≤+−−−−
+<≤−
<
=
.,
,,)()()()(
,,)()()(
,,)()(
,,)(
,,
)()(
40
43321
3221
211
1
0
3
6
13
4
13
6
13
24
1
3
4
13
6
13
24
1
3613
241
3
24
1
3
mt
mt mmt mt mt mt
mt mmt mt mt
mt mmt mt
mt mmt
mt
t Bm
The normalized 3rd
order B-spline is then given by )()()( )()( t Bmmt mm33
4 −+=β
⎪⎪⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
+≥
+<≤+−−−−−+−−−−
+<≤+−−+−−−−
+<≤+−−−−
+<≤−
<
==
.,
,,)()()()(
,,)()()(
,,)()(
,,)(
,,
)()(
40
43321
3221
211
1
0
4
3
3
233
3
23
6
1
33
3
23
6
1
3323
61
3
6
1
3
mt
mt mmt mt mt mt
mt mmt mt mt
mt mmt mt
mt mmt
mt
t Bm
Substituting and evaluating forα+= 1t )()( t m3β ,,, 2101−=m we have
,)()(
6
1
226
2331 +
α−
α+
α−=αβ− ,)(
)(
3
2
2
23
30 +α+
α=αβ
,)()(61
222
23
31 +α+α+α−=αβ .)()(6
3
32 α=αβ
Substituting back in Eq. (13.91), we have ][)(][)(
k n xn y
k
k +αβ= ∑
−=
2
1
3
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][][][ 16
1
2223
2
21
6
1
226
232
323
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +
α+
α+
α−+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +α+
α+−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +
α−
α+
α−= n xn xn x
][ 26
3
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ α+ n x α⎟
⎠
⎞⎜⎝
⎛ ++−−+⎟ ⎠
⎞⎜⎝
⎛ ++−= ][][][][][ 12
11
2
1
6
1
3
21
6
1n xn xn xn xn x
.][][][][][][][ 32 26
11
2
1
2
11
6
11
2
11
2
1α⎟
⎠
⎞⎜⎝
⎛ +++−+−−+α⎟ ⎠
⎞⎜⎝
⎛ ++−−+ n xn xn xn xn xn xn x
In the z -domain, the input-output relation is thus given by
where,)()()()()( 30
2123 α+α+α+= z H z H z H z H z Y
,)( 210
6
1
2
1
2
1
6
1 z z z z H +−+−= − ,)( z z z H
2
11
2
1 11 +−= − ,)( z z z H
2
1
2
1 12 +−= −
.)( z z z H 6
1
3
2
6
1 13 ++= − The corresponding Farrow structure is shown on top of the
next page:
y[n]
x [n]
H 3( z) H 1( z) H 2( z)
α α
H 0( z)
α
13.35 For the factor-of-4/3 interpolator design, if we use cubic B-spline with uniformlyspaced knots at the problem reduces exactly to the design given in the solution of
Problem 13.34.
13.36 From Eq. (13.94) with 0k we have Substituting the
expression for on the left-hand side Eq. (13.97) we get INCOMPLETE
).()(1
1
L L
i
ii z E z z H ∑
=
α
13.37 For a half-band zero-phase lowpass filter, the transfer function is of the form
where,]2[]0[)(
0
21 ∑∞
≠ ∞
nn
n znh zh z H .2
1]0[ =h If the half-band filter has a zero at
then or,1 z ,0]2[]0[)1(
0
=∑
∞
≠ ∞
nn
nhh H .]2[]0[
0
∑
∞
≠ ∞
=
nn
nhh
13.38 From Eq. (13.99), a zero-phase half-band filter satisfies the condition
a constant.
)( z H
=)()( z H z H
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(a) The zero-phase equivalent here is given by .2)( 11
z z z H Hence,
.422)()( 1111 = z z z z z H z H A plot of the scaled magnitude
response of is given below:)(1 z H
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
H1(z)
(b) The zero-phase equivalent here is given by .9169)( 3132
z z z z z H
Hence,
.3291699169)()( 31331322 = z z z z z z z z z H z H
A plot of the scaled magnitude response of is given below:)(2 z H
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
H2(z)
(c) The zero-phase equivalent here is given by
.319326193)( 3133
z z z z z H Hence,
.643193219331932193)()( 31331333 = z z z z z z z z z H z H
A plot of the scaled magnitude response of is given below:)(3 z H
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
H3(z)
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(d) The zero-phase equivalent here is given by
.325150256150253)( 531354
z z z z z z z H Hence, )()( 44 z H z H 13553135 150256150253325150256150253 z z z z z z z z z z
.512325 53 = z z A plot of the scaled magnitude response of is given
below:
)(4 z H
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
H4(z)
13.39 (a) A function of has p-th order zero at a frequency)(ω F iω=ω if
.)(
0=ω
ω
ω=ω i p
p
d
F d The function has p zeros at)( ω je H 1=ωcos , i.e., at π=ω .
Hence, .)(
0=ω π=ω
ω
p
j p
d
e H d Moreover, the order of the highest power of
is As a result,21 /)cos( ω− .1− p .)(
0
01
1
=
ω =ω−
ω−
p
j p
d
e H d
(b) Now,
2
2
1
2
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +=
ω+ ω−ω j
j eecos
and .cos
2
2
1
2
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −−=
ω− ω−ω j
j ee
Substituting these expressions in Eq. (13.120) we arrive at
.)()(
l
l
l
l
l
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ −−⎟
⎠ ⎞
⎜⎝ ⎛ +−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +=
ω−−
=
ω−ω−
ωω ∑ 2
11
2
11
0
j p
j
p j
jp j ee
peee H Replacing with
in z the above we get the expression for the zero-phase transfer function asω je
.)()(
l
l
l
l
ll
⎟⎟ ⎠ ⎞
⎜⎜⎝ ⎛ −−⎟
⎠ ⎞⎜
⎝ ⎛ +−
⎟⎟ ⎠ ⎞
⎜⎜⎝ ⎛ += −
−− ∑ 2
112
1 1
1
1 z z p z z z H
p
p
13.40 (a)
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+
2
2
22
2
2_1 z
_1 z
1 z_ _1
_11 + z
0 H ( z)
1 H ( z)
1 H ( z)0 H ( z) +
z z__2 _1
X ( z) Y ( z)V ( z)1
V ( z)0
V ( z)2
W ( z)0
W ( z)1
W ( z)2
U ( z)0
U ( z)1
U ( z)2
R ( z)0
R ( z)1
R ( z)2
Analysis of the above structure yields ),(1
1
)(
)(
)(
)( 1
1
2
1
0
z X z
z
zV
zV
zV
z
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡=
V
),(1
1)(1
1
)(
)(
)(
)( 2 / 12 / 1
2 / 1
2 / 12 / 1
2 / 1
2
1
0
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡= z X z
z
z X z
z
zW
zW
zW
zW
),(
)(
)()()1(
)(
)(
)(
)()()1(
)(
)(
)(
)(
)( 2 / 1
1
102 / 1
02 / 1
2 / 1
1
102 / 1
02 / 1
2
1
0 z X
z H
z H z H z
z H z
z X
z H
z H z H z
z H z
zU
zU
zU
z ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡=
U
).(
)(
)()()1(
)(
)(
)(
)()()1(
)(
)(
)(
)(
)(2
1
21
20
10
1
21
21
20
1
20
1
2
1
0
z X
z H
z H z H z
z H z
z X
z H
z H z H z
z H z
z R
z R
z R
z ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡=
R
)()()()()1()( 212
11
01 z R z z z R z z R z zY
] )()()()()()1()()1( 21
1221
20
1120
11 z X z H z z z H z H z z z H z z
] )()()()()()1()()1( 21
1221
20
1120
11 z X z H z z z H z H z z z H z z
[ ] ).()()(2)()(2)(2 ][ 21
120
121
220
1 z X z H z z H z z X z H z z H z Hence,
].[ )()(2)( 21
120
1 z H z z H z zT
(b) ).(2)(2
1)(
2
1)(
2
1)(
2
12)( 11 z H z z H z H z H z H z zT =⎥
⎦
⎤⎢⎣
⎡
(c) Length of and length ofK z H =)(0 .)(1 K z H =
(d) The total computational complexity of the above structure is2
3 T F K
multiplications per second, where is the sampling frequency in Hz. On the other hand,a direct implementation of requires)( z H T KF 2 multiplications per second.
13.41
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+
2
2
2
2
_1 z_1 z
R( z) X ( z)1
X ( z)2
Y ( z)1
Y ( z)2
Analysis yields ),()()( 22
21
1 z X z X z z R )()()( 2 / 12 / 11 z R z R zY
),(2)]()([)]()([ 2212 / 1
212 / 1 z X z X z X z z X z X z =
)()()( 2 / 12 / 12 / 12 / 12 z R z z R z zY
).(2)]()([)]()([ 11
22 / 1
11
22 / 1
11 z X z z X z z X z z X z z X z = Thus, the output
is a scaled replica of the input while the output is a scaled replica of
the delayed input
][1 n y ][2 n x ][2 n y
].1[1 n x
13.42
22
H ( z2
)
22
z1 –
z1 –
X ( z)1
X ( z)2 Y ( z)2
Y ( z)1
W ( z) V ( z)
Analysis yields ),()()( 22
121 z X z z X zW
),()()()()( 22
2121
2 z X z H z z X z H zV
),()()()(2
1)( 1
2 / 12 / 11 z X z H zV zV zY =
).()()()(2
1)(
2
12 / 12 / 12 / 12 / 1
2 z X z H z zV z zV z zY =
Therefore, ),()(
)(
1
1 z H z X
zY = and ).(
)(
)( 1
2
2 z H z z X
zY = Hence, the system is time-invariant.
13.43
22 H(z2)
22
z1 –
z1 –
1 X ( z)
2 X ( z) 2Y ( z)
1Y ( z)
From the solution of Problem 13.42, ),()(
)(
1
1 z H z X
zY = and ).(
)(
)( 1
2
2 z H z z X
zY = Here now,
and hence,),()( 12 zY z X = ).()()()()()()( 121
11
21
2 z X z H z zY z H z z X z H z zY =
Thus, ).()(
)( 21
1
2 z H z z X
zY = Hence, the system is time-invariant.
13.44
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33 H(z3)
3
33
3
–2
3
z 1 –
z1 –
z1 –
z 1 – z (C 1) – +
1 X ( z)
2 X ( z)
3 X ( z)
W ( z) V ( z)
Y ( z)
Y ( z)1
2Y ( z)
Y ( z)3
Analysis yields ),()()()( 33
231
131 z X z zY z z X zW
),()()()()()()()()( 33
3232
3131
33 z X z H z z X z H z z X z H zW z H zV
[ ])()()(3
1)( 3 / 43 / 13 / 23 / 13 / 1
1π π j j e zV e zV zV zY
[ ])()()()()()(3
1 4
1
42
1
2
1
π π π π j j j j
e z X e z H e z X e z H z X z H
[ ])()()()()()(3
1 42
43 / 43 / 122
23 / 23 / 12
3 / 1 π π π π π π j j j j j j e z X e z H e ze z X e z H e z z X z H z
[ ])()()()()()(3
1 83
83 / 83 / 243
43 / 43 / 23
3 / 2 π π π π π π j j j j j j e z X e z H e ze z X e z H e z z X z H z
[ ])()()()()()(3
1111 z X z H z X z H z X z H
[ ])()()()()()(3
12
3 / 43 / 12
3 / 23 / 12
3 / 1 z X z H e z z X z H e z z X z H z j j π π
[ ])()()()()()(
3
13
3 / 83 / 23
3 / 43 / 23
3 / 2 z X z H e z z X z H e z z X z H z j j π π
),()( 1 z X z H
[ ])()()(3
1)( 3 / 43 / 13 / 43 / 13 / 23 / 13 / 23 / 13 / 13 / 1
2π π π π j j j j e zV e ze zV e z zV z zY
),()( 31 z X z H z =
[ ])()()(3
1)( 3 / 43 / 13 / 83 / 13 / 23 / 13 / 43 / 13 / 13 / 1
3π π π π j j j j e zV e ze zV e z zV z zY
).()( 22 z X z H z =
Now, and)()( 33 zY z X = ).()( 12 zY z X = Hence,
)()()()()( 31
31
2 zY z H z z X z H z zY
= )()()()()( 122211 ][ zY z H z z X z H z z H z
=
Therefore,).()( 132 z X z H z = )(3)(2)( 3
)1(2 zY z zY zY C
)()(2)(3)()(3)()(2 1322
121)1(
132 ][ z X z H z H z z X z H z z z X z H z C for
Thus, the transfer function of the system of Figure P13.12 is.0C
.)(2)(3)( ][ 322 z H z H z zG
M13.1 (a) (i)
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0 20 40 60 80 100-2
-1
0
1
2Input Sequence
Time index n
A m p l i t u d e
0 20 40 60 80 100
-1
-0.5
0
0.5
1
1.5
2Output sequence up-sampled by5
Time index n
m p
u
e
(ii)
0 10 20 30 40 500
0.2
0.4
0.6
0.8
1Input Sequence
Time index n
A m p l i t u d e
0 10 20 30 40 50
0
0.05
0.1
0.15
0.2Output sequence up-sampled by5
Time index n
M13.2 (a) (i)
0 10 20 30 40 50-2
-1
0
1
2Input Sequence
Time index n
A m p l i t u d e
0 10 20 30 40 50
-2
-1
0
1
2Output sequence down-sampled by 5
Time index n
A m p l i t u d e
(ii)
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0 10 20 30 40 50
0
0.2
0.4
0.6
0.8
1Input Sequence
Time index n
A m p l i t u d e
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1Output sequence down-sampled by 5
Time index n
A m p l i t u d e
M13.3 (a)
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Input spectrum
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Output spectrum
(b)
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Input spectrum
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Output spectrum
M13.4 (a)
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Input spectrum
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Output spectrum
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(b)
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t
u d e
Input spectrum
0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t
u d e
Output spectrum
M13.5 (a)
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Input spectrum
0 0.2 0.4 0.6 0.8 1
0
0.05
0.1
0.15
0.2
0.25
ω / π
Output spectrum
(b)
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
Input spectrum
0 0.2 0.4 0.6 0.8 1
0.04
0.06
0.08
0.1
0.12
0.14
ω / π
Output spectrum
M13.6 (a)
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0 20 40 60 80 100 120
-2
-1
0
1
2Input sequence
Time index n
A m
p l i t u d e
0 10 20 30 40
-2
-1
0
1
2Output sequence
Time index n
A m
p l i t u d e
M13.7 (a)
0 10 20 30 40 50-2
-1
0
1
2Input sequence
Time index n
A m p l i t u
d e
0 50 100 150
-2
-1
0
1
2Output sequence
Time index n
A m p l i t u
d e
M13.8 (a)
0 10 20 30 40-2
-1
0
1
2Input sequence
Time index n
A m p l i t u d e
0 5 10 15 20 25-2
-1
0
1
2Output sequence
Time index n
A m p l i t u d e
M13.9 Using Program 13_9.m we arrive at the transfer function of the desired elliptic half-
band lowpass filter: [ ],)()(
2
1)( 2
112
00 z z z z H A A where
642
6422
00192.03903.02456.11
2456.13903.00192.0)(
=
z z z
z z z zA and
.1206.08884.07442.11
7442.18884.01206.0)(
642
6422
1 =
z z z
z z z zA The power-complentary half-
band highpass transfer function is given by [ ].)()(2
1)( 2
112
00 z z z z H A A A plot
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of the magnitude responses of the above half-band lowpass and highpass filters is
shown below:
0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a g n i t u d e
H0(z) H1(z)
M13.10 (a) A digital lowpass half-band filter can be designed by applying a bilinear
transformation to an analog lowpass Butterworth transfer function with a 3-dB cutoff
frequency at 1 rad/sec. The 3-dB cutoff frequency of the digital lowpass Butterworth
half-band filter is therefore at .5.0 / )1(tan2 1 = π ωc
To design a 3rd
-order digital lowpass Butterworth half-band filter we use the MATLAB
statement [num,den] = butter(3,0.5); which yields
.3333.01
)1(1667.0)(
2
31
=
z
z z H As can be seen from the pole-zero plot of given below all
poles are on the imaginary axis:
-1.5 -1 -0.5 0 0.5 1 1.5
-1
-0.5
0
0.5
1
3
Real Part
m a g n a r y
a r
Using the MATLAB statement[d1,d2] = tf2ca([1 3 3 1]/6, [1 0 1/3 0]);
we arrive at the parallel allpass decomposition of as)( z H
)],()([)( 21
1202
1 z A z z A z H where
2
3
1
2
3
1
20
1)(
=
z
z z A and Hence,
the power-complementary highpass transfer function is given by
.1)( 21 = z A
.3333.01
)331(1667.0)]()([)(
2
3212
112
02
1
=
z
z z z z A z z A zG
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0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a
g n i t u d e
H(z)G(z)
(b) To design a 5th-order digital lowpass Butterworth half-band filter we use the
MATLAB statement [num,den] = butter(5,0.5); which yields
.0557.06334.01
)1(0528.0)(
42
51
=
z z
z z H As can be seen from the pole-zero plot of given
below all poles are on the imaginary axis:
-1 -0.5 0 0.5 1
-1
-0.5
0
0.5
1
Real Part
I m a g i n a r y P a r t
Using the MATLAB statement[d1,d2] =
tf2ca(0.0528*[1 5 10 10 5 1], [1 0 0.6334 0 0.0557]);
we arrive at the parallel allpass decomposition of as)( z H
)],()([)( 21
1202
1 z A z z A z H where
2
22
01056.01
1056.0)(
=
z
z z A and
.5279.01
5279.0)(
2
22
1 =
z
z z A Hence, the power-complementary highpass transfer function
is given by .0557.06334.01
)1(0528.0)]()([)(
42
512
112
02
1
=
z z
z z A z z A zG
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0 0.2 0.4 0.6 0.8 10
0.2
0.4
0.6
0.8
1
ω / π
M a
g n i t u d e
H(z)G(z)
M13.11
M13.12
0 0.2 0.4 0.6 0.8 1
-80
-60
-40
-20
0
ω / π
G a i n ,
d B
L-th band Nyquist Filter, L = 5