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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


4 I ( z) 5

40 kHz 10 kHz 2 kHz

F ( z)

10 kHz40 kHz


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 ∆


15165.150

40000

80006.14

13002.0001.0log20 10 ⇒

⎟⎠

⎞⎜⎝

⎛

=F N

1569.14

40000

82006.14

13002.0001.0log20 10 ⇒

⎟⎠

⎞⎜⎝

⎛

= I N


000,3045

10000)1151(, =F M R mps and 000,160

4

40000)115(, = I M R mps.


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


5 I ( z) 4

40 kHz 8 kHz 2 kHz

F ( z)

8 kHz40 kHz


4000:)( = pF zF Hz, 5000sF Hz, ,002.0,001.0 =s p δδ and thus,

.40000

1000= f ∆


6200= f ∆


12152.120

40000

10006.14

13002.0001.0log20 10 ⇒

⎟⎠

⎞⎜⎝

⎛

=F N

2044.19

40000

62006.14

13002.0001.0log20 10 ⇒

⎟⎠

⎞⎜⎝

⎛

= I N


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


10 I ( z) 2

40 kHz 4 kHz 2 kHz

F ( z)

4 kHz40 kHz


8000:)( = pF zF Hz, Hz,000,10sF ,002.0,001.0 =s p δδ and thus, .40000

2000= f ∆


2200= f ∆


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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


000,1242

4000)161(, =F M R mps and 000,224

10

40000)155(, = I M R mps.


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 =


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 ∆


14562.144

24000

5006.14

13002.0001.0log20 10 ⇒

⎟⎠

⎞⎜⎝

⎛

=F N

3771.36

24000

19706.14

13002.0001.0log20 10 ⇒

⎟⎠

⎞⎜⎝

⎛

= I N


080,705

2400)1145(, =F M R mps and 200,91

10

24000)137(, = I M R mps.


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



−−−−−−

−−−−−−−=

.))()((

))()((

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



−−−−−−−−−−−−

=

.))()((

))()((

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



−−−−−−

−−−−−−−=

.))()(())()((

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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8/13/2019 Chapter13 SM[1]


][][][ 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

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Chapter13 SM[1]