• Author / Uploaded
• Rax Rey

Citation preview

CONTENTS Chapter 1

1

Chapter 2

11

Chapter 3

51

Chapter 4

89

Chapter 5

191

Chapter 6

211

Chapter 7

233

Chapter 8

293 319

Chapter 9 1\

Chapter 10

333

Chapter 11

355

Chapter 12

375

iii

Chapter 1

1.1 (a) (b) (c) (d) (e)

One dimensional, multichannel, discrete time, and digital. Multi dimensional, single channel, continuous-time, analog. One dimensional, single channel, continuous-time, analog. One dimensional, single channel, continuous-time, analog. One dimensional, multichannel, discrete-time, digital.

1.2 (a) (b) (c) (d) (e)

f = 0;;" = 2~O ~ periodic with Np = 200. f = ;~~ (21§) = ~ ~ ~erio.dic with Np = 7. f = 2J = 2 ~ perlOdlc with Np = 2. f = .,~ ~ non-periodic. f = ~1;(2~) = ~~ ~ periodic with N p = 10.

1.3

=

(a) Periodic with period Tp 25". (b) f .,5 ~ non-periodic. (c) f non-periodic. (d) cos( ~) is non-periodic; cos( "8") is periodic; Their product is non-periodic. (e) cos( 1(2") is periodic with period Np =4 sine 1(:) is periodic with period N p =16 cos( "; + i) is periodic with period N p =8 Therefore, x(n) is periodic with period N p =16. (16 is the least common multiple of 4,8,16).

= = ;r" : :;.

1.4 (a) w

= 2;/

implies that

f

= -Ii.

Let Q

= GCD of (k, N), Le., k

=k'

Q,

N

=N'

Q.

Then,

f

= N"Ie'

W h'ICh'Imp l'les

N'= N. Q

1

t hat

(b)

N k GCD(k,N) Np

= = = =

7 01234567 7 1 111 117 17777771

(c)

N k GCD(k, N) Np

= = =

16 0 1 2 3 4 5 6 7 8 9 10 11 12 16 1 2 1 4 1 2 1 8 121 4

=

1 68 16 4 16 8 16 2 16 8 164 .. , 1

16 16

1.5 (a) Refer to fig 1.1 (b)

5

10

15

20

->1("'5)

Figure 1.1:

z(n)

= = =

f =

=

za(nT) za(n/F.) 3sin(1rr:/3) => 1 1r 21r('3) 1

6,Np 2

=6

25

30

(c)Refer to fig 1.2

x(n)

= {o, ~, ~, 0, -~, -~} ,Np = 6.

3

,

,,

, ,,

\

,,

\ \

\

I I

,

--tfr---...I._--a-_~-~--r----f:I~....-t

1 ,

\

,

(ms)

,'20

I

\

\

,,

, ,,

I

,

·3

Figure 1.2: (d) Yes.

x(l) = 3

. 100?r = 3stn( T) ~ F, = 200 samples/sec.

1.6 (a)

x(n)

= =

Acos(27rFon/F,+9) Acos(27l'(T/Tp)n + 9)

=

But T /Tp f ~ x( n) is periodic if f is rational. (b) If x(n) is periodic, then f=k/N where N is the period. Then,

.

Td

= (~T) f

= k( Tp)T = kTp. T

Thus, it takes k periods (kTp ) of the analog signal to make 1 period (Td ) of the discrete signal. (c) Td kTp ~ NT kTp ~ f = kiN = T/Tp ~ f is rational => x(n) is periodic.

=

=

1.7 (a) Fmax = 10kHz ~ F, 2: 2Fmax = 20kHz. (b) For F, = 8kHz, FroId = F,/2 = 4kHz => 5kHz will alias to 3kHz. (c) F=9kHz will alias to 1kHz.

1.8 (a) Fmax (b) FroId

= 100kHz, F, 2: 2Fmax = 200H z. = If = 125Hz. 3

1.9 (a) Fmax (b) Ffold

(c)

= 360Hz, FN = 2Fmax = 720Hz. = tt = 300Hz. z(n)

z(n)

= = = = =

za(nT) za(n/ F,) sin(4801rn/600) + 3sin(7207rn/600) sin( 47rn/5) - 3sin(47rn/5) -2sin(47rn/5).

=

Therefore, w 41r/5. (d) Ya(t) z(F,t) -2sin(4807li).

=

=

1.10 (a)

= 10921024 = 10.

Number of bits/sample

F,

FroId

=

[10,000 bits/sec] [10 bits/sample] 1000 samples/sec.

=

500Hz.

=

(b)

Fmax FN

= = =

18001r 21r 900Hz; 2Fmax

= 1800Hz.

( 0.5 :::} /2

= O.l.

3cos[(21r)(0.3)n) + 2cos[(27r)(0.1)n]

10 = 5-(-5) 1023 = 1023'

1.11 z(n)

= =

za(nT) 3cos (lOO1rn) 200 4

+ 2sin (2507l'"n) 200

T'

yc(t)

Crn)

(~n) T - 2sin 4

=

3cos

= = =

1 1000 => yc(t)

= z(t/T')

(rlOOO/)

C.-WOOl)

3cos - - - 2· szn 2 4 3cos(500rl) - 2sin(750n)

1.12 (a) For Fs

= 300Hz, zen)

= 3cos (1r n) + 10sin(1rn) - cos (1r n) 6

= 3cos (b) xr(t)

= 3cos(10000irt/6) -

n

(7(6

) - 3cos

3

('7r n) 3

cos(100001rt/3)

1.13 (a)

Range m

= =

Zmax - zmin 1 + range

=12.7.

~-

= 127 + 1 = 128 => 1092(128)

= (b) m

7 bits.

= 1 + 5.~~ = 636 => /092(636) => 10 bit A/D.

1.14 R

Ffold Resolution

=

(20samples)x(8~ ) sec

sample

= 160 bits sec F, = "'2 = 10Hz. 1volt = 28 -1 = 0.004.

1.15 (a) Refer to fig 1.3. With a sampling frequency of 5kHz, the maximum frequency that can be represented is 2.5kHz. Therefore, a frequency of 4.5kHz is aliased to 500Hz and the frequency of 3kHz is aliased to 2kHz. (b) Refer to fig 1.4. y(n) is a sinusoidal signal. By taking the even numbered samples, the sampling frequency is reduced to half i.e., 25kHz which is still greater than the nyquist rate. The frequency of the downsampled signal is 2kHz. 5

Fa • 5KHz, FO-SOOHz

Fa • 5KHz, FO-2OOOHz

0.5 0

-0.5

-1

0

50

100

Fa. 5KHz, FO-3OOOHZ

Fa.5KHz.~

0.5

o -0.5

Figure 1.3:

1.16

= =

(a) for levels 64, using truncation refer to fig 1.5. 128, using truncation refer to fig 1.6. for levels for levels 256, using truncation refer to fig 1.7. (b) for levels 64, using rounding refer to fig 1.8. for levels = 128, using rounding refer to fig 1.9. for levels 256, using rounding refer to fig 1.10. (c) The sqnr with rounding is greater than with truncation. number of quantization levels are increased. (d) levels 64 128 43.9000 49.9200 theoretical sqnr sqnr with truncation 31.3341 37.359 sqnr with rounding 32.754 39.2008

= =

=

But the sqnr improves as the

256 55.9400 43.7739 44.0353

The theoretical ·sqnr is given in the table above. It can be seen that theoretical sqnr is much higher than those obtained by simulations. The decrease in the sqnr is because of the truncation and rounding.

6

FO • 2KHz, Fs-SOkHz

FO • 2KHz, Fs-25kHZ

Figure 1.4:

....,.. •

~,

UIing 1Nncdon, SQNR • 31.33'1 dB -__.("---.r--,

1...---

...

'E" If III

'E"

"

0

A

I

I I

I

-0.5

-1

0

50

100 -:lon

150

50

200

0 -0.01

'F

! -0.02 I I

-0.03

100 -:lon

Figure 1.5:

7

100 -:lon

150

200

levels .. 128, USIng truncalion, SaNR .. 37.359dB 1 l"""II'I-.........- - . . _ - . _ . - _ _ .

g

g JC

cr JC

0

"

"

I I

I

I

-0.5 -1

0

50

100 ->n

150

200

50

100 ->n

150

200

100 ->n

200

0 -0.005

.s.

• "

~.O,

I I

-0.015 ~.020

Figure 1.6:

=

levels 256. using truncation, SONR-437739dB 1

g

c)C' o "

i

"

I

I I

I

~.5

-1

~-.Ji'--_..K._-.Ji'--_..M-I

o

50

100 ->n

150

200

50

X 10-4

0;..;.::.-------.------. -2

-8 0

50

100 ->n

150

200

Figure 1.7:

8

100 ->n

150

200

• • • 64, using raunclng. SQNR.32. 7S&dB

11"'7'r--'7'l"""---,r---""'l"I""-'"

~ o II. I I

-0.5 -1 0~-.I"-5O----I"-100---"1-15O---"l..J200

->n

50

100 ->n

150

50

100 ->n

150

200

50

100 ->n

150

200

200

Figure 1.8:

1---------_.....-..,

. . . 128, Uling raunclng, SQIIRa39.2OO8cE

'C'

l:l" A

0

I I

-0.5 -1

0

50

100 ->n

150

200

50

100 ->n

150

200

0.02 0.01 'C'

¥

II.

I -0.01 -0.02 0

Figure 1.9:

9

levels:: 256. usmg rouncing. SONR=44.0353dB 1

.s.

~ o A

fr

,

A

I I

I

-0.5 50

100 ->n

1SO

100 ->n

200

0.01..-----------.

g

• A I I

-0.01

0

50

100 ->n

1SO

200

Figure 1.10:

10

1SO

200

Chapter 2

2.1 (a)

Refer to fig 2.1. (b) After folding s(n) we have

21.3

-3

-2

-I

0

2

3

4

Figure 2.1:

After delaying the folded signal by 4 samples, we have

z(-n +4)

={... O,~, 1, 1, 1, 1,~,~, 0, .. -}.

On the other hand, if we delay x(n) by 4 samples we have

z(n - 4) = {- .. ~,O,~,~, 1, 1,1,1,0, .. -}. Now, if we fold z(n - 4) we have

z(-n - 4) = {- .. 0,1,1, 1, 1,~,~, o,~, ...} 11

(c) z(-n +4)

={-- .~, 1,1,1, 1,~,~, O, ...}

(d) To obtain z( -n + k), first we fold zen). This yields z( -n). Then, we shift z(-n) by k samples to the right if k > 0, or k samples to the left if k < 0. (e) Yes. 1 2 zen) = 3c5(n - 2) + 3c5(n + 1) + u(n) - u(n - 4)

2.2

(a) z(n - 2) = { ... o,~, 1, 1, 1,1,~,~, o, ...}

(b) z(4-n)= {- ..

O'~'~'l'l'l'l'O'.,,}

(see 2.1(d»

(c) z(n + 2)

={... 0,1,1,

i,d, ~,o, ...}

(d) z(n)u(2 - n)

={-- .0,1+ 1,1,0,0, .. -}

(e) z(n -1)6(n - 3)

= {. ,,~,O, l,O, ...}

(f)

z( n2 )

= {

0, z( 4), z(I), z(O), z(l), z( 4),0, ...}

= { O,~,l,t,l,~,O, ...} (g)

z(-n)

12

(b)

=

:t(n)-x(-n) 2

= {... 0, -~, -~, -~, 0,0,0, 4,~,~, O, .. .} 2.3 (a) u(n) - u(n - 1)

(b)

t

=6(n) = {

6(1:) = u(n) =

1:=-00

n n

0,

n>

{~:

:

n~O

2.4 Let

:te(n)

1 = 2[:t(n)+ :t(-n)],

xo(n)

= ![:t(n) 2

:t(-n)].

Since

= xe(n)

and

zoe -n) = -:to(n), it follows that

zen)

= :teen) + zo(n).

The decomposition is unique. For

we have

and

13

=0

~~

n k = -. 365

Therefore, the total solution is

yen) = yp(n) To determine

Cl

and

C2,

8 1 1

+ Yh(n) = S(2")u(n) + cl(2)"u(n) + Ci( 3)"u(n).

assume that y( -2) = y( -1) = O. Then,

yeO)

= 1 and 27

=-65 y( 0) + 2 =-176

y( 1) Thus, 8

=--53 17 11 :::;> 3Cl + =-6 5

5 + Cl + C2 = 1

16

1 :::;>

1

5" + 2C1 + 3C2

=

Cl

+ C2

2C2

and, therefore, 2

= -1, = 5'

Cl

C2

The total solution is 8 1 yen) = [ -(2)" - (-)" 5 2

2 _)n 1 ] u(n) + -( 5 3

2.27 yen) - 3y(n - 1) - 4y(n - 2) = zen) + 2z(n - 1) The characteristic equation is A2

Hence, A

=4, -1 and Yh ( n)

-

3A - 4 = O.

= Cl ( n )4n + C2 ( -1 )" .

Since 4 is a characteristic root and the excitation is zen) = 4n u(n),

we assume a particular solution of the form

Yp(n) = kn4 n u(n). Then

kn4 n u(n) - 3k(n - 1)4n - 1 u(n - 1) - 4k(n - 2)4 n - 2 u(n - 2)

= 4n u(n) + 2(4)n-l u (n . For n

1)

= 2, k(32 - 12)

= 42 + 8 =24

-+

k

6 =-. 5

The total solution is

y(n)

= Yp(n) + Yh(n) =

To solve for

Cl

and

C2,

[~n4n + c14

we assume that y( -1)

n

+C2(-lt] urn)

= y( -2) = O. Then,

y(O) = 1 and y(1)

=3y(0) + 4 + 2 = 9

Hence, Cl

+ C2 = 1 and 28

24

"5 + 4C1 4C1 -

C2

C2

=9

=-215

Therefore, C1

The total solution is

26 25

= - and

C2

= [~n4n + 26 4" 5 25

y(n)

1 25

=-..!..(-It] u(n) 25

2.28 From 2.27, the characteristic values are A Yh(n)

When x(n)

=6(N), we find that

= ~ and

C2

=-

i.

C14"

=

+ C2 1 and Therefore,

Cl C1

=

+ c2(-I)n

yeO) = 1 and y(l) - 3y(O) = 2 or y(l) = 5.

Hence, This yields,

= 4, -1. Hence

h(n)

=

4Cl -

C2

=5

[~4n - ~(-lt] u(n)

2.29

=

(a) L 1 N 1 + M 1 and L2 = N 2 (b) Partial overlap from left:

+ M2

low N 1 + M 1

high N 1 + M 2

Full overlap: low N 1 + M2 Partial overlap from right:

-

high N2

(c)

t, 1,1,1,1 }

zen)

=

{ 1,1,

hen)

=

{2,?,2,2}

N1 N2

= = = =

M1 M2

-2, 4, -1,

2,

Partial overlap from left: n = -3 Full overlap: n = 0 Partial overlap from right:n = 4 29

n= -1 n=3 n=6

1

+ M1

2.30 (a) yen) - 0.6y(n - 1) + 0.08y(n - 2)

= zen).

The characteristic equation is A2

A

0.6A

-

= 0.2,0.4 Hence,

1'1

Yh (n) With zen)

= 6(n),

+ 0.08 = O.

= Cl 5

2"

+ C2 5 .

the initial conditions are

= 1, = o => y(l) = 0.6. Hence,cl + C2 = 1 and 1 2 -Cl +- = 0.6 => Cl = -1,c2

yeO) y(l) - 0.6y(0)

5

5

Therefore h( n)

=

=3.

[-(~)" +2(~)"] u(n)

The step response is

sen)

"

=

Lh(n-k),n~O e=O

= ~ [2(~)"-· - (~)"-.] {O.~2 [(r+l - 1] - O.~6 [(

= (b)

r+l -I])

yen) - 0.7y(n - 1) + O.ly(n - 2) = 2z(n) - zen - 2).

The characteristic equation is A2

-

0.7A + 0.1

= O.

;\ = ~, ~ Hence, With zen)

u(n)

= 6(n), we have yeO) y(l) - 0.7y(0) Hence,Cl

1

+ C2 1

= 2, = o => y(l) = 1.4. = 2 and = 1.4 = -57

2C1 + 5 2 => Cl + SC2 = These equations yield 10 4 Cl -,C2 = --.

=3

3

hen)

30

14

5

The step response is n

sen)

= L

hen - k),

k=O

=

10

t(!t- k - ~ t(!t- k

3 k=O 2

= =

3 k=O 5

10(!t~2k_~(!)n~5k LJ

32

35

k=O

LJ

k=O

10 (! n (2n + 1 _ l)u(n) _ !(! n (5n+1 _ l)u(n) 3 2 3 5

2.31 hen)

=

yen)

=

z(O)h(O) 1 2z(O)

{d,~,~, :6}

h'

2, 2.5, 3,3, 3, 2,1,

o}

= yeO) => z(O) = 1 3

+ z(l) = y(l) => z(l) = 2

By continuing this process, we obtain 3 3 7 3

z(n)= { 1'2'2'4'2' ...

}

2.32 (a) h(n) (b)

=hI(n) * [h

2 (n)

ha(n) h2 (n) - ha(n)

- ha(n)

* h4 (n)]

* h 4 (n) = * h4 (n) = hI(n)

Hence hen)

(n-1)u(n-2) 2u(n) - 6(n) 1 1 1 2 6(n) + 46(n - 1) + 26(n ~ 2)

= = [~6(n) + ~6(n 5 = '12 6(n) + 46(n -

1) +

~6(n -

2)J

5 1) + 26(n - 2) + 2u(n - 3)

(c)

?,

,,(n)

=

{I, 0, 3, 0, -4}

yen)

=

25 13 } I 5 { 2'4'f'4'2,5,2,O,o, ... 31

* [2u(n) - 6(n)]

2.33 First, we determine

= =

sen) sen)

u(n)*h(n) 00

L

u(k)h(n - k)

k=O n

=

Lh(n - k) k=O 00

= = For zen)

= u(n + 5) -

La n - k k=O

an +1 - 1 ,n a-I

~

0

u(n - 10), we have the response

a n +6 _ 1 an - 9 - 1 u(n+5)u(n-IO) a-I a-I

s(n+5)-s(n-l0)= From figure P2. 33,

yen)

=

Hence, yen)

=

zen) * hen) - zen) * hen - 2) a n +6 _ 1 an - 9 - I a-I u(n+5)- a-I u(n-IO)

an

+4 _

1

a-I u(n + 3) +

-

a n - ll -

1

a-I

u(n - 12)

2.34 hen) sen)

= =

[u(n) - u(n - M)] 1M 00

L

u(k)h(n - k)

1=-00

=

{til 17'

n

~h(n-k) =

n h(n)

= ay(n - 1) + bz(n)

=

banu(n)

=

--=1

(Xl

Lh(n)

"=0 =>b

b I-a

= 1- a. 37

y(n)

x(n)

+

z-I -1

z-I -2

z-I

-3

,

Figure 2.5:

\

(a) n

sen)

8(00)

:>6 (c) b

=1 -

=

Eh(n - k)

«=0

[1- an+!J

=

b

= =

--=1 I-a 1- a.

1- a

u(n)

6

a in both cases.

2.46 (a) yen) yen) - O.8y(n - 1) The characteristic equation is A - 0.8 A

Yh(n)

= = = =

O.8y(n - 1) + 2r(n)

2r(n) + 3r(n - 1) 0 0.8.

= c(O.8t 38

+ 3x(n -

1)

Let us first consider the response of the sytem

yen) - O.8y(n - 1)

= zen)

to x(n) = 6(n). Since y(O) = 1, it folows that c = 1. Then, the impulse response of the original system is

hen) = 2(0.8)"u(n) + 3(0.8)"-lu(n - 1) = 2c5(n) + 4.6(0.8)"-lu(n - 1) (b) The inverse system is characterized by the difference equation 1

zen) = -1.5z(n - 1) + 2y(n) - O.4y(n - 1) Refer to fig 2.6

y(n)

0.5

z -1 -0.4

-1.5 Figure 2.6:

2.47

=

y(n) 0.9y(n - 1) + x(n) + 2x(n - 1) + 3x(n - 2) (a)For x(n) 6(n), we have

=

y(l) y(2) y(3) y(4)

= = = = =

y(5)

=

yeO)

1, 2.9, 5.61, 5.049, 4.544, 4.090, ...

39

x(n)

(b)

= = 5(2) = 8(3) =

y(O) = 1, yeO) + y(l) = 3.91 y( 0) + y( 1) + y( 2) = 9.51 y(0) + y(l) + y(2) + y(3) = 14.56

s(4)

Ly(n) = 19.10

s(O) s(l)

4

=

0 IS

= Ly(n) = 23.19

5(5)

0

(c) hen)

= =

+ 2(0.9)"-lu(n - 1) + 3(0.9)"-2u (n c5(n) + 2.9c5(n - 1) + 5.61(0.9)"-2u (n - 2) (0.9tu(n)

2)

2.48 (a) 1

1

= 3z(n) + 3z(n - 3) + yen - 1) for zen) = c5(n), we have yen)

hen)

= U,~,~,~,~,~,~,··-}

I

(b)

= 21 y( n - 1) + 8"1 y( n - 2) + 21z (n = c5(n), and = y( -2) = 0, we obtain

yen) with zen)

y(-I)

=

hen)

2)

{O , 0'2'! !4' ~16' !8' ~ ~ ~ } 128' 256' 1024' ...

(c)

yen) with zen)

y( -1) hen)

= 1.4y(n - 1) - OA8y(n = c5(n), and = y( -2) = 0, we obtain

=

2) + zen)

{I, 104, 1048, lA,1.2496,1.0774,O.9086, ...}

(d) All three systems are HR.

(e) yen) The characteristic equation is ~2 _ lA~

+ 0.48 ~

Yh(n)

=

1.4y(n - 1) - OA8y(n - 2) + z(n)

=

0 Hence

=

0.8,0.6. and

=

cI(0.8)" 40

+ c2(O.6)"For

z(n)

=6(n). We have,

\

Cl 0.8Cl

= = = = =

+ C2

+ 0. 6C 2 ~ cl

C2

h(n)

1 and 1.4 4,

-3. Therefore [4(0.8)" - 3(0.6)"] u(n)

2.49 (a)

hI (n) h 2(n) h 3 (n)

= = =

+ c 1 6(n -

co6(n) b26(n)

1) + c26(n - 2)

1) + b0 6(n - 2) + aoa2)6(n - 1) + alQ26(n -

+ b16(n -

ao6(n) +

(al

2)

(b) The only question is whether ?

h3 (n)

= = = = =

Let ao a1

+ a2CO a2 a l

C2

-a2 + a2 CO ~

cOa22 -

Cl02

h 2(n)

Cl

+ C2

= hl(n)

Co, Cl, C2.

Hence

0 0

For Co :/; 0, the quadratic has a real solution if and only if

2.50 (a)

y(n) For y(n) -

't

2y (n -

1)

1

=

2y (n -

=

6( n), the solution is

1) + .r(n)

+ z(n \

h(n)

2.51 (a)

convolution: YI(n) ,correlation:

'Y1 (n)

= 41

h.

3, 7. 7, 7,6,4}

{I, 3,7, 7,!, 6, 4}

1)

(b)

•

Note that Y2(n)

convolution: y,(n)

=

{4,?,~, -2,4, -6, -~, -2}

correlation: 'l'1(n)

=

{4,?,~, -2,4, -6, -~, -2}

=1'2(n), because h2(-n) = h2(n) (c) = { 1,11,20,30,20, 11,4}

convolution: 1I3(n)

=

correlation: 1'l(n)

{1,4, 10, 10,25,24,16}

(c)

= { ~,4, 10, 20,25,24,16}

convolution: y.. (n)

Note that h 3(-n) hence, 1'3( n) and h.. ( -n) ~

{ 4,11, 20, ~O, 20, 11,4}

= = = = =

correlation: 1'.. (n)

1'.. (n)

+ 3), y.. (n + 3) h 3(n + 3), Y3(n + 3) h.. (n

, \

2.52 Obviously, the length of h(n) is 2, i.e.

=

h(n)

= = =

ho

3h o + hi ~ ho

{h o, hd

1 4 1, hi

=1

2.53 N

(2.5.6)

y(n)

= - L: Qky(n -

M

k)

+

L bkz(n -

k)

k=O

k=1

N

(2.5.9)

w(n)

= - L: Ql;w(n -

k)

+ z(n)

k)

(A)

1;=1

M

(2.5.10)

y(n)

= L: bl:w(n -

k)

1:=0

From (2.5.9) we obtain N

z(n)

= w(n) + L: Qkw(n k=1

By substituting (2.5.10) for y(n) and (A) into (2.5.6), we obtain L.B.S 42

= R.H.S.

2.54

.... yen) - 4y(n - 1) + 4y(n - 2) The characteristic equation is

=

zen) - zen - 1)

4A + 4

= = =

0

A2

-

A Yh(n) The particular solution is

Yp(n)

2,2. Hence, c 1 2" + C2 n2"

= k(-I)"u(n).

Substituting this solution into the difference equation, we obtain

k(-1)"u(n) - 4k(-I)"-lu(n -1) For n

= 2, k(1 + 4 + 4) = 2 ~ k = i.

+ 4k(-I)"-2 u(n - 2) = (-l)"u(n) The total solution is

[CI2" + C2n2" + ~(-1)"] u(n) From the initial condtions, we obtain yeO) = 1, y( 1) = 2. Then, y(n) =

2

+-9

Cl

~

2Cl

Cl 2 9

+ 2C2 -

~ C2

= = = =

1

7

9' 2 1

3'

2.55 From problem 2.54,

\\"ith yeO)

= 1, y(l) = 3, we have Cl

= =

1 3 1 2

Thus hen)

=

[2" + ~n2"] u(n)

2.56 zen)

= = =

zen) * 6(n) zen) * [u(n) - u(n - 1)] [z(n)-z(n-l)]*u(n) 00

=

E

[z(k) - z(k - 1)] u(n - k)

A:=-oo

43

(-I)"-lu(n - 1)

2.57 Let h( n) be the impulse response of the system k

s(k)

L

=

h(m)

m=-oo

=> h(k) = s(k) - s(k - 1) 00

y(n)

L

=

h(k)z(n - k)

k=-oo 00

L

=

[s(k) - s(k - 1)] z(n - k)

k=-oo

2.58 z( n)

= { 1, 0,

y(n)

no - N. ~ n otherwise

-N

1,

~ no + N

~ n ~

= { 0, otherwise

N

00

1'zz(l)

=L

z(n)z(n -I)

n=-oo

The range of non-zero values of 1'zz (I) is determined by

no - N ~ n ~ no

+N

no - N ~ n - I ~ no

+N

which implies

-2N

~

I

~

2N

For a given shift I, the number of terms in the summation for which both z(n) and z(n - I) are non-zero is 2N + 1 - III, and the value of each term is 1. Hence, zz (I)

l'

For

1'zy (I)

={ 2N + 1 -

III, -2N ~ I S 2N

0,

otherwise

we have

1'zy

2N + 1 -1/- nol, ( I) -_ { 0,

no - 2N ~ I ~ no otherwise

2.59 (a) 00

1'zz(l)

=

L

z(n)z(n -I)

n=-oo

1'zz( -3) 1'zz(-2)

= z(0)z(3) = 1 = z(0)z(2) + z(1)z(3) 44

=3

+ 2N

"Y:u(-l)

x(O)x(l)

::

+ x(1)x(2) +

x(2)x(3)

=5

3

"Yzz(O)

L x2 (n)::7

::

n=O

=

Also "Yzz (-I) Therefore 1'zz(l)

"Yzz(l)

{1,3,5, p,3,l}

::

(b) 00

= n=-oo L y(n)y(n -

"Yyy (I)

I)

We obtain

=

"Yyy (I)

{1,3,5,7,5,3,1}

=

we observe that y(n) x( -n + 3), which is equivalent to reversing the sequence x(n). This has not changed the autocorrelation sequence.

2.60 00

1'zz(l)::

E

x(n)x(n -I)

n=-oo

1'zz(O) Therefore, the normalized autocorrelation is

+ 1-

::

{2N 0,

::

2N

::

2N

::

0, otherwise

kd

+ "Y2 s (n

III,

-2N ~ I ~ 2N otherwise

+1 1

+ 1 (2N + 1 - 1/1), -2N

~ I ~ 2N

2.61 (a) 00

"Yzr(l)

::

L

x(n)x(n - I)

n=-oo 00

=

L

[5(n)

+ "Yl s (n -

- k 2)]·

n=-oo

[5(n -I) + 1'1s(n -1- kd + 1'25(n -1- k2)] ::

(1 +1'? + 1'~)1',,(/) +1'1 [1',,(1 + kd + 1',,(1- kd] + "Y2 [-y" (1 + k2) + 1'" (I - k 2)] +"Yl1'2 (-y,,(1

+ k1 - k 2) + 1',,(1 + k2 - kdl

=

(b) "Yrr(l) has peaks at 1 0, ±k1 , ±k2 and ±(k1 + k 2 ). Suppose that k 1 < k 2 . Then, we can determine "Yl and k 1 • The problem is to determine 1'2 and k'2 from the other peaks. (c) If 12 0, the peaks occur at I 0 and I ±k1 . Then, it is easy to obtain "Yl and *'1'

=

=

=

45

2.62 (a) (b) variance (b) Delay D

= 0.01. Refer to fig 2.7. = 20. Refer to fig 2.7. 1.5

0.5

~ 0.5

~

lr

~I

0

A I I

0

'-0.5 -0.5 -1

-1

0

50

100 ->n

150

-1.5 0

200

50

100 ->n

150

200

15.----------, 10

~ ~

5

A I

0

I

~ ,."lMrUt.r-" ..... ~..~---_4

-------J 20

_5l..-......- -......

-20

o

->1

Figure 2.7: variance

= =

(c) variance 0.1. Delay D (d) Variance 1. delay D

= 0.01

= 20. Refer to fig 2.8. = 20. Refer to fig 2.9. 1.5.---------..,

0.5 ~

I:i'

0

A I I

-0.5 -1 -1 UL

o

..J

50

100 ->n

150

-1.5"----------..J o 50 100 150 200 ->n

200

20---------.. . 15

~10 I:!

i

5

_5L.---------...I -20 0 20 Figure 2.8: variance

= 0.1

(e) x(n) = {-I, -1, -I, + 1, + 1, + 1, + 1, -1, + 1, -1, + 1, + 1, -1, -1, + I}. Refer to fig 2.10. (f) Refer to fig 2.11.

46

3

2

0.5

..... 1

'C ";" A I I

c

~O I

0

I

-0.5

-1

-2 -1

0

50

100 ->n

150

-3

0

200

50

100 ->n

150

200

15

20

Figure 2.9: variance = 1

2.63 (a) Refer to fig 2.12. (b) Refer to fig 2.13. (c) Refer to fig 2.14. (d) The step responses in fig 2.13 and fig 2.14 are similar except for the steady state value after n=20.

2.64 Refer to fig 2.15'.

47

1\

0.5

0.5

C' It

C'

1;:

0

A

A

I

1-0.5

I

-0.5 -1

-1

0

50

100 ->n

150

-1.5 0

200

50

100 --> n

150

200

20 15

~ t! A-

10 5

I I

-5 -10

-20

0 ->n

20

Figure 2.10:

1.5.....- - - - - - - - - - ,

g lC

AI I

50

100 ->n

150

50

200

2O--po--------.., 15

Ii:

10

t! A-

I

-20

o

20

->n

Figure 2.11:

48

100 ->n

150

200

,,"ptA. response h(n) ollne system

05

c

~ 1\ I

I

o

20

25 ->n

30

50

Figure 2.12:

U~--r----"T--or--""'--"--"T"'""---r--r---r---.,

1.5 1.4

1.3

...

'E"12 A

I 1.1

0.8 0.8

10

15

20

25 ->n

Figure 2.13:

49

30

50

1.1~--r--"""'-""""""-T""""-~---r--.,---r--....,..--...

1.5 1.•

1.3

...

'C" 1.2 It.

I 1.1

0.8 0.8

5

10

15

20

25

->"

30

Figure 2.14:

__- __--.---_---.

7~--.----r--...--~--...,.....-

8 5

'E" 3

:r A

I

2

o -1

-2 L-.._..L-_........_ _..I..-_...._ - - I......_ o 10 20 30 40 50

->n

Figure 2.15:

50

......_--A._ _"'--_~_.....J

10

70

10

10

100

Chapter 3

3.1 (a)

X(z)

= Lx(n)z-n

=

3z 5 + 6 + Z-l - 4z- 2 ROC: 0 < IZI
/al

n=O

00

and

1

=

(1

n=O

Hence, X(z)

- G'z

1 l-az- 1

=

1

ROC:

1 -1)2

Izi > -II a

1 l_lz-l

--~+---:-G

2-(a+~)z-1

=

(1 -

az- 1 )(1

ROC:

- ~Z-l)

(e)

X(z)

(0) X(z)

(e) 00

X(z)

=L

nancoswonz- n

n=O

=

L 00

nan

[eilJ/On + e-ilJ/on] 2

n=O

52

z-n

Izi >

=

1 [

'2

aeiwo Z-1 (1 - aeiwo z- 1 )2

+

ae- jwo Z-1 ] (1 - ae- iwo z- 1 f2

[az- + (az- )3] sinwo - 2a 2 z- 2 (1 - 2acoswoz- 1 + a2 z- 2 )2 1

=

1

Izi > a

(f) 00

X(Z)

=

A

2: rncos(won + 4»Z-n n=O

= A 2: r n=O = -A[ 00

2

=

[ejWOnej~ + e-iWOne-iiP]

n

2

ej , . 1 - re JWo z-1

-

4>

)Z-I]

+ r 2 z- 2

1 - 2rcoswoz- 1

_ n

e-j~] . 1 - re- Jwo z- 1

+

A [eos4> - reos( WQ

z

Izi > r '

(g)

X(z) 1 2: n(3")n-l z-l 00

But

n=1

f:

n2(

~ t- 1

n=1

z-n

Therefore, X (z)

(h)

X(z)

f:(~tz-n = n=O

= =

- f: (~tz-n n=10

1

(~)lOz-10

l - 12 z- 1

1- 12 z- 1

1_(~z-I)10

1 - 1 z -1

1

Izl> '2

2

The pole-zero patterns are as follows: (a) Double pole at z 1 and a zero at z O. (b) Poles at z = a and z =~. Zeros at z = 0 and z !(a+ ~). (e) Pole at z = -~ and zero at z = o. . (d) Double poles at z = ae JWo and z = ae- JWo and zeros at z = 0, z = ±a. (e) Double poles at z = aeiwo and z = ae- jwo and zeros are obtained by solving the quadratic

=

=

acoswoz 2

=

-

=

2a 2 z + a3coswQ

= O.

(f) Poles at z re jwo and: = ae- jwo and zeros at z = 0, and z = rcos(wo - ¢J)/eos¢. (g) Triple pole at : = ~ and zeros at z 0 and z = ~. Hence there is a pole-zero cancellation so

=

53

i

that in reality there is only a double pole at z = and a zero at z = O. (h) X(z) has a pole of order 9 at z = O. For nine zeros which we find from the roots of

1 - ( ~ z-1 )10

= =

2

Or,eqUivalently,(~)10-z10

0

1 .l.l2

= '2 f

Bence, Zn Note the pole-zero cancellation at z

0

10

,n=I,2, ... ,k.

= !.

3.3 (a) 0

00

L

n=O

n=-oo

= 1:(~)"z-n +

(~)"Z-n-l

00

1

,,( l)n n 1 _ 1 z -1 + ~ '2 z - 1 3 n=O 1 1 1 1 + --1- -I, 1- 3z1 - '2z

= =

5

6 (1- 1z-1)(1- ~z)

= The ROC is (b)

i

< Izi < 2.

=

E(~)"z-n- E2 n

n=O

n=O 1 1 - 2z- 1 '

1 1- 1 z -1 3

=

Z-

5 -1

-3 Z

= The ROC is

Izi > 2.

(c) 00

X3(Z)

=

L

Z1(n

+ 4)z-n

n=-oo

The ROC is

i

< Izi < 2.

(d) 00

X 4(Z)

= 1: n=-oo 54

Z1(-n)Z-n

n

m=-oo

=

Xdz- 1 ) 5

= The ROC is ~

6

(1- ~z)(l- ~z-l)

< Izi < 3.

3.4 (a) 00

n(_I)n z -n = E n=O

X(z)

f:(

=

-z.!!dz

=

-Z:' [1 +1Z_1]

=

-(I+z-1)2,lzl> 1

_1)n z-n

n=O

z-1

(b)

X(z)

=

00

E n z-n 2

n=O

=

d2

00

z2 dz 2 Ez-n n=O

= =

=

Z2::2

[1 _1Z_1]

Z-1 2z- 1 - (1-z- 1 )2 +---(1-z- 1 )3 z-l(1 + z-l) (l-z- 1 )3 ,lzl> 1

(c) -1

X(Z)

=

E

_nanz- n

n=-oo

=

d

-1

- z - ~ a(n)Z-n dz L..J n=-oo

(d) 00

X(Z)

= L(-I)nCoSinz-n n=O

55

From formula (9) in table 3.3 with a

=

-1,

=

X(z)

1 + 2z- 1coSi 1+ 1 z- 1

=

1 + z-

t+ z-

+ Z-2

2'

ROC:

Izi > 1

(e) 00

X(z)

= L(-l)n z-n n=O 1

=

Izl> 1

1 + z-1 '

(f)

.1'(n)

=

h,O, -1,0,1, -I}

X(z)

=

1 - z-2

+ Z-4

z-5,

_

z;iO

3.5 Right-sided sequence :.1',.(n)

X,.(z)

= =

0, n

< no

-1

00

L

.1',.(n)z-n

+

n=no

L .1',.(n)z-n n=O

The term L~~no .1',.(n)z-n converges for all z except z = 00. The term L~=o .1',.(n)z-n converges for all Izl > ro where some ro. Hence X,.(z) converges for ro < 1=1 < 00 when no < 0 and Izi > ro for no > 0 Left-sided sequence :.1',(n)

=

0, n

> no no

o

XI(Z)

= L

.1',(n)z-n +

n=-oo

L .1',(n)z-n n=1

The first term converges for some Izi < r,. The second term converges for all z, except z Hence, X,(z) converges for 0 < Izi < r, when no > 0, and for Izi < r, when no < O.

=

Finite-Duration Two-sided sequence :.1'(n)

0, n

> noandn
n1

no

=

X(z)

L

.1'(n)z-n

n=nl n=no

-1

=

L

.1'(n)z-n

n=nl

=

The first term converges everywhere except z The second term converges everywhere except z

+

L

r(n)z-n

n=O

00.

= O. Therefore, X(z) converges for 0 < Izi
y(n) - y(n - 1) Hence,Y(z) - Y(Z)Z-l Y(Z)

= = =

x(n)

X(Z) X(z) 1 - z-l

3.7

00

=

X 1 (Z)

=

L(~tz-n + n=O

1

-1 L (~)-nZ-n n=-oo 1

+---1 1-~z

l-lz-l

5

=

6

(1 -lZ-l)(1 - ~Z)

= f:(~)nZ-n

X 2 (Z)

n=O

= Then,Y(z)

=

Hence,y(n)

=

1 1 1 _1'22

3.19 (a)

rnsinwonu(n), 0 1

3.31

=

=

From the definition of the Fibonacci sequence, y(n) y(n - 1) + y(n - 2), y(O) 1. This is equivalent to a system described by the difference equation y(n) y(n - 1) + y(n - 2) + z(n), where z(n) = 6(n) and y(n) = 0, n < O. The z-transform of this difference equation is Y(z) = z-ly(z) + z- 2 y(z) X(z) Hence, for X(z) 1, we have

=

=

Y(z)

=

Y(z)

=

1

69

=

where A Hence, yen)

=

.;5+1 v'5-1 I-v'5 2v'5' B = 2V5 = - 2.;5

=

.;5 + 1 (.;5 + 1 )"u(n) _ 1 - v'5 (1 - v'5 )"u(n) 2v'5 2 2V5 2

=

_1 [( 1 + .;5 )"+1 _ (1 - V5 )"+1] u(n) .;5 2 2

3.32 (a)

(a)

=

Hence, Y+(z)

= Therefore, y( n )

=

1Z-

1

1 _

04

4

1 + 12 z - 1 - 14 z -2 0.154 0.404 1 1-0.31z1+0.81z- 1 [0.154(0.31)" - 0.404(0.81)"] u(n)

(b)

Therefore, yen)

= = = =

1.5 - 0.5z- 1 1 - 1.5z- 1 + 0.5z- 2 2 0.5 l-z- 1 1-0.5z- 1 [2 - 0.5(0.5 )"] u( n) [2 - (0.5)"+1] u(n)

(c)

1

= =

1 - 1 z -1 3

1.5-i z -

1

(1 - ~z-I)(1 - 0.5z- 1 )

2

!

=

1-

2

0.5z- 1

1 - ~z-1

= [~(O.5). - 2( ~t] u(n)

Hence, yen)

(d)

=

= = Hence, yen)

=

1 1 - z-1 ~

4

_ 1 z-

1

1

(1 - z-I)(I- 1z-2) 4

1-

-3

7

3+T+24 z-1

1- 1 Z -1 2

1 + 1z 2

[~~(!)" + !-(_!)"] u(n) 3 8 2 24 2 70

1

3.33 (a)

Y(z)

[1 -

0.2z- 1]

=

Y(z) X(z)

=

[1 -

0.3z- 1 - 0.02z- 2 ] (1- 0.lz- 1 )(1 - 0.2z- 1 ) 1 - 0.2z- 1 1 - 0.lz- 1 X(z)

= (b)

= =

Y(z) Y(z) X(z)

X(z)[1-0.1z- 1 ] 1 - 0.lz- 1

Therefore, (a) and (b) are equivalent systems.

3.34 X(z) ~

zl(n) orz2(n)

1

= = =

1 - az- 1

a"1'(n) -a"1'( -n - 1)

Both xl(n) and x2(n) have the same autocorrelation sequence. Another sequence is obtained from X(z-l) = _1_ l-az X(Z-I)

Hence x3(n)

1

= =

1- az I-

1 I - 1G Z -1

1 6(n) - (-)"1'(n) a

=

\\'e observe that z3(n) has the same autocorrelation as zl(n) and z2(n)

3.35 H(z)

= =

X(z) Y(z)

= = =

E

3" z-"

"=-1

1-

f:( ~)" z-" n=O

-1 3 1 z-

1

+ 1- 5z2 l'

2 ROC: - < Iz I < 3 5

1 1-

Z-1

H(z)X(z) _13 z-1 5

(1- z-I)(I- 3z- 1 )(I-lz- 1 )' 13

=

+

3

1 - z-

2

'2

'6 1

1 - 3z -

71

1

3 1- 1 z- 1 5

ROC: 1
3; ROC 2 : Izi < 3; ROC 3 : < Izi < 2; ROC 4 : 2 < Izi < 3;

l

3.50 z( n) is causal. (a) 00

X(z)

=

limz_ooX(z)

= 84

L z(n)z-n n=O

z(O)

= ~;=t;: ::} limz_ooX(z) = oo::} x(n) is not causal. X() (l-!z-')' ::} I'zmz_ooX(z) = 1 Hence X(z) can • Z = 1- !Z-l

(b)(i) X(z)

(11")

sequence, (iii) X(z) sequence.

= (z-P' (Z-2)3

=> limz_ooX(z)

,

.

be assocIated wlh a causal

.

= O.

Hence X(z) can be associated wih a causal

3.51

= a"u(n) ::} H 1(z) = l-~z-i' lal < 1. This system However when h 2(n) = a"u(n + 3) ::} H 2(z) = t_-o.3zz~i the system is stable

The answer is no. For the given system h 1(n) is causal and stable. but is not causal.

3.52 Initial value theorem for anticausal signals: If x(n) is anticausal, then x(O) = limz_oX(z) Proof: X (z) = 2:~=-oo x( n )z-" = x(O) + x( -l)z + x( -2)z2 + ... Then limz_oX (z) x(O)

=

3.53 1

s(n) = (3)"

_ 2 u(n+2)

(a) h(n)

=

s( n) - s( n - 1)

=

(~)"-2u(n + 2) - (~)"-3u(n + 1)

=

1 34 6(n+2)-546(n+ 1)-18(3)"u(n)

3

3

2

= 81 z -

H(z)

=

54 z

+

-18 1

1- 3z

_ I

81z(Z-1 ) 1 - 13 z - 1

=

i.

H(z) has zeros at z 0,1 and a pole at z = (b) h(n) 816(n + 2) - 546(n + 1) - 18(~)"u(n) (c) The system is not causal, but it is stable since the pole is inside the unit circle.

=

3.54 (a)

.1'(n)

= -1

21rj

i e

z"-1 dz 1- !Z-1

= _1_1~dz 21rj

for n

~

O,x(n)

Ie z -

= (~)" 2 85

~

for n < 0, .1'( -1)

=

.1'( -2)

= =

1 1 dz 21rj!e z(z - ~) 1 1 - - 1 Iz=o + -Iz=! z - 2' z _1_

1

_1_

1

!)

21rj!e z2(z -

=

(_1-

dd z

z - 21

By continuing this process, we find that x(n)

=0

dz

1:=0 + ~Iz=t z

)

=0

= 0 for n < O.

(b)

For

n~

1

1

X(z)

=

_I,l z lO = J~ Ae- o'e-;2.F'dt = =

IXo(F)1

A e- CO+;20Fj,] DO -a - j27rF 0 A a + j27rF A 2 + (21r F)2

= Ja

90

2

~ )

1 21r F -tan- ( - ) a

= Refer to fig 4.2 (b)

A = 2,

a =4

2r----------..---------

0.5,...------,,.-.----------.

-x

0.4

1

u.

ca

::0.3 :..

'5 0

s

CD

~

~0.2

(/)

ca

~

Q.

0.1

0 -10

-5

0 --> F

-2 -10

10

5

-1

-5

Figure 4.2:

:=

=

= = =

J~ A.··.-j2.F·dt + J~ A.-··.-j2.F·dt A

A . F + )21r

. F+ a a - )21r 2aA a2

+ 21rF 2

2aA a 2 + (21r F)2 0

Refer to fig 4.3

4.3 (a) Refer to fig 4.4.

Z(t)={

I-t;l, Itl~T

0,

otherwise

Alternatively, we may find the fourier transform of

y(t) = x'(t) = {

I' T'

91

-T

F

5

10

A.2, •• 6 0.7

0.6

0.5

0.4

rE

I 0.3 0.2

0.1

00

5

10

15

20

->F

Figure 4.3: j

\

1(t)

IX(F)I

Figure 4.4:

92

Then,

= f~T y(t)e- i2 • F ' dt

Y(F)

= fO

!e- i2 • Ft dt

+ fT (.:.!. )e- i2 • F 'dt

TOT 2sin 2 7f'FT j7f'FT 1 j27f'F Y (F) -T

= =

and X(F)

r( Sin1fFT .. Fr) r( Sin1fFT .. Fr)

= IX(F)I LXo(F)

= =

2

2

0

(b)

Ck

f

= -1

Tp

T

,/2

= .2.. [fO Tp

=

x(t)e-j2dt/T'dt

-T,/2

fT

+!

)e-j2dt/T'dt + (1 - ! )e- j2dt/T'dt] TOT

(l

-T

.!.... [Sin1flcT /Tp ] Tp

2

1rlcT/Tp

(c) From (a) and (b), we have

Ck

= t;Xo(f;)

4.4 (a)

x(n) = {- .. , 1, 0, 1, 2, N

=

C.

= 6

r'

2, 1, 0, 1, ...}

6 1 5

L z(n)e-

. J2 ,.,kn/6

n=O

-"... + e -1'].... • +e = [3 + 2e -72... 3

+ 2e -'160".]

= 6'1 [ 3 + 4COST 7r/c + 2cos27f'/c] 3 Hence, Co

=

9 6,CI

4 1 = 6,C2 = O,C3 = 6,C4 = O,cs = 64

(b)

1

Pt

5

= 6' L

Iz(n)1

2

n=O

= !6 (3 2 + 2 2 + 12 + 0 2 + 12 + 22 ) 93

19 16

=

-

s

= L Ic(n)1 2

PJ

n=O

Thus, P1

=

[ ( _9 )2

= = =

PJ

16 19 16

4 )2 + 02 + (_ 1 )2 + 02 + (_ 4 )2) ] + (_

6

6

6

19 16

-

4.5 x(n)

1 = 2 + 2cos1rn/4 + cos1rn/2 + 2cos31rn/4, => N = 8

(a) 1

Ck

= 8L 7

"

x(n)e- p

"kn/4

n=O

x(n) Hence, Co

3 3 3 3} = { -II 2 + -../2 1 2 - -../2 -I 2 - -../2 1 2 + -../2 2 4 " 4 ' 2' 4 4 I

=

2,Cl

I'

= = l,c2 = C6 = 21 = = 4I = 0 -,C3

C7

Cs

-,C4

(b) 7

P

= L

lc(i)1

2

i=O

53 8

= 4.6 (a)

x(n)

Ct

= =

. 1r( n - 2) 4sIn 3 . 21r(n - 2) 4sIn 6

1 5 = 6L

"

x(n)e-2Jdn/6

n=O 5

. 2",( n LJ sIn 6 = 64 "'"'

2)

e

-2jdn/6

n=O

=

~

[_e- j2 11'k/3 _ e- j d / 3

94

+ e-jd/3 + e- j2d / 3 ]

1\

_ j2) [Sin 21fk + sin 1fk] e= _1_( vJ 6 3 j2 = O1C 1- - -}'2e- 1(/3 ,C2 -- C3-- C4-- 0,c5 -- c·1 j2d / 3

Co Icd = Icsl = Hence,

and

LC1 Lcs

Lco

= = =

2:

ICol = IC21 = 1c31 = IC41 = 0 7r

27r

S7r

2

3

6

7r+---=-S7r 6 LC2

=

LC3

=

=0

LC4

(b)

.1'(n)

where

Clk

=

21fn cos--

=

Clk

3

. 27rn

+ san-- => S

+ C2k

is the DTFS coefficients of cos 2;n and

27rn cos - 3

is the DTFS coefficients of sin 2~n. But

-1;--" ) =-21 (~ e -----r- + e

Hence, Clk

C2k

N = IS

={

~, 0,

=

k

S,10 otherwise

Similarly,

27rn sin-S

1 J.12 = 2j(e 5

e

-

-12n .&

).

Hence,

k=3 k 12

=

otherwise Therefore, 1 2J'

k=3 k=S k = 10 k = 12

1

c. = ca + C2.1 l'

2' -1

2)'

0,

otherwise

21(n· 211'n l ' 161(n - '2sznl5' 1 · 411'n Hence, N -- IS . F0 11owmg ' t he same me th 0 d .1' ( n ) = cos-szn= '2szn-rg3 S as in (b) above, we find that k = 2,7 k 8,13 otherwise (c)

=

(d)

N Ck

= S . 1E -,:I.... = -S n=O .1'(n)e 5

= =

1 [ -,2".

-,....

-I'..

- e .& + 2e 5 - 2e 5 S 27rk - 2szn(-) . 41fk] -2j [ -sin(-) 5 5 5

95

-

e

-,.... ] 5

Therefore,

Co

= 0,

Cl

=

. -2" ) + 2sln( . -4"]) -2j [-sm( 5 5 5

C2

=

. -2,,]) -2j [. sln( -4" ) - 2szn( 5 5 5

C3

= =

C4

-C2 -Cl

(e)

Therefore,

N

= 6

Ck

1 = -6

Co

Cl C2

=

L z(n)e -I~r'" 1 [ -J... e - 1 + 2e

=

d )6I [ 1 + 4cos( '3

=

1 2 2

6

n=O

3

6

-I~ .. II

-

3

e

-

-I.'" 3

C4

Cs

=

-5 6

j

\

0

2 3

N

= 5

Ct

=

1 -,~ x(n)e s 5 n=O

=

-1 [1 + e ~] 3

L

..

4

= Therefore,

=1

Ck

3

2d )] 2cos( 3

(f)

(g) N

+ 2e -,s... ]

= -3 = 0 = =

C3

5

Co

=

5 2 1rk _,rll -cos(-)e 3 5 5 2

5 2

=

-cos( -)e 5 5

3

=

2

21r

C2

-cos( -)e 5 5

-I~"

=

2

31r

-cos( -)e 5 5

-IJr

C3 C4

=

-cos( -)e 5 5

2

= z(O) = lorco = 1

(h) N

1r:.L!.

Cl

= 2 96

4",

s

3

-I." 3

II

1

2L

=

Ck

1

.

x(n)e-J'rnk

n=O

=>

!(1 -

= =

Co

2

e- j d )

=

0,C1

1

4.7 (a) 7

z(n)

= =

Note that if Ck

~

LCt e ~

e',

~

7

,2 .. ""

1:=0

L n=O

•

p =-n

0,

p;f. -n

1 [~

-,2... ] 1 [~ -,.... ] -2 e • +e' +2j e • -e • 4h(n + 1) + 4h(n - 1) - 4jh(n + 3) + 4jh(n - 3), -3

=

We have x(n)

f

8,

=

Since Ck

then

,2"(,,+,,).

~

=

Le • e •

•

1=0

7 ~

Z:h,.,.

~ n ~

(b)

Co x(n)

= O,Cl

=

v'3 V3 V3 h = 2,C2 = 2,C3 = O,C4 = -2'cS = -2,C6 = = 0 C7

7 ~

LCt e

12i"·

1:=0

= =

] -.j3 [~ e 4 +e ~ 4 -e ~ 4 -e I~"" 4

2

r;: [1Tn 1Tn ] ~ v3 sin"2+sin4 e •

(c) 4

x(n)

=

~

L

Cke

,2..... •

k=-3

= =

2+e

~ 4

+e

1Tn 2 + 2cosT

-,.... 4

1

+ -e

~

2

2

-I....

2

1

7rn

1

+ -e

2

1

+ -e 4

31Tn

+ COST + 2cos-4-

4.8 (a)

If k N-l

L

eJ2dn/N

= 0, ±N, ±2N, ...

=

n=O

97

sum~';011

=N

~

•

1

-13 ....

+ -e • 4

5

Irk

#;

0,±N,±2N,...

N-l

L

1 - ej2d 1 - e j2d / N

= =

ej2dn/N

n=O

°

(b) Refer to fig 4.5. (c) 1e=1

1e=3

1e=2

*1) 5, 2(4)

s{

, '

'1(0)

s(4)

, (5)

'2(0) , (3)

~(2)

2

3

(I)

(0) 3 , 3 (2)

•

•

, (3)

3 s (5)

s (4) 3

3

s (5) 2

1e=4 ,

Ie =6

1e=5

(5)

5- *'5(0) 4

s 5(5)

, (4)

, (0)

4

s (1) 4

, (3)

,

4

s (2) 5

4

Figure 4.5: N-l

N-l

L 5A:(n)5;(n) = L

ei 2dn / N e-j2~in/N

n=O

n=O

=

N-l

L

ei2~(A:-i)n/N

n=O

=

=

N,k = j 0, k #; i

Therefore, the {5A: (n)} are orthogonal.

4.9 (a)

x(n)

=

u(n) - u(n - 6)

98

•

(0)

6

(1)

5

(4)

5

s

'6(5)

00

X(w)

=

x(n)e- jwn

L "=-00 5

=

e- jwn

L "=0

1 - e- j6w 1 - e- jw

= (b)

= 2"u( -n)

x(n)

0

2"e- jW "

L

=

X(w)

"=-00 00

jw L(T)n

=

m=O

2

=

2 - ejw

(c)

z(n) X(w)

1

= (4)n u(n+4)

1: (~)"e-jW" = (1: (~)me-jWm)44ei4w

=

n=-4

m=O

=

44 ej4w jw 1 - le4

(d)

x(n)

=

X(w)

= = = =

(e) x(n)

=

10'1" sinwOn, 10'1 < 1

00

Note that

00

L

\x(n)1

=

n=-oo

Suppose that Wo

= ~, so that Isinwon\ 00

L n=-oo

99

L

la\"lsinwonl

"=-00 .

=

l. 00

10'1"

= L Iz(n)ln=-oo

00.

Therefore, the fourier transform does not exist.

(f)

x(n)={ 2_(~)n, Inl~4 0,

otherwise

4

X(w)

=

L

z(n)e- jwn

n=-4

=

nt)-(~t]

=

2e j4w 1 - e- jw

.-j"n

_! [_4ei o4W + 4e- j4

=

2 2ei o4w

t&1 _

••

---:-.+) JW 1 - e-

3ei 3w

+ e- j3w _

2ei 2w

+ 2e- j2w

_

e jw

• • • + 3sm3w + 2s1n2w + smw]

[4s1n4w

(g) 00

L

=

X(w)

= =

z(n)e- jwn

n=-oo _2ei 2w _

+ ei w + 2e- j2w -2j [2sin2w + sinw] ei w

(h)

x(n)

= { A(2M + 1 -Inl), 0,

Inl ~ 1..1 Inl > M

M

X(w)

= L

z(n)e- jwn

n=-M M

=

A

L

InDe- jwn

(2M + 1 -

n=-M

=

M

(2M

+ l)A + A L(2M + 1 -

k)(e- jwk

1=1 M

= (2M + I)A + 2A L(2M + 1 -

k)coswk

k=1

4.10 (a)

zen)

=

1 -2 1f

1

x(O) 100

ft

-ft

X(w)eiwndw

+ eJwk )

+ e- jw ]

=

wo

1r 1r

=

1 . _eJwnl-wo jn -lr

=

-=-(e-Jwon _ e- Jlrn )

=

1·

In

-.!..-eiwnl lr jn Wo

= ~(eilrn In

_ eiwon)

sinnw o , n#;O

=

Hence, z(n)

.

n~

(b)

X(w)

z(n)

= cos 2(w) w + !e- iW )2 = (!ei 2 2 2W + 2 + e- i2w ) = !(ei 4 [ X(w)ei",ndw = 1.. 21r -lr = =

1 - [21r6(n + 2) + 41r6(n) + 21r6(n - 2)] 81r 1 :4 [6(n + 2) + 26(n) + 6(n - 2)]

(c)

2..1'"

=

x(n)

21r _'"

X(w)eiwndw

f.wo+!t . eJwndw 21r wo-lf-

= -1

=

~6w (Sin(n6w/2») einwo 1r

n6w/2

(d)

x(n)

7lr/ lr/ {f.lr/S 2ei wn dw + J.3 8eiwndw + /. 8eiwndw + fIr eiwndW} 21r 0 "/8 6,,/8 17"/8 3 8 = .!. [f."'/8 2C05wndw + J. "/8 c05wndw + /.7"/ coswndw + ff( 2coswndw]

=

-.!..-Re

1r

= -n1r1

lr/B

0

[ . 71rn

Sln8

. 61rn + 51n8

6,,/S

. 31rn . 1rn] Sln- - Sln8 8

4.11

101

17f(/B

z(n) - z( -n)

= 2 = U,0,-2,~,2,0,n

zo(n)

3

Then, XR(W)

= L ze(n)e- jwn n=-3

jXJ(w)

= L zo(n)e- iwn n=-3

3

= =

Now, Y(w)

y(n)

XJ(w) + XR(W)ei 2w . Therefore, F- 1 {XJ(w)} + F- 1 {XR(W)ei 2W }

(g) 00

= n=-oo L z(n)e- jwn = _2ei 2w _ ei w + ei w + 2e- i2w = -2j [2sin2w + sinw]

X(w)

(h)

_ { A(2M + 1 x (n ) 0, M

X(w)

=

L

In!), Inl ~ M Inl > M

x(n)e- jwn

n=-M

=

M

A

L

(2M

+ 1 -Inl)e- iwn

n=-M M

= (2M + l)A + A L(2M + 1 - t)(e- jwk + ei wk ) k=1

=

M

(2M

+ l)A + 2A L(2M + 1 -

k)coswk

k=1

4.12 (a)

z(n)

=

1

-2 1r

1 ft

X(w)eiwndw

-ft

z(O)

= For n f- 0, l~wO eJwndw

=

7t'

1 jvm I-wo

-:-f

)n

102

_'"

1· = -:-(e-JUIon _ e- J.1I'n) )n

~eJUlnl1l'

=

jn

1£10

= ~(ei1l'n _ eiwon) In

= -

Hence, zen)

sinnwQ n1r

,n ¢ 0

(b)

X(w)

z(n)

= cos 2 (w) Ul + !e- iUl )2 = (!ei 2 2 2W + 2 + e- j2U1 ) = !(ei 4 [ X(w)eiwndw = ~ 21r _11'

1

=

- [21r6(n 81r

=

- [6(n

1

4

+ 2) + 471'6(n) + 21r6(n -

+ 2) + 26(n) + 6(n -

2)]

2)]

(c)

z(n)

= 2~

L

= _1

J.wo+~ ~wndw .

21r

=

X(w)eiwndw

1£10-

if

~6w (sin(n6w/2») eJnUlo 1r

n6w/2

(d)

x(n)

=

~Re { 271'

{'1/8 2eJ wn dw Jo

+ /.37(/8 eJwndw + /.77(/8 ~U1ndw + Ill' 7(/8

67(/8

eiwndW}

J77(/8

= ~ [ 17(/8 2cosw"ndw + /.311'/8 coswndw + /.711'/8 coswndw + Ill' 2coswndW] 1r Jo 11'/8 611 /8 J7../8 . 771'n . 61rn . 31rn . 1rn] = -n1r1 [ szn+ szn- szn- - sm8 8 8 8 4.13 ze(n)

zo(n)

=

z(n)

+ z( -n) 2

=

g,O,l,?,l,O,n

=

z(n) - z( -n) 2

= g,o,-2'V,2,O,n 103

3

~ xe(n)e- jwn

=

Then, XR(W)

n=-3

3

L

=

jXJ(w)

xo(n)e- jwn

n=-3

= Xr(w) + XR(W)ei 2w . Therefore, y(n) = F- 1 {Xr(w)} + F- 1 {XR(W)ei 2w } = -jx o(n)+ze(n+2)

Now, Y(w)

I

j

j

1

.

j}

= { 2,O,I-2,2,1+2'~'2-)2,O'2

4.14 (a)

x(n)

= -1

[1.

9 ,../10 eiwndw 8,../10

21F'

=

2... 21F'

+

1-

S ,../10 eiwndw -9"/10

+2

J."

eiwndw + 2

9'1(/10

1-'I(

[~(ei9,..n/l0 _ e- j9'fln/l0 _ ei S'fln/l0 + e-jS,..n/lO) }n

+~( _e1 9'fln/l0 + e- j9 ,..n/l0 + e1'fln _ e-j'fln)] )n

= _1_ [sin1l"n _ sin81rn/l0 - sin911"n/lO] n1l" = __n1l"1_ [sin411"n/5 + sin911"n/lO] (b)

x(n)

= -211"1

= -211"1 = =

1

1 0

_,..

X(w)e1 wn dw

1 (- + 0

w

_,..

1r

[w

0

e 0 ] 1-,.. + -.-1-,.. )n

·wn,..

)n1r

L'" X(w)ejwndw

. l)eJwndw + -1 21r

1 1I"n . /2 Jn -sin-e,.. 1rn

211"

jwn

-2 -.-eJ 1r

+ -1

2

(c)

x(n)

104

L'" -eJwndw w . 0

1r

9 wodw " ' /e11 0]

4.15 0:5n:5Af

I

x1(n)= { 0:

otherwise

M

= ~ e- jwn n=O

= x2(n)

= {OI "

1 - e- jw (M+1) 1-

e- jw

-M:5 n :5 -1 otherwise

-1

X 2(W)

=

~ e- jwn n=-M

n=l

jwM

= X(w)

=

1 - e . _el.w ___

1 - elW X 1 (w) + X 2 (w)

=

1 + e jw - ejw _ 1 _ e- jw (M+1) _ e jw (M+1) 2 - e- jw - e jw

= =

2coswM - 2cosw(M + 1) 2 - 2cosw 2sin( Wi\! + ~ )cos~ 2sin 2 1'

=

+ e jwM + e- jwM

sin(.\1+~)W sin( ~)

4.16

Ln

= x(n) = -1 (b) LX(w) = 7r for all w 1 (c) x(O) 2• J~" X(w)dw Hence, (d)

(a) X(O)

=

X(1r)

=

J:" X(w)dw = 21rX(0) = -61r

00

L

z(n)e- jn "

= L(-I)"x(n) =

n=-oo

4.17 (a)

X(w)

=

Lx(n)e-jwn

105

-3 - 4 - 2 =-9

=

X(O)

Lx(n) n

L nx(n)e-jwnlw=o = -iL: nx(n)

d~~w) Iw=o

=

- j

n

n

.dX(w) I

Therefore, c (b) See fig 4.6

X(O)

=

J dw

w=o

X(O)

= 1 Therefore, c = ~ =o.

dX(w)

dw

-..w

_ _ _ _ _ _ _ _+--

I

Figure 4.6:

4.18 Xl

(n)

Now, suppose that

xk(n)

F

anu(n) 1 1 - ae-jw

=

(n+k-l)! n ( ) n.'(k _ 1)'. a u n

-

(1 - ae-;w)A:

F

1

holds. Then 106

\

1 . dX I: ( W ) dw

X ( )

=

"k J

=

ae- iw (l-ae- jw )l:+l

+

I: W

1

+ (l-ae- jW )k

4.19 (a) n

n

(b)

=

Lx·(-n)e- jwn

00

x·(n)e1 wn

L

= X·(w)

n=-oo

n

(c)

n

n

Y(w)

n

X(w) + X(w)e- jw (1 - e-iw)X(w)

=

= (d)

n

y(n)

Hence, X(w)

::;. Y(w)

= ~ x(k) 1:=-00 = y( n) - y( n - 1) = z(n) = (1 - e-iw)y(w) X(w) = 1 - e- jw

(e)

Y(w)

=

LZ(2n)e- iwn

=

Lz(n)e-i,n

=

X(~)

= = =

Lz(n . '2 )e-,wn

n

n

2

(f)

Y(w)

n

L

z(n)e- j2wn

n

X(2w) 107

4.20 (a)

X}(w)

L z(n)e- jwn

=

n

= ei 2w + ei Wl + 1 + e- jw + e- j2w = 1 + 2cosw + 2cos2w (b)

X 2 (w)

L z2(n)e- jwn

=

n

j2w

= ei 4w + ei 2w + 1 + e= 1 + 2cos2w + 2cos4w

+ e- j4w

(c)

X 3{w)

L z3(n)e- jwn

=

n

= ei 6w + ei 3w + 1 + e- j3w + e- j6w = 1 + 2cos3w + 2cos6w (d) X 2 (w) (e) If

=X

1

(2w) and X 3 {w) = X 1 (3w). Refer to fig 4.7

.

\

z.(n)

an integer ={ ~~rJ, Iotherwise

Then,

XA;(W)

= n, f

zA;(n)e- jwn L an integer

=

L: z(n)e-jA;wn

=

n X{kw)

4.21 (a)

z} (n)

X1{w)

=

~(eill"n/4 + e- j ll"n/4)z{n) 2

l[

1r 1r] X(w--)+X(w+-) = -2 44

(b) z2(n)

=

1 2j (eill"n/2 + e- j ll"n/2):r(n)

X 2 (w)

=

2j X(w-'2)+X(w+'2)

l[

1r

108

1r]

~(w)

--~

-+-~---~----w

~(w)

Figure 4.7:

I

(c) Z3(n)

X 3 (w)

= =

~(eil'n/2 + e- j l'n/2)z(n)

2 1r 1r] -1[ X(w--)+X(w+-) 222

(d) z4(n)

X 4(w)

1·

.

+ e-Jl'n)z(n) = _(e'l'n 2 1 = 2 [X(w - 11') + X(w + 1r)] = X(w - 11')

4.22

ct =

1 N

N-l

.

I: y(n)e- J2

I'kn/N

n=O

=

~ ~ Ltco z(n- IN)]

= 1~

00

N-I-IN

I: I:

1=-00 m=-IN

109

.-j2dn/N

z(m)e- j2d (m+IN)/N

\

00

But

N-l-IN

L L 1:-00

=

z(m)e-jw(m+1N)

X(w)

m:-1N

Therefore, c~

4.23

,..n z( n )w( n) sinwen 7rn

=

where z(n)

= =

w(n)

Then

sinwen

= =

Let zN(n)

-

sinwen 7rn

F

-oo~n~oo

~

n:5 N

1,

- N

0,

otherwise

X(w)

= 1, Iwl:5 = o otherwise = X(w). W(w) = { X(e)W(w We

I

XN(W)

j

=

we: -We:

e)de

sin(2N + I)(w - e)/2 de sin(w - e)/2

4.24 (a)

X1(w)

= Ln

z(2n + I)e- jwn

= LI: =

z(k)e- jw l:/2e1 w / 2

X( ~ )e1 W / 2 2

ei w / 2

=

1 - aeJw / 2

(b)

X 2 (w)

=

L z(n +

2)ern/2e-jwn

n

= - L z(k)e-

j

l:(w+ j r/2)ei 2w

I:

= -X(w + j1r )e1 2w 2

(c) X 3 (w)

=

Lz(-2n)en

110

jwn

L z(k)e-

= -

jiw / 2 )

k

=

X(-~) 2

(d) X 4 (w)

= L

~(eio.a1l'n + e-jo.a1l'n)r(n)e-jwn

n

~ Lz(n)

=

[e- j (w-o.a1l')n

+ e- i (w+oa1l')n]

n

1

= 2 [X(w - 0.311') + X(w + 0.311')] (e) Xs(w) = X(w) [X(w)e- jw ] = X2(w)e- jw (f)

= =

X6(W)

X(w)X( -w) 1

(1 - ae- jW )(I- ae jW ) 1 (1 - 2acosw + a 2 )

= 4.25

(a) Ydu:) = Ln Yl(n)e- jwn = Ln,n even z(n)e- jwn The fourier transform Yl(W) can easily be obtained by combining the results of (b) and (c). (b)

Y2(n)

=

z(2n)

Y2(W)

=

LY2(n)e-

=

L

=

L

=

X(~)

jwn

n

r(2n)e- jwn

n

z(m)e- jwm /2

m

2

Refer to fig 4.8. (c)

(n) 113

= {r(n/2),

Ya(w)

0,

= =

L

n even. otherWIse

Y3(n)e-

jwn

n

L even

n

r(n/2)e- iwn

r(m)e- i2wm

=

L

=

X(2w)

m

111

\

~(w)

-n -n/2

o

n/2

n

2n

3rt/2

Figure 4.8:

We now return to part(a). Note that YI(n) may be expressed as

Hence, Y1(w)

= Y2 (2w).

Refer to fig 4.9. j

\

4.26 (a) Because the range of n is (-00,00), the fourier transforms of z(n) and y(n) do not exist. However, the relationship implied by the forms of x( n) and y( n) is y( n) = z3( n). In this case, the system HI is non-linear. (b) In this case,

X(w)

=

Yew) = Hence, H(w)

= = ~

1 1W 1- !e2

'

1 jw 1 - !eIi

'

Yew)

X(w) jw 1- !e2 jw 1 - !e8

System is LTI

Note however that the system may also be nonlinear, e.g., yen) = z3(n). (c) and (d). Clearly, there is an LTI system that produces yen) when excited by zen), e.g. H(w) 3, for all w, or

=

H(~)=3.

(e) If this system is LTI, the period of the output signal would be the same as the period of the input signal, i.e., N 1 N 2 . Since this is not the case, the system is nonlinear.

=

112

~(w)

-nl8

0

n/8

-n - 37t/4 -n/2 -n/4

0

n/4

-n -77t/8

n/2 3n/4

1t

Figure 4.9:

4.27 (a)

WR(W)

=

M

L WR{n)e-

jwn

n=O M

= E e- jwn n=O

= __

(b) Let wT(n)

= hR{n) * hR{n -

1-

e

e-j(M+l)w

1 - e- jw . (M.±!.) _jMw/2 sln 2 W .

w

sm 2

1),

hR{n)

={

1, 0,

Ow

-2'----........----'----"'----~-----"--- .......- - . . . j

o

0.5

1.5

2

->w

Figure 4.23:

124

2.5

3

3.5

0.8

f~0.6 ~ 0." I

0.2 0 0

0.5

1.5

2

2.5

3

3.5

->w

..

3.5

0.5 ->w

Figure 4.24:

Refer to fig 4.25.

(b)

H('!:")

=

H(!)

= =

(2cos'3 )e- J

=

(6cos'3 )sin( '3

10

3

y(n)

0 ·u

57r' 57r'

.

3

1r

(6cos'3 )san( '3 57r'

'"

1r

51r

'3)

+

10 -

-

47", 30 )

(c)

H(O) = H( 4"') 10

y(n)

2

=

2

=

20 + 10cosT

2",n

'"

+ '2

4.31 h(n) H(w)

Steady State Response: H(

== 6(n) + 26(n - 2) + 6(n - 4)

= 1 + 2e- j2w + e- j4w = (1 + e- i2w )2 = 4(cosw)2e-j2w

'" = '2)

Therefore, Yu(n)

=

0 0, (n ~ 4)

Transient Response: 125

3.5

->w 2 r - - - - - - , - - - - , . - -__- -__--...,.---.,...---..,

A I 1_

1

--....a..--. . 2. . . 1.5

_2L..----Io--......... o 0.5

--....1-.--~----J

2.5

3.5

3

->w

Figure 4.25:

= =

Yt,,(n)

+ 20e .("~-~l u(n - 2) + IDe .("~-41 u(n 106(n) + jl06(n - 1) + 106(n - 2) + jl06(n - 3) 10e T u(n)

4)

4.32 (a) yen)

=

zen)

+ zen -

Yew)

= (1 + e-

H( w)

=

j4w

4) )X(w)

(2cos2w)e-j 2w

Refer to fig 4.26. (b) yen) 1r

But cos2"(n - 4)

1r

and cos-(n - 4) 4

Therefore, y( n)

= = = = = =

(c) ~ote that H ( ~) = 2 and H (~)

1r

1r

1r

cos-n + cos-n + cos-(n - 4) 242 1r

cos2'ncos27r

•

1r

+ cos-(n 4

4)

1r

+ san2'nsin27r

7r

cos2'n 1r • 1r . cos-nCOS1r - san-nsan1r

4

4

'4"

-cos-n 1r 2cos2'n

= O.

Therefore, the filter does not pass the signal cos( ~n). 126

2r"""'1'::----,----r--~~-_r_--_r_--."..-___.

'.5

'i

r, A I I

0.5

----11.0-....a.---'.. .5 . - -.....2~--l1&...-lo2.5--~3----'3.5

°O~----I0.-5

->w

2,------,r------,.----,..---r---_r_--..,.....-----,

-2 ' - - - - . 1.-- - - ' - - - - '.......5 - -.....2~--.l2.5--~3----'3.5 0 05 ->w

Figure 4.26:

4.33 y(n)

=

1 2

- [z(n) - z(n - 2)] I

Y(w)

= ~(1 2

H(w)

=

- e- j2w )X(w)

~(l 2

e- j2w )

j = (sinw )e ( i -w) 11" H(O) = O,H(-)=l 2 11" Hence, y",(n) = 3cos("2n + 60°)

Ytr( n)

=

0

4.34

=

z(n) Acos~n (a) y(n) = z(2n) (b) y(n) x 2 (n) (c)

=

= Acos~n ~ w = ~ = A 2cos 2 ~n = ~A1 + ~A2cos~n. Hence, y(n)

Hence,

W

= =

W

= 0 and w = ~

z(n)cos1I"n ~

4'

Acos ncos1I"n

A 571' A 371' = -cos-n + -cos-n 2 4 2 4 311" 511' = -4 and w =-4 127

•

4.35 (a) y(n)

=

1

2" [z(n) + x(n

-

1)]

Y(w)

= ~(1 + e-iw)X(w)

H(w)

+ e- JW ) = -(1 2

1

=

.

cos( -w )e-'." 2

Refer to fig 4.27. (b)

0.'

1°·6 ~ 0.4 I

0.2 0 0

1.5

0.5

2

2.5

3

3.5

->w

0 I

_-0.5

•

I

-1

A I 1_ .

5

'

-2

0

1.5

0.5

2

2.5

3

->w

Figure 4.27: y(n)

1

= --2 [z(n) -

z(n - 1)]

Y(w)

=

1 . --(1- e-JUI)X(w) 2

IH(w)1

=

sin2

9(w)

w

= e'w

Figure 4.42: when Wo

at w

= 1r/2, H(w) = 1 - ei 2w

= 1r/3, H(1r/3) = y(n) =

=

ei 2ff / 3 = lei,,/3 1r IH( 1r /3)13cos( in + 30° -

1-

1r

3cos( in

-

30°)

4.44 (a)

y( n ) H(w)

= x (n) - x (n - 4)

= =

1 - e- j4w 2e-j2wejff/2sin2w

Refer to fig 4.43.

(b) x(n)

y(n)

1r

1r

1r

= cos'2 n + cos'4 n ,

=

1r 2cos'4 n ,

(c) The filter blocks the frequency at w

1r

H( '4)

H( '2n)

= 2,

= ~.

4.45 y(n)

=

1

2 [x(n) -

..r(n - 2)] 139

=0

1r LH("4)

=0

60°)

'.5

'i

1:, A I I

0.5

0.5

....•

'.5

2

2.5

3

3.5

2

2.5

3

3.5

->w

'

10 A I I

-, -2

0

0.5

'.5

->w

Figure 4.43:

z(n)

= = =

H(O)

= 0,

y(n)

=

H(w)

!(l - e- j2w ) 2

e-jwei~/2sinw 7r

5 + 3sin( 2'n

+ 60°) + 4sin(""n + 45°)

H(2') "" = 1, .

H(",,)=O

7r

3szn(2'n+600)

4.46 (a)

y(n) Y(w)

= =

~

z(2n)

L

This is a linear, time-varying system

y(n)e- jwn

n=-oo"" 00

= L

z(2n)e- iwn

n=-oo

= = =

X(~) 2

7r

1,

Iwl:5 2'

0,

- < 2 -

7r

Iwl -< 7r

(b)

y(n) Y(w)

=

z2 (n)

=

1 2""X(w)

~ This is a non-linear, time-invariant system

* X(w) 140

Refer to fig 4.44. (c)

Y(w) 1/4

o

-rt/2

rt/2

w

I

Figure 4.44:

y(n) Y(u:)

= (COS1rn )x( n) ~ This is a time-varying system 1 = 21r [1r6(w - 1r) + 1rc5(w + 1r)] * X(w) 1 = 2 [X(w - 11') + X(w + 11')] 311' = 0, Iwl:5 4" 1 311' = 2' 4" :5 Iwl :5 11'

4.47

(a)

H(z)

= =

1 - !COS!'Z-1 4 4 1 - 2(*)COS~Z-1 11-

£1 z8

£1 -1 + 4 z 141

+ (~)2z-2

1

J.. .. -2 16'"

\

(b) Yes. Refer to fig 4.45 (c) Poles at z ~e±jt, zeros at z =~.

=

x(n)

t------..-------... y(n) z -I

z -I

z -1/16

Figure 4.45:

zen) X(z)

=

Y(z)

=

1 - lz-l 4

X(z)H(z)

yen)

4.48 y( n)

= x ( n) - z (n - 10) 142

-I

!

,.. ~"

I

o801....---0...S--~--'''-S--,£",2--2'£"'S--'£"'3- - - - 1 35 ->w

Figure 4.46:

(a)

H (w)

=

1 - e- j j5w

lOw

eii sin5w

=

2e-

IH(w)1

=

21sin5wL

e(W)

=

~

- 5w,

= '1r2 -

5w

for sin5w

+ 1r,

>0

for sin5w

w 2 , . - - - _ - -__---r---...---....----..---,

-' •

fo A

1_, I

-2'---.-I.---""""---"""'---"'------I~--..I.--~

o

0.5

1.5

2

2.5

3

->w

Figure 4.47:

= -1

[ Sln-n . 31r - Sln-n . '1l' ]

8

1rn

=

8

2 . 1r 1r -Sln-ncos-n 1rn 8 4

(b) Let

Then,

{2,

H (w) 1

Iwl ~ i i < Iwl < 1r

0,

-

and

hen)

1r

= h 1 (n)cos-n 4

4.50

Y(z)

1 = 21 y (n - 1) + zen) + 2z(n - 1) 1 _ 1 _ = 2z ly(z) + X(z) + 2z I X(z)

H(z)

=

yen)

=

Y(z) X(z) 1 + 12 Z -

1

(a)

H(z)

=

2 - 1 1 - 12 z - 1

h(n)

=

2(2)"u(n) - 6(n)

1

144

3.5

(b) 00

H(w)

= L h(n)e-

jwn

n=O

2 - 1 jw !e2

= .1-

jw 1 + !e2

= =

jw 1 - !e2

H(z)lz=e''''

(C)

=

H(?:) 2

= = =

Hence, y(n)

1 + !e-ji 2 1

.

r

1 - 2'e- J 'i' 1 - }2' 'I 1 + j~ le-j2tan-l! 1r

cos(-n 2

+ -7r 4

11

2tan- -) 2

4.51 Refer to fig 4.48.

IXCw)1 'or (a)

IXCw)1 'or (b)

4

I

3

I

t.

-8

,,

1

2

I

2 2

3

°0

~

IX(w)11or (c)

2

3

~

3

~

IXCw)1 'or (eI) 12

'.5

10

t:

I, 2

I

~

2 2

3

°0

~

Figure 4.48:

4.52

145

2

= H(w) =

1-

J2z-1 + Z-2

1 - .J2e- jw

.

+ e- 2jw

Vi

= 2e-JW(cosw - -2) y(n)

= x(n) - J2x(n -

1)

+ x(n -

2)

= sin~u(n) y(O) = x(O) = 0

for z(n)

y(l)

V2 = x(I)- ,f22z(0)+x(-I)=2

y(2)

+0 = 0 = x(2) - v'2x(l) + z(O) = 1 - yliVi 2

y(3)

= z(3) -

y(4)

= 0

v'22x(2) + x( 1) = -Vi - h + -V2 = 0 2 2

4.53 (a) H(z) (b)

= kl~O~~~i'

Refer to fig 4.49.

j

\

Figure 4.49:

H(w)

1 - e- jw

= k ---1 + O.ge-}W 146

IH(w)1

=

8(w)

=

=

2lsin~1

/c

Vl.81

=

tan-

1

+ 1.8cosw sinw + tan- 1

1 - cosw

=

0.9sinw 1 + 0.9cosw

= 210

(c) H(1r) k1~O.~~~~" /c 0 1 20k 1 ~ k (d) y(n) = -O.9y(n - 1) + 210 [z(n) - z(n - 1)] (e)

=

2

H(~) = O.014e j6 (i) 6

y(n)

=

4.54 (a) H(z) = (b) For a

1+bz -

1

bo~.

Refer to fig 4.50.

= 0.5, b = -0.6,

H (z) = bo ~~g~;

=: . Since the pole is inside the unit circle and the

Direct form I: y(n)

·a

Direct form II : y(n)

Figure 4.50: filter is causal. it is also stable. Refer to fig 4.51. (c)

H(z)

= b 1 + 0.5z- 1 1 _ 0.5z- 1 24'5 + cosw

0

= 147

bo-=-s--4 - cosw

z-plane

Figure 4.51: The maximum occurs at w

= O.

Hence, 9

H(w)lw=o

~

(d) Refer to fig 4.52. (e) Refer to fig 4.53. obviously, this is a highpass filter. highpass filter is improved.

bo

24 boT

= 4 = 9b~ = 1 = ±!3

By selecting b

= -1,

the frequency response of the

4.55 IH(w)1

2

d 1 dw IH(w)l2

A

= [1 + r 2 - 2rcos(w - e)] [1 + r 2 - 2rcos(w + e)) = ~ [2rsin(w - 8)(1 + r 2 - 2rcos(w + e)) +2rsin(w

(1

+ r 2 )(sin(w -

8)

+ e)(1 + r 2 -

2rcos(w - e))]

= 0 + sin(w + 6)) = 2r [sin(w - 8)cos(w + 8) + sin(w + 8)cos(w 148

8)]

1H(w)1

0.8 :;:0.6

~0.4 0.2

°°

0.5

1.5

• pt-.

2

2.5

3

3.5

1-

04 . -0.6

-0.8

- 'OL..-----Io S --~--'...5---21-----&2.--5--~3---J3.5 O.-

•

Figure 4.52:

z-plane

Figure 4.53:

149

+ r2 )2sinwcos6 =

(1

= Therefore, cosw

w,.

= =

2rsin2w 4rsinwcosw 1 + r2 --cose 2r

[1

2

+r ] cos- 1 ~cose

4.56 y(n) H(w)

IH(w)1 8(w)

= = = = = =

1 -z(n)

1 1 + -z(n - 1) + -x(n - 2) 4 2 4 1 -jw 1 -j2w -41 + -e + -e 2 4 jw ( 1 + e- )2 2

. 2w e- Jw cos -

cos

2

w

2

-

2

LH(w) =-w

Refer to fig 4.54

j

0.8

"

1

0.8

1, o.~ 0.2 0 -4

-3

-2

-1

2

3

~

2

3

~

->w ~

__ 2

f• ,

0

A I

-2 -4 -4

-3

-2

-1

0 ->w

Figure 4.54:

4.57 (a)

z(n) 150

X(z) Hence, H(z)

= = =

1

-1 1_ z-

+

1 - 1 Z -1

1

l'

ROC:

4 < Iz I
0 for 1 + 2cosw < 0

Refer to fig 4.55.

4.59 (a) y(n)

=

j

1 M 2.\1 + 1 k~M z(n - k) 151

0.8

EO.6 ~

,

~ o.~

0.2 °O'-----O.l-5---""---.... '.S-----'2::::a....::::--.....----'-3----1 .! 3 2S ->w

__----,

2r---__r---.....,-----r-----r--~--

-3 L - - - - l.I-!- - " " " - - -.... '.!---"""'2---2..L..- -.....--~3.! 0 O 3 5 -->w

Figure 4.55: M

H(w)

=

'"'" e- jwk 1 LJ

1

2M

+

k=-M

1[1 + 2t

1 2M

=

+

COSWk]

k=l

(b) y(n)

1 4M z (n

=

+ At) +

1 2M

1

M-l

L

x(n - k)

+ 4M z (n -

A-f)

k=-M+l

H(w)

L

1 1 [ M-l ] 2McosMw + 2M 1 + 2 coswk

=

k=l

The filter in (b) provides somewhat better smoothing because of its sharper attenuation at the high frequencies.

4.60 (a)

XU)

=

1:

z(t)e-

j20F

'dt

1:[1: = 1: [1:

=

zl(I - T)Z,(T)dT] e-j,·F'dt

Z,(T)

1:

=

XI(F)

=

XdF)X 2 (F)

zl(I - T)e-j'·Fl'-')dt] e-j,·F'dr

z,(T)e-j'·F, dr

152

(b)

=

x(t)

=

,. +t

1'-,--,.

Id.x=T+t,-T~t~O

1 t

-t+t

x(t) = {

T -

Id.x=T-t,-T>t~O It I,

0,

ItI $

T .

otherWIse

Refer to fig 4.56. (c)

x(t)

o

-'t Figure 4.56:

4.61 H(z)

= =

1 + z + z2 1 - z9 1 - z-l

153

+ ... + z8

't

t

=

H(w)

=

1 - e- j9w 1 - e- jw e- j9w / 2 sin9w /2

e- jw / 2 sinw/2

= e =

IH(w)1

-j4w

sin9w/2 . /2 smw

I sin9w/2, sinw/2 -4w, when sin9w/2

e(w)

>0 = = -4w + 7r, when sin9w /2 < 0

H(w)

= 0, at w = 9 ' Ie = 1, 2, ... , 8

27rk

The corresponding analog frequencies are

!.f,

Ie = 1,2,3,4, or ~kHz, ~kHz, ~kHz, ~kHz.

4.62 Refer to fig 4.57.

Figure 4.57:

(1 - ej311'/4 z -1)(1 _ e- j3 11'/"z-1)

H(z)

=

G~----~-~----

H(w)

=

H(z)lz=e J •

(1 - ~z-1)2

154

H(O)

=

IH(w)1

=

12

=

G

=

G

(1 - ej3T / 4 )(1 _ e- j3w / 4 ) 1 (1 - 2)2

12 l=>G T =1 4 2+V2 1 = 0.073 4(2 + y'2)

4.63

=

Hz(w)

l-rd'e- jw 1 - reos(w - 9)

=

+ jrsin(w -

9)

(a)

= HI - reos(w -

IHz(w)1

= [1 + r 20log 1o IH z (w)1

2

9)]2 + [rsin(w - 9)f}1 2reos( W - 9)]!

-

101og 1o [1 - 2reos(w - 9) + r 2 ]

=

Hence proved. (b)

=

tan

=

tan

-1

imago part real part

-1

rsin(w-9) 1 - reos(w - 9)

Hence proved. (c)

= =

1 r

"~~lnJ~w-B)

[l-,.co~

+ 2

1+r

-

2 -

w-BH 5

rcos( w - 9) 2rcos(w - 9)

Hence proved. (d) Refer to fig 4.58.

4.64 Hp(w)

= ,1 ., 1 - re eJ

r

JW


W=1r

1+r2

-

2rcos(w - wo)

1r

3

IH(1r)1

2

C

= =

4C 2 ( 1 -(1 3

2

1+r

+r2

)2

=1

+ r + r 2)

(d) Refer to fig 4.62.

(e) 160

] [ 2( 1 - rcos(w + "'0)) ] 1 + r 2 - 2rcos(w + wo)

Figure 4.61:

j

1

x(n) y(n)

Figure 4.62: 161

= In the vicinity of W =

G2

11 - eiwoe-iwI211_ e-jWoe-jWI2 11 - rejwoe-jw 12 11 _ re-jwoe-jw 12

we have

Wo,

.

2

~

G

= G2 cos(W - wo)

')

11 - e) wOe - ) w 1* .. 2 Jw I 11 - re)WOe2( 1 - cos(W - wo)) ] _ ~ 1 + r 2 - 2rcos(W - wQ) - 2

[

=

1 + r 2 - 4G 2

=

WQ ±

= =

2

2r - 4G2

cos

1 COS- (

-1 (

1 + r2

- 4G2 2r _ 4G2 )

2

1 + r - 4G2) 2r _ 4G2

r - 1 2cos-1(1 _ ( .j2 )2)

= 2J2(1-r), .j2 = 2.;r::r 4.70 For the sampling frequency F, = 500samples/sec., the rejected frequency should b~' WI 6 21l"( 160°0) 2 5 7r. The filter should have unity gain at W2 21l"( ~~~) £1l". Hence,

=

=

=

6 H(-1l") 25 4 and H( -1l") 5 H(w)

= 0

= 1 jW = G(I- ein-e- )(1 - e-jn-e-)W)

= 4 H(-S1l")

=

Hence, G

=

. 61l" Ge- JW [2cosw - 2cos-] 25 4 6 2Gl[cos( 571') - cos( 25 71')]1 = 1 1

'2

cos

6 2 5 1l"

- cos t 1l"

4.71 From (5.4.22) we have,

H(w)

= =

Hence, bo =

1 - e- j2w o b (1 _ re j (wo-w»)(1 _ re-j(wo-w»)

11 -

2

e- j2w 1 b =1 o (1- r)2[(1 - rcos2wo)2 + (rsi n2wo)2) ';(1 - r)2(1 - 2rcos2wQ + r 2) 21 sinw ol 2

162

=

4.72 From

;3

=

(n - I)wo

1)wo

=

with y( n)

=

coswon, it follows that

+ cost:lIJ =

+ I)wo + cos(n y(n

= (n + I)wo

o+fJ 0-13 2cos--cos-- we obtain 2 2' 2cosnwocoswo

and coso

cos(n

0

+ I) + y(n

= =

- 1) y(n)

2coswoy(n) or equivalently, 2coswoy(n - 1) - y(n - 2)

4.73 sino

+ sinl3 when

sinnwo

0

+ sin(n -

2)wo If y(n) y(n) Initial conditions: y( -I)

= = =

0 13 2. 0-13 ' san-+-cos- , we 0 b tam 2 2 nWQ and 13 (n - 2)wo, we obtain

=

Asinwon, then

= =

2coswoy(n - 1) - y(n - 2)

=

2sin( n - 1)wQCOSWO

-Asinwo,y(-2)

= -Asin2wo

4.74 For h(n) H(z) Hence, y(n)

Fo'r h(n) H(z) Hence, y(n)

= =

= = = =

Acoswonu(n) 1 - z-lcoswo 1 - 2coswoz- 1 + z-2 2coswoy(n - 1) - y(n - 2)

+ Az(n) -

AcoswQz(n - 1)

Asinnwou(n) z-lsinwo A ------:~--=1 - 2coswQz-l + z-2 2coswoy(n - 1) + y(n - 2)

+ Az(n) -

AsinwQz(n - 1)

A

4.75 Refer to fig 4.63. ydn)

= AcosnwQu(n), Y2(n) = AsinnwQu(n)

4.76 (a) Replace z by z8. We need 8 zeros at the frequencies W =

H(z)

=

= Hence, y(n)

=

O,±i'±I,±3;,1r Hence,

1- z-8

1 - oz-8 Y(z) X(z) a y( n - 8)

163

+ z (n)

- z (n - 8)

x(n) +------i~

Y (n) 1

-AcoswO

z -1

Asin"t

-1

Figure 4.63: (b) Zeros at 1,e±jT,e:l:: j 'i,e:l:: j ¥,-1 11:1::'" 1:1::'" .l:l::'.l!. Poles at al,a1e J.,a'e Jl",a1e J. ,-I. Refer to fig 4.64. (c) IH(w)1 21cos4wl 2acos8w + a 2

=

LH( ) W

={

"'1 -

-tan-1 4,inBUI l-oeo,B~' 1r _ tan- 1 Olln~UI

1- oeo,BUI '

cos4w cos4w

>0 bo = 0.089

12

(2je- jw sinw)(2e-jW)(coSW - cOS 3 1r )

41sinwllcosw - cos 3; I _ 1.6cos 2; e- jw + 0.64e- j2w lll _ 1.6cos 4;

e- jw

+ O.64e-J 2W) + 0.64e- j2w l

(b) H(z) as given above. (c) Refer to fig 4.66. The filter designed is not a good approximation of the desired response.

4.82 Y(w)

= e- jw x(w) + d~~w)

(a)

For x(n)

=

b(n), X(w) 167

= 1.

Hence,

dX(w)

~

h(n)

=

0, and Y(w)

=

-1

= e- jw

jll' Y(w)ejwndw ~ j'" eiw(n-l)dw 27r 27r

=

-11'

"

-11'

= 27rj(n1 - 1) ei w (n-l)llI'-11' sin7l'(n - 1) 7r(n - 1)

= (b) yen)

= zen - 1) - jnz(n). the system is unstable and time-variant.

4.83 00

H(w)

=

h(n)e- jwn

L

n=-oo

= 1, = 0,

Iwl:5 We 00

G(w)

=

L

We

< Iwl7r

g(n)e- jwn

n=-oo 00

=

L

n=-oo

j

",

h( ~)e-jwn 2

00

=

h(m)e- j2wm

L

m=-oo

=

H(2w)

Hence, G(w)

1, = { 0, Iwl :5 T and Iwl ~ T < Iwl < 7r- T

11' -

T

4.84

=

=

yen) zen) - zen) * hen) [eS(n) - hen)] * zen) The overall system function is 1- H(z) and the frequency response is 1 - H(w). Refer to fig 4.67.

4.85 (a) Since X(w) and Yew) are periodic, it is observed that Yew) y(n) eJlI'n z (n) (-I)nz(n)

=

=

(b) zen) = (-I)ny(n).

4.86 yen)

= O.9y(n 168

1) + O.Iz(n)

= X(w - "J

Therefore,

H(w)

I-H(w)

1

--t-----.w. . . - --- w

w

c

c

w

I-H(w)

H(w)

11------,

1

o

n

C

- -....... ---....Ioo.-----w (

w

n

w

Figure 4.67: (a)

H(z)

= =

0.1 1 - 0.9z- 1 0.1 1 - 0.ge-.i(w-i) 0.1 1 - jO.ge- jw

(b) h(n) = 0.1(0.ge.i i )"u(n) (c) Since the impulse response is complex, a real input signal produces a complex-valued output signal. For the output to be real, the bandpass filter should have a complex conjugate pole.

4.87 (a)

Let g(n)

=

nh(n)

Then, G(w)

=

J--

.dH(w) dw

00

D =

L

Ig(n)1

2

"=-00

If

= -27r _... \G(w)\2 dw 169

But dH(w)

dw

11

r

=

211'

=

2..111' [jdH(W)(_j)(dH(W»)-] dw 211' -11' dw dw

=

[dH(W) dw

-r

G(w)G*(w)dw

+ jIH(w)1 de(W)]

ei8(w)

dw

Therefore,

D

= 2..1 211'

r

-r

{[dH(W)]2 +IH(W)\2[de(W)]2}dW dw dw

(b) D consists of two terms, both of which are positive. For IH(w)1 0, in which case the second term becomes zero. selecting 6( w)

=

#=

0, D is minimized by

4.88 y( n)

= a y( n

- 1) + 6z (n ), 0 < a < 1

6 H(z) = 1 - az- 1 (a)

H(w) IH(O)I

6

= 1 - ae= A=1 I-a

jw

6 =

±(l - a)

(b)

~ 2b 2

= =

cosw

=

IH(w)!2

= W3

=

62 1 + a 2 - 2acosw 1 + a 2 - 2acosw

1

= -2

2 -1 [ 1+ a - 2( 1 - a) 2] 2a 1 2 -(4a-l-a) 2a 2 1 4a - 1 - a cos- ( ) 2a

(c)

W3 Let /(a) Then /'(a)

= = = =

Therefore I(a) is maximum at a W3 increases as a - 0.

(a - 1)2 ) 2a (a - 1)2 12a a2 - 1

cos-1(l-

-~

1- a 2 ~>O

= 1 and decreases monotonically as a - 0. Consequently, 170

(d)

b Wa

= =

±(l-a) 1

COS- (

4a - 1 - a 2 2a )

o.

The 3-dB bandwidth increases as a -

4.89 y(n) H(w) IH(w)1 8(w)

= z(n) + az(n - M), = 1 + oe-;wM = Jl + 2acoswM + -asinwM = tan- 1+ ocoswM

0

>0

0 2

1

Refer to fig 4.68. .-,0. . . . 0.1 __-..,._-..__-.,......___,

2r----,~--r-~-.....,...-"""T""-

1.5

c:

~, It.

I

0.5 00

O.De

0.1

0.1S

0.2

0.25 ->1

0.3

o.atS

0..

0.04S

D.S

O.De

0.1

0.1S

0.2

0.25 ->,

0.3

0.35

0."

0.04S

0.5

0.5

I

0

A.

. I

I

-0.5 -10

Figure 4.68:

4.90 (a)

Y(z) H(z)

= =

~

= =

~(I+z-l)

[X(z) + z-1 X(z)]

Y(z) X(z)

Z

+

1

2z 171

Zero at z

= -1 and a pole at z = O. The system is stable.

(b) 1

Y(z)

=

H(z)

= X(z)

- [-X(z) 2

+ z-1 X(z)]

Y(z)

= -21 ( -1 + z -1) z -1 --2z

= Zero at z

= 1 and a pole at z =O.

The system is stable.

(e)

Y(z)

Three zeros at z

= -1 and three poles at z =O.

The system is stable.

4.91 Y(z)

=

H(z)

= =

For 6

= 1, H(w) = = IH(w)1 LH(w)

+ 6z- 2 X(z) + z-"X(z)

X(z) Y(z)

X(z) 1 + 6z- 2 1 + ei

2w

+ z-4 + e- j4w

+ 2cosw)e- jw

(1

= 11 + 2coswl 1 + 2cosw 1 + 2cosw

-w, ={ ."..-w,

~

0

w 2

I

~ -1 I

-3 0

0.5

1.5

->w

Figure 4.69:

I

3

2.5 :::::: 2

!- 1.5 A I I

°0

0.5

1.5

2

2.5

3

3.5

2

2.5

3

3.5

->w

3

I A I I

-2

0

0.5

1.5

->w

Figure 4.70:

173

1

(a)

Y(z) H(z)

Z6

z 6th order pole at z

=O.

= = =

X(z)(1 - 0.95z- 6 ) (l - 0.95z- 6 ) Z6 - 0.95 %6

= 0.95

=

(0.95)tei 2d / 6 , k

= 0,1, ... ,5

Refer to fig 4.71.

(b)Refer to fig 4.72.

r=(O.95Y/6

Figure 4.71:

=

(c) Hin(z) zi~~.9S' (d)Refer to fig 4.74.

T

= (0.95)!.

Refer to fig 4.73.

4.93 (a)

H(z)

=

%-1

1 - z-1 - z-2 z-1

174

e:

~ 1 1\ I I

0.05

0.1

0.15

0.2

0.25

0.3

0.35

-->f

Figure 4.72:

r=(O.9Sj/6

Figure 4.73:

175

0.4

0.45

0.5

--r__-__--_-_----.

20 r - - - - - - r - - - , - - - - - - - r - - - r - - -__ 15

€

~10 1\ I I

5

00

0.05

0.1

0.15

0.2

0.25 -->1

0.3

0.35

0.4

0.45

0.5

Figure 4.74: 1

=

If ROC

. v'5-1 2


1- - 2 -

hen)

=

[_1 (1 + v'5)" v'5 2

IS

-~

ROC, then

__ 1 (1 - v'5)"] v'5

u(n)

2

then

I,.

hen)

11-../5" = - v'5(-2-) u(n) -

11+../5" y"5(-2-) u(-n -1)

If Izi

< 1 - - 2 - is ROC, then

h(n)

= [__ 1 (1 + ../5)" + _1 (1 -

../5 - 1

../5

V5

2

v'5)"] u(-n _ 1) 2

From H (z ), the difference equation is

= y( n

y( n )

- 1) + y( n - 2)

+ z (n -

1)

(b)

H(z)

1

=

The difference equation is

y(n)

= e- 4G y(n - 1) + zen)

H(z)

=

1 (1- e- a z- I )(I- e.1i e- a z - 1 )(1

=

4 l-e- G z- 1

1

1

+

4 I-je- G z- 1

+ e- a z- 1 )(1 + je-az- I ) 1

+

4 l+e- G z- 1

+

1

4 l+je- a z- 1

If ROC is Izi > 1, then hen)

= ~ [1 + (j)" + (-1)" + (-j)"] e-a"u(n) 4

< 1, then h(n) = -~ [1 + (j)" + (_1)" + (-j)"] e-a"u(-n -

If ROC is Izi

4

176

1)

4.94 Y(z) X(z) Therefore H ( z) l

h(n)

=

= = = = =

+ 3z- 2 _ z-3 + 6z- 4 (1 + z-1 + 2z- 2 )(1 - 2z- 1 + 3z- 2 ) 1 + z-1 + 2z- 2 1 - z-1

Y(z) X(z) 1 - 2z- 1 + 3z- 2

h-2,3}

4.95 y(n)

= 21 y (n -

z(n)

=

(4)"u(n)

H(z)

=

Y(z)

= X(z)

=

Y(z)

=

Hence, rhh(n) Rr~(z)

1

X(z) 1

1 - 1 Z -1 2

1 1 (1 - ~z-I)(1 - ~z-l)

=

X(z)X(z-l)

=

(1- ~z-l)(l- ~z)

=

Hence, rrr(n)

1) + z(n)

= = =

1

-4z- 1 (1 - tz-1)(1 - 4z- 1 ) 16 1 16 1 151- 1 z- 1 151-4z- 1

"

16(!)n u (n) 15 4

+

16(4)n u (_n - 1) 15

H(z)H(z-l) 1

= (1-

~z-l)(l- !z)

= (1-

-2z- 1 !z-I)(I- 2z- 1 ) 1 4 1

4

= = =

X(z)Y(z-l)

=

1 (1 - ~z-I)(1 - iZ)(1 - !z)

31- 12 z- 1 -31-2z- 1

~(! )nu(n) + ~(2)"u( -n 3 2

3

177

1)

Hence, rry(n)

Ryy(z)

1

16

1

128 1 1051- 1 z- 1

-171-2z- 1

=

16 ( )n ( ) 16 )n 128 1 n 17 2 u -n-l - 15(4 u(-n-l)+ 105(4) u(n)

+ 151-4z- 1 +

4

= Y(z)Y(z-l)

=

= Hence, ryy(n)

16

=

1

(1- ~z-1)(1-lz-1)(1-~z)(l-lz) 64 1 128 1 64 1 - 21 1 - 2z- 1 + 105 1 - 4z- 1 + 21 1 - 12 z-l 64 ( 2 )n (

= 21

-

128 1 105 1 - 1 z -1 4

)

128 ( )n ( ) 64 ( 1)n () 128 1 n u -n-l - 105 4 u -n-1 + 212 un - 105(4) u(n)

4.96

=

(a) hen) {lOr,9,-7,-8,0,5,3} The roots(zeros) are 0.8084 ± jO.3370, -0.3750 ± jO.6074, -1.0, -0.7667 All the roots of H(z) are inside the unit circle. Hence, the system is minimum phase. (b) hen) {5, 4,-3, -4,0,2, I} H(z) 5 + 4z- 1 - 3z- 2 - 4z- 3 + 2z- s + z-6 The roots(zeros) are 0.7753 ± jO.2963, -0.4219 ± jO.5503, -0'.7534 ± jO.1900 All the roots of H(z) are inside the unit circle. Hence, the system is minimum phase.

=

=

4.97 The impulse response satisfies the difference equation

I,

N

L a;h(n -

k)

=

o(n), ao

=1

n=O'~La&:h(-k)

=

aoh(O) = 1

&:=0 N

;=0

ao n

= 1, ~ aoh(l) + aoh(O)

= =

1

h(O) 0

-aoh( 1) h(O)

n

= N, ~ aoh(N) + alh(N - 1) + ... + aNh(O)

~

-he 1)

= h (0) 2

yields aN

It is apparent that the coefficients {an} can be determined if we know the order N and the values h(O), h(l), ... , heN). If we do not know the filter order N, we cannot determine the {an}.

4.98 hen) = boo(n) + b1 0(n - D) + b2 0(n - 2D) (a) If the input to the system is zen), the output is yen) = bo.:r(n)+b 1 x(n-D)+b 2 x(n-2D). Hence, the output consists of x(n), which is the input signal. and the delayed signals x(n - D) and x(n - 2D). The latter may be thought of as echoes of x(n).

178

(b)

= = = =

H(w)

IH(w)1 6(w) (c) If

Ibo + b2 1 « Ib 1 1,

+ b1e- jwD + b2e-j2wD bo + blcoswD + b2 cos2wD - j(b1sinwD + b2 sin2wD) bo

+ b12 + b2 2 + 2b 1(b o + b2 )coswD + 2bob 2 cos2wD -tan- 1 blsinwD + b2 sin2wD bo + blcoswD + b2 cos2wD b0 2

then the dominant term is b1e- jwD and

IH(w)\

= Jb

0

2

+ b1 2 + b2 2 + 2b 1 (b o + b2 )coswD

and IH(w)1 has maxima and minima at w = ±fJ1r,k = 0,1,2, ... (d) The phase e(w) is approximately linear with a slope of -D. Refer to fig 4.75.

11

€

~1

"I

09 080

005

0.1

0 15

0.2

0.25

0.3

035

O~

0.45

0.5

0.3

0.35

o.~

045

0.5

-~I

~o

005

01

0 15

0.2

025 ->1

Figure 4.75:

4.99 1

H(z) = B(z) = 1 + bZ = A(z) l+az 1

~ h(n)z-n

f:'o

(a)

H(z) Hence, h(O) h(l)

h(2) h(3) h(4)

= = = = = =

1 + (b - a)z-l

+ (a 2 - ab)z-2 + (a 2 b - a3 )z-3 + (a 4 - a3b)z-s + ...

1,

b- a, a 2 - ab, a2 b - a3 , a 4 _ a3 b 179

(b) y(n)

+ ay(n -

= = =

1)

For x(n) hen) + ah(n - 1) Multiply both sides by h(n) and sum. Then

= = = =

rhh(O) + arhh(1) rhh(l) + arhh(O) rhh(2) + arhh(l) rhh(3) + arhh(2)

x(n)

+ bx(n - 1)

6(n),

6( n)

+ b6 (n - 1)

h(O)

+ bh(l)

bh(O)

0 0

By solving these equations recursively, we obtain 2ab + 1 1 - a2 (ab - 1)(a - b) 1 - a2 (a6-1){a-b) -a -----~--'1 - a2 2 (a6 - 1)( a - 6) a 1- a2 b2

=

=

-

4.100 x(n) is a real-valued, minimum-phase sequence. The sequence yen) must satisfy the codd,tions, yeO) = x(O), ly(n)1 = Ix(n)l, and must be minimum phase. The solution that satisfies the condition is y( n) (-1)" x( n). The proof that y( n) is minimum phase proceeds as follows:

=

Y(z)

= E y(n)z-" "

=

E(-I)"x(n)z-"

= =

Ex(n)(-z-I)"

"

"

X(-z)

This preserves the minimum phase property since a factor (1- oz-l) -

(1

+ oz-l)

4.101

=

Consider the system with real and even impulse response hen) {~, 1, ~} and frequency response H(w) 1 + ~cosw. Then H(z) z-I(~z2 + Z + ~). The system has zeros at z -2 ± -/i We observe that the system is stable, and its frequency response is real and even. However the inverse system is unstable. Therefore, the stability of the inverse system is not guaranteed.

=

=

=

I

4.102 (a)

g(n)

=

fen)

= h(n). g(-n) => F(w) =

hen)

* x(n) => G(w) = H(w)X(w) 180

H(w)G(-w)

Y(w)

Then, Y(w)

= = =

F(-w) H(-w)G(w) H( -w)H(w)X(w)

=

H-(w)H(w)X(w)

=

IH(w)1 X(w)

2

2

But Ho(w) == IH(w)1 is a zero-phase system. (b) G(w) F(w) Y(w)

But Hb(W)

= = = = = =

H(w)X(w) H(w)X(-w)

+ F(-w) + H(-w)X(w) X(w)(H(w) + H-(-w»

G(w)

H(w)X(w)

2X(w)Re(H(w»

=2Re {H(w)} is a zero-phase system.

4.103 (a) Correct. The zeros of the resulting system are the combination of the zeros of the two systems. Hence, the resulting system is minimum phase if the inividual system are minimum phase. (b) Incorrect. For example, consider the two minimum-phase systems.

z 1 - 21-1 1 - 1 z -1 3

-2(1 + !Z-I)

iz-1

1-

-1 - 16 z -

1 -1

1- 3z

1

'

which is not minimum phase.

4.104 (a)

Hence, H(z)H(z-l)

(b)

=

2(1-0 2 ) 1 + 0 2 - 20cosw 2( 1 - a2 )

1+0 2 -0(z+Z-I) 181

H(z)H(z-l) Hence, H(z) or H(z)

=

2(1 + a)(1 - a) (1 - az- I )( 1 - az)

=

J2(1 - a2) 1 - az- I

=

J2(1 - a2) 1- az

4.105 H(z)

= (1- 0.8ei,../2 z -1)(I- 0.8e-;,../2 z -I)(I- 1.5ei,../4 z -1)(1- 1.5e-;,../4 z -1) = (1 + 0.64z- 2 )(1 - ~z-1 + 2.25z- 2 )

(a) There are four different FIR systems with real coefficients:

H 1(z)

=

(1

+ 0.64z- 2 )(1 - ~z-1 + 2.25z- 2 )

H2(Z)

=

(1

+ 0.64z- 2 )(1

H 3 (z)

=

(1

+ 0.64z- 2 )(1 - ~z-1 + 2.25z- 2)

H 4 (z)

=

(1

+ 0.64z- 2 )(1 - ~z-I + 2.25z- 2 )

-

~z-1 + 2.25z- 2)

H (z) is the minimum-phase system. (b)

Ht{z)

= 1 - ~z-1 + 2.89z- 2 _ 1;; z-3 + 1.44z- 4

hI (n)

3 -1.92 } = { ~,- y'2' 2.89, y'2 ,1.44

H 2 (z)

= 0.64z - y'2 z + 2.44 - y'2z

h 2 (n)

3 } = {-1.92 0.64, v'2 2.{4, - v'2' 2.25

H3(z)

=

h 3 (n)

= { 2.25, -3 v'2' 2.{4, - 1.92} v'2 ,0.64

H 4 (z)

=

1.44z -

h 4 (n)

=

{-1.92

1.92

2

3

-1

+ 2.25z -2

I

~ 2 2.2~z

4

1.44,

3 1.92 -1 - .j2z + 2.44 - .J2 z

1.92

3

v'2 z + 2.89z

v'2

2

-

+ 0.64z -2

3 z+1 v'2

3}

,2.89,- v'2't'

(c) E 1 (n) E2(n) £3(n) E 4 (n)

= = = =

{I, 5.5, 13.85, 15.70, 17.77} {0.64, 2.48,8.44,12.94, l8.0} {2.25, 6.75, 12.70, 14.55, 14.96} {1.44, 3.28,11.64,16.14,17.14}

Clearly, h 3 ( n) is minimum phase and h 2 ( n) is maximum phase. 182

4.106 H (z)

(a) The new system function is H'(z)

=

1 --N":-;----

1+

L:k=l QkZ-k

= H().-lz)

'( Hz)

1 = --~N-:----1 + L:t=l aeAk z-t

If Pk is a pole of H(z), then >'PI: is a pole of H'(z). Hence, A < Ip~ul is selected then IptAI < 1 for all k and, hence the system is stable.

(b) y(n)

=- 2:f=l QeAky(n -

k) = zen)

4.107 (a) The impulse response is given in pr10fig 4.76. (b) Reverberator 1: refer to fig 4.77.

12r----.....,.---

---...,....---......-----,

1SJO

1000

2000

->n

Figure 4.76: Reverberator 2: refer to fig 4.77. (c) Unit 2 is a better reverberator. (d) For prime number of D1 , D2, D3 , the reverberations of the signal in the different sections do not overlap which results in the impulse response of the unit being more dense. (e) Refer to fig 4.78. (f) Refer to fig 4.79 for the delays being prime numbers.

4.108 (a) Refer to fig 4.80. (b) Refer to fig 4.81.

183

mpulh reapo,.. tor unlt1

L ~ I

600

500

1.5.__---.....---"'1...----"'1...-----.,.-----.-----,

300 ->n

600

Figure 4.77:

3.....--..,.----...--....,.---,.....--...,...----,..----, 2 :

1

{ A

OWlr.,1I11I.:llln/l'

1-1 -2 -3

0

2

3

6

7

->w

7 ->w

Figure 4.78:

184

1.2,------,-----,.....----...,...----..,.---_

0.8

02

500

1500

1000

2000

Figure 4.79:

~

,

i

10

E

0

/II I

I

-10

---------------------'----..J 1.5 2 2.5 3 3.5

_20'----....... o 0.5

-~w(1'Ild)

_2.5'-----'----.Io---......- -......---~-----_.-J o 0.5 1.5 2 2.5 a 3.5 -~w(1'Ild)

Figure 4.80:

185

100

-8

i

E

50

0

A

1-50 -100 0

0.5

1.5

2.5

2

3

->w(rM}

..

-4'----.......- _ - - a ._ _

o

0.5

~

_ __ L ._ _......._ _

1.5

2.5

2

~

_

_.J

3

3.5

->w(""

Figure 4.81:

4.109 (a)

= = = = = = =

B F, ZI

%2

Z3 Z4

H(z)

10kHz

j

\

20kHz

= 0.5 7.778k = 0.3889 20k 8.889k = 0.44 5 20k 4 6.667k = 0.3334 10k 20k

20k (z - 0.5)(z - 0.3889)(z - 0.4445)(z - 0.3334)

(b) Refer to fig 4.82. (c) It satisfies the objectives but this filter is not recommended in a practical application because in a speech application linear phase for the filter is desired and this filter does not provide linear phase for all frequencies.

4.110 Refer to fig 4.83. Practical: r r r

=0.99 =0.6

= 0.9

w" = i w" f w" =0

=

Bandw~dth : !~8 ~ 0.0245 Bandwldth - 32 - 0.49 Bandwidth 1.1536

=

Theoretical: 186

0

~

i~ ~

-100

I

-HiO

0

0.5

1.5

2

3.5

3

2.5

->w(~

• -2 --OI.---O...l..5~---'---""1.5--""""2--....l2.~5- -.... 3--....J.5 3 ->w(~

Figure 4.82:

r

r

= 0.99, = 0.9,

Bandwidth Bandwidth

= 2(1 -

r)

= 0.02

=2(1 - r) = 0.2

J

For r very close to I, the theoretical and practical values match.

4.111 H(z) H(z)

= =

Hmin(z)

= = = = =

Hap(z)

= =

Let B 1 (z)

B2(Z)

A(z)

=

(1- O.9ei° 41r z- 1 )(1- O.ge-iO.41rz-l)(1_1.5e.i°.s"z-1)(1_1.5e-iO.S1rz-1)

B(z) A(z) (z - 0.gei041r )(z - 0.ge- jO .4 ,,)(z - 1.5eiO .h )(z - 1.5e- jO .61r ) z4

(z - 0.9e.1°. 4 ,,)(z - 0.ge- iO .4 ,,) (z - 1.5e.1°. 6,,)(z - 1.5e- j O. 6,,) Z4

Bdz) B 2 (z) A(z) (z - O.9e.1°.4,,)(z - O.ge- j O. 4 ")(z-1 %4

B 2 (%) B 2(z-1 ) (z - 1.5e j061f )(z - 1.5e- j06 ,,) (Z-1 _ 1.5eJ061f)(z-1 - 1.5e- j061f )

187

-

1.5ejO .611')(z-1

-

1.5e- j 0

6 ,,)

)

4IO.,.....---,--....,.--_r--.......-......-...,....--T---.......---, 30

.... '.0.11 -'.0.1 _,.0.1 .

'20

.10 i

0

-10 _2O ......- - " - -......- - . . & - - - - ' - - - - ' - - - . a . - - - & . . . . - - - J ... -3 -2 -1 0 2 3

->w(rad)

,.,.....---,--....,.---r--__--...,---T---.,.....---,

.... '.o.n -,.0.1

_,.0.'

...

........._ - "_ _......_ _......._ _......._ _-'-_ _.a.-_ _

-3

-2

2

&...._~

3

Figure 4.83:

Hap(z) has a flat magnitude response. To get a flat magnitude response for the system, connect a system which is the inverse of Hmin(z), i.e., 1

I

= (b) Refer to fig 4.84 and fig 4.85.

188

\

. . . . .IM)

pIDII tar ~nMled .~em

101.5

1'X\...O-~--

1

210

Figure 4.84:

I

mag tor Hc(Z)

mag 01 compensated system

10

•

5

s.

0

~

~1\

I

2

j

1

t

0

-5

1\

1_1

-10 -15

-4

-2

0

2

->w(rad) pf'IaM tor Hc(z)

•

->w(rad) ptIaH ot c:ompersat tid s)'Stlm

I

0

,..

A

0:

1~5

I I

-2 -4 -4

2

0

-2

1.5

2

i

-2

•

-1

-2

0 ->w(rld)

2

•

-1:'

Figure 4.85:

189

-2

0 ->w(rad)

2

•

\

Chapter 5

5.1 Since x(n) is real, the real part of the DFT is even, imaginary part odd. Thus, the remaining points are {0.125 + jO.0518, 0, 0.125+ jO.3018}

5.2 (a) %2(1)

x2(/)

=

= =

x2(1), 0n

-->w

cit

dcII(n)

3

4

0.'

i

~0.5

I -0.2 0

10

20

° 0

30

10

20

->n

Figure 5.1:

(c) Refer to fit 5.l. (d) Refer to fit 5.l. (e) Refer to fit 5.2. (f) N=15. Refer to fit 5.3.

I

5.29 Refer to fig 5.4. The time domain aliasing is clearly evident when N=20.

5.30 Refer to fig 5.5. (e)

Zam(n)

=

z( n )cos(21r Ie n) N-l

Xam(w)

=

2: z(n)cos(21rfen)e-j21t/n

n=O

=

~

N-l

2: z(n) [e- j21t (J-Jc)n + e- j21t(J+Jc)n]

n=O

Xam(w)

=

1

2[X(w - we) + X(w

+ we)]

5.31 _{2 12 1 (a) Ck;,-;'3i"'-2.,..·

.. }

(b) Refer to fig 5.6. The DFT of z(n) with N = 128 has a better resolution compared to one

with N

= 64. 205

\

x(n)

06

.

s

A05

II

OA

03 02 01

->n

Figure 5.2:

5.32 (a)

Y(jn)

where sincz

z

=

Y(jn) (b) woP

sin z

A

. (To(n -

Tosanc

= 27rk for an integer k or

no)) e_'

J

Tp(n-ng)

:I

= ~ 11'

Wo

l

2

(c) N-l

Y(w)

L

=

eiwone-jwn

n=O

=

sin!f(w - wo) e-j¥(w-wo) sin w- w g 2

Larger N ~ narrower main lobe of IY(w)l. To in Y(jn) has the same effect.

(d)

Y(k)

= =

IY(k)1

=

Y(w)lw=;;A:

sin1r(k - I) e- j . 7I'(A:-1} sm N Isin1r(k - 1)1

Ism . 71'(1-/)1 N

= N6(k-/) 206

¥7I'(A:-l)

I(n)

x(w)

20

08

15

~06

"104 I

02 0

-20

-10

10

0

-5

20

2

0

->n

3

4

5

10

->w xtilde(n)

ctt 15

1.

i: 1.2 oK

i

u

,

"

1

M

I

" I

'08

~50

5

10

O~~O

15

-->w

-5

0 -->n

Figure 5.3: (e) The frequency samples ~ k fallon the zeros of Y (w). By increasing the sampling by a factor of two, for example, we will obtain a frequency sample between the nulls.

Y(w)lw=mk=*c,

k=O,1, .. ,2N-l I

207

"

X(w)

x(n)

10

1.5

--

i

c

E 5 1\

1

)(

1\

I I

10.5 0 0

2 X(w) w" N=20 6

0 0

8

10

3000

1~9 with ~9aO

1.5

--

i

c

E 5 1\

1

)(

1\

I I

j

\

10.5 0 0

6 2x(w) wi#, N=100

0 0

8

10

fOb

30

100

150

xlR) with N=

1.5

--

0)

c

ca

E 5 1\

1

)(

1\

I I

: 0.5

2

4 -->w

6

8

50

-->n Figure 5.4:

208

x(n)

xc(n)

-V

-

r::

..5r 0 )( -2

Io...--

.-....:.

.

&

.

_

-1

~

o

.............~...........~...................:..L:_..u...._---I

~~

o

300 with ~~8 40..-------------,

300 100 xam(n) 200 2..---------.....---.,

-E

0

~pw)

r::

~

,

\

0

-2~-~------~

300 30..---------.....----,

0 0 150 xJ,w

Figure 7.6:

H(z)

= = =

2(1 - :-1)(1 + -/2z-1 + z-2) (1 + 0.5z- 1)(1 - 0.9z- 1 + 0.8z- 2 ) 2 + (2-/2 - 2)z-1 + (2 - 2-/2)z-2 - 2z- 3 ) 1 - 0.4z- 1 + 0.36z- 2 + 0.405z- 3 A B+Cz- 1 1 + O. 5z- 1 + -1---0-.-9-z--I-+-0-.8-z---1

Refer to fig 7.10 (e)

H(z)

=

1 - 1 Z -1 2

= =

-

1 Z -2 4

1 + Z-1 (1- 0.81z- 1 )(1 + 0.31z- 1) 1.62 -0.62 -1---0-.-8-1z---1 + 1 + 0.31z- 1

Refer to fig 7.11 l-z-1tz-~ (f) H(z) l-z- 1 +O.5z- 2 ~ Complex valued poles and zeros.Refer to fig 7.12 All the above systems are stable.

=

7.10 Refer to fig 7.13

(1)

(2) (3)

H(z)

=

V(z) W(z) y(z)

=

= =

1 1 - 2rcoswoz- 1 + r 2 z- 2 X(z) - rsinwQz-ly(z)

V(z) - rcoswQz- 1 W(z) rcoswoz- 1y(z) - Tsinwoz-lW(z) 239

I

\

Dileet form II:

Direct form I:

'"----y(0)

-'I\)

y(0)

x(O)

)'(0)

:1(0)

)'(0)

Figure 7.7:

By combining (1) and (2) (4)

we

W(z)

I

obtain

=

1 X(z) _ rsinwoz-l Y(z) 1 - rcoswoz- 1 1 - rcoswoz- 1

Use (4) to eliminate W(z) in (3). Thus,

+ r2sin2woz-2] 1 2 2 Y(z)[l - 2rcoswoz- + (r cos wo + r2sin2wo)z-2]

=

Y(z) X{z)

=

Y(z)[(l - rcoswoz- 1)2

7.11 Ao{z) A 1(z)

81{Z) A 2{z)

= =

= = = =

B 2(z) A3(z)

= =

Bo{z) = 1 Ao(z) + k 1B o(z)z-t 1 + !z-l

2

-1 + z-1 2 Al(Z) + k 2BtCz) 1 + 0.3z- l + 0.6z- 2 0.6 + 0.3z- 1 + z-2

A2(z)

+ k 3 B 2 (z) 240

X(z)

= X(z)

\

Dinct form I:

Di:nlet form D:

.......----y(0)1l(0)

y(o)

.......1:

c:..:.dIc: 1l(0)

y(D)

1l(0)

Figure 7.8:

I

1 - 0.12z- 1 + 0.39z- 2

= =

-0.7 + 0.39z- 1

-

-

0.7z- 3

0.12z- 2 + z-3

+ k4B a(z) 2 = 1- ~z-1 + 0.52z- 150 A3(Z)

=

Therefore, H (z)

53 C(l - 150 z-1

0.74z- 3

7.12 Refer to fig 7.14

=

=

bOle

+ buZ-1 + b2 le Z - 2

1 + ouz-1 + 02A:Z- 2 botzl:(n) + wu(n - 1)

YA:(n) wu(n)

= bax(n) - oU:Yk(n) + W21:(n - 1)

W2A:( n)

=

b2k x(n) - 02kYl:(n) 241

!Z-4

3

+ 0.52z- 2 - 0.74z- 3 + ~Z-4)

where C is a constant

HA:(z)

+

\

Direct fonD B:

Direct fonD I:

1(D)X(D)

y(D)

XeD)

y(D)

Figure 7.9:

7.13

=

YJMl G * XIN DO 20 J=l,K YJ =B(J ,0) * XIN + Wl(J) Wl(J) B(J,l)*XIN - A(J,l)*YJ W2(J) = B(J,2)*XIN - A(J,2)*YJ YJMl = YJMl + YJ 20 CONTINUE· YOUT YJMl

=

=

+ W2(J) RETURN

7.14

=

YJMl XIN DO 20 J=l,K W -A(J,l) * WOLDl - A(J,2) * WOLD2 + YMJl YJ = W + B(J,l)*WOLDl + B(J,2)*WOLD2 WOLD2 = WOLD 1 WOLDl = W YJMl = YJ 20 CONTINUE YOl:T = YJ RETURN

=

242

l>irecl farm 1:

I>int:t farm 0: )----~y(0)

y(o)

lI(O)

Figure 7.10:

7.15 H(z)

= A 2(z) = 1 +2z- 1 + !z-2 3 1 B 2(z) = ! + 2z- + z-2 3 1 le 2 = 3 A2(Z) - k 2B 2(z) A 1(z) = 1- Ie~ lei

= =

1+

~z-1 2

3

2

7.16 (a)

1 + 13 z - 1 - 13 z - 2 _13 + 13".. -1 + ...,.-2

243

]

Dinct form I:

Dinlct form D:

~----y(D)X(D)

y(D)

Y(D)

Figure 7.11: J

zeros at z (b)

\

= -l,e±ii = A2(Z) - z-1 B 2(z) = 1 + ~z-1 - ~z-2 _ z-3 The zeros are z

=

3 3 1 -5 ± jv'IT ,

=

6

=

(c) If the magnitude of the last coefficient IkNI 1, i.e., kN ±1, all the zeros lie on the unit circle. (d) Refer to fig 7.15. We observe that the filters are linear phase filters with phase jumps at the zeros of H(z).

7.17 (a) Refer to fig 7.16

244

Direct form I:

Direct form II, cascade, parallel:

yen) y(n)x(n)

Figure 7.12:

j

\

z(n)

Idn) 91(n) h(n) 92(n) h(n) = 13(n)

= = = = = = = = =

6(n) 6(n)

+ 0.656(n -

1)

0.656(n) + 6(n - 1)

hen) - 0.3491(n - 1) 6(n) + 0.4296(n - 1) - 0.346(n - 2)

+ 91(n - 1) -0.346(n) + 0.4296(n - 1) + 6(n - 2) 12(n) + 0.892(n - 1) -0.34h(n)

6(n)

1) + 0.00326(n - 2)

+ 0.1576(n -

+ 0.86(n -

(b) H(z) = 1 + 0.157z- 1 + 0.0032z- 2 + 0.8z- 3 . Refer to fig 7.17

7.18 (a)

H(z)

=

A3(Z) 8 3 (:) k3

=

= =

C3(Z) A3(Z) 1 + 0.9z- 1 - 0.8z- 2 + 0.5z- 3 0.5 - 0.8z- 1 + 0.9:- 2 + z-3 0.5

245

3)

Figure 7.13:

A 2(z)

B 2(z) k2

Al (z)

B 1 (z) k1

C 3(z) D3(z) k3

C 2(z)

= = = = = = = = = = = = =

D 2(z)

=

k2

=

C 1(z)

A3(Z) - k3B3(Z) 1 - k~ 1 + 1.73z- 1 - 1.67 z-2 -1.67+ 1.73z- 1 + z-2 -1.67 A2(Z) - k 2B 2 (z) 1- k~ 1+1.62z- 1 1.62 + Z-1 1.62 1 + 2z- 1 + 3z- 2 + 2z- 3 2 + 3z- 1 + 2z- 2 + z-3 2 C3 (z) - k 3D3(Z) 1- k§ 1 + -z 4 -1 + -z 1 -2 3 3 4 -1 + z -2 -1 + -z 3 3 1

= =

D 1 (z)

= 246

3

C2(Z) - k 2D 2(z) 1 - k~ 1 + !z-1 4 3 -1 -+z 4

~ (n)

-a2k

Figure 7.14:

k1 C 3 (z)

= = =

3

-

4

+ VIDl(Z) + V2D2(Z) + V3D3(Z) 1 + 2z- 1 + 3z- 2 + 2z- 3

VQ

From the equations, we obtain 107

VQ

VI V2 V3

= 48 13 = --4 = -1 = 2

The equivalent lattice-ladder structure is: Refer to fig 7.18 (b) A3(Z) 1 + 0.9z- 1 - 0.8z- 2 + 0.5z- 3 , Ikd> 1 and Ik 2 1 > 1 ~ the system is unstable,

=

7.19 Refer to fig 7.19

Y(z) C(z)

= [rsineX(z) + rcosey(z) - rsineC(z)] Z-1 = [-rcoseX(z) + rsiney(z) + rcoseC(z)] z-1

H(z)

=

Y(z) X(z)

247

j "

.

2

: :"1 :J:

'0

I 0

t

~ -1 I I

-2 0

0,05

0.1

0.15

0.2S

0.2

0.3

0.35

0."

O.~

0.5

0.3

0.35

0."

O.~

0.5

-~~)

..

..

!2 'a

to ~ -2 I I

....0

0.05

0.1

0.2S

0.2

0.15

-~ '"ec(Hz)

Figure 7.15:

rsinez- l 1 - 2rcos(3z-l + r 2z- 2 r n sin(8n)u(n) rsinez(n - 1) + 2rcosey(n - 1) - r 2 y(n - 2)

=

= = The system has a zero at z = 0 and poles at z = re*i e , Hence, h(n) and y(n)

I

7.20 H(:)

= =

S(z) s(n) p ~ Ql

Q2

A ~

ql

q2 vl(n + 1) v2(n)

= = =

= = = = =

= = = =

1 1 - 2rcoswoz- l 1+

+ r 2z- 2

rcosw - j 0

reo,2wp 2.. nwo

Z - (rcoswo + jrsinwo) rcosw - J' reo.,2wp

o

2"nwo

z - (rcoswo + jrsinwo) vl(n) + jV2(n) Q1

+

rcosw + j 0

z - (rcoswo - jrsinwo)

X() Z

+ j Q2

rcoswo rsinwo

+ jq2 rcoswo -rcoswQ 2sinwo Qlvl(n) - Q2v2(n) + qlz(n) rcoswOv1(n) - r,inwov2(n) + rcoswQz(n) Q2vt{n) + Qlv2(n) + q2 z (n) , -rcoswo rsznWQvl(n) + rcoswOv2(n) + 2' x(n) sznwo

ql

248

,.eo,2wp 211 nw o

1

f n) = y(n)

&3 (n)

Figure 7.16: or, equivalently, v(n -

+ 1) = [ rc~swQ

rsznwQ

-rSinWQ] v (n) rcoswQ -

=

y( n)

=

+

[rCOSWQ] rco~wo 2, inw o

Z

(n)

+ s· (n) + z (n) 2vI(n) + zen) n)

S(

or, equivalently,

= [2

yen)

O]~(n)

+ zen)

where V(n) -

_

-

[ VI (n) V2

() n

]

7.21 (a)

k1 Al (z)

B 1(z) A 2 (z) B 2{z) A3(z)

H(z)

=

B3(Z) A 4 (z)

= = = = =

0.6 1 + 0.6z- 1 0.6 + z-1

Adz) + k 2 Bt{z)z-1 1 + 0.78z- 1 + 0.3z- 2 0.3 + 0.78z- 1 + z-2 A2(z) + k3B 2 (z)z-1 1 + 0.93z- 1 + 0.69z- 2 + 0.5z- 3 0.5 + 0.69z- 1 + 0.93z- 2 + z-3 = A3(Z) + t4 B3(Z)z-1 3 2 1 = 1 + 1.38z- + 1.311z- + 1.337z-

= = = =

249

+ 0.9z- 4

len)

......- - y ( n )

Figure 7.17: (b) Refer to fig 7.20

,

7.22

\

(a)

From (7.3.38) we have

+ k2 )

= = = =

k1

+

y(n) But, y(n) Hence, k 2 and, k 1 (1

-k 1(l + k 2 )y(n - 1) - k 2 y(n - 2) + z(n) 2rcoswoy(n - 1) - r 2 y(n - 2) + z(n) r2 -2rcoswo 2rcoswo ---1 + r2

Refer to fig 7.21 (b) \\7hen r = I, the system becomes an oscillator.

7.23 H(z)

=

=

1 - 0.8:- 1 + 0.15:- 2 1 + 0.1:- 1 - 0.72:- 2 B(:) A(z)

1 For the all-pole system A(z)' we have

k1(l

+ k2 ) k2

250

=

0.1

= -0.72

Figure 7.18: ~

k1 k2 For the all-zero system, C 2 (z) A2(Z) B 2(z) k2

= = = = = =

Al(Z)

=

= B 1 (z) = k1 = Ao(z) =

0.357 0.72 1 - 0.8z- 1 + 0.15z- 2 1 - 0.8z- 1 + 0.15z- 2 0.15 - 0.8z- 1 + z-2 0.15 A2(Z) - k2B2(Z) 1- k~

1 - 0.696z- 1 -0.696 + z-1 -0.696

Bo(z) = 1 2

C 2(z)

= L

vmBm(z)

m=Q

= =

VQ + VIBl(Z) + V2B2(Z) 1 - 0.8z- 1 + 0.15:- 2

= = = = = =

0.15

The solution is:

V2 V1 - 0.1 8v 2 VQ - 0.696v1 + 0.15v2 ~ VQ V1 V2 Thus the lattice-ladder structure is: Refer to fig 7.22

251

-0.8 1 1.5

-0.68

0.15

rcos a

n)

r sina

x(n)

-rcos a rcos

a

Figure 7.19:

7.24 M

y(n)

L: blcz(n -

=

It)

lc=O

+ 1) = Vi +1 ( n ), vM(n + 1) = z(M)

i

Let vi(n

= 1, 2, ... , M

- 1

M

Then, y(n)

boz(n) +

=

L: b vM+l-i(n) i

i= 1

Then 0 0

Vl(n + 1) V2(n + 1)

1

0

0

1

0 0

0 0

= VM-l(n + 1) vM(n + 1)

y(n)

0

+ 1

0 0

= [bM

. ., .

"

0

vM-I(n) vM(n)

VI(n) V2(n) bM-I

...

bl

0 0

VI (n) V2(n)

z(n) 0

1

1

]:

+boz(n)

[ VM(n) General representation:

= FV(n)+!z(n) y(n) = f'V(n) + boz(n)

!!(n+l)

The general state-space realization for this system is shown in figure 7.36 where d 252

= bo.

Direct form:

s(n)

)

.

o

. . 0

+

z-I

0.5

+

z-I

0.9

+

z·1

z-1

+

Figure 7.20:

7.25 = y( n

y( n ) a2

= =

bo

=

al

- 1)

[O~l

:],

2)

+ z (n )

-1

-0.11 1

Type 1 representation:

£=

+ 0.11 y( n -

F

[

n'

[= [

~.ll ] ,

d= 1

Type 2 representation:

,F = -

[01 0.11] 1 '

_ [ 0.11 ]

i-

1

d=1

I

Refer to fig i .23

7.26 = y( n - 1) + y( n - 2) + z (n ) = -1 = 1

yen) al

=

a2

60 Type 1 representation:

:]. F[n· F=[~ :] F [:],

F=[~

f=[:]'

d=l

F[n,

d=1

Type 2 representation:

,

253

I-----,...-------'y(n)

Figure 7.21: Diagonal form (see example 7.5.2) . F-

- -

[~ 2 0

0

ft

I2

] I

7.27

1) ] = [01 0.11] [ vl(n) ] + [0.11 ] z(n) 1 v2(n) 0

vl(n+ [ v2(n + 1)

y(n) zV1 (z) z V2 ( z) Y(z)

= [0

1] [

vl(n) ] v2(n)

= O.llV2 (z) + O.11X(z)

= =

~ H(z) = =

VI ( z)

+ V2 ( z) + X (z )

V2 (z)

+ X(z)

Y(z) X(z) 1 1 - z-1 - 0.l1z- 2 11

12 1-1.1z- 1

h(n)

+ z(n)

+

1

12 1+0.lz- 1

= = g~(l.l)n + 112(-O.lt] u(n)

7.28 v(n

-

+ 1)

= [0 1 -0 2]£(n) + [ 9291 02

01

254

] z(n)

VI = -0.68

v 2= 0.15

+

y(n)

Figure 7.22:

1'1-11 A

-_I

A - Qt -02

Q2

A_

0t

I

I= (' A

-

°t

)2

-

(

°2

)2

=0

7.29 Refer to fig 7.24 To obtain the transpose structure, we reverse the directions of all branch transmittances and interchange the input and output. Thus we obtain the structure shown below: Refer to fig 7.25

7.30 (a) H(z) = (b)

1_::J;jZ-l+ 0 .25z -2 1-0.8r 1 +064z- 5 .

Refer to fig 7.26

al

=

a2

= 0.64

bo

=

-0.8

255

1

\

Type I Realizatioa:

y(a)

Ty.. 2 "'lizetina·

(a)

Figure 7.23: I

Type 1 representation:

Type 2 representation: -0.39

i = [ 0.8- ~

]

d=1 I

Refer to fig 7.27 (c)

H(z)

=

1 - 0.707z- 1 + 0.25z- 2 1 - 0.8z- 1 + 0.64z- 2

= 25 + 64

0.609(1- coS(j)z-l) 1 - 2(0.8cosj)z-1 + (0.8)2 z -2

-0.208sin( i )z-1 1 - 2(0.8cosi)z-1 + (0.8)2 z -2

+--~-~-~--~~

Therefore, h( n)

1r

= 0.3916(n) + [0.609(0.8tcos( 'in) 256

.

7r

0.208(0.8)n szn ( '3 n ))u(n)

\

1

a _x(_n.-..)---+--~-"""'-_.M.-------7L..-------------.&----.

y(n)

a

Figure 7.24: State-space realization of coupled-form structure

I

From the state-space representation, we have h(n)

= [-0.39

0.8 -

~]

0 0.81] [ 0.64

n-l [

0] 1 u(n -

1) + 6(n)

(d) Coupled-form state space realization:

H(z) Hence, Ai: Pi:

= 1+

= =

0.0464 + jO.255

z - (0.4 + jO.4V3) 0.0464 + jO.255 9u 0.4 + jO.4v'3

au

0.0464 - jO.255

+ -----~

z - (0.4 - jO.4V3) 9i:2 0.255

= 0.0464

= 0.4

ai:2

=

= 0.4V3

Therefore,

F

-=

[

0.4 -0.4V3

-0.4V3 ] 0.4 '

i

=[

(e)

(a)H(z) = (b)

1

I--t

Z-I+o 25z- 2 2z- I +O.64z- 2

Refer to fig 7.28

(c)

257

0.0464 ] 0.255 '

d=l

l

1 a.

yen)

~---+-_....&..-_~

x(n)

_ _L-_~J.-~_----I.

a.

Figure 7.25:

h( n)

= [-0.39

0.3v'2]

1] 0 0.8-/2 [ -0.64

0] 1 u( n -

1) + 6{ n)

g=[~],

d=!

n-l [

(d)

F _ [0.4-/2 -0.4../2] 0.4-/2 0.4../2

I

q

=[

0.212 ] 0.133

I

7.31 _

(a) H(z) ( b) H(z) =

1+z- 1 1 r::J=T'1-0.8J'2z-1+0.64z-;'

i:- +

231 1 1-

-131+296z-

1

1-0.BJ2z-1+0.64z-~

Refer to fig 7.29

(c) 1 + z-1

a3

= 1 _ 1.63z- 1 + 1.21z- 2 - 0.32z-3 = -1.63 = 1.21 = -0.32

bo b1

= =

H(z) al a2

1 1

258

Direct form II :

Transposed form :

y(n)

x(n

where 8=0.8. b

x(n) . . , . . . - - -..

. . . . . - - -.......- - v ( n )

b

•

c

d

=if}. n.. c =0.25 and d =-0.64

Figure 7.26: Type 1 representation:

F=

[~ ~ ~], 0.32 -1.21 1.63

i=

U],

f=

[0.32 ], -1.21

d=l

2.63

Type 2 representation:

E.. =

O~ [

0

0 1

0.32] -1.21 , 1.63

i

= [0.32 -1.21 2.63

],

1=

[~l

d=l

7.32 H(z) where H 1 (z)

= =

0.5H 1(z)H2(Z) 1 + 0.4z- 1 - 1.2z- 2 + 2z- 3 1 1 - 0.6z- 1 + 0.2z- 2 + 0.5z- 3

H (z) can be realized as a cascade or two lattice filters (an all·zero lattice and an aU· pole lattice) or as a lattice-ladder filter. Let us choose a cascade configuration:

For H 1 (z) we have

A 3 (z) B3(Z) 259

= 1 + 0.4z- 1 - 1.2z- 2 + 2z- 3 = 2 - 1.2z- 1 + 0.4z- 2 + z-3

Type I Ra.lizabaa:

.0.39

)'(D)

.(D) OJJ..V=;

Figure 7.27:

k3 A2(Z)

B 2(z)

k2 A 1(z)

1: 1

= = = = = = = =

2 A3(Z) - k3B 3(z) 1 - k~ 14 -1 2_ 1- -z +-z 2 15 3 2 14 -1 -2 - - -z + Z 3 15 2 3 A2(Z) - k 2B 2(z) 1 - ki

= = = =

1 - 0.6z- 1 + O.2z- 2 + 0.5z- 3 0.5 + 0.2z- 1 - 0.6z- 2 + z-3

1 _ 14 z-1 15 14 15

For H2(Z) we have

F3(Z) G3(Z) 1:3 F2(Z)

= 260

0.5 F3(Z) - k3G 3(Z) 1 - k5 14 -1 2_ 2 1 - 15 z + 3z

Direct form II :

Transposed form :

n)

~---.,..._-_.y(n)

x ( n ) - - - - -..

b

a

c

d

where a=c.m. b = • Vi 12. c = 0.25 and d = -0.64

Figure 7.28:

j

\

G2(Z)

=

t2 = F 1(z)

= =

t1 =

14 -1 15

2 3 2

- - -z 3

F2(Z) - t2G2(Z) 1- t~ 1 _ 14 z-1 25 14 25

The all-zero lattice has reflection coefficients

t3 = 2 2 t2 =

3 14 15

tl = The all-pole lattice has reflection coefficients

t3 = t2 = tl = 261

+ z -2

1 2 -23 14 25

Parallel:

cascade: x(n)

(n) x(n)

Figure 7.29: Refer to fig 7.30

7.33 (a) ,

IAI-FI=

A2

-

I 0.81A

-1 A-I

I=A(A-l)+0.81=0

A + 0.81 = 0 ~ A1,2 = 0.5 ± i 1. 50

(b)

~(z)

=

Z( z

~ll(Z)

I - E) -1

=

Z2 -

=

Z

Z z2 _

Z

+ 0.81 262

1

+ 0.81

[Z-l 1

X'D)

.1

-1__H_iZ_)

I

HI (z)

1_/2_-..-....y(n)

Figure 7.30: 1-

Z-1

=

1-

11>11 (n)

=

(0.9)" [008(0.98n) -

41 12 (z)

41 22 (z)

= = = = =

4>22(n)

=

11>12(n)

41 21 (z) 4>21 (n)

z2 -

z-1

+ 0.81z- 2

~8in(0.98n)] u( n)

-0.8z Z + 0.81

~(O.9t sin(0.98n )u( n) z

z2 - Z + 0.81 -1.08(0.9)" sin(0.98n)u( n) z2

z2 -

Z

+ 0.81

(0.9)" [008(0.98n) +

~8in(0.98n)] u(n)

(c)

H(z)

h(n)

= = = = = = =

-1 0.81 1 0 -1 1 - Z-1 + 0.81z- 2 (0.9)"cos(0.98n)u(n)

+ [-0.Scos(0.98(n -

1)) - sin(0.98(n - 1))] (0.9)"-l u(n - 1)

(d) For the zercrinput repsonse Yzi(n) we have

z[Y(z) - ![(O)]

= FV(z) 263

Yzi (:) = f~l(Z)

= =

f(zl- F)-l z\/(O)

Yzi(n)

=

.:-1

V(z)

(:1- F)-l zV(O)

{Yzi(Z)}

For the zercrstate repsonse Yz ~ (n) we have

zV(z) Yu(z) Yu(z)

X(z) yu(n) y(n)

= = = = = = =

FV(z) + iX(z) !V(z) + X(z) 9.(zl- F)-l i X(z) + X(z), Z-1

1 - z-1 z-1 {Yu(z)} Yzi(n)

+ yz,(n)

(O.9t [2cos(O.98n) + 1.1sin(O.98n)] u( n)

(e) Refer to fig 7.31

......-..---.. y(n)

Figure 7.31:

7.34 Using the same method as in 7.33 we obtain the following answers: (a)

264

roots: z

= -2,-1

(b)

¢ll(n)

= ¢12(n) = ¢21(n) = ¢22(n) =

[-1(-2)" [-1(-2)"

+ 2(-I)"]u(n) + 2(-I)"]u(n)

-2¢12(n)

[2(-2)" - (-I)"]u(n)

(c)

H(z)

h(n)

= =

1

z2 + 3z + 2 [_(_2)"-1 + (-I)"]u(n - 1)

(d)

y(n) = h(n)

=

+ Y.u(n) [_(_2)" + (-I)"]u(n) + [~(-2)" - !(-1)" + !]u(n)

Yzi(n)

3

2

6

(e) Refer to fig 7.32

t--:----,.----... y(n)

-3 Figure 7.32:

7.35 Using the same method for the solution as in 7.33 1 we obtain the following answers: (a)

C(z)

= z2 + 0.6z - 0.07 265

roots: z

=

0.1, -0.7

(b)

=4>22(n) = 4>12(n) = 4>21(n) =

4>l1(n)

1 [-(0.1)"

2

1 + -(-0.7)"]u(n)

2

[!(O.l)" - !(-0.7)"]u(n) 2 2

(c)

H(z) h(n) (d) y(n)

= (O.I)"u(n) + [1: -

= =

1 z _ 0.1

6(n)

+1

+ (0.1)"-1 u(n -

1)

Igo(O.l)"]u(n) Refer to fig 7.33

-0.3

x(n) --..,.--.---.....

-0.3

Figure 7.33:

7.36 (a)

al

= =

1 + z-1 1 _ 0.9z- 1 + 0.08z-2 -0.9

a2

=

0.08

H(z)

266

bo b1

= 1 = 1

Type 1 representation:

F= [ -~08 0~9 ] ,

i=

[n,

- [ -0.08] 1.9 '

! -

d=l

Type 2 representation:

F=[~

-0.08 ] 0.9

2. =

I

[ -0.08] 1.9

f=

I

[n,

d=1

(b) Refer to fig 7.34

Parallel:

j

\

x(n)

y(n)

Figure 7.34:

H(z)

=

1 + z-1 (1- 0.lz- 1 )(l- 0.8z- 1 ) 11

=

-7"

1-

0.lz- 1

+

18

7"

1 - 0.8z- 1

Cascade:

H(z)

=

1 + z-1 (1- 0.8z- 1 )(1- 0.lz- 1 ) 267

= F1

F2

H 1 (z)H 2 (z)

= [0.8), = [0.1],

i 1 = [1), ~h = [1],

9..1

= [1.8),

[2

= [0.1],

(c) Method 1: 11

H(z)

=

h(n)

=

.!!

+

-7

1 - 0.1:- 1

7

1 - 0.8z- 1

+

[_ 1; (0.1)"

18 (0.8)"] u(n) 7

Method 2: From (a) we have, !L(n

= = = =

+ 1) y(n) Y(z)

Hence, H(z)

+ iz(n) £'!L(n) + dz(n) Fv(n)

9..'(z1 - F)-l i X(z) 9..'(z1 - E.)-l i

r

+ dX(z),

X(z) = 1

+d

[Oi~8 [0.~8 Z~~.9] [~] + 1

H(z)=

=

H (z)

1 + z-1 -1---0-.9-z---1-+-O-.-O-8z---2

which is the same system function as in (a).

7.37

I "

(c)

~(z)=z(z1-F)-1=

z2 -

: -z + 4

1

-l!l

[z-~ -~] 1 2

1_!z-1

! (1- ! z

( 1- z -

_

- [

_ [

f(n) -

1 )( 1- I

r

1)

z-

1)

z-1 • -1 )( 1- {

-( ~)nu(n) + 2(i)"u(n) _( ~)"+lu(n) + !(~)"u(n)

(d) 1 [ -j

13] 12

= (z -

~)(z -

H(z)= _ 1+1,-1 H(z) - 1-il-l+ll-~ (e) det{zL - F) = z2 -!z + Hence, the system is stable.

i

[zi

4(~)nu(n) - 4(i)"u(n) ] 2(~)"u(n) - (~)"u(n)

-1] [0]

z-!

1

+1

i) = o. z = !' i are both inside the 268

unit circle.

7.38 H(z) = ~t(zL - F)-Ii + d H ( z)

= [0 1 J [ ~1 ~~ \1 ] [ ~.11 H(z)

Therefore, h( n )

]

+1

1

=

1-

=

12 1 - l.lz-l

Z-l 11

O.11z-2

+

1

12 1 + O.lz-l

= [~~(l.lt + 1~(-O.lt] u(n)

7.39 (a) Refer to fig 7.35 (b)

x(n)

y(n)

Figure 7.35:

1:( I)

= [ _0"6 ~ 1 ] 1(0) + [ ~ ] z(O) = [ ~ ] 269

y(n)

h(O) h(l)

= = =

11 [18 2]lJ.( n)

+ 2z( n)

11 [18 2]lJ.(0)

+ 2z(0) = 2

11 [18 2]lJ.(1)

+ 2z(l) = 2

proceeding in the same way, we obtain

h(2) h(3) h(4) h(5)

5 = --8 = 0 25 = 128 25 = 128

(c)

H(z)

Therefore, y(n)

= = =

H..t(zI - F)-Ii + d 2 + 4z- 1 + 2z- 2 1 + z-1 + '!'z-2 16 5 -y(n - 1) - 16 y (n - 2)

+ 2z(n) + 4z(n -

1) + 2z(n - 2)

i

I

\

=

(d) For z(n) 6(n), we obtain y(O) (e). Refer to fig 7.36

= 2,y(1) =2,y(2) = -i, etc.

2

x(n D

4 D

Figure 7.36:

270

These results check with (b)

~~y(n)

7.40 (a) Refer to fig 7.37

Figure 7.37: All-zero lattice

I

+ 1) = x(n) V2 (n + 1) = k IX ( n) + v I ( n ) vdn

v3(n + 1)

vN(n

=

+ 1) = y(n) =

k 2 x(n) + klk2vl(n)

+ k1kN-IvI(n) + k 2kN - I v2(n) + ... + vN-l(n) x(n) + k1vl(n) + k2v2(n) + ... + kNvN(n) kN_Ix(n)

0 0

0 0

0

1

F=

+ v2(n)

k l k2

1

k 1 k3

k 2k3

t1kN-l

t2 t N-I

0

1

0

0

i=

k1 t2

kN- I

1 0

(b) If we proceed as in (a) we obtain 271

\

F=

-k 1

-k 2

1 - kr

-k 2

0

1 - k~

-kN

1

k1 k2

i=

-kg

1 - k~_1

kN-l

-kN

7.41 (a)

For positive numbers, range is 01. 00 ... 0 z2 100I

01. 11 ... 1 x20 111 ~ 11 2.5596875x 10 2

~

11 or 7.8125z10- 3

negitive numbers 10. 11 ... 1 z2 100I

10.00 ... 0 x20 111 ~ 11 -2.56z10 2

~

11 or - 7.8163z10- 3

(b) For positive numbers, range is 01. 00 ... 0 Z210000001

01. 11 ... 1 Z201111111

---......-. 23

~

23

or 5.8774717z10-

39

3.4028234z 10 38

negitive numbers 10. 11 ... 1 z210000001

10.00 ... 0 X201111111

~

~

23

23

or - 5.8774724z10- 39

-3.4028236z 1038

7.42 (a) Refer to fig 7.38

= (1+a1z-1+02Z-2)-1 poles

Zpl,2

= -al ± Jai 2

for stability

(i)a~ - 4a2 -al - Ja~ - 4a2 2 ~ )""a-2 ---4-a-2 1 272

~

0

>

-1


"1 = ml

= =

=> Iz1l

>

2'

=> IZ 31

>

1, overflow

or

(Til = 0, ml = 1, C = 0)

or

(10 ml 0 ...

(i)

(ii)(Til = 1, ml = 0, C = 0) _ 1 => (10"1 0 ... Ohol > "2

1

IZ 21> 2

Ohol

1

and IClol

>

=> IZ 31

>

2

1, overflow

(b)

Z1 Z2 Z3

+ Z2 Z1 + Z2 + Z3 Z1

= 0100 = 0110 = = =

-0110= 1010 1 0 1 0 overflow o 1 00, correct result I

276

>

~

7.45 (a)

=

H(z)

-a

+ z-l

1 - az- 1

-a + e- jw

2

= 11 - ae- jw I (-a + cosw)2 + (-sinw)2

=

= (b) Refer to fig 7.42 (c) If Iill 1- ill, where

=

a means

(1- acosw)2 + (asinw)2 a 2 - 2acosw + 1 -----~=1 Vw 1 - 2acosw + a 2

the quantized value of a, then the filter remains all-pass.

y n) z -1 a

Figure 7.42: (d) Refer to fig 7.43 (e) Yes, it is still all-pass.

7.46 (a) y(n) = [2(~)n - (~)n] u(n) (b) Quantization table

x

>

1 1- 32 277

z=1

x(n)

.....----~y(n)

---~""""'"---"'-'

a +

Figure 7.43: 31

->z 32 29

>

29 32 27 32

15 16 14 z=16

z=-

z=-

->z 32 -

>

->z

1 32 -

>

1 32

z


n

Figure 7.50:

=

y(i) c*x(i) + yl(i) end (d) Refer to fig 7.56.

+ y2(i);

288

60

7D

80

90

100

v 0=0.7829

VI =0.4336

v 2=-0.1563

+

y(n)

Figure 7.53:

fR tor part a

IR tor part b

0.5

0.5 C

'2

"E

"E

Oil

0

-05 0

50

-05 0

100

IR for part c

50

100

IR tor part f

08 06 05

O.

i-

.e. .s:.

02

011

~

Or -0.2

0

50 ->n

100

-0.5 0

Figure 7.54:

290

50 ->n

100

,..----1 c

Z(o.. } -+-_ _-1

(o)

L.----f ~ (z)

PIrallcI form ItructIIrc aia. 2Drknter oouplecHona ~ ICtiDaI

x(n)

Figure 7.55:

Parallel form lITlpulse response

3

· ·

25

'2

2

~1

5

I I

· 05~ 0 0

10

20

30

.a

SO ->n

60

70

80

go

100

Parallel structure CO~ed term Impulse response

.

08 ~06 A

10. 02 0 0

.

10

20

30

40

SO ->n

Figure 7.56:

291

60

70

80

go

100

Chapter 8

8.1 (a) To obtain the desired length of 25, a delay of

hd(n)

= = =

Then, h(n)

= =

Hd(W)

= 12 is incorporated into Hd(W). Hence,

25;1

11'

le- i 12w,

0< - Iwi -< -6

otherwise

0,

it

Hd(w)e-Jwndw 211' -f sini(n - 12) 1I'(n - 12) hd(n)w(n)

=

where w( n) is a rectangular window of length N 25. L~~o h(n)e- iwn ~ plot IH(w)\ and LH(w). Refer to fig 8.1. (c) Hamming window:

(b)H(w)

=

5Or---..---'T'""""-""I""'""-_--r--....,.--...,..---r-~-__,

i

0

,

-50

II I

I

1_100

-1500

0.05

0.1

0.15

0.2 0.25 0.3 - > FNq(HI)

0.35

0..

O.~

0.5

• __-..---'T'""""-""I""'""-_--r--........-...,..---r-~-__,

~O

0.05

0.1

0.15

0.2

0.25

0.3

- > FNq(HI)

Figure 8.1:

w(n)

=

n1l'

(0.54 - 0.46COST2 293

0.35

OA

O.~

0.5

h(n)

=

hd(n)w(n)

for 0 $ n $ 24

Refer to fig 8.2. (d) Bartlett window: so.......--r--~--,..---,--r---r--..,....---r---,r----,

,

0

I

-so

A I I

1_100

-1SO---""------...-----&.--.............--...---'-~ o O.OS 0.1 0.15 0.2 0.2S 0.3 0.35 0.4 0.45 0.5 - > fNq(Hz)

4r---..,.---r--..,..-__--r----r---r--..,..---,--.,

-'---""------....---'-_&.--......._ ....... o O.OS 0.1 0.15 0.2 0.2S 0.3 0.35

_--'----'-~

0.4

0.45

0.5

- > fNq(Hz)

Figure 8.2:

w(n)

=

1 _ 2(n - 12) 24

0$ n $ 24

Refer to fig 8.3. O....-==::z:::---r---r-.....,--,----r---r---r---,---,

-soO

0.05

0.1

0.15

0.2 0.25 0.3 - > Frec(Hz)

0..

0.4

0.45

0.5

--r--_-..,....--...-.. . . --.,

4.---..,.--..,...--...-.......

-'0

0.05

0.1

0.15

0.2

0.25

0.3

->~)

Figure 8.3:

294

0.35

0.4

0.45

0.5

8.2 (a) le- i

= = =

0

Iwl ~

12W,

~ < Iwl < ~ - 3

'6 -

11"

i,

w

-3 < Iwl < -

7r

.

-2 Htl(w)e-Jwndw 7r -W' 6(n) _ sini(n - 12) + sini(n - 12) w(n - 12) 7r(n - 12)

(b) Rectangular window:

=

w( n)

1, 0 ~ n ~ 24 0, otherwise

= Refer to fig 8.4. (c) Hamming window:

10~-,--,--,--or---or---.,.---.,----r---..,..----.

o~-~

ilIi -10

1-20 ": -30 I

~

-500

0.05

0.1

0.15

0.2 0.25 0.3 - > Fre4(Hz)

0.35

0.4

0.45

0.5

4,.--,--or---.,.---....---.,.---.,.---.,---..,..--..,..---,

-4

0

0.05

0.1

0.15

0.25

0.2

0.3

0.35

0.•

->F~)

Figure 8.4:

nw

w(n)

= (0.54 - O.46cos12

h(n)

= hd(n)w(n) 24

H(w)

=

E h(n)e-iwn n=O

Refer to fig 8.5. (d) Bartlett window: w(n)

=

(n - 12)

1-

12 295

'

o ~ n ~ 24

0.45

0.5

5.---~-~-..,.---r--.....---r--

il5 I

__---.--__---.

-5

'0 1I I

-15 _20L.--~-""'-"""'--'---"""'---'--"""'-~---'--~

o

0.05

0.1

0.15

0.2 0.25 0.3 - > FnqHz)

0.35

o.~

O.~

0.5

__---.--__---.

~~-~-~-..,.---r--.....---r--

-4L.-_~_..&-.._"""'-""""-~-"""'-"""'-~---'--~

o

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

O.~

0.5

->~Hz)

Figure 8.5: Refer to fig 8.6. Note that the magnitude responses in (c) and (d) are poor because the transition region is wide. To obtain sharper cut-off, we must increase the length N of the filter. J

8.3

=

(a) Hanning window: wen) ~(1 - eos~;), 0 ~ n ~ 24. Refer to fig 8.7. (b) Blackman window: w(n) = 0.42 - O.Seos ~; + 0.08eos "6n . Refer to fig 8.8.

8.4 (a) Hanning window: Refer to fig 8.9. (b) Blackman window: Refer to fig 8.10. The results are still relatively poor for these window functions.

8.5 M

=

4,

Hr(O)

= L

= 1,

1f'

B r ("2)

"-1

Br(w)

2

M-l - n)] h(n)eos[w(-2-

n=O

1

= 2L n=O 1

At w

3 h(n)eos[w(2 - n))

=0, Br(O) = 1 = 2 L h(n)cos[O) n=O 2[h(O) + h(l)] = 1 (1 ) 296

1

=2

\

Oc===--r---r--,----r-~....-

i

f

.....-

....._ - _ - _

-s

A

II -10 _15'-----"-.......---'----'--..&...._'""-_'--___"_---"'-_--' o 0.0& 0.1 0.15 0.2 0.2S 0.3 0.35 0.. O.~ 0.5 ->m«Hz)

~'-----"-.......---'--.....a...-..&....-__- ' - - - - - " -.......- - - '

o

0.05

0.1

0.15

0.2 0.2S 0.3 - > Freq(Hz)

0.35

0..

o.~

0.5

Figure 8.6:

At w =

7f'

7f'

1

1

2' H r ("2) = '2 =

-h(O) + h(1) Solving (1) and (2), we get h(O)

2

L

3

h(n)cos[~( - - n)] 2 2 (2)

n=O

=

0.354

= = h(2) = h(3) = Hence, hen) =

0.073 and 0.427

h(1)

h(l) h(O) {0.073, 0.427, 0.427, 0.073}

8.6 M = 15.Hr (27rk) = 15 .

{I,0,

k=

0, 1,2,3

k=4,5,6,7

M-3

Br(w) hen) hen)

= = =

M-l ~ M-l h ( - - ) + 2 L- h(n)cosw(-- - n) 2 n=O 2 h(M-l-n) h(14-n) 6

Br(w)

= h(7) + 2

L h(n )cosw(7 -

n)

v.-'

n=O

Solving the above eqn yields, h( n)

=

{0.3189, 0.0341, -0.1079, -0.0365,0.0667,0.0412, -0.0498,0.4667 0.4667, -0.0498,0.0412,0.0667, -0.0365, -0.1079, 0.0341, 0.3189} 297

i

0

~

I

-SO

A I I

1_100

-1500

0.05

0.1

0.15

0.2 0.25 0.3 - > Freq(Hz)

0.35

004

0.65

0.5

.........~--",,_--a._--a.._--'-_~_"""'_-'-_"""'---J 0.05 0.1 0.15 0.2 0.25 0.3 0.35 o.~ 0.65 0.5

~L--

o

- > Freq(Hz)

Figure 8.7:

8.7 1,

k

0.4,

k

0,

k

= 0,1,2,3

=4

= 5,6,7

M-3

Hr(w)

=

M-l ~ M-l h(-2-) + 2 LJ h(n)cosw(-2- - n) n=O

h( n)

= h( M - 1 - n)

h(n)

=

h(14 - n) 6

Hr(w)

=

h(7)

+2L

h(n)cosw(7 - n)

n=O

Solving the above eqn yields,

h(n)

=

{0.3133, -0.0181, -0.0914,0.0122,0.0400, -0.0019,-0.0141,0.52, 0.52, -0.0141, -0.0019, 0.0400,0.0122, -0.0914, -0.0181, 0.3133}

8.8 (a)

yo(t)

=

dzo(t) dt

= ~[ei21rFtl = Hence, H(F)

= 298

dt j21rFei 271Ft j27rF

i

'IE

-50

A

: -100 I

_1SOL--.l...--.-._.-._.-._.-._.&.-_...L-_...L-_........- - - J o 0.05 0.1 0.15 0.2 0.25 0.3 0.3& 0.. O.~ 0.5 - > Frec(Hz)

•,..--,..--__-r---__-......--......--or--or--T-----w

~

0

O.OS

0.1

0.15

0.25 0.3 0.2 - > Frec(Hz)

0.36

0.•

O.~

0.5

Figure 8.8: (b)

IH(F)I LH(F)

= = =

211'F

= = = =

jw,

11'

F

2' 11' -2'

>0 1\

FO

2' 11' -2'

w Freq(Hz)

o.~

0.35

O..s

0.5

Figure 8.9: to the differentiator in (c). (e) The value H(wo) is obtained from (d) above. Then y(n) = AIH(wo)lcos(won

+8+ j -

~)

8.9 I

Hd(W)

hd(n)

hd(n)

j1Ow

= = =

we, _we- j1Ow ,

= = = =

COSlr( n (n 0, n COSlr( n (n 0, n

-1 [ 211'

\

O~W~lr -lr~w F,.qHZ)

Figure 8.10:

Hence, H(z)

=

1 - rz- 1

-------~2~~2' WQ

1 - 2rcoswQz-1

+r

z-

= 0.3 r = 0.99

From the impulse invariance method we obtain

H(s)

1] = 21[1 s+0.1-j3 + s+0.1+j3

H(z)

=

~ [1- e-O.l~.i3Tz-l

=

1 - rcoswQz-1 1 - 2rcoswoz- 1 + r 2 z-2

+

1-

e-O.l;.-i3Tz-l]

The poles are the same, but the zero is different.

8.11 Ha(s)

=

s

=

S2 -

(Ou - nds + nuo,

2 1 - z-1

T 1 + z-1 2

H(z)

where a

(1- z-1)(1

+ z-l)

T = (n u - n) I (t)2(1- Z-1)2 + (n u _ 0')(1')(1- z-l)(l + z-l) + n u n,(l + z-1)2 2(a - 13)( 1 - z-2) = [4 + 2( a - 13) + a13] - 2(4 - (13)z-l + [4 - 2( Q - 13) + ap]z- 2 = nuT, 13 = n,T

In order to compare the result with example 8.3.2, let Wu

= nuT = 3n5 301

o 6r------.----,.---r---.---....---.--.....,...--,.--~_.,

j

1

~.08 ~06

-O.a.

-008

-Oa.

0.04

006

008

0,

004

006

008

01

i

-h0

" I I

-1

i ,

-0.06

-0.02

0

-~

002

Freq(Hz)

Figure 8.11:

WI

Then, H(z) In our case, we have

211'

=

n,T=5

=

0.245{1 - z-2) 1 + 0.509z- 2

( example 8.3.2 )

W

Q

13

= 2.753 WI 2tan"2 = 1.453

= nu T = 2tanTu

= niT

=

By substituting into the equation above, we obtain H(z)

= =

2.599( 1 - z-2) 10.599 + 5.401z- 2 0.245(1 - z-2) 1 + 0.509z- 2

8.12

=

Let T 2 (a) H ( z) (b)

= ~! ~ =~ ~ y( n) = y( n -

1) + z( n)

+ z (n -

1)

(c) O~W~1r

-11'

w

__-~......_-__r'--_r_-_..,

2r--__r'--_r_-~--

A I I

-1

-3

-2

o

-1

2

-->w

Figure 8.13:

Zs

= Z3

ZS

=

Z;

=1 = 2. Z7 and H(z)

=

z-SH(z-l)

Therefore, H (w) is linear phase. (c) Refer to fig 8.16

8.15 From the design specifications we obtain l

fJ

fp

f, Assume t

= 1. Then, np and

= = = = =

0.509

=

2tanwfp w, 2tanT

n, =

= 77

99.995 4 1 24 6 24

=6 1

=4

2tan w p 2

= 1.155

2tan7r I, = 2

fJ = - = 196.5 l

304

3

k Butterworth filter: Nmin Chebyshev filter: Nmin Elliptic filter: sino

sin(3 Nmin

= nn,p = 1.732 10g7] = 9.613 ~ >

=

= -= = -7] = ~

k(sino) k(cos(3) . {3 k(cOSQ )' k( sIn)

8.16 From the design specifica.tions we ha.ve (

lJ

Ip

=

0.349

= 99.995 1.2 0 5 = 8= .1 2

I, = 8 = 0.25

np = n, = 7]

k

= =

=

=> N 10 logk h- 1 cos 7] 5.212 => N 6 cosh- 1 k 1 0.577 => Q 35.3 0 k 1 0.577 => {3 0.3 0

w 2tan ; = 1.019 w, 2tan- = 2 2 lJ - = 286.5 (

n, = 1.963 n, 305

=

= =

= 3.78

=> N

=4

x(o)

y(o)

o

-112

~

y(o)

2

"""B""" · L._ y~

l

l

_}

Figure 8.15: Butterworth filter: Nmin

~

Chebyshev filter: Nmin

~

Elliptic filter: N min

~

8.17

=

=

Passband ripple IdB => l 0.509 Stopband attenuation 60dB :;. 6 1000

=

=

= 0.3", W, = 0.35",W n = 2tan ; = 1.019 W, n, = 2tan2' = 1.226 6 '7 = - = 1965.226 l n, k = n = 1.203 I' cosh-If] = 8.277 = 13.2 Nmin ~ WI'

p

cosh-1k

0.627

306

=> N

= 14

x(n)

+

-413 • 314 = ·25/12

(4/39- + (3/4l + 4 COS 2(1.)0 = 481/144

Figure 8.16:

Special software package, such as MATLAB or PC-DSP may be used to obtain the filter coefficients. Hand computation of these coefficients for N = 14 is very tedious.

8.18 Passband ripple = 0.5dB ~ ( = 0.349 Stopband attenuation = 50dB Wp

=

W,

Op

= 0.35,," W = 2tan-;f = 0.792

0,

= 2tanT =

0.247r

w,

1.226

Ie

=

-lJ =906.1 ~' = 1.547

Nmin

~

cosh-lT] cosh- l Ie

T]

=

(

"

7.502 1.003

=--=

7 8 .4

~

N =8

Vse a computer software package to determine the filter coefficients. 307

8.19 (a) MATLAB of length N normalized to (b)c1 = 0.02,

is used to design the FIR filter using the Remez algorithm. We find that a filter 37 meets the specifications. We note that in MATLAB, the frequency scale is ~ of the sampling frequency. Refer to fig 8.17. 62 = 0.01, ~f = 12000 - 11;0 = 0.05

=

08

04

02

00

01

0.2

0.3

0."

0.5

->'

0.8

07

08

oe

Figure 8.17:

With equation (8.2.94) we obtain

M

-20!og10( ~ - 13

=

14.6~f

+

1 ~ 34

With equation (8.2.95) we obtain

Doo (C1 02) f(01 02) and

= =

1.7371 11.166

D oo (01 C2) - f(C162)(~f)2

M =

~f

Note (8.2.95) is a better approximation of M. (c) Refer to fig 8.18. Note that this filter does not satisfy the specifications. (d)The elliptic filter satisfies the specifications. Refer to fig 8.19. (e)

FIR order storage No. of multo

37 19 19 308

IIR 5 16 16

+ 1 ~ 36

0.1 0.8 0.7

0.'

Ia.s 0.' 0.3 02 0.1 0 0

0.1

0.2

0.3

0.'

0.'

0.7

0.8

0.1

Figure 8.18: M=37 FIR filter designed by window method with Hamming window

8.20 (a) h(n)

=

{?'

H(z)

=

L

1,2,3,4,5,4,3,2,1,0, ...}

10

h(n)z-n

"=0

+ 2z- 2 + 3z- 3 + 4z- 4 + 5z- 5 + 4z- 6 + 3z- 7 + 2z- 8 + z-9 H(u') = e-j 9W[2cos4w + 4cos3w + 6cos2w + 8cosw + 5] = 12co~4w + 4cos3w + 6cos2w + 8cosw + 51. Refer to fig 8.20. =

(b)IH(w)1 (c)

z-1

bo

=

h( 1) = 1 9

rhh(O) =

L h (n) = 85 2

"=1 8

rhh(l) =

L

h(n)h(n

+ 1) =

80

"=1 7

rhh(2) =

L h(n)h(n + 2) = 68 "=1

8.21 (a) de gain: Ha(O)

309

=

1

o.g~

·

0.8

·

0.7

0.'

0.• 0.3 0.2 0.1

. O.oe

0.1

0.15

0.2

0.25 ->f

0.3

0.35

\ 0 .•

· -

O.~

0.5

Figure 8.19:

3dB frequency: IH o(jO)1 2

1 2 1 2

=

Q2

or Q2

+ O~ =

=> Oc = Q For all 0, only H(joo) = 0 =! = !hll(O) e e => e- o ' = e- 1 1 =>r = Q ho(r)

(b)

h(n) H(z) H(w) H(O)

3dB frequency: IH(w c )1 2 (1 - QToCoswe)2

+ (e- oT sinw e)2 Hence,

We

Since IH(w)1 2 310

= = = = = = = = = =

ho(nT) e- onT u(n) 1 1 - e- oT z-1 1 1 - e-oTe- jw

H(w)lw:o 1 1 - e- oT !IH(0)1 2 2 2(1 - e- aT )2

2sin- 1 (sinh

QT

T )

1 1 - 2e- aT cosw

+ e- 2aT

20

115 iE

"I 10

o~~~----'L.--~~~.......c;;""'-.L..-.._.L..-..-"""";;;:-~-=:;;''''-----J

o

0.0lS

0.1

0.15

0.2

0.25 0.3 -> Fre«Hz)

0.35

0.4

0.45

0.5

Figure 8.20:

it oscillates between

1 and 1 (1 - e- oT )2 (1 + e- oT )2 but never reaches zero h(T)

=> T

is the smallest integer that is larger than

T

=

e-o'rT

~

f

= e- 1

1

aT

(c)

H(z)

= = =

DC Gain: H(z)lz=1 At z

= -1(w = 11"), H(z) since IH aUn e )1 2 We

Let a Then H(z)

h(n)

= =

a 2 l-z-

1

+a aT(1 + z-l) 2(1 - z-l) + aT(1 + Z-I) aT{1 + z-l) 2+aT+(aT-2)z-1 T

1+:-1

1

o 1

= -2' we have n e = a

=

= = = =

2tan- 1 ne T 2 2tan- 1a1'2 2-aT 2+aT I-a (l+a)Z-1] 2 1 - az- 1 1- a -2[6(n) + (1 + a)a n - 1 u(n - 1) ]

[1+

311

l

\

h(O) h(n) h(O) ~ (1

+ 0)0"-1 n

I-a = -2 1 = e 1 = e In1 = 1+0 Ina Ine-:T) - 1 = I n e-2+oT oT ) o -

-

8.22 (a)

(b) I

Let hl(n)

=

hd(n)w(n),

Then, h(n)

=

hl(n - 100) will be the impulse of the filter for 0 ~ n ~ 200

(c)

- 100 ~ n ~ 100(M

I

\

= 101)

0, O 0.0411'. (b) The assumed spectral amplitude normalization in fig 10.1 implies that the analog FT (magnitude spectrum) of zo(t) is (refer to fig 10.4). The given sample rate is identical to Fy above, F y = 250Hz. The DTFT of samples taken at this rate is Y(O) Xo(O - qn y ) where

= +; L,

=

fly 211'Fy. On a scaled frequency axis w" = nTy = /);' Y(w") = Consequently y( n) y( n ).

=

10.2 (a) X(w)

= (l-a~-J"')

(b) After decimation Y(w')

= !X(~') = 2(1-ae-~) 1

(c) DTFT {x(2n)}

= L

x(2n)e- i \ll2n

n

= L x(2n)e- i = Y(w') n

333

\ll'n

.

+; Lq Xo(w" -

q21r).

IX(W ' / Assumes peak

of

'b(,) normalized to unity

O.~

-n

X (w') shifted to 0.& b

-

period 2n

o

-o.8n

w'

0.8n

I-

~.16n

-I

Figure 10.1:

10.3 (a)Refer to fig 10.5 (b)

Let

Wi

Y(w")

=

n

"n

2

z( ~ )e- iw " n even 2

+

'

= L n

=

Wi

w = F" =

Fz

L n

![z( n - 1) odd 2 2

+ z( n + 1 )]e-jw"n

E z(p)e-iW"2P + ~ L[z(q) + z(q + 1)]eP

2 jw "(2 9 +1)

9

+ !e- iW "[X(2w") + ei 2W " X(2w")] 2 = X(2w")[1 + cosw"]

=

X(2w")

X(w ' )

= {ol ',

X(2w")

0:5 Iw/l :5 0.2r otherwise

={ 1, 0,

0:5 12w"l :5 0.211" otherwise

_ { 1, 0:5 Iwit I :5 0.111" 0, otherwise

Y(w")

=

I + cosw", { 0,

334

o :5 Iw"l :5 0.111" otherwise

lW(w')1

,.

H(w') 0.5

-------~

I I I I I I I

0.5

-n

-o.4n

i-

-t

~

0.4n

a.In

period 2n

w

n

Figure 10.2:

(c) Refer to fig 10.6

Y(w")

={

O.35;r ~ Iw"l ~ 0.45;r or 0.55;r ~ Iw"l :5 0.65;r otherwise

1 + cosw",

0,

10.4

= p;,

o/!-.

(a) Let w' w" = Refer to fig 10.7 Let x"(n) be the downsampled sequence.

= =

z"(n)

X"(w")

z(nD) .!.X( w")

D

D

As long as Dw:n :5 11', X(w') [hence z(n)] can be recovered from X"(w")[z"(n) using interpolation by a factor D:

X(w')

= z(Dn)]

= DX"(Dw')

=

=

The given sampling frequency is w~ ~. The condition Dw:n :5 ;r - 2w:n :5 ~ w~ (b) Let za(t) be the ral analog signal from which samples z(n) were taken at rate Fz . There exists a signal, say z~(t'), such that z~(t') = X a ( z(n) may be considered to be the samples of z'(t')

*).

f!: 7f tJ·

taken at rate Iz = 1. Likewise z"(n) = z(nD) are samples of z'(t') taken at rate = = From sampling theory, we know that z'(t') can be reconstructed from its samples x"(n) as long as it is bandlimited to 1m :5 or W m :5 which is the case here. The reconstruction formula

2h,

iJ,

IS

z'(t')

= Lz"(k)hr(t' k

where

335

kD)

IV(w')1

period 2 n

•

n

O.08n

w'

lY(w")1 = O.IIV(O.! w")1

O.8n

7t

w"

Figure 10.3:

Refer to fig 10.8 Actually the bandwidth of the reconstruction filter may be made as small as 2'" - w , , so hr may be as 1) m

h (t') = sin( w~t') (w~t')

r

where

w:n ~ w~ $

jf -

w:n.

In particular x( n)

x(n)

= x' (t' = n) so

=E x(kD)hr(n -

kD)

k

(c) Clearly if we define

v(p) = { x(p), 0,

if p is an integer multiple of D other p

then, we may write 10.4 as

x(n)

=

E v(p)hr(n - p) p

so z(n) is reconstructed as (see fig 10.9)

336

w:n, or as large

...........................

~·/2···················

-D.e

Figure 10.4:

10.5 (a)

Let w

=

X.I(w)

=

0 F~'

w " =20 F~

L z.I(n)e-;wn n

=

E z(2m)e-

Jw2m

m

12: X(w - -q) 211' = -2 2 f

= ~ EX(w- ,..q) f

To recover z(n) from (b)

Z.l (n):

see fig 10.10 Recall w'

=

Xd(W')

=

2w LZd(n)e-;w'n n

= L n

even

z.I(n)e-;w'i

= L z.I(n)e-;w'i n

337

x(n)

'fe~y_(m_)F

z

x

-1

Figure 10.5: since x,(n)

I

\

= 0 when n odd =

W'

X,(-) 2

see fig 10.11 No information is lost since the decimated sample rate still exceeds twice the bandlimit of the original signal.

10.6 A filter of length 30 meets the specification. The cutoff frequency is are given below: h(l) h(2) h(3) h(4) h(5) h(6) h(7) h(8) h(9) h(10)

= = = = = = = = = =

= 0.006399 h(29) = -0.01476 h(28) = -0.001089 h(27) = -0.002871 h(26) = 0.01049 h(25) = 0.02148 h(24) = 0.01948 h(23) = -0.0003107 h(22) = -0.03005 h(21) = -0.04988 h(30)

338

We

=i

and the coefficients

X(.)

o o

I

I I

O.7n 0.35n

l.In

O.9n n O.45n

1.3n O.65n

O.55n

2n

w'

n

W"

Figure 10.6: h(ll) h(12) h(13)

h(14) h(15)

pt(n)

= = = = = =

h(20) = -0.03737 h(19) = 0.01848 h(18) = 0.1075 h(17) = 0.1995 h( 16) = 0.2579 h(n+k),

k = 0,1, ...

corresponding polyphase filter structure (see fig 10.12)

10.7 A filter of length 30 meets the specification. The cutoff frequency is are given below:

h(2)

= =

h(3)

=

h(4) h(6)

= = =

h(7)

= h(24)

= 0.01598

h(8)

= = =

= -0.01562

h(l)

h(5)

h(9) h(10)

h(30) = 0.006026

=

h(29) -0.01282 h(28) = -0.002858 h(27) = 0.01366 h(26) = -0.004669 h(25) = -0.01970 h(23) = 0.02138 h(22) = -0.03498 h(21)

= h(20) = 0.06401 h(12) = h( 19) = -0.007345

h( 11)

339

We

=

i

and the coefficients

IX(w')1

-w'm

w' m

n

w'

n

w"

IX"(w")1

-Ow' m

Figure 10.7: h(13) h(14) h(15) pt(n)

= = = =

=

h(18) -0.1187 h(17)=0.09805

=

h(16) 0.4923 h(2n+k), k=O,l;n=O,1, ... ,14

corresponding polyphase filter structure (see fig 10.13)

10.8 The FIR filter that meets the specifications of this problem is exactly the same as that in problO.6. Its bandwidth is t. Its coefficients are g(n, m)

g(O, m) g(l,m)

g(14,m)

= = = = =

h(nI

+ (mD)I)

h(nI

+ mD -

=

{h(28), h(29)}

mD

[-1-]1)

5m h(2n + 5m - 2[2]) {h(O), h(l)} {h(2), h(3)}

340

IX(w ')1 period2 nlD

Hr (w ')

,

,,

, ,,

,

, ,,

, , ,, ,, ,

\/",

, " ,,

,

w'

m

nlD

,,

, ,,

,,

2nlD· w'

m

2nID

Figure 10.8: A polyphase filter would employ two subfilters, each of length 15

=

pa( n) PI (n)

{h(O), h(2),

, h(28)}

= {h( 1), h(3),

,h(29)}

10.9 (a)

x(n) D = I = 2. Decimation first Z2(n)

= {xa, Xl, X2, ...}

= {xa, X2, X4, ... } Y2(n) = {Xc, 0, X2, 0, X4, 0, ...}

Interpolation first

= {xc, 0, Xl, 0, Z2, O, ...} Yl(n) = {xc, Xl, Z2,"'} zl(n)

so Y2(n)

(b) suppose D

= die

¢

Yl(n)

and I = ile and d, i are relatively prime.

x(n) Decimation first

=

{ZO,Xt,X2,"'}

z2(n)

=

{ZO,Zdk,Z2dk, ... }

Y2(n)

=

{ZO'~!Zd&'~'Z2dk" ik-1

Interpolation first 341

i&-l

,,

.. }

,,

,,

,,

w'

x..(n)=x(kn)

~I,---_+D

~_v..;..(n..;;...)~~I

x(n) h~n)

----..

I

Figure 10.9:

Zl(n)

= {xo,0, ... ,O,Xl,0, ... '0IX2,01 ... ,0, ... } ~~~

it-l

=

Yl(n)

ik-l

{XO'~'Xd'~""} d-l

Thus Y2(n)

ik-l

d-l

= ydn) iff d = dk or k = 1 which means that D and I are relatively prime.

10.10 (a) Refer to fig 10.14

Yl (n)

= h(n) * wl(n) = h(n) * x(nD) = E h(k)x[(n 00

k)D]

1:=0

H(zD) H(zD)

=

-

... h(O)zo

+ h( 1)zD + h(2)z2D + ...

h(n)

342

xs (w)

w

I

x! (n) .,

---

where

yen)

.Ik__hen) ..a..-_f----.,;x(:.....:n)~. [

J;J -Tt/2

w

Tt/2

Figure 10.10:

j

\

=

ho,~, Hl'~' h(2)' ... }

{

D-l

D-l

nD-l

so w2(n)

= L

h(k)x(n - k)

t=o n

= L h(kD)x(n t=o

kD)

n

= L h(k)x(n k=O Y2(n)

= =

kD)

w2(nD) n

L h(k)x(nD -

kD)

k=O n

So yt{n)

= =

E h(k)x[(n -

k)D]

k=O

Y2(n)

(b) 00

wl(n) Yl (n)

= =

E h(k)x(n t=o

wdp),

n

343

k)

=pI(p an integer)

period .,2n

-n

2n/3

w'

7t

Figure 10.11:

= U'2(n)

=

=

other n

0,

z(p),

n

= pI

other n

0,

Let h(n) be the IR corresponding to H(zl) 00

Y2(n)

=

2: h(k)w2(n -

k)

k=O 00

=

2: h(kI) w2(n - kI) k=O 00

=

2: h(k)w2(n -

kI)

k=O

for n

= pI 00

Y2(n)

= E h(k)W2((p -

k)I)

k=O 00

= = for n

2: h(k)z(p -

.=0

WI (p)(

Y2(n)

see above )

:F pI 00

=

2: h(k).O = 0 k=O

344

k)

+---~y(n)

l\(n)

I .~

x(n)

F

o~---

P4 (n)

x

11-----1

Figure 10.12: so we

=

conclude Yl (n)

Y2( n)

10.11 (a)

H(z)

=

L h(2n)z-2n + L h(2n + 1)z-2n-l L h(2n)(z2)-n + z-1 L h(2n + 1)(z2)-n n

n

= = Therefore H 0 ( z) = Hdz)

n H o(z2)

n

+ z-1 H 1(z2)

L h(2n)z-n n = Ln h(2n + 1)z-n

(b)

H(:)

= Ln h(nD)z-nD + L +L

h(nD + l)z-nD-l + ...

n

h(nD + D - l)z-nD-D+i

n

=

D-l

L z-c L h(nD + k)(zD)-n

c:o

n

345

x(n) y(n)

F

x

Figure 10.13: I

Therefore HI: (z)

=

2: h(nD + k)z-n n

(c)

H(z)

=

1 1 - az- 1

=

2: anz- n

00

n=O 00

Ho(z)

=

L a2n z- n n=O

1

=

l-a 2 z- 1

=

L a2n + z- n

00

HI(z)

1

n=O

=

a

l-a 2 z- 1

10.12 D1

=

Fa

= 10 kHz, 346

25,

D2

=4 Fa

F1 = -

D1

= 400 Hz

\

x(n)

-I

x(n)

_I

,_D_

win)

.I,~~_

~_~- Y1 (n)

win)

-1__ tD_

H(z)

Figure 10.14:

Passband a ~ F ~ 50 Transition band 50 < F :5 345 Stopband 345 < F

~

5000

Passband a ~ F ~ 50 Transition band 50 < F :5 55 Stopband 55 < F :5 200 For filter 1, 61

=

0.1 "2 =0.05,

~f

=

345-50 _ -2 10,000 - 2.95z10

M1 = For filter 2, 61 ~f

-101096162 - 13

14.6~f 62

= 0.05, - 50 = 55400

M2 =

62

= 10

-3

~

+ 1 = 11

= 10- 3

= 7.5z10- 3

-101096162 - 13

14.6~f

+ 1 ~ 275

The coefficients of the two filters can be obtained using a number of DSP software packages. 347

10.13 To avoid aliasing Fie

:5 ft>. Thus D =/ = 50. Single stage 61

:::;

l:11

:::;

All

:::;

62 :::; 10- 3 0.1, 65 - 60 _ 5 -4 10,000 - .rIO -1010g6 1 62 - 13 14.6 l:11

+ 1 :::::: 3700

Two stages

Dl stage l:Fl

25,

:::;

.= 2

D2

It =2,

/2

= 25

10,000 = 400 25

=

Passband 0 :5 F

:5 60 :5 335 Stopband 335 < F :5 5000

Transition band 60 < F

61

10- 3 62 = - 4 2.75z10- 2 M1 = 84

= 0.1,

6/ = stage 2:F2

400 2

=

= 200

:5 F ~ 60 Transition band 60 < F :5 65 Stopband 65 < F ~ 100 Passband 0

61

, \

=

61 =

62

0.1,

10- 3

=-4

M2 = 13

0.1875

Use DSP software to obtain filter coefficients.

10.14 b+(n) is nonzero for 0 :5 n :5 2N - 2 with N even. Let c(n) for -(N - 1) :5 n :5 N - 1. From (10.9.43)

= b+[n -

+ (_1)N-1 B+(w - 11') B+(z) + (_1)N-1 B+(-z)

B+(w)

or Therefore, C(z )z-(N-1) + (-1 )N-1 C( -z)( -z )-(N-1) or C(z)

+ C( -z)

c(n)+c(-n)

when n

(N -1)]. So c(n) is nonzero

= = = =

oe- jw (N-1)

=

06(n)

#- Oc(n) =

oz-(N -1) oz-(N-1)

0 -c( -n)

when n is odd c(n)

=

-c( -n)

when n is even but n ~ 0, c(n)

=

0

(half- band filter) when n 348

= 0, c(n)

=

0

2

10.15 one stage:

61 ~f

, -IO/og,St 62 - 13 M = 14.6~f

62 = 10- 3 100 - 90 = 10- 3 10,000

= 0.01, =

+ 1 :::::: 2536 = 2z10 5 Hz 12 = 2 = 1,

two stages: Fa

II

= 2x10 = -Fo II

F1

4

Hz

Passband 0 ~ F ~ 90 Transition band 90 < F ~ 19,900 Therefore

and 611

.\1 1

=

-10Iog6 1 62 -13 ~f 14.6

19,900 - 90 2x 10 5

= =

~f

61

2'

612

= 0.09905

= 62

+ 1 :::::: 29 F2

= -F/2 = 1xl04 Hz 1

Passband 0 ~ F ~ 90 Transition band 90 < F ~ 9,900 Therefore

~f

and 621

. M2

=

-101ogC1C2 - 13 14.6 b.f

I

=

9,900 - 90 2z10 4

=

2'

61

= 0.4905 C22 = 62

+ 1:::::: 7

10.16 Suppose the output of the analysis section is xao(m) and Zal(m). After interpolation by 2, they become Yo(m) and Y1(m). Thus _

Yk ( m ) -

{

X ok

0,

(!f),

m even Ie m odd

= 0, 1

The final output is z(m) when m is even, say m z(m)

= =

= z(2j) =

Yo(m) 2j,

* 2h(m) + Yl(m) * [-2(-1)mh(m)]

2Yo(m)

* h(m) -

2Yl(m). h(m)

= 2 LYo(k)h(m - Ie) - 2 LY1(Ie)h(m - Ie) k

=

2L

Yo(21)h(2j - 2/) - 2 LY1 (2/)h(2j - 21) I

349

k

I

\

= 2

L x o(l)h[2(j -

I)] - 2

a

L x dl)h[2(j -

I

a

I)]

I

= =

2[x a o(j) - Xal (j)] 2[x aO(j) - Xal (i)]

=

2[zao(j) + X o l(j)] • Pl(i)

* h(2j) * Po(j)

In the same manner, it can be shown that

+ 1)

z(2j

10.17 Refer to fig 10.15, where hi(n) is a lowpass filter with cutoff freq. ];. After transposition (refer

I~----------~---------I

--H

H

hi

I

~

-

-

- -

-

-

-

- -

-

--H

1-

hfn) - -

-

-

-

H

*\.

IL

I

- - -

1

~n) I -1i

1I

- _I

Interpolator L

Interpolator 1

Figure 10.15: I

=

=1 1

1 2 . ..

=

ILL-stage interpolator

=

to fig 10.16). As D I, let Di h+l-i, then D D1 D2 ... DL. Refer to fig 10.17 Obviously, this is equivalent to the transposed form above.

10.18

=

Suppose that output is y(n). Then Ty fTr . Fy filter is h( n) of length M kI (see fig 10.18)

=

350

= ill = ft = fF

r .

Assume that the lowpass

r-------------------I I

H

h[n)

I 1

,-------------------,

-H

I

H·\.

~

bln)

I

I

.J

H·\

'

10.19 (a)

for any n

= II + j

(O~j~I-l)

/-1

LPc(n-k)

= LPc(lI + j - k) k=O

k=O

= = = =

Pi (II) Pi (I) h(j + 1I) h(n) 1-1

Therefore, h( n)

= LP«(n - k) «=0

(b) z-transform both sides /-1

H(z)

=L

,

J

Figure 10.16:

1-1

~.

z-l:Pl:(z)

1:=0

(c)

351

- - -- - - - - - - - - - - - - - - -

-.

--~-----------------

I

•I~

Figure 10.17: L-stage decimator

j

\

I-I

= ]- L:L h(k + mI)eJ 27i'/m z-m

=L

m

1=0

h(k

+ mI)z-m

m

m

10.20 (a) Refer to fig 10.19. (b)

Bandwidth cut off freq sampling freq of z(n) sampling freq for the desired band of frequencies Therefore, D (c) Refer to fig 10.20. (d) Refer to fig 10.21.

352

= =

=

3

7r 2 27r 27r 2

= - =7r = 27r = 2 2

• • •I"

.. -.. ~.~~~!!~!~~-_._-----_

.

n=O.l, .... K-l

8(0.1)

I

g(nJ-l)

I

..~ ~

input buffer

n=O,l, ..., K-l

I

length K

I

1

~

0=0,1, ...• K-l

I I

II I

•

._ •••

~

buffer length K

K-l ~

n=O output buffer 'enlth I

(n)

=

F (Ilk) y

Ii

Figure 10.18:

spectrum of x(n)

a>

c :]

spectrum of y(n)

1

0.8

0.8

.g 0.7 ~

E0.6

~

c:

~0.6 E

C)

m

~ 0.4 I I

A I I

0.2

a

0

2

4 -->w

6

0.5 0.4 0

8 Figure 10.19:

353

2

4 -->W

6

8

x(n)

IX(w)1

40

-s

1200 1000

30

1800 'e

f600

~20 I I

~400

10

I

200

OlllWl

0 0

500

1000 -->n

0

1500

1000

500

1500

Figure 10.20:

1000

"T

100

800 700

A

I .we 300

·

200

· ·

100

~ ~ ~ ~~ ~ ~ ~ 200

\4 '-l ~ '-- ~ ~ ~ ~ '-'~ 100

Figure 10.21:

354

100

~

~ 1000

1200

Chapter 11

11.1 (a)

25 (1 - :-1

H(z) and O'~ so z( n)

= = =

(b) The whitening filter is H-l(z)

11.2 (a) r (z)= rr

+ ~z-2)(I

- z-1

+

~z-2)

1 1 - :-1

+ t:- 2

25 1

z(n-I)-'2 z (n-2)+w(n)

=1 -

z-1

+ ~z-2

27(l-!Zl)(1-!Z) 2 (1 - ~ ZI )(1- ~ Z)

For a stable filter, denominator (1 - tZ1) must be chose. However, either numerator factor . (I_l z 1) ~ may be used. H(z) (1 _ ~Zl) or (T~fZ)

=

~

[minplc.]

(b)

~lust

invert the min. pk. filter to obtain a stable whitening filter.

11.3 (a) 1 + 0.9:- 1 1 - 1.6z- 1 + 0.63z- 2 1 - 1.6:- 1 + 0.63z- 2 1 + 0.9z- 1

H(z)

=

whitening filter, H- 1 (z)

=

zeros: z

=

0.7 and 0.9

pole: z

=

-0.9

355

(b) f:rr(W)

=

u~H(w)H(-w)

=

u·

')

11 + O.ge- jw 12

-:---.:--~-_.:--~~

11 -

w

1.6e- jw

+ O.63e- 2jw I 2

11.4 A(z)

=

+ -z 5 -2 + -z 1 -3

13 -1 1 + -z 24

8

3

1 3

~ + ~z-1 + 13 z-2 + z-3

=

3

8

24

1 2

3 -1 + z -2 = -21 + -z 8

Adz)

=

A2(Z) - 1c 2B 2(z)

=

1+~Z-1

1 - Ic~ 4

11.5

= =

1+2z-1+~z-2 3

~+2z-1+z-2 3 1 3

A2(Z) - t2B2(Z) 1- t~

=

3_ 1 1 +-z

2

3

2

11.6 (a)

Al (z)

=

B 1(z)

=

A2(Z)

=

= 356

1+

~z-1

2 1 + - z -1 2 A 1(z) + t2 Bdz)z-1 1 1 - -~1 2 1 + -z3 3·

B 2 (z)

1

1

1

= -3 + 3z - +

2 Z-

= A2(Z) + k aB 2 (z)Z-1 = 1 + Z-a The zeros are at z = -1, e±i i" H(z) = A 3 (z)

Refer to fig 11.1 (b)

Figure 11.1:

= Aa(z)

= = =

The zeros are at z

=

If k 3

H(z)

(c) If

Ikpl =

1, the zeros of H(z)

= Ap(z)

-1, we have

A2(z) - B 2 (z)z-1 2 -1 --z 2 -2 -z -3 1+-z 3 3 5 .v'IT -1, -'6 ±} -6-

are on the unit circle. Refer to fig 11.2.

11.7 = 1+0.6z- 1 Bdz) = 0.6 + :-1

AI(z)

357

Figure 11.2: A 2(z)

B 2(z) A3(Z) B3(Z) H3(Z)

h(n)

= = = = = = = = =

A 1(z) + k 2B 1(z)z-1 1 + 0.78z- 1 + 0.3z- 2 0.3 + 0.78z- 1 + z-2 A2(Z) + 0.52B2(Z)Z-1 1 + 0.93z- 1 + 0.69z- 2 + 0.5z- 3 0.5 + 0.69z- 1 + 0.93z- 2 + z-3 A3(Z) + 0.9B3(z)z-1 1 + 1.38z- 1 + 1.311z- 2 + 1.337z- 3 + 0.9z- 4 { 1,1.38,1.311,1.337,0.9,0, ...}

11.8 Let y(m) = z(2n - p - m). Then, the backward prediction of z(n - p) becomes the forward prediction of y( n). Bence, its linear prediction error filter is just the noise whitening filter of the corresponding anticausal AR(p) process.

11.9 p

x(n

+ m)

= - L Qp(k)z(n k=1

358

k)

e(n)

= z(n+m)-i(n+m) p

=

z(n

+ m) + L

ap(k)z(n - k)

k=l

E[e(n)z*(n -I)]

= 0,

1=1,2, ... ,p

p

~

L: ap(kh'zz(k -I)

= -1zz(l + m),

1= 1,2, ... ,p

k=l

The minimum error is

E{le(n)1 2 }

= =

E[e(n)z·(n

+ m)]

p

'1'zz(O)

+

L ap(k)1zz(m + k) k=l

Refer to fig 11.3.

x(n+m)

e(n)

+

forward

........

z -m-l

A

x(n+m)

linear

predictor

Figure 11.3:

11.10 1'-1

i(n - p - m)

= - L: bp(k)z(n -

k)

k=O

e(n)

= r(n-p-m)-x(n-p-m) 359

,

\

p-l

= z(n - p - m) +

2: bp(k)z(n -

k)

k=O

E[e(n)z·(n-/)] p-l

~Lbp(khzz(l-k) c=o

= 0,

=

I=O,2, ... ,p-1

-"/zz(l-p-m),

I=O,2, ... ,p-1

The minimum error is

E{le(n)1 2 }

= E[e(n)z·(n =

p- m)] p-l "/zz(O) + L bp(k),,/zz(p + m - k) k=O

Refer to fig 11.4.

Backward linear predictor

A

x(n-p-m)

I

Z

-px(n-p-m)

+

Figure 11.4:

11.11 The Levinson-Durbin algorithm for the forward filter coefficients is "/zz(m)

+ t:_l~m-l

Eln Qm(k)

= am-l(k) + kma~_l(m - k), k= l,2, ... ,m-l; m= l,2, ... ,p

360

1

but bm(k) or am(k) Therefore, b;'" (0) == km

b;"'(m - k) Equivalently, bm(0)

= k~ bm(k)

= = = = = =

k

a:n(m - k), b:n(m - k) 'Yrr( m)

= 0,2, ... , m

+ in-I Q:n-l Ebm

b:n_l(m - 1- k)

'Y;r(m)

+ kmbm-l(k)

+ 2:-1~-1 Ebm

bm-1(k - 1) + k:nb;"'_l (m - k)

This is the Levinson-Durbin algorithm for the backward filter.

11.12 Let

b

_m

= [ ~-l 0

]

+

[dm-l ] bm(m)

Then, Ln-l [ 'Y bt ~-l Hence, 6·

Ln-l~-l

+ Lm-ldm-l + bm (m)2m_l l~-l~-l + l~_lrb-l + bm(m)"Yrr(O) But Ln-l~-l ~ Ln-l4m-l

Hence, rb-l 1

6·

Also, Ga-l1m_1 Therefore, bm(m)l~_lf!~-l + bm(m)"Yrr(O)

= = = = = = =

~-l

cm(m) rm-l 6-

-bm(m)lm_ 1 1

6·

-bm(m)L;-l!.m_1 6-

f!m-l cm(m) - tr:-l~-l

solving for bm(m), we obtain cm(m) - ~_l~-l bt

6·

'Yrr (0) + 1m-I f!m-l

=

cm(m) - ~_IQm-1

E!n-l

we also obtain the recursion

bm{k)

11.13 Equations for the forward linear predictor:

361

=

bm_1{k)

+ bm(m)a:n_l{m -

k), k=1,2, ... ,m-1

where the elements of f m are "Yu(l + m),

I

= 1,2, ...

I

p. The solution of gm is

cm(m) -l~_lgm-l

=

"Yr r ( 0)

+ 1~ _1g~ - 1

cm(m) -l~_lgm-l

=

E!n-1

am(k)

=

am-l(k) + am(m)o:n_dm - k), k=1,2, ... ,m-l; m=l,2, ... ,p

where Om is the solution to LnQ:m

=

2m

The coefficients for the m-step backward predictor are

km

=g~.

11.14

1 - [ k~

-k m -1

] [

1

k~

=(1- Ikml') [~ ~1] ={1-lkml'),l 11.15 (a), m

E(fm(n)z(n - i)]

= E[L am(k)z(n -

k)z(n - i)]

k=O

=

0, by the orthogonality property

=

L a:n(k)E[z(n - m + k)z(n -

(b) m

E[gm(n)z(n - i)]

i)]

k=O m

= L a:n(k)"Yzr(k =

k=O

0,

i

m

+ i)

= 0,1, ... , m -

1

(c)

E[fm(n)z(n)]

=

m

E{fm(n)[fm(n) -

L Qm(k)x(n k=l

362

k)]}

= E{IIm(n)1 2 }

=

Em m-l

L

= E{gm(n)[gm(n) -

E[gm(n)x(n - m)]

bm(k)x(n - k»)}

k=O

= E{19m(n)1 2 } = Em (d) i

E [Ii (n ) Ii (n )]

where i

= E {Ii (n )[ x( n) +

L OJ ( k )x (n -

k)]}

c=l

=

E{/.(n)x(n)}

= =

Ei Emax(i,j)

> j has been assumed

(e)

=

E [Ii (n) Ii (n - t)]

i

E {Ii (n )[ x (n - t)

+ L OJ ( k)z (n -

t - k»)}

k=l when 0 ~ t ::; i - j, x(n - t - 1), x(n - t - 2), ... , x(n - t - j) are just a subset of x(n - 1), x(n2), ... x(n - i) Hence, from the orthogonality principle, I

Also, when -1

2: t 2:

i - j holds, via the same method we have

E(f. (n)1; (n - t)]

=0

(f) i-I

E[gi(n)gj(n - t)]

=

E{gi(n)[x(n - t - j) +

L bj(k)x(n -

t - k)]}

k=O

when 0 ~ t ::; i - j, {x(n - t), x(n - t -1), ... , x(n - t - j)) is a subset of {x(n), ... , x(n - i + I)} Hence, from the orthogonality principle,

Also, when 0

2: t 2:

i - j

for i = j, E{/i{n

+ 1 we obtain the same result + i)li(n + j» =

= for i ¢

i,

suppose that i

> j.

2

E{/. (n

(g)

+ i»

Ei

Then j

E {Ii (n

+ i) f; (n + j)} =

=

E {Ii (n

+ i)( x (n + j) + L OJ ( k) x (n + j c=l

0 363

- k)]}

(h) suppose i > j

E{gj(n

j-l E{gj(n + i)[x(n) + Lh](k)x(n + j - k)]}

+ i)gj(n + j)} =

k=O

= E[gi(n + i)z(n)] = Ei (i) for i ~ j

E {/i(n)gj (n)}

for i < j,

E {/j(n )gj (n)}

= = = = = =

j-l E{/i(n)[x(n - j) + L hj (k)x(n - k)]} E=O

E {Ii (n )[hj (O)x( n)]} kjE[/i(n)z(n)]

kjEi

i E{gj(n)[z(n) + L Qj(k)x(n - k)]} 1=1

0

(j) j

\

i-I

E{/i(n)gj(n - I)}

= = = = =

E{/j(n)[x(n - 1 - j)

+L

hi(k)x(n - 1 - k)]}

1=0

E[Ji(n)x(n - 1 - i)]

i

E{/i(n)[gj+l(n) -

L b + (k)x(n i

1

k)]}

1=0

-E[/i (n )6 i+ 1 (O)z(n)] -ki+l E i

(k)

= = = in =

E{gi(n - I)z(n)}

E{/i(n

+ I)z(n -

i+l E{gi(n -l)[fi+l(n) - LQi+l(k)z(n - k)]} A:=1

-E[gi(n - I)Qi+l(i + I)z(n - I - i)] -ki+l E i i E{/i(n + 1)[/i(n - i) - L Qi(k)x(n - i - k)]} A:=1

(1) suppose i > j ;-1

E{Ii(n)gj(n - I)}

= E{/i(n)[z(n-I-j)+ Lbj(k)x(n-I- k)]} A:=O

=

0 364

~

Now, let i

j. then i

=

E{/i(n)gj(n - I)}

E{gj(n - I)[x(n) +

L ai(k)x(n -

k)]}

k=l

=

E{gj(n - 1)x(n)}

=

-k;+lE;

from Cd)

11.16

= 0,

I ~ i ~ m 0~ i ~m - 1 (c) E[/m(n)x-(n)] = E[gm(n)x·(n - m)] = Em (d) E[/i(n)I;-(n)] = Emar(i,j) (e)

(a) E[/m(n)x·(n - i)]

(b) E[gm(n)z·(n - i)]

= 0,

•(

E [ Ii (n) I j n - t

)]

= 0,

r

lor

(f)

{I ~ ?t ~t ?

i - j, i _ j,

-1

{O0 >~_ tt >~_ ii --

•(

)]

Ef!.(n

+ i)f;(n + j)l = {;'"

E [gi ( n )gj n - t

= 0,

r

lor

j, J"+ 1 ,

(g)

(h) E[g,(n (i)

:; ~

+ i)gj(n + j)] = Ema:r(i,j) i?j i 16

~

=[

0 0

0 0

0 0

.£.] => 1~2 3

1: 3

2

1: 2 = -

3

47 =-128 I "

(b)Refer to fig U.S

x(n

Figure 11.5:

368

11.21 ex>

fzz(f)

L 'Yzz(k)e-j21rjle = L am(p)e-j21rjp = L a~(q)ej21rjq =

Ie=-ex> m

Am(f)

,,=0 n

A~(f)

G=O

1~ r..

(f) Am

(f)A~ (f)d/ = L L L i'zz(k)a",(p)a~(q) j! e-i2w /('+P+')d/

~

Ie

=

-1

q

" ex>

m

n

L L L 'Yzr(k)am(p)a~(q)6(q - P -

k)

Ie=-ex> 1'=09=0 m

n

L L 'Yrr(q - p)am(p)a~(q) = L L E[z(l + q)Z-(I + p)]am(p)a~(q)

=

,,=09=0

" = E {~a",(p)Z"(1 + p) ~ a~(q)z(I + P)} 9

= =

E {fm(l

+ m)J:(1 + n)}

E m 6mn

where the last step follows from prob. 11.16 property (g)

11.22

H(z)

= = = = = = = = = = =

A2(Z)

=

Adz) Bd z ) A2(Z) B 2(z) A3(Z) B3(Z) A.. (z)

1 + 0.6z- 1

0.6 + z-1

A 1(z) + k 2B 1(z)z-1 1 + 0.78z- 1 + 0.3z- 2 0.3 + 0.78z- 1 + z-2 A2(Z) + k 3B 2(z)z-1 1 + 0.93z- 1 + 0.69z- 2 + 0.5z- 3 0.5 + 0.69z- 1 + 0.93z- 2 + z-3 A3(Z) + k.. B3(Z)z-1 1 + 1.38z- 1 + 1.311z- 2 + 1.337 z-3 + 0.9z- 4 1 A.. (z)

11.23 1 + 0.lz- 1 - 0.72z- 2

369

I

"

= -0.72 = -0.72 + 0.lz- 1 + z-2 Adz) = A2(Z) - k 2B 2(z)

k2 B 2 (z)

= = = Ao(z) = C 2(z) = = = Hence, /30 = /31 = /32 = k1 Bdz)

1 - k~

1 + 0.357z- 1 0.357 0.357 + z-1

=

Bo(z) 1 130 Bo(z) + 131 Bdz) + 132 B 3(Z) 130 + 131 (0.357 + Z-I) + 132 ( -0.72 + 0.lz- 1 + Z-2) 1 - 0.8z- 1 + 0.15z- 2 1.399 -0.815 0.15

Refer to fig 11.6

input

1.399

-0.815

0.15

++--------~

Figure 11.6:

11.24 Refer to fig 11.7 ht(n) mininizes E[e 2 (n)] (wiener filter) length M r",(w)

=

cr;\H(w)1 2 370

= 2 (a)

+ +--..... output

s(n)

x(n)

h ~n)

y(n)

~""(n)

FIR

Figure 11.7: 0.49

r,,(z)

= \1- 0.8e- jw I2 0.49 = (1 - 0.8z- 1 )(1 -

0.8z)

We can either formally invert this z-transform, or use the following idea: The inverse z· transform of 11.1 will have the form

..,.,,(m) = "'(,,(0)(0.8)lm l From the AR model for s( n) it is easy to show

= = and '"'(" (1) = = ""11(0)

solve for '"'(" (0)

=

so..,.,,(m)

=

Now..,." (m)

0.8'"'(,,(1) + '"'('11(0) 0.8'"'(,,(1) + 6 2 0.8'"'(" (0) + "'(III (1) 0.8"'(,,(0) 49 36 49 (~)Iml 36 5

-

= E[x(n)x(n - m)) = E{[s(n) + w(n))[s(n - m) + wen - m)l} = '"'(,,(m) + C1;6(m) = 49 (~)Iml + 6(m) 36 5

(b)

'"'(dr(l)

=

den) '"'(13;(1)

= = = =

sen) E[s( n )x( n - I)) E{s(n)[s(n -I) 1,,(1) 371

+ wen -

I)]}

So the normal equations are

h'(O)

= 0.462,

h'(l)

(c) ~ = MMSE 2 = ~~ - 0.462 x :~ - 0.248 x :: x

= 0.248

t = 0.462

11.25 f dZ(Z)

= r..,(z) =

f zz(z)

= =

G(z)

[ r u(z) ]

G(z-l) +

(1-0.45z- 1 ) (1-0.8z- 1 )

=

[

= = =

h7(n) = (,: = MMSE oo

(1 - 0.8z- 1 )(1 - 0.8z) r..,(z) + 1 1. 78( 1 - 0.45z- 1 )(1 - 0.45z) (1 - 0.8z- 1 )(1 - 0.8z)

=

=

H:(z)

0.49

0.49

]

(1-0.8z- 1 )(1-0.45z) +

[0.766 0.345z ] 1 - 0.8z- 1 + 1 - 0.45z + 0.766 1 - 0.8z- 1 1 1 - 0.8z- 1 0.766 1 1.781 - 0.45z- 1 - 0.8z- 1 0.43 1 - 0.45z- 1 0.43(0.45 )"u( n)

=

~ f. [f" (z) 21r} e

=

-

1 f.

21rj

e

He ( Z ) r" (z -

0.28 (z - 0.45)( 1 - 0.8z)

d

1 )] Z - 1 d Z

Z

= 0.438

11.26 Using quantities in prob. 11-25,

0.275

=

(1- 0.45z- 1 )(1- 0.45z)

=

1 2 . f.[fc~d(Z) 7r}

e

372

H~A:)rdz(z-l)]z-ldz

=

_1_ 1 0.275 dz 21rj Ie (z - 0.45)(1- 0.45z- 1 ) 0.345

= 11.27

1',,(m)

~3

= MMSE3 = 1 -

= (0.6)lm l

0.455 - 0.15 x 0.6 - 0.055 x 0.36

= 0.435

Increasing the length of the filter decreases the MMSE.

11.28 I

1'rz (0)

= 2.5641,

1'%%( 1)

=1.6026,1'%r(2) =0.064

For m < 0, 1'rr(m) = 1'rr( -m)

11.29 r,,(z)

r,,(z) 373

\

r,,(z)

= =

r,,(z) + O'~

0'; + O'~(.)(.) (.)(. )

z(n)is ARMA(p,p). Suppose

r,n·(z)

=

(L:~=o b,,(k)z-I:)(L::-o bp(k)zl:) (L~=o ap(k)z-I:)(Lt=o ap(k)zl:)

Comparing parameters of the two numerators p

tT~ + O'~

p

L a;(k) = tT~ L b;(k) 1:=0

tT~

1:=0

P-f

P-f

1:=0

1:=0

L ap(k)ap(k + q) = O'~ L bp(k)bp(k + q)

q

= 1,2, ..

'1

There are p+ 1 equations in p+ 1 unknown parameters O'~, bp(I), ... , bp(p). Note that bp(O)

374

P

= 1.

Chapter 12

12.1 (a)

(b)

=

1 N

N-l

L

z(n + m)z·(n)

n=O N

N

L

"'rzz(m)e-

j27 m /

=

m=-N

L

~

m=-N N-l

1

N-l

L

z(n + m)z·(n)e- i27 / m

n=O

n+N

L N L = n=O

z(l)z·(n)e- j27 /(I-n)

I=n-N

=

1 N

N-IN-l

LL

n=O N-l

.

z(l)z·(n)e-j27/lei27/n

1=0

= ..!..I ~ z(n)e-j21r/nI2 NLJ n=O

375

12.2 E[brr(m)1 2 ]

=

~2

N-lml-1 N-lml-1

L

=

~2 L

L

.

n=O

L

n

E[x·(n)x(n

+ m)x(n')x·(n' + m)]

n'=O

+ m)]E[x(n')z·(n' + m)]

{E[x·(n)x(n

n'

+E[x·(n)x(n')]E[x·(n' + m)x(n + m)]

+ m)]E[z(n')x(n + m»)}

+E[x·(n)x·(n'

=

~2 L

Lh';z(m)

n

Let P = E[I;;rr(m)1 2 ]

=

+ i;z(n -

n')

n'

+i;r(n' + m - nh'zz(n n - n'. Then m

2 [ N -N 1 ' ] ' izz(m)

+m-

1 + N2

L

n')]

L[izz(P)irr(P 2 ° - mhzr(P + m)]

n

=

IE[;zz(m)]1

2

+ ~2

p

L Lb;z(P)i;r(P n

mhzz(p + m)]

p

Therefore, var[;rr( m)]

=

~2 L

Eb;z(p);;z(p - m)irz(P + m)]

n

~

~

p

L

-

i

00

j

\

[;;z (P)i;z (p - m)irz(P + m)]

p=-oo

12.3 (a)

E(-y..( mh;.(m'l]

=

E ([

~ N-~-1 zO(nlz(n + m) .J

[~ N~~O-1 z(n'lzO(n' + m'lJ } = ~2 L L E{x·(n)z(n + m)x(n')z·(n' + m')} n

n'

= ~2 L L n

{E[z·(n)x(n

+ m)]E[z(n')z·(n' + m')]

n'

+ m')z(n + m)] + m')]E[z(n')x(n + m»)}

+E[z·(n)x(n')]E[x·(n' +E[x·(n)z·(n'

=

(14

NZ2 L

L[6(m)6(m')

n

+6(n' Hence, E[prr(fdPrz(!2)]

=

+ m' -

N-1

L

+ 6(n -

n')6(m - m')

n'

n)6(n

+m-

n')]

N-1

L

m=-(N-1) m'=-(N -1)

376

Eb.rr(m};rr (m')]e- j2 1l'm h e- j2 1l'm' h

=

;2 LLLL[6(m)6(m/ ) + 6(n 0'4

m

ml

+ m' -

+6(n'

=

n

n')6(m - m')

n'

n)6(n

+m-

n')]e- j2 11'm/Ie- j2 11'm'h

4{1 + [sin1r(fl+h)N]2 [Sin1r(fl-h)N]2} N sin1r(fl + h) + N sin7r(fl - h)

O'z (b)

(c)

var[pzz(f)]

=

COv[prr(ft )Pzz(h)]lh=h=J

=

cr: [1+ (~:~:::,

r]

12.4 Assume that z(n) is the output of a linear system excited by white noise input w(n), where = 1. Then Pu(f) r rz(f)Pww(f). From prob. 12.3, (a), (b) and (c), we have

=

0';

E(prr(fdPu(h)]

cov(prr(ft )Pzr (h)]

f u(fdfzz(h)E[pww(fdPww(h)]

=

fzz(fdfzz(h){1+

[Sin~(!l +h)N]2 + [Sin~(!l N Sln1r(!l

+ h)

-h)N]2} N Sln1f(ft - h)

= r zz(fl)fzz(!2 )cov[pww(ft )Pww(h)] =

var (P:rz (f)]

I

=

rzr(ft)fzr(!2){[Sin~(!1 +!2)N]2 + [sin~(fl-h)N]2} N sln1r(ft

+ h)

= cov(prr(!dPzr(!2)]Ih=h=J

=

r J. rz

[1 + (Sin21r! N )2] Nsin2-rr!

12.5 Let Yt(n)

=

=

z(n) N-l L

* ht(n) z(m)e

J'''.(''-''') N

m=O

=

N-l e ";,," L z(m)e -1'..0." m=O

Yt(n)ln=N

=

N-l L z(m)e -1:319 .", m=O

=

X(k) 377

N sln-rr(fl - h)

"

Note that this is just the Goertzel algorithm for computing the OFT. Then, N-l

IYA:(n)1 2

= IX(k)1 2 = L I

z(m)e -'~N"'"

2

1

rn=O

12.6 From (12.2.18) we have M-l

W(f)

=

_1_ 1

MU

=

~U

L

w(n)e- i2 I' Jn I2

n=O M-l M-l

L L: w(n)w·(n')e-

j21f

!(n-n')

n=O n'=O

j-!! W(f)df

= ~U L: ,L w(n)w·(n') j!_~ e- j2 1'!(n-n')df n

n

.,

= ~U L: L: w(n)w·(n')6(n n

=

n')

n'

t [~ % l !'] = !w(n

1

by the definition of U in (12.2.12)

12.7 (a) (1) Divide z(n) into subsequences of length ~ and overlapped by 50% to produce 4k subsequences. Each subsequence is padded with zeros. (2) Compute the M-point OFT of each frame or subsequence. (3) Compute the magnitude square of each OFT. (4) Average the 4k M-point OFT's. (5) Perform the 10FT to obtain an estimate of the autocorrelation sequence. (b)

¥

M-l

L: z3(m)e- J~M'"

m=O

=

~l l3i:'" + ~l L..J zl(m)eL m=O

=

z2(m -

M 1~i1'" 2' )e-

m=¥

M-l

M-l

m=O

m'=O

L zl(m)e- 1~X:'" + e- jd L

z2(m')e- f~wt...,

(c) Instead of zercrpadding, we can combine two subsequences to produce a single M-point subsequence and thus reduce the number of sequences form 4k to 2k. Then, we use the relation in (b) for the DFT.

378

12.8

=

=

=

(a) Obviously, ~f 0.01. From (12.2.52), M ~~ 90. (b) From (12.2.53), the quality factor is QB = 1.1N~f. This expression does not depend on M: hence, there is no advantage to increasing the value of M beyond 90.

12.9 (a) From table 12.1, we have

QB = ~~f

1.11N l::J.f 1 QB 1.11N 111

=

Qw = 1.39N ~f Qw

1

-== 1.39N 139 QBT = 2.34N~f

~l::J.f

1 QBT - =234 = 2.34N

~~f

(b) For the Bartlett estimate,

QB =

=

~M

N

M N

- = 100

QB

For the \Velch estimate with 50% overlap,

Qw =

=

~M

16N

M 16N

- = 178

Qw

For the Blackman-Tukey estimate,

QBT = ~M

=

1.5N

M 1.5N _

---1

50

QBT

12.10 (a) Suppose P~)(f) is the periodogram based on the Bartlett method. Then, M-l

I: zi(m)e-

p~)(f)

=

~I

p~~)(f)

=

0

=

1 ~W

j21f

i=0,1, ... ,k-1

!nI2,

n=O

p~;)(f)

M-l

I

I: zl(m)e-

j21f

2

!n 1

n=O

p;;)(f)

= =

(1 - w)p11 )(f)

wp;~)(f)

+ (1

1

- w)P1 )(f)

3i9

j

I

2 = (I - w)[wp11){f) + P1 \f)] P~':){f)

m = (I - w) L, mw -1: P11:\f)

1:=1

Therefore, E{P~:f){f)}

= (I - w) L, Mw m -1: E[p11:){f)] 1:=1

var{P~~)(f)} var{ p~~)(f)}

M

jt_! r u(e»

[Si~7r(f -

e»M] 2 de> sln7r{f - a)

=

(1 - w) 1 - w 1 1- w M

=

(1-

= =

E{[P~~)(f)F} - [E{P~:'){f)}f

wM)~ jt

_!

M

ru(e»

[Si~7r(f -

e»Mr de> sln7r{f - 0)

E{[(1- w) ~ Mw m -1: P11:)(f)]2} 1:=1

L, Mw m -1: P11:\f)]}2

-{E[(1 - w)

k=l

W)2l~ Mw 2(M-i)E{p1i )(fW - {E[P1i \f)W]

= =

(1 -

=

- W 2 [ 1+ (1-w) 2 11-w 2 rzz{f)

=

(1 - w 2w ) 1 - w r 2 (f) [1 + (Sin27r f M 1 + w zz Msm27rf

(1 - w)2

L Mw 2(M-1:)var[P11:)(f)] 1:=1

2M

e

(b)

where W(f)

= =

E{P~~)(f)}

=

_1_ 1

.L

j:il-! r zz(a)W{f - o)da M-l w{n)e-i2rJnI2

"

MU L.i

n=O

=

M

(1 - w)2

L, w2(M-k)var[P~~(f)] 1:=1

=

(1- w

2M

)

C~ :) r~.(f)

12.11 Let R~i) be defined as follows:

r~i)(l) r~i)(O)

380

r] r]

in27r fM Msin21rf

I

,

Then,

E·t(/)R~i1E(/)

=

M-l

M-l

LL

~ r~i1(k -

k')e- i27r (Ic-A:')J

Ic=O k'=O

= ~

= Therefore,

P;:) (I)

= =

M-l

E

E

r~i)(m)e-j21fm!

k=O m=k-(M-l) (M-I) ""' (MimI) (i)(

L.J

M

r;u

)

me

-j21fm!

-(M-I)

P;~(/)

l E E·t(/)R~k) K

E(/)

k=l

12.12 To prove the recursive relation in (12.3.19) we make use of the following relations: N-l

= = = = =

Em where fm(n)

gm(n) and

Em- l

L [Ifm(n)1

2

+ Igm(n -

n=m 1m-len)

+ kmgm-l(n k:nfm-d n ) + gm-l(n -

I)1 2e]

(1)

1)

(2)

1)

I

N-I

L

[Ifm_l(n)1

2

+ 19m-l(n -

2 1)1 ]

n=m-l 2 2 I/m-l(m - 1)1 + Igm-l(m - 2)1 N-I 2 2 + [lfm_l(n)1 + 19m-l(n - 1)1 ] n=m

E

N.-l Also,

L [1m-le n ) + 9:n-tC n -

n=m

1)]

=

1-

-

-"2kmEm-l

We substitute for fm(n) and 9m(n - 1) from (2) into (1), and we expand the expressions. Then, use the relations for Em - l and km to reduce the result.

12.13 zen) E[z(n)] since E[w(n)]

= 2'1 z( n - 1) + w( n) - w( n - 1) = ~E[z(n - 1)] + E[w(n)] - E[w(n = 0, it follows that E[z( n)] = 0

To determine the autocorrelation, we have

1

h(O)

=

2'h(-I)

h(I)

=

2'h(O)

381

1

+ 6(0) -

+ 6(1) -

6(-1)

6(0) =

= -1 1

-2'

1)]

\

p

=

q

=

1,

1

a

= -2'

bo

= 1, = = = = = = = =

Hence, "Yzz,(O) '"Yn·(I)

and "Yzz(O)

"Yzz(l) "Yzz(m)

"Yzz(m)

1 2"Yzz(l)

+ O'w(1 + 2)

1

2 + O'w(-I)

2"Yzz(0)

4

1

2

2

30'w

1

--0'

3

2 w

-al"Yzz(m - 1) _!(!)m-l 2 3 2 O'w' "Yzz(-m)

m

_!(!)-m+l 2 3 2 O'w'

>

m

1


r rz(f) =

rn(f)

(! )lm1e- i 2'J: j m L 2 m=-oo

=

0.75 1.25 - cos21r f r rz(f)rh h (I)

=

3cos 21r1 (1.64 - 1.6cos27f f)( 1.25 - cos27r f)

=

(b)

It

54

1.64 - 1.6cos21r 1

1.25 - c0821r 1

9

=

150

1

2Z - 1)(1- 3z - 1)

25

1.64 - 1.6cos27f f

3

_ 50

= 150(0 .8)lm l - 50(!2 )Im l 383

4

1.25 - cos27f f

H 2(z)

=

(c) O'~ = "Yrr(O) = 150 - 50 = 100

12.18 proof is by contradiction. _ (a) Assume the Ikml > 1. Since Em E m - 1 < O. Hence, O'~ < 0, and

= (1 -lkm I

is not positive definite. (b) From the Schur-Cohn test, Ap(z) is stable if the unit circle.

2

)Em_ 1 , this implies that either Em < 0 or

IkmI
3, by the above recursion. Thus,

"Yrz(O) 1'zz(l ) 1'zz(2) "Yzr(3) "Yzz(4) 1'rz(5)

= = = = = =

= 0, 1,2,3. Then we can obtain "Yrr(m)

4.93 4.32 4.2 3.85 3.65 3.46

(c)

= =

14 -1 1 - -z 24 1 24 1 9 -1 24 - 24 z 384

9 24

-z

-2

1_ 3 + -z 24

14 -2 - 24 z +

-3 Z

A2(Z)

A3(z) - k 3B 3(z) 1- k§ 1 - 0.569z- 1 - 0.351z- 2 -0.351 -0.351 - 0.569z- 1 + z-2 A2(Z) - k 2B 2(z) 1- k~ 1 - 0.877z- 1

= =

k2 = B 2(z) =

A 1(z)

=

= k 1 = -0.877

12.20 (a)

[

"'frr(O) "'frr(-I) ;rr(-2)

;Z:(2)] [

;zz(l) ;::(0) ;::(-1)

Hence,

;::(1) ;::(0)

;::(m) ;zz(m) 2 u tAl

1 ] 0 -0.81

= [ u~0

= O.81;zz(m - 2),

]

0

m~

3

= {2.91, 0, 2.36, 0, 1.91,0, 1.55,0, ...} q

The values of the parameters dm

MA(2): dm MA(4): dm MA(8) : dm

L

=

bA;bA;+m

are as follows:

.=0

= {2.91 , 0, 2, 36} = {2.91, 0, 2, 36, 0, 1.91} = {2.91, 0, 2, 36, 0,1.91,0,1.55, o}

(b) The A-f A(2), M A( 4 )andAI A(8) models have spectra that contain negative values. On the other hand, the spectrum of the AR process is shown below. Clearly, the MA models do not provide good approximations to the AR process. Refer to fig 12.1.

12.21

=

{1.656u~, 0, 0.81u~, 0, ...}. For AR(2) process:

;zz(m)

o

0.81u; ] [

o

1.656u;

The solution is

1.656u~

° 9 01

02

For the AR( 4) process, we obtain g g,

= = =

0

=

1.07 and

1] = [ °

gu; ]

01

02

0

1.12 -0.489

= {I, 0, -0.643,O,O.314} 385

l

\

6r----,-----r--~--...,.._--_r__--_-___r--_r--__r_-__,

Q)

5

"C

.a4

'2

~3 E

1'2 I I

1

........- -........--...I----~--~--

O~--""'----..Io.---

o

0.05

0.1

0.15

0.2 0.25 0.3 ---> frequency(Hz)

0.35

. . . . - -.0.45 ....

---J

0.4

Figure 12.1: For the AR(8) process, we obtain 9 ~

= =

1.024 and {I, 0, -0.75,0,0.536,0, -0.345, 0, 0.169}

Refer to fig 12.2.

12.22 (a)

Th~

=

40'~ (2 - z-I)(2 - z) -9- (3 - z-I)(3 - z)

=

u~H(z)H(z-1

minimum-phase system function H (z) is

H(z)

2 2 - z-1

= =

33- Z-1 4 1 - !z-1 91- 13 z- 1

(b) The mixed-phase stable system has a system function

H(z)

= =

21-2z- 1 3 3 - z-1 2 1- 2z- 1 91-i z- 1

12.23 (a)

386

0.5

MA(2)

AR(2)

2

2.5

-! 1.5

~

:)

:eCD ~

2

.! CD

1

~ 1.5

It. I

It. I I I

10.5 0 0

0.2 0.4 ---> frequency(Hz) AR(4)

1 0.5 0

0.6

2

2

.":: 1.5 c:

~ 1.5 ~

~

r

f It.

0.2 0.4 --> trequency(Hz) AR(8)

0.6

0.2 0.4 --> trequency(Hz)

0.6

1

It. I

I

I I

I 0.5

0.5 0

0.2 0.4 ---> trequency(Hz)

0 0

0.6

Figure 12.2:

= = Adz)

= =

Hence, k 1 (b) As r -

l,k 2 -1 and k 1

-

=

j

\

2

r r 2 _ 2rcosez- 1

+ z-2

A2(z) - 1: 28 2(z) 1- k~ 2rcose 1 1z1+ r 2 2rcose 1+ r 2

-cose

12.24 (a)

= =

01 (I) Hence, A3(Z)

=

=

-1.25, 02(2) 1.25, 03(3) -1 1 - 1.25z- 1 + 1.25z- 2 - z-3

First, we determine the reflection coefficients. Clearly, 1: 3 = -1, whcih implies that the roots of A3( z) are on the unit circle. We may factor out one root. Thus, .

= (1- z-l)(l- ~z-l + z-2) 4

where

0

= (1- z-l)(I- O"z-l)(l- O·z-l) 1 + j.;63

=

8 387

=

Hence, the roots of A 3 (z) are z 1,0, and 0-, (b) The autocorrelation function satisfies the equations

=

=

(c) Note that since k3 -1, the recursion E/n E~_1(1-lkmI2) implies that E~ implies that the 4x4 correlation matrix r zz is singular. Since E~ 0, then O'~ 0

=

=

= O.

This

12.25 "Yzz(O) "Yzz(1) "Yzz(2) "Yzz(3)

= = = =

1 -0.5 0.625 -0.6875

Use the Levinson-Durbin algorithm 1

"Yzz(l)

= - - - = -2 "Yzz(O)

I 1

= 1 + !z-1 2

E1

=

(1 - o~(I)hzz(O)

= 43

"Yzz(2) + ot{l)-yzz(1) E1 1

= 01(1) + 02(2)01(1) = 4 Therefore, A2 (z) ~

=

1

1

1

1 + -z- - -z4 2

1

k 2 =-2

388

2

1

=-'2

12.26 (a) (1)

H(w) rrr(w) rrr(w)

= = =

1 - e- jw 1 + 0.81e- jw

IH(w)1 20';, 1 - e- jw 2 2 11 + 0.81e- jw I O'w

(2)

=

H(w)

r rr(W) =

=

(1- e- j2w )

IH(w)1 20'; 40'2w sin 2 w

(3)

H(w)

=

1

1 - 0.81e- jw

0'2w 1.6561 - 1.62cosw (b) Refer to fig 12.3. (c) For (2),

o~ m ~ m> 2 m< 0

since bo

1rr (0) 1'rr(2) 1rr( -2) 1rr(m)

= = = = =

I,

bl

1

[ 0.81

0,

0 1.81 0

1rr(0) "Yrr(m) "Yrr(m)

2

we have

20'2w _0'2 w _0'2w m #; O,±2

For (3), the AR process has coefficients ao

o

= 0 and b = -1,

2

= = =

= 1, al = 0 and a2 =0.81.

0.81 ] [ 1'rr(O) ]

o

1'rr(l) 1'rr(2)

1

=

[ 0'; ] 0 0

2.90'~ 0, m odd 2.9(O.9)lmI0'~ ,

389

m even

(1 )

(2)

8r---------------

2r-----~-----~ 1.5 ~

r

1

"I 0.5 I

0.2 0.4 ---> trequency(Hz)

0.2

0.6

0.4

--> trequency(Hz)

(3)

6-----------G)

5

~4 'c

~3

~2 I

I

1

0.2 0.4 ---> trequency(Hz)

0.6

Figure 12.3:

12.27 (a) 00

r rz(z) =

L 1'zz(m)z-m -00

=

1

-1

~(4)-mz-m

00

0

-00

~z

1 1 1 1- 4 z-

=

--1-+

=

16 (1 - ~z)(1 - ~z-l)

1- 4 z

15

since

r rz(z) = H(z)

=

(12

H(z)H(z-I),

0.968 1 - 14 z -

1

is the minimum-phase solution. The difference equation is

x(n)

1 = -x(n 4

1) + O.968w(n)

390

1

+ ~(4)mz-m

0.6

where w(n) is a white noise sequence with zero mean and unit variance. (b) If we choose

H(z)

= =

then, x(n)

= =

1 1 - ~z Z-1 z-1 _ 1 4

4z- 1 1 - 4z- 1 4x(n - 1) - 4xO.968w(n - 1)

12.28 1'zz(O) 1'zz(l) 1'zz(2) 1'zz(3) al (1)

Al(Z)

=> k 1 = 0 E1 a2(2)

= = = = = =

1 0 _a 2 0

_ 1'zz(l) 1'zz(O) 1

=0

= (l-ar(1»1'zz(O) = 1 1'zz(2) + al(I)"Yzz{l) =a = E = al{l) 2+ a2(2)al(I) = 0 = 1 + a z-2

I

2

1

a2(1) Therefore,A2(z)

=> k 2 = a 2 £2

aa(3)

a3(2) aa(l)

= = = = =

= A2(Z) => k 3 = 0 E 3 = E2 =

Therefore,A3(z)

=

4 (l - a~(2»EI 1- a _ "Yrz(3) + a2(l)"Yrz(2) + a2(2hzz{1) £2 a2(2) + aa(3)a2(l) a 2

= a2(l) + aa(3)a2(2) = 0 1 + a 2z- 2 1 _a 4

12.29 (a) For the Bartlett estimate,

M

(b).~

391

= = =

0.9

t::.!

=

0.9 90 0.01 0.9 0.02 = 45

=0

)

(c)for (a), QB

for (b), QB

= = = =

N M 2400 90 N M 2400 45

= 26.67 = 53.33

12.30 Ap(z)

= A p_ (z) + kpB p_ (z)z-1 1

1

where Bp_1(z) is the reverse polynomial of Ap-dz). For Ik p I < I, we have all the roots inside the unit circle as previously shown. For Ikpl 1, Ap(z) is symmetric, which implies that all the roots are on the unit circle. For Ikpl > 1, Ap(z) = A.(z) + lBp_dz)z-I, where A,(z) is the symmetric polynomial with all the roots on the unit circle and Bp_1(z) has all the roots outside the unit circle. Therefore, Ap(z) will have all its roots outside the unit circle.

=

12.31

(a)

= =

Therefore, H(z)

z-0.9 z-1-0.9 UI z2 + 0.81 z-2 + 0.81 z - 0.9 z2 + 0.81 z-l(l- 0.9z- 1 ) 1 + 0.81z- 2 2

(7

= (b) The inverse system is

1 + 0.81z- 2

1

H(z)

= z-I(1- 0.9z- 1 )

This is a stable system.

12.32 X(k) ~

N-l

2: z(n)e -'~"..

n=O

(a)

E[X(k)]

= L

E[z(n)]e -iljd

n

E[lX(k)1 2 ]

=

=0

L L E[z(n)z·(m)]e -l~d,*"-"'l L L 0';6(n - m)e -'~dA"-Prl) n

m

n

m

392

N-l

= U; L = Nu;

1

n=O

(b)

n

= u;

=

n'

L L 6(n -

n')e -Z3","'''' e -'3d~"-"'1

n n' 2 Z3'N''''

uze

=

Nu;,

=

0,

m=pN otherwise

p

= 0, ±l, ±2, ...

12.33

=

"'rt!t!(m)

E[v·(n)v(n 9

9

= L L

.'=0.=0

+ m)]

b~bk,E[w·(n - k)w(n + m - k')]

9

9

= (1; L L

.'=0.=0

b;b.,6(m + k - k')

9

= (1; L b~bA:+m .=0

=

(1;d m

--

(1:.... "" L,.;

q

Then , r \it! (f)

dme-j2ff/m

m=-q

12.34 l'zz(m)

=

E[z·(n)z(n

+ m)]

= =

A E{cos(wln

=

Tcoswln

2

+