50999264FGX0_Digital Signal Processing Applications_Soution Manual

8/18/2019 50999264FGX0_Digital Signal Processing Applications_Soution Manual

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Solutions of Examples for Practice

Example 1.4.7

Solution : The frequency of the sinusoid is,

W = 100 Hz

The sampling frequency is, f s = 240 Hz.

Here f W s 2 i.e. 240 > 200. Hence there is no aliasing.

The minimum sampling rate is given as,

f s 2W

We know that f T s s

1

and W T

1

. Then above equation becomes,

1T s

2T

Here T is the period of x t( )

T s T 2Thus sampling period should be less than or equal to half period of the signal.

Example 1.4.8

Solution : i) x t e j t ( ) 25 500

Here = 500

2 f = 500 f 250 Hz

Nyquist rate = 500 Hz

ii) x t t t ( ) ( ) 1 0.1 sin 200 cos 2000 = cos . sin cos2000 01 200 2000 t t t

= cos . sin( ) sin( )2000 0 12

1800 2200 t t t

= cos . sin . sin2000 0 005 1800 0 05 2200 t t t

Here max = 2200

f max = 1100 Hz

Nyquist rate = 2200 Hz

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iii) x t t ( ) 10 sinc 5

= 105

5

sin

t

t

Here = 5

2 f = 5 f 52

Nyquist rate = 5 Hziv) x(t) = 2 sinc (50 t ) sin ( 5000 t )

= 2 5050

5000sin

sin

t

t t = 1

2512

4950 5050 t t t cos cos

Here max = 5050

2 f max = 5050 f max 2525 Hz

Nyquist rate = 2 25 25 5050 HzExample 1.4.9

Solution : i) Here 1 2000 and 2 4000

f 1 =

1

2

2000

2 = 1000 Hz and f 2 =

2

2

4000

2 = 2000 Hz

W = f f max 2 = 2000 Hz

Nyquist rate = 2 2 2000 4000W samples/sec

Nyquist interval = 12

14000W

= 0.25 msec

ii) Here x t( ) =sin ( )

( )

( cos ) /

( )

2

2 2

4000 1 8000 2

t

t

t

t

= 80002

8000

2

f W = 4000 Hz

Nyquist rate = 2 2 4000W = 8000 samples/sec

Nyquist interval = 12

18000W

= 0.125 msec.

Example 1.4.10

Solution : i) g t 1 ( ) = sinc (200 t )

=sin 200

200

t

t

Here max = 200

f max = 100 Hz

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Nyquist rate = 2 × f max = 200 Hz

and Nyquist interval = 1

2 f max=

1200

= 0.005 sec

ii) g t 2 ( ) = sinc2

(200 t )

=sin 200

200

2

t

t =( cos ) /

( )

1 400 2

200 2

t

t

=1 400

2 200 2 cos

( )

t

t

Here max = 400 f max = 200

Nyquist rate = 2 f max = 2 × 200 = 400 Hz

And Nyquist interval = 1

2 f max=

1400

= 2.5 msec

iii) g t 3 ( ) = sinc (200 t) + sinc3 (200 t)

=sin sin200

200200

200

2

t

tt

t

=sin cos

( )

200

2001 400

2 200 2

t

tt

t

Here 1 = 200 and 2 = 400

f 1 = 100 Hz and f 2 = 200 Hz

f max = 200 Hz

Nyquist rate = 2 f max = 2 200 = 400 Hz

and Nyquist rate = 1

2 f max=

1400

= 2.5 msec

Example 1.5.6

Solution : i) ( ) ( )

1 2kk

t k

The given signal can be expressed as,

( ) ( )

1 2k k

t k =

( )

( )

t 2k

t 2k

for odd values of k

for even values of k

The above signal is shown in the Fig. 1.1. Observe that the signal is

periodic with period T = 4




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ii) x n( ) = ( )1 n

= 1 1 1 1 1 10 1 2 3 4 5( ) ( ) ( ) ( ) ( ) ( )

, , , , , , .....n n n n n n

This signal is periodic with period N = 2 samples

iii) x t ( ) = w t kk

( )

25

5

From above Fig. 1.2 (a), observe that x t( ) exists from k = – 11 to 11 only. Hence it is

non periodic signal.

iv) x(t) = w t kk

( )

3

Above signal is shown in the Fig. 1.2 (b). Observe that the signal is

periodic with period 3.



(t + 4)k = –2

–k(t + 2)

= –1

–6

–5 –4 –3 –1

–2

(t)= 0k

–k(t – 2)

= 1

(t – 4)= 2k

–k(t – 6)

= 3

2

0 1 3 4 5

6

7 8 t

Fig. 1.1 Sketch of ( ) ( )

1 2k k

t k

1 2 3 4 5 6 7 8 9 10 11 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11

1 2 3 4 5 6 7 8 9 10 11 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12 –13 12 13

x(t) = w (t –2k)

k = –5

5

x(t) = w (t –3k)

k = –

k

k

(b)

(a)

Fig. 1.2 Sketches of example 1.5.6


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Example 1.5.7

Solution : i) x(n) = lm e jn 4

= lm n

j n

cos sin 4 4

, since e

j = cos sin j

= sin

n 4

Compare this equation with x n f n( ) sin 2 , hence 24

f nn

f 18

cycles/sample. Since

f k

N

18

. There will be 8 samples in one period of DT sine wave.

Fig. 1.3 shows the waveform of x n( ) = sin n

4 and its even and odd parts are also shown.

Even and odd parts are given as,

Even part, x ne ( ) = 1

2{ ( ) ( )}x n x n and

Odd part, x no ( ) = 1

2 { ( ) ( )}x n x n

ii) x(t) =t t

t t

0 1

2 1 2

This is a triangular pulse as shown in the Fig. 1.4. It's odd and even parts are also shown

in the Fig. 1.4.



x (n) =e [x(n) + x(–n)] =012

x (n) =o [x(n) – x(–n)]12

Zero even part

Odd part

0 1 2 3 4

5 6 7 8n

x(n)

1 N = 8

n

x(n)

n

n

n

x(–n)x(–n)

sin n

4

Even part of x(n) Odd part of x(n)

n

Fig. 1.3 Even and odd parts of x n( )


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iii) x(t) = cos22

t

x t( ) = 1

2cos t

, since cos2 = 1 2

2cos

= 1

212

cos( ) t



t

x(t)

10

1

2

t

x(–t)

–1 0

1

–2

1 2 t

Even part

–1 0 –2

0.5

x (t) =e12

[x(t) + x(–t)]

Odd part

t –1

0

0.5

0.5

1 2

–2

x (t) =o12

[x(t) – x(–t)]

Fig. 1.4 Odd and even parts of traingular pulse


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Here 2 ft = t f = 12

or T = 2. The waveform x t( ) and its odd and even parts are

shown in the Fig. 1.5.

Example 1.5.8

Solution : i) x(n) =cos( n) for n

otherwise

4 4

0

This signal is given only for |n| 4. Hence it is not periodic. Let us calculate its energydirectly,



t

[Zero odd part]

x (t) =o [x(t) – x(–t)] = 012

DC shift of 12

t

t0

0

0

0

t

x(t)

x(–t)

1

1

1

T = 2

12

cos t

x (t) =e [x(t) + x(–t)12

Even part

Fig. 1.5 Odd and even parts of x(t ) = cos22

t


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E = x(n)n –

2

= cos ( n)2n –4

4

Here cos2 ( n) = 1 for any value of n. Hence,

E = 1

n 4

4

= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9


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E =

| ( )|x t dt2

Let us change the limits of integration as T T 2 2

, and take limT

. This will not change

meaning of above equation.i.e.,

E = lim | ( )|T

x t dtT

T

2

22 = lim | ( )|

T T

T x t dt

T

T

1

2

22 by rearranging

= lim lim | ( )|T

T T T

x t dtT

T

1

2

22 = lim

T T P

since quantity inside brackets is P.

= By taking limits as T

Thus, energy of the power signal is infinite over an infinite time.

Example 1.5.10

Solution : i) x(n) = 1 cos2

n

This signal is periodic, and it has a DC shift. It's frequency is,

2 f n = n

2 f

14

k N

N 4

Power is, P = 1

2N 1 | x(n)|

n N

N 2

Here 1 cos

n2

= 2 cos n

42

, hence above equation becomes,

P = 1

9 2 cos

n4

n 4

42

2

=

49

cos n

4n 4

44

= 49

cos cos 3

4 cos

2 cos

44 4 4 4

1 cos

4 cos

2 cos

34

cos4 4 4 4

Since cos x = cos (x),

P = 49

1 2 cos 2 cos 3

4 2 cos

2 2 cos

44 4 4 4

= 4

9 1 2

12

0 1

2169




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Example 1.7.3

Solution : i) x n( )3 1

This waveform is obtained by precedence rule of time shifting and scaling. Fig. 1.6 (a)

shows the sketch of x n( ).

Shifting : First x n( ) is shifted/delayed by '1' sample to obtain x n( )1 . This signal isshown in the Fig. 1.6 (b).

Scaling : The time shifted signal x n( )1 of the Fig. 1.6 (a) is time scaled by 3 samples.

This means it is compressed by 3 samples. This signal gives x n( )3 1 . It is shown in theFig. 1.6 (c).

ii) y n( )1

Fig. 1.6 (d) shows y n( ) and (e) shows y n( ) . Then y n( )1 = y n ( )1 is obtained bydelaying y n( ) by '1' sample. It is shown in the Fig. 1.6 (f).



1 2 3 –1 –2 n

1

23

1

23

(b)

x (n–1)

40

1 2 3n

–3

–2

–1

2 (e)y (–n)

1

3

–3 –2 –1

1 2 3 –1 –2 n

2 (c)

x (3n–1)

0 4

1

1 2 3 –1 –2 –3 n

1

2

3

1

2

3

(a)

x(n)

1 2 3

–1 –2 –3n

1

2

3

–1

–2

–3

(d)

y (n)

1

2 3n

–1

–2

–3

2 (f)

y (1–n)

1

3

4

–2 –1

Fig. 1.6 Sketches of x n( )3 1 , y n( )1


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iii) x n y n( ) ( )2 4 Fig. 1.7 (a) shows x n( )2 . Note that x n( )2

is obtained by time compression of x n( )

by the factor of '2'. Fig. 1.7 (b) shows

y n( ) 4 . The addition x n y n( ) ( )2 4 isshown in the Fig. 1.7 (c).

Example 1.8.4

Solution :

i) The system is memoryless, since y(n) depends upon x(n) i.e. present sample value

only.

ii) Since sine function is multivalued, it is not invertible. Hence system is noninvertible.

iii) System is causal, since y(n) depend on x(n) only.

iv) When x(n) 0, y(n) = sin 00

= 1 by L'Hospital's rule.

Hence it is stable system.

v) Response of system to delayed input by 'k ' samples will be,

y (n, k) =

T x n k

x n k

x n k

sin



4n

–1

–2

–3

(b)y (n–4)

5 6 7

321

3

2

1

1

3

–1 n

3

2

(c)x (2n) + y (n–4)

4 5 6 7

2

1

2

1

–2

–3

1 –1 n

22 (a)

x (2n)

Fig. 1.7 Sketched of x n y n( ) ( )2 4


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Let us delay the output by replacing 'n' by 'n – k',

y n k =

sin x n k

x n k

since y n, k = y n k system is time invariant

vi) The output of the system to inputs x n1( ) and x n2 ( ) will be,

y n1 =

T x nx n

x n11

1

sin... (1)

y n2 =

T x nx n

x n22

2

sin... (2)

The linear combination of two outputs will be, y (n)3 = a y (n) a y ( n)1 1 2 2

= ax (n)

x (n) a

x (n)

x (n)11

1 2

2

2

sin sin By equations (1) and (2) ... (3)

Now the output due to linear combination of two inputs will be,

y n3 = T a x n a x n1 1 2 2 =

sin a x n a x n

a x n a x n1 1 2 2

1 1 2 2

... (4)

Here from equations (3) and (4),

y n3 ( ) y n3 ( ), the system is nonlinear.

The presence of 'sine' function makes the system nonlinear.

Example 1.8.5

Solution : y(n) = x n( ) Output is magnitude of present input, hence the system is static or memoryless.

Magnitude operation is not linear. Hence the system is nonlinear.

Delaying the input by ' 'k samples, output will be,

y n k ( , ) = T x n k ( ) = x n k ( )And the delayed output will be,

y n k ( ) = x n k ( )

Since y n k y n k ( , ) ( ) , the system is shift invariant.

Output is magnitude of present input. Hence the system is causal.

As long as x n( ) is bounded, its magnitude will also be bounded. Hence the system is

stable.

Example 1.8.6

Solution : i) Dynamic system, since output depend upon previous inputs.

ii) Causal system, since output depends upon present and past inputs.

iii) Stable system, since y n( ) is bounded as long as x n( ) is bounded.

iv) Time invariant, since y n( ) is not direct function of time 'n'.




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Example 1.8.7

Solution :

i) Memory : It is dynamic (requires memory) since the term x(n+1) is present.

ii) Stability : Value of cosine function is between (–1, 1). Hence y(n) is bounded as

long as x(n) is bounded. Hence the system is stable.

iii) Causality : The system is noncausal since x(n+1) requires future input.

iv) Linearity : Presence of 'cosine' function makes the system nonlinear.

v) Time invariance : System is time invariant, since there is no time index

manipulation.

Example 1.8.8

Solution : i) y t ( ) = sin [ ( )] x t 2

This system is not memoryless, noncausal, stable. y (t)1 = sin [x (t 2)]1

y (t)2 = sin [x (t 2)]2 Linear combination of two outputs,

y t3 ( ) = a y t a y t1 1 2 2( ) ( ) = a x t a x t1 1 2 22 2sin [ ( )] sin [ ( )] Response to linear combination of inputs,

y t3 ( ) = f a x t a x t[ ( ) ( )]1 1 2 2 = sin [ ( ) ( )]a x t a x t1 1 2 22 2 Here y t3 ( ) y t3 ( ) hence system is nonlinear.

Response to delayed input will be,

y t t( , )1 = f x t t[ ( )] 1 = sin [ ( )]x t t 2 1Let us delay an output itself,

y t t( ) 1 = sin [ ( )]x t t 2 1

Here y t t( , )1 = y t t( ) 1 , hence this is time invariant system.

ii) y n( ) = x n[ ]2This system is not memoryless, noncausal and stable.

y n1( ) = x n1 2[ ]

y n2 ( ) = x n2 2[ ]

Linear combination of the two outputs, y n3 ( ) = a y n a y n1 1 2 2( ) ( ) = a x n a x n1 1 2 22 2( ) ( )

Response to linear combination of inputs,

y n3 ( ) = f a x n a x n[ ( ) ( )]1 1 2 2 = a x n a x n1 1 2 22 2( ) ( )

Here y n3 ( ) = y n3 ( ) hence the system is linear.

Response to delayed input by 'k ' samples will be,

y n k ( , ) = f x n k [ ( )] = x n k ( )2




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Let us delay an output itself by 'k ' samples,

y n k ( ) = x n k [ ( )]2 = x n k [ ]2 Here y n k ( , ) y n k ( ) , hence this system is time variant.

Example 1.10.8

Solution :x n1( ) = 1 1 1

x n2 ( ) = 1 1 1

_________________

1 1 1

1 1 1

1 1 1 _________________

x n x n1 2( ) ( ) 1 2 3 2 1

Example 1.10.9

Solution : i) y n u n * u n 3n

y(n) = k

h k x n k

= k

n k u k u n k

3

Here u k 3 = 1 for k 3 and u n k = 1 for n k , hence

u k u n k 3 = 1 for 3 k n .

y(n) =k

nn k

k

nn k

3 3

=

n

k

n k 1

3

=

n

n 11 1

1 1

3

sincek N

N 2k

N N

1

1 2 1

1

=

nn

32

1

1 1

1

1




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ii) y n u n * n p p

0

4

y(n) = u n n p p

*

0

4 = p

u n n p

0

4*

We know that x n n n x n n0 0* from properties of impulse function

Therefore u n n p u n p* 4 4 . Then above equations will be,

y(n) = p 0

u n p

4

Example 1.10.10

Solution :

Convolution is given as,

x(n) y n x k y n k k

Here nxl = – 2 and n yl 1

similarly nxh = 1 and n hy 2The index of summation will be,

n n n n nxl yl xh yh (– 2 – 1) n (1 + 2) i.e. – 3 n 3

x(n) y(n) = x k y n k andk

( ) n3 3

The value of n is varied from 0 to 3 in Fig. 1.8 (c) (d) (e) and (f ) respectively. The

corresponding output is shown in Fig. 1.8 (j), (k), (l) and (m). For n 4, there is no overlap between x(k) and y (n – k) Hence convolution becomes zero.

The index is varied from n = – 1 to – 3 in Fig. 1.8 (g), (h) and (i) respectively. The

corresponding output is shown in Fig. 1.8 (n) (o) and (p). For n – 4, there is no overlap between x(h) and y (n – k). Hence convolution becomes zero.

Thus the convolution of two sequences becomes, x(n) y(n) = { , , , , – , – , – }1 2 1 0 1 2 1




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Example 1.12.5

Solution : i) Stability

Here h(n) = ( ) u(n )n0.99 3 = ( )n0.99 for n3

Consider |h(k)|k =

= ( )0.993

k

k =



x(k)

–2 –1 0 1k

(a)

y(k)

–1 0

1 2 k

n = 0 y(–k)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)

(n)

(o)

(p)

n = 1

n = 2

–1 –2

0 1 k

k

k

k

k

k

k

y(1–k)

y(2–k)

–1

0 1 2

y(3–k)n = 3

n = –1

n = –2

n = –3

y(–1–k)

y(–2–k)

y(–3–k)

01

2 3

2

3

10

4

0

–1

–2 –3

0

–2

–3

–1

–4

0 –1 –2 –3

–4 –5

x(k) y(–k)

x(k) y(1–k)

x(k) y(2–k)

x(k) y(3–k)

x(k) y(–1–k)

x(k) y(–2–k)

x(k) y(–3–k)

–2 –1

0 1 k

k

k

k

k

k

k

–4 –3 –2 –1

–4 –3 –2 –1 –5

x(k)y(–k)= –1 –1 + 1 + 1= 0

x(k)y(1–k)= 0 –1 – 1 + 1 + 0= –1

x(k)y(2–k)= 0 + 0 –1 – 1 + 0 + 0= –2

x(k)y(3–k)= 0 + 0 + 0 – 1 + 0 + 0 + 0= –1

x(k)y(–1–k)= 0 – 1 +1 + 1 + 0= 1

x(k)y(–2–k)= 0 + 0 + 1+ 1 + 0 + 0= 2

x(k)y(–3–k)= 0 + 0 + 0+ 1 + 0 + 0 + 0= 1

Fig. 1.8 Convolution of x (n) and y (n) using graphical method


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= (0.99) (0.99) (0.99) 0.993

2 10

( ) k

k =

= (0.99) (0.99) (0.99) 1

1 0.993

2 1 , since ak

k = 0

= 11 a

= 103 0, upper limit of 'k ' will be '0'. If n < 0, upper limit of 'k ' will be 'n'

Case - I : For n > 0, y n( ) = 1

2

0

k

k = –

= 1

20

k

k =

By adjusting the sign

= 20

k

k =

=




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Case - II : For n < 0 y n( ) = 1

2

k n

k =

= 1

2

k

nk =

By adjusting the sign

= 2 k

nk =

=

Thus the step response is "infinity" since the system is unstable.




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Example 2.1.7

Solution : The given difference equation is,

y n( ) =

1

2

1

2 1x n x n( ) ( ) ... (1)The linear convolution of unit sample response h n( ) and input x n( ) gives output y n( ). i.e.,

y n( ) =k

h k x n k

( ) ( )

This equation can be expanded as,

y n( ) = .... + h x n h x n h x n h x n( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .... 1 1 0 1 1 2 2

On comparing above equation with equation (1) we find that,

h ( )0 = 1

2 and h ( )1 = 12

And all the other terms are absent, hence they are considered zero. Consider,

H ( ) =k

j k h k e

( )

This is Fourier transform of unit sample response and it is called transfer function. Putting

for h h( ) and ( )0 1 ,

H ( ) = h e h e j j( ) ( )0 10 1

=

1

2

1

2

e j

= 1

2 1

1

2 1

e j j

cos sin

= 12

1 1

2 cos sin j ... (2)

Real part of H H R( ) ( ) ( cos ) 12

1 = cos22

Imaginary part of H H I ( ) ( ) sin 12

= sin cos 2 2

Magnitude of H H H H R I ( ) ( ) ( ) ( ) 2 2


2 DTFT, DFT and FFT


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= cos sin cos42

2 2 2

= cos sin cos4 2 2

2 2 2

= cos cos sin2 2 22 2 2

= cos 2

2

= cos 2

... (3)

Phase of H ( ) = H tan H H

I

R( ) ( )

( )

1

= tansin cos

cos

1

2

2 2

2

= tan tan

1

2

= 2

... (4)

Table 2.1 shows the calculations of magnitude and phase of H ( ) for few values of .

Magnitude

H ( ) cos

2

Phase H ( )

2

0 2

23 0.5

3

2

1

2

4

3

32

6

0 1 0

3

32

6

2

1

2

4

23 0.5

3

0

2

Table 2.1 Calculation of H H ( ) and ( )

Fig. 2.1 (a) shows the magnitude and Fig. 2.1 (b) shows the phase plot of transfer function

based on calculations in above table.

Here note that transfer function H ( ) is the continuous function of ' ' in the range of . Hence magnitude and phase plots in this figure are continuous.

Digital Signal Processing Applications 2 - 2 DTFT, DFT and FFT



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Comments on magnitude and phase plots of H ( ) :

1. The magnitude and phase plots are continuous function of ' ' for nonperiodicsequences.

2. The magnitude plot is even symmetric around 0, where as phase plot has oddsymmetry. i.e.,

H

H

( )

and ( )

H

H

( )

( )

... (5)

3. The magnitude and phase plots are periodic with period 2 . Readers can verify thisstatement by actually calculating magnitude and phase transfer functions of the

preceding example.

Example 2.2.8

Solution : To obtain DFT

X 4 = [ ]W x4 4

X

X

X

X

( )

( )

( )

( )

0

1

2

3

=

1 1 1 1

1 1

1 1 1 1

1 1

j j

j j

1

1

0

0

=

2

1

0

1

j

j



0

0

–

–

1

–

(a) Magnitude plotof the transfer function

(b) Phase plotof the transfer function

|H( )|

H( )

Fig. 2.1 Magnitude and phase plots of the transfer function H e j ( ) 12

1


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To obtain IDFT

xN = 1N

W X N N

x4 = 14 4 4

W X

x

x

x

x

( )

( )

( )

( )

0

1

2

3

= 1

4

1 1 1 1

1 1

1 1 1 1

1 1

j j

j j

2

1

0

1

j

j

= 1

4

4

4

0

0

=

1

1

0

0

Thus the original sequence is obtained back.

Example 2.2.9

Solution : Here bandwidth, W = 4 kHz

Resolution = 50 Hz

i) To obtain minimum sampling rate

By Nyquist theorem,

Fs = 2 2 4 8 W kHz kHz

ii) To obtain minimum number of DFT samples

Let the DFT samples be N . These samples are equally spread over the frequency range of

0 to 2 . In analog domain 2 corresponds to f s 8000 Hz. In other words 'N ' samples will be equally spread over 0 to 8000 Hz. Since minimum spacing between the samples

(Resolution) should be 50 Hz,

50 8000 8000

50 160

N or N samples.

It is given that N = 2m point DFT is used. Hence, let us calculate value of m for N 160.Thus,

m = loglog

log

log

log .2

10

10

10

102

160

2 7 322N

N

Here 'm' must be taken nearest higher integer.

m 8

N = 2 2 2568m

samples.Thus the DFT must have 256 samples.

iii) To obtain minimum duration

Since 8000 samples are taken in 1 sec. 256 samples will require,

Minimum duration = 2568000

32 msec.




25/38

Example 2.4.8

Solution : X (0) = 0.25

X(1) = 0.125 – j 0.3018

X(2) = 0

X(3) = 0.125 – j 0.0518

X(4) = 0

For real valued sequence, X(N – k) = X k ( )

X k ( )8 = X k

X ( )5 = X ( )8 3 = X ( )3 = 0.125 + j 0.0518

X(6) = X ( )8 2 = X ( )2 = 0

X ( )7 = X ( )8 1 = X ( )1 = 0.125 + j 0.3018

Example 2.5.3

Solution : Here, h n = 2 2 1, , i.e. M 3

Input segments length is N = 8.

Since N = M L 1

8 = 3 1 L L 6Let us form the segments of input sequence as follows :

x n1 =

3 0 2 0 2 1 0 01

, , , , , , ,' '

L samples of x n M zeros

x n2 = 0 2 1 0 0 0, , , , ,

' '

Thesetwozerosare apperedtomake L sampl

es

L samples of x n M zeros

' '

, ,0 0

1

And h n = 2 2 1 0 0 0 0 0

1

, , , , , , ,

' ' M samples L zeros

Now let us calculate

y n1 = x n h n1 8

y n2 = x n h n2 8




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y

y

y

y

y

y

y

y

1

1

1

1

1

1

1

1

0

1

2

3

4

5

6

7

=

3 0 0 1 2 0 2 0

0 3 0 0 1 2 0 2

2 0 3 0 0 1 2 0

0 2 0 3 0 0 1 2

2 0 2 0 3 0 0 1

1 2 0 2

0 3 0 0

0 1 2 0 2 0 3 0

0 0 1 2 0 2 0 3

2

2

1

0

0

0

0

0

6

6

1

4

2

6

4

1

Similarly,

y n2 = 0 2 1 0 0 0 0 0, – , – , , , , , 8 2 2 1 0 0 0 0 0, , , , , , ,

= 0 4 6 4 1 0 0 0, – , – , – , – , , ,Last M – 1 samples of y n1 are added to first M – 1 sample of y n2 .

y n1 6 6 –1 – 4 2 6 4 1

y n2 0 – 4 – 6 – 4 –1 0 0 0

y n 6 6 –1 – 4 2 6 4 – 3 – 6 – 4 –1 0 0 0

Actually y n should contain samples of x n + samples of h n – 1 12 . The last two zeroin above y n are obtained due to two zeros appended in x n2 . Hence last two zeros can

be discarded.

Thus the output will be,

y n = 6 6 1 4 2 6 4 3 6 4 1 0, , – , – , , , , – , – , – , – ,




27/38


Example 3.1.7 Kept this unsolved example for student's practice.

Example 3.1.8

Solution : i) y n u nn

1

2

We know that a u naz

z an z

1

1 1–,

–

12

1

1 1

2

121

n z

u n z

z–

,–

or Y z = 1

1

1

2

121

–

,–

z

z

ii) y n n x n

We have x n z

z z

z

2

2 164

–,

Differentiation in z-domain property states : n x n z d

dzX z

z –

n x n z d

dz

z

z

z

–

–

2

2 16

z

z z

z

– .–

–

32

162 2

, Here d

dx

uv

=

v dudx

u dvdx

v

–

2

z z

z

32

16

2

2 2–

or Y z =

32

164

2

2 2

z

z z

–,


3 z-Transform


28/38

Example 3.2.9

Solution : x(n) =

1

5 5

1

2 1

n n

u(n) u n( )

Rearranging the given x(n),

x(n) = 15 5 2 11n nu(n) u( n ) , Here 1

2 2 1

= ( 0.2) 5(2) ( 1) n nu(n) u n

= ( 0.2) 5[ 2 1)] n nu n u n( ) (

By using standard z-transform pairs,

X(z) = 1

1 0.2

5

1 21 1

z zfor| | z > 0.2 and| | z < 2

or ROC : 0.2


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= z

z j z j

1

12

12 1

212

= 12

32

j

and A2

= z jX z

z z j

1

2

1

2 12

12

( )

= z

z j z j

1

12

12 1

212

= 12

32

j

Key Point Here note that A 1 = A 2* . This is always true. Hence value of only A1 is

sufficient to determine A2 .

X z

z

( )=

12

32

12

12

12

32

12

12

j

z j

j

z j

Step 3 : X ( z) =

12

32

1 1

212

12

32

1 1

212

1 1

j

j z

j

j z

Step 4 : Here causal sequences are assumed. Taking inverse z-transform of above equation,

x n( ) =

1

2

3

2

1

2

1

2

1

2

3

2

1

2

1

2

j j u n j j

n

( )

n

u n( )

Simplification of above equations : Convert all the term to their polar form. It can be

converted with the help of P-R function on calculator. i.e.,

x(n) = (1.58 – 71.56º) (0.707 45º)n u(n) + (1.58 71.56º) (0.707 – 45º)n u(n)

= 1 58 0 707 1 58 0 70771 56 45 71 56. ( . ) ( ) . ( .. .e e u n e e j j n j j n u n45 ) ( )

= [ . ( . ) . ( . ). .1 58 0 707 1 58 0 70771 56 45 71 56e e e e j n j n j n j 45n u n] ( )

= 1 58 0 707 45 71 56 45 71 56. ( . ) [ ] ( )( . ) ( . )n j n j ne e u n

= 1 58 0 707 2 45 71 56. ( . ) cos( . ) ( )n n u n By Euler's identity

= 3 16 0 707 45 71 56. ( . ) cos( . ) ( )n n u n

Example 3.5.14

Solution : i) y n y n y n x n x n 1 0 5 2 1.

Y z = z Y z z Y z X z z X z 1 2 10 5.

H z =

Y z

X z

z

z z

1

1 0 5

1

1 2.

Digital Signal Processing Applications 3 - 3 z-Transform



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= z z

z z

2

2 0 5

.

=

z z

z j z j

1

0 5 0 5 0 5 0 5. . . .

Zeros : z z1 20 1 ,

Poles : p j p j1 20 5 0 5 0 5 0 5 . . , . .

Fig. 3.1 shows the pole zero plot.

ii) y n y n y n x n x n 0 7 1 0 1 2 2 2. .

Y z = 07 01 21 2 2. . z Y z z Y z X z z X z

H z =

Y z

X z

z

z z

2

1 0 7 0 1

2

1 2. .

=2 1

0 7 01

2 1

2

0 7 01

2

2

2

2

z

z z

z

z z

. . . .

=

z z

z z

1

2

1

2

0 5 0 2. .

Zeros : z z1 21

2

1

2 ,

Poles : p p1 20 5 0 2 . , .

Fig. 3.2 shows the pole-zero plot.

Example 3.5.15

Solution : y z = z X z

x n = 3 2 1 0 1 2 3, , , , , ,

X z = 3 2 0 2 33 2 1 2 3 z z z z z z

Y z( ) = z z z z z z z{ }3 2 0 2 33 2 1 2 3

= 3 2 0 1 2 34 3 2 1 2 z z z z z

y n = 3 2 1 0 1 2 3, , , , , ,



Im(z)

Re(z)

z2

–1

z1

p2

p1

0.5 10

Fig. 3.1 Pole zero plot of (i)

Re (z)

Im(z)

z2

–12 12

0

p2 p1 z1

0.2 0.5

Fig. 3.2 Pole zero plot of (ii)


31/38

Example 3.5.16

Solution : First let us convert powers of z in H z( ) to positive values. i.e.,

H z( ) = z

z

z

z z

2

2

1

1 2

3 4

1

3.5 1.5

= z z

z z

( )3 42

3.5 1.5

H z

z

( )=

3 42

z

z z

3.5 1.5... (1)

The denominator polynomial is quadratic (i.e. second order). Hence it will have two roots.

They are given as,

p p1 2, = b b ac

a

2 4

2

Here b c a 3.5 1.5, and 1.

p p1 2, =3.5 (3.5) 4 (1) (1.5)

2 (1)

2

= 3.5 2.5

2

= 3, 0.5

Thus p p1 23 0 5 and . . Hence we can write equation (1) as,

H z

z

( )=

3 4

3

z

z z

( ) ( )0.5

= A

z

A

z1 2

3 0.5 A A1 2and can be obtained as follows :

A1 = z H z

z z

3

3

( )

= 3 4

23

z z z

0.5

and A2 = z

H z

z z 0.5 0.5

( )

= 3 4

3 1

z z z

0.5

Thus the partial fraction expansion is, H z

z

( )=

23

1 z z 0.5

The system has two poles and they are at p j p j1 23 0 0 5 0 and . . Thus both thepoles have real part only. We can further rearrange above equation as,




32/38

H z( ) =2

3

z

z z

z 0.5

= 2

1 3

1

11 1

z z0.5... (2)

i) To determine ROC and h n

( ) for stable system :Since the system is stable, the ROC of H z( ) must include the unit circle. The system has

one pole at p1 3 and second pole at p2 0 5 . . Hence the ROC must be

ROC : 0.5 || z 3 which includes|| z 1 i.e. unit circle.

Thus the above specified ROC of 0.5 || z 3 includes the unit circle of|| z 1 but does notinclude the poles at p p1 23 and 0.5. Since the ROC is the ring or annular region, theunit sample response h n( ) will be two sided. Since 0 5. || z or || . z 0 5, the second term of equation (2) will correspond to causal part of h n( ). Hence,

IZT z

1

1 1

0.5= ( ) ( )0.5 n u n for ROC :|| z 0.5

Similarly since || z 3, the first term of equation (2) will correspond to anticausal part of h n( ). Hence,

IZT z

1

1 3 1

= ( ) ( )3 1n u n for ROC :|| z 3

Hence the unit sample response can be obtained as inverse z-transform of H z( ) of

equation (2). i.e.,

h n( ) =

IZT H z( )

= 2 3 1( ) ( ) ( ) ( )n nu n u n0.5

This is required unit sample response for stable system.

ii) Causal system :

The poles of this system are at p p1 23 and 0.5. For the causal system the ROC is theexterior of the circle. The ROC should not contain any poles. Hence ROC will be || z 3.The unit sample response can be obtained by taking inverse z-transform of equation (2)

i.e.,

h n( ) = IZT H z IZT z

IZT z

( )

2

1 3

1

11 10.5

Since ROC is || z 3 and causal system we can obtain the inverse z-transform of aboveequation as,

h n( ) = 2 3( ) ( ) ( ) ( )n nu n u n 0.5

Here since ROC is || z 3, it does not include unit circle of || z 1. Hence this system isunstable.




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iii) Anticausal system :

The poles of the system are at p p1 23 0 5 and . . We know that ROC is the interior of thecircle for anticausal system. Hence the ROC will be || z 0.5. Here note that ROC won't be

|| z 3, since it is also the exterior of || z 0.5. Therefore ROC of | | z < Magnitude of polenearest to origin is considered. The unit sample response h n( ) can be obtained by taking

inverse z-transform of equation (2). i.e.,

h n( ) = IZT H z IZT z

IZT z

( )

2

1 3

1

11 10.5

For ROC of|| z 0.5 we obtain inverse z - transform of above equation as follows,h n( ) = 2 3 1 0 5 1( ) ( ) ( . ) ( )n nu n u n

Since the ROC of this system is for| | . z 0 5, it does not include unit circle at || z 1. Hencethis system is unstable.

Example 3.5.17

Solution : The given difference equation is,

y n y n 12

1 = 2 1 0x n y, Whenever initial conditions are given we must use unilateral z-transform.

Here since y 1 0, we can use z-transform also. Taking unilateral z-transform of aboveequation,

Y z z Y z y X z 12

1 21

Putting y 1 0, above equation becomes same as z-transform, Y z z Y z

12

1 = 2X z

H z =

Y z

X z

= 2

1 1

21 z

Here we use IZT 1

1 1

p z p u n

k k

n

, hence we get,

h n = 2 1

2

n

u n

This is the unit sample response of the given difference equation.




34/38




Example 4.4.6

Solution :

i) y n . (x(n) + x(n ) 0 5 1 = 0 5 0 5 1. x n . x n –

Fig. 4.1 (a) shows direct form - I and Fig. 4.1 (b) shows direct form - II structure. Here

b0 0.5 and b 1 0.5

ii) 3y n – 2y n – 1 y n – 2 4x n – 3x n – 1 2x n – 2

3 y n = 2y n – 1 – y n – 2 4x n – 3x n – 1 2x n – 2

y n = 23 1

13 2

43 1

23 2 y n y n x n x n x n– – – – – –

Here a123

– , a213

, b043

, b1 1 – , b223

.

Fig. 4.2 shows the direct form-I and direct form-II


4 Digital Filter Structures

z –1

+

x(n)

0.5 0.5

y(n)

(a)

z

–1

+x(n)b = 0.50

y(n)

(b)

b = 0.51

Fig. 4.1 Direct form - I and II implementation of example 4.4.6 (i)


35/38

iii) y n – y n – x n( ) ( ) ( )5 1 7 y n = 5 1 7 y n x n–

Fig. 4.3 shows direct form I and II.

Example 4.4.7

Solution :

i) y n y n – – y n – 2 x n x n – 1 0.5 1 0.25 0.4

Digital Signal Processing Applications 4 - 2 Digital Filter Structures


(b) Direct form - II

+

b =04

—3

x(n) y(n)

–a =1 2—3

+

+

z –1

b = –11

–a = –21

—3

b =22

—3

+

z –1

Fig. 4.2 Direct form-I and II implementation of example 4.4.6 (ii)

+

z –1

b =04

—3

b = –11

z –1

x(n)

+

+

z –1 –a =1

z –1

+

–a = –21

—3

2—3

y(n)

b =22

—3

(a) Direct form - I

+

z –1

y(n)x(n)7

5

(a) Direct form - I

+

z –1

y(n)x(n)

5

7

(b) Direct form - II

Fig. 4.3 Direct form - I and II realization of example 4.4.6 (iii)


36/38

ii) y n y n – y n x n x n– 0.1 + 0.2 – 2 3 3.6 – 1 0.6 – 2 1 x n



+

z –1

+

z –1

+

0.5

– 0.25

0.4

1y(n)x(n)

Direct form - II

Fig. 4.4 Direct form I and II realizations of 4.4.7 (i)

+ +

z –1

z –1

+

z –1

3.6

3

– 0.1

0.2

x(n) y(n)

Direct form - I+

z –1

0.6

++

z –1

+3.6

3

– 0.1

0.2

x(n) y(n)

Direct form - II+

z –1

0.6

Fig. 4.5 Direct form-I and II realizations of 4.4.7 (ii)

+ +

z –1

z –1

+

z –1

0.4

1

0.5

– 0.25

x(n) y(n)

Direct form - I


37/38

iii) y n – y n – + y n – x n x n – 0.1 0.72 0.7 0.2521 2 2

Example 4.4.8

Solution :Direct form - I and II

Here b 0 0.7, b 1 0 , b 2 0.2 and a1 01 . , a2 0.72



+

z –1

+

z –1

0.7

– 0.1

0.72

x(n) y(n)

Direct form - II

– 0.252

+

Fig. 4.6 Direct form-I and II realizations of 4.4.7 (iii)

+ +

z –1

z –1

+

z –1

0.7

– 0.1

0.72

x(n) y(n)

Direct form - I

z –1

– 0.252

+

+

y (n)x (n) +

z –1

z –1

z –1

z –1

0.7

– 0.2

– 0.1

– 0.72

Fig. 4.7 Direct form - I realization


38/38

Cascade structure Y z = 0 1 0 72 0 7 0 21 2 2. . . . z Y z z Y z X z z X z

H z Y zX z

= 0 7 0 21 0 1

2

1 2. .

.

z

z z0. 72= 0.7z 0.2

z 0.1 z 0.72

22

= 0.7 z 0.285

z 0.1z 0.72

2

2

= 0.7 0 . 53 0 . 53

0 . 1 0.722 z z

z z

=

0.7 1 0 . 53z 1 0 . 53z1 0.1z 0.72 z

1 1

1 2

= 0.7 1 0 . 53

1 0 . 1 0 . 721 0 . 53

1

1 2

z

z z z 1 = H z H z1 2

Here, H z1 = 0.7 1 0 53

1 0 1 0 72

1

1 2 2

.

. .

z

z z

H zand = 1 + 0.53 z–1

Fig. 4.8 shows the cascade realization of H z1 and H z2 .


+ y (n)+

+

z –1

z –1

0.7

– 0.1

– 0.72 – 0.2

Fig. 4.7 (a) Direct form - II realization

+ + y (n)x (n) +

+

z –1

z –1

z –1

1 1

– 0.1 – 0.53 0.53

– 0.72

0.7

Fig. 4.8 Cascade realization

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