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Z Transform

Transcript of 3 Z Transform

• Signals and Systems

Z-Transform

ECE - 2006

1. If the region of convergence of

x1[n] +x2[n] is 1/3 < |z| e-0.05

(D) 5z

ze5, |z| > e-5

ECE - 2009

5. The ROC of Z-Transform of the discrete time sequence ,

x(z)

z transform Sampler

(fs = 10Hz)

200k +

5V

10F

S x[n]

• Signals and Systems

x[n] = (1

3)

nu[n] (

1

2)

nu[n 1] is

(A) 1

3< |z| 1

2

(C) |z| 1

2

S2: The system is stable but not causal for ROC: |z| 2. . Then, x [2] is_______.

EE - 2007

1. The discrete time signal

x[n] X (Z) = 3n

2+nn=0 z

2n,

Where denotes a transform pair relationship, is orthogonal to the signal

(A) y1[ n ] y1 ( z ) = (2

3)

nn=0 z

n

(B) y2[ n ] y2 ( z )

= (5n n)n=0 z(2n+1)

(C) y3[ n ] y3 ( z ) = 2|n|

n= zn

(D) y4[ n ] y4 ( z ) =2z4 + 3z2 +1

EE - 2008

2. Given X (z) = Z

(za)2 with |Z|> a, the residue of X(z) zn1 at z = a for n 0 will be

(A) an-1

(B) an

(C) nan

(D) nan-1

3. H (z) is a transfer function of a real system. (When a signal x[n] = (1 + j)n is the input to

such a system, the output is zero. Further, the region of convergence (ROC) of (1

• Signals and Systems

1

2z1)H(z) is the entire Z-plane (Except z = 0). It can then be inferred that H (z) can have a

minimum of

(A) one pole and one zero

(B) one pole and two zero

(C) two poles and one zero

(D) two poles and two zeros

EE - 2014

4. Let X(z) =1

1z3 be the Z-transform of a causal signal x[n]. Then, the values of x[2] and

x[3] are

(A) 0 and 0

(B) 0 and 1

(C) 1 and 0

(D) 1 and 1

IN - 2008

1. The region of convergence of the z-transform of the discrete-time signal

x[n] = 2nu[n] will be

(A) |Z|>2

(B) |Z|< 2

(C) |Z|>1

2

(D) |Z|1

3

(1

2)

n

(n 1) z

z 12

, |z| 1

9

(Right sided sequence, ROC in

exterior of circle of radius 1

9

)

Thus overall ROC in1

9< |z| a

Let Y(z) = zn1 X(z) =zn

(za)2

Residue of Y(z)= 1

1!

d

dz[(z a)2Y(z)]|

z=a

=d

dz[zn]|

z=a

= n zn1|z=a = n an1 =

1

anan

Note that the Question asked is to find x(n) given

X(z) = z

(za)2 with |z|>a , for which the answer is known through the following fundamental

ZT pairs and properties.

u(n) z

z 1, |z| > 1

n u(n) zd

dz(

z

z 1) =

z

(z 1)2,

|z| > 1

n an u(n)

za

(za 1)

2 , |z

a| > 1

=az

(z a)2 , |z| > |a|

n an1u(n) z

(z a)2 , |z| > |a|

3. [Ans. A]

(11

2z1). H(z)has ROC as entire Z-plane which implies that it is polynomial.

H(z) has one pole

4. [Ans. B]

X(z) =1

1 z3

1 + z3 + z6

1 z3 1

1 z3

z3

z3 z6

z6

• Signals and Systems

z6 z9

.

x(z) = 1 + z3 + z6+. .

x(n) = {1, 0, 0, 1, 0, 0, 1, }

x(2) = 0 and x(3) = 1

IN

1. [Ans. A]

Given,

x[n] = 2nu(n)

x(z) = x[n]zn

n=

= 2n u[n]zn

n=

= 2n. zn

n=0

= (2

z)

n

=1

1 2z1

n=0

ROC |2z1| < 1 |z| > 2

2. [Ans. B]

y[n] 1

3y[n 1] = x[n]

Y(z)

X(z)= H(z) =

1

1 13 z

1

For input x[n] = (1

2)

n

u[n]

X(z) =z

(z 12)

, |z| >1

2

Y(z) = H(z)X(z) =1

(1 13 z

1)

z

(z 12)

= z

(1 13) (z

12)

Y(z)

z=

z

(1 13) (z

12)

=2

(1 13)

+3

(z 12)

Y(z) =2z

z 13

+3z

z 12

y(n) = [2 (1

3)

n

+ 3 (1

2)

n

] u(n)

3. [Ans. C]

• Signals and Systems

For a system to be stable, it should contain Z=1 circle in its ROC

For ROC, (1 1

4Z1) > 0

Then, Z >1

4

4. [Ans. C]

Only this option gives poles inside unit circle.