Signals and Systems

Z-Transform

ECE - 2006

1. If the region of convergence of

x1[n] +x2[n] is 1/3 < |z|<2/3, then the region of convergence of x1[n] – x2[n] includes

(A) 1

3< |z|< 3

(B) 2

3< |z|< 3

(C) 2

3< |z|< 3

(D) 1

3< |z|<

2

3

ECE - 2007

2. The z-transform x[z] of a sequence x[n] is given by x[z] = 0.5

1−2z−1. It is given that the region

of convergence of x[z] includes the unit circle. The value of x[0] is

(A) −0.5

(B) 0

(C) 0.25

(D) 0.5

ECE - 2008

Statement for linked question 3 and 4:

In the following network, the switch is closed at t = 0− and the sampling starts from t = 0.

The sampling frequency is 10Hz.

3. The samples x [n] (n = 0, 1, 2, ...) are given by

(A) 5 (1 – 𝑒−0.05𝑛)

(B) 5𝑒−0.05𝑛

(C) 5 (1 – 𝑒−5𝑛)

(D) 5𝑒−5𝑛

4. The expression and the region of convergence of the z-transform of the sampled signal

are

(A) 5z

z−e−5 , |z| < e-5

(B) 5z

z−e−0.05, | z | < e-0.05

(C) 5z

z-e-0.05 , | z | > e-0.05

(D) 5z

z−e−5, |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

10μF

S x[n]

Signals and Systems

x[n] = (1

3)

nu[n] − (

1

2)

nu[−n − 1] is

(A) 1

3< |z| <

1

2

(B) |z| >1

2

(C) |z| <1

3

(D) 2 < |z| < 3

ECE - 2010

6. Consider the z-transform

X(z) = 5z2 + 4z−1 + 3; 0 < |z| < ∞. The inverse z-transform x[n] is

(A) 5 δ[n + 2] + 3 δ[n] + 4 δ[n − 1]

(B) 5 δ[n − 2] + 3 δ[n] + 4 δ[n + 1]

(C) 5 u[n + 2] + 3 u[n] + 4 u[n − 1]

(D) 5 u[n − 2] + 3 u[n] + 4 u[n + 1]

7. The transfer function of a discrete time LTI system is given by H(z) =2−

3

4z−1

1−3

4z−1+

1

8z−2

Consider the following statements:

S1: The system is stable and causal for ROC: |z| >1

2

S2: The system is stable but not causal for ROC: |z| <1

4

S3: The system is neither stable nor causal for ROC: 1

4< |z| <

1

2

Which one of the following statements is valid?

(A) Both S1 and S2 are true

(B) Both S2 and S3 are true

(C) Both S1 and S3 are true

(D) S1, S2 and S3 are all true

ECE - 2011

8. Two systems H1(z) and H2(z) are connected in cascade as shown below. The overall

output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the

second system H2(z) is

(A) (1−0.6z−1)

z−1(1−0.4z−1)

(B) z−1(1−0.6z−1)

(1−0.4z−1)

(C) z−1(1−0.4z−1)

(1−0.6z−1)

(D) (1−0.4z−1)

z−1(1−0.6z−1)

ECE/EE/IN - 2012

9. If x[n] = (1 3)|n|⁄ − (1 2)nu[n]⁄ , then the region of convergence (ROC) of its Z-transform in

the Z-plane will be

x(n) H1(z) =

(1−0.4z−1)

(1−0.6z−1)

H2(z) y(n)

Signals and Systems

(A) 1

3< |Z| < 3

(B) 1

3< |Z| <

1

2

(C) 1

2< |Z| < 3

(D) 1

3< |Z|

ECE - 2014

10. Let x[n] = (−1

9)

nu(n) − (−

1

3)

nu(−n − 1). The region of convergence (ROC) of the z-

transform of x[n]

(A) is |z| >1

9

(B) is |z| <1

3

(C) is1

3> |z| >

1

9

(D) Does not exist

11. Let x[𝑛] = 𝑥[−𝑛]. Let X(z) be the 𝑧-transform of [𝑛]. If 0.5+j0.25 is a zero of

X(z), which one of the following must also be a zero of X(z).

(A) 0.5 − j 0.25

(B) 1/(0.5 + j 0.25)

(C) 1/(0.5 − j 0.25)

(D) 2 + j 4

12. The z-transform of the sequence x[n] is given by X(z) =1

(1−2z−1)2, with the region of

convergence|z| > 2. . Then, x [2] is_______.

EE - 2007

1. The discrete – time signal

x[n] ↔ X (Z) = ∑3n

2+n∞n=0 z2n,

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

(A) y1[ n ] ↔ y1 ( z ) = ∑ (2

3)

n∞n=0 z−n

(B) y2[ n ] ↔ y2 ( z )

= ∑ (5n − n)∞n=0 z−(2n+1)

(C) y3[ n ] ↔ y3 ( z ) = ∑ 2−|n|∞n=−∞ z−n

(D) y4[ n ] ↔ y4 ( z ) =2z−4 + 3z−2 +1

EE - 2008

2. Given X (z) = Z

(z−a)2 with |Z|> a, the residue of X(z) zn−1 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

2z−1)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

1−z−3 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

2

IN - 2011

2. Consider the difference equation

y[n] −1

3y[n − 1] = x[n] and suppose that x[n] = (

1

2)

nu[n]. Assuming the condition of initial

rest, the solution for y[n], n ≥ 0 is

(A) 3 (1

3)

n− 2 (

1

2)

n

(B) −2 (1

3)

n+ 3 (

1

2)

n

(C) 2

3(

1

3)

n+

1

3(

1

2)

n

(D) 1

3(

1

3)

n+

2

3(

1

2)

n

IN - 2014

3. The system function of an LTI system is given by

Hz = 1 −

13 z−1

1 −14 z−1

The above system can have stable inverse if the region of convergence of H(z)is defined as

(A) |𝑧|<1

4

(B) |𝑧|<1

12

(C) |𝑧|>1

4

Signals and Systems

(D) |𝑧|<1

3

4. The transfer function of a digital system is given by: b0

1−z−1+a2Z−2 ; where a2 is real.

The transfer function is BIBO stable if the value of a2 is:

(A) −1.5

(B) −0.75

(C) 0.5

(D) 1.5

Answer Keys and Explanations

ECE

1. [Ans. D]

ROC {x1 + x2} = ROC{x1 − x2}(if there are no pole-zero cancellation)

2. [Ans. B]

ROC is |z| < 2

∴x[n] = − 0.5∗ 2nu[−n − 1] ⇒ x[0] = 0

3. [Ans. B]

VR(s) = (200 × 103

200 × 103 +1

10 × 10−6s

)5

s

=5 × 2 × 105 × 10 × 10−6

2 × 105 × 10−5s + 1

=10

2s + 1=

5

s + 0.5

VR(t) = 5e−0.5t

Therefore samples

X(n) = 5e− 0.5n 10⁄

= 5e−0.05n

4. [Ans. C]

X(z) = ∑ 5e−0.05nz−n

∞

n=0

= 5 ∑(e−0.05z−1)n

∞

n=0

For |e−0.05z−1| < 1 or |z| > e−0.05

X(z) =5z

z − e−0.05

Signals and Systems

5. [Ans. A]

Recall the z – transform pairs:

For the Right sided sequence: anu(n) →z

z−a , |z| > |a|

For Left sided sequence:

−anu(−n − 1) →z

z−a , |z| < |a|

∴ (1

3)

n

u(n) →z

z −13

, |z| >1

3

− (1

2)

n

(−n − 1) →z

z −12

, |z| <

1

2

∴ For x(n) containing both the above terms the ROC is the overlapping region:

1

3< |z| <

1

2 .

6. [Ans. A]

For the given X(z) , note that the ROC is the entire z – plane except

z = 0 and z = ∞.

Recall the z – transform pair: δ(n) → 1

Recall the time shifting property:

If g(n) → G(z)

δ(n ∓ n0) → z∓n0G(z)

∴ δ(n ∓ n0) → z∓n0

∴ For given X(z) ,

x(n)=5 δ(n + 2) + 4 δ(n − 1) + 3δ(n)

7. [Ans. C]

Given : H(z)= 2−

3

4z−1

1−3

4z−1+

1

8z−2

Express H(z) in +ve powers of z and find the zeros and poles.

H(z) =z (−

34 + 2z)

(z2 −34 z +

18)

=2z (z −

38)

(z −14) (z −

12)

Poles are at z = ¼ and z = ½ lying inside the unit circle in the z – plane.

Consider the three possible ROC’s:

(i)|z| >1

2

For a casual system h(n) =0, n< 0 and h(n) is a Right sided sequence and ROC is the region

outside the circle passing through the right most pole z = ½

∴ The system is Causal.

For stability of LTI system unit circle should be inside the ROC

∴ system is stable S1 is valid statement.

Signals and Systems

(ii)|z| <1

4 ,

H(n) is a Left sided sequence and the system is not causal.

It is anticausal : h(n) ≠ 0, n < 0

And h(n) =0 , for n≥ 0

As unit circle is not inside the ROC , the system is unstable.

∴ S2 is not a – valid statement.

(iii)1

4< |z| <

1

2

The ROC is the annular region between two circles bounded by the poles:

At z =1

4 and z =

1

2

H(n) is a two sided signal existing for

n < 0 as well as for n ≥ 0 and System is not causal. As unit circle is not inside the ROC , the

system is unstable.

∴S3 is a valid statement

∴ Both S1 & S3 are true and S2 is not true.

8. [Ans. B]

y[n] = x[n − 1]

Taking z – transform of both sides

Y(z) = z−1X(z)

Y(z)

X(z)= z−1

For cascaded system

H(z) = H1(z). H2(z)

z−1 =(1 − 0.4z−1)

(1 − 0.6z−1)H2(z)

∴ H2(z) =z−1(1 − 0.6z−1)

(1 − 0.4z−1)

9. [Ans. C]

x(n) = (1

3)

|n|

− (1

2)

n

u[n]

= x1(n) + x2(n)say

x1(Z) = ∑ x1(n)z−n

∞

n=−∞

= ∑ x1 (1

3)

|n|

z−n

∞

n=−∞

Signals and Systems

= ∑ x1 (1

3)

−n

z−n

∞

n=−∞

+ ∑ x1 (1

3)

n

z−n

∞

n=−∞

= ∑ x1 (1

3z)

l∞

n=−∞

+ ∑ (1

3z−1)

n∞

n=−∞

= {1

3z + (

1

3z)

2

+ ⋯ . }

+ {(1

3z−1)

0

+ (1

3z−1)

1

+ ⋯ }

ROC is

|1

3z| < 1 |

1

3z−1| < 1

⇒ |z| < 3 |z| > |1

3|

x2(n) = − (1

2)

n

u(n)

x2(n) = 1

1 −12 Z−1

ROC is |1

2z−1| < 1

or|z| >1

2

We draw the three ROC on the Z plane

So,

The over lapping region is given by

1

2< |z| < 3

10. [Ans. C]

Given x[n]

= (−1

9)

n

u[n] − (−1

3)

n

u[−n − 1]

For (−1

9)

n

u[n] ROC in |z| >1

9

(

Right sided sequence, ROC in

exterior of circle of radius 1

9

)

Thus overall ROC in1

9< |z| <

1

3

11. [Ans. B]

x(n) = x(−n)

⇒ x(z) = x(z−1)

∴ 0.5 + j0.25 is a zero of x(z)

∴1

0.5 + j0.25 is also zero of x(z)as

Signals and Systems

x(z) = x(z−1)

12. [Ans. *] Range 11.9 to 12.1

x(z) =1

(1 − 2z−1)2

=1

(1 − 2𝑧−1)(

1

1 − 2𝑧−1)

taking Inverse z-transform using

Frequency domain multiplication

=time domain convolution

x(n) = 2nu(n) ∗ 2nu(n)

= 2ku(k) ∗ 2n−ku(n − k)

= ∑ 2k

n

k=0

2n−k

= ∑ 2n

n

k=0

= 2𝑛 ∑ 1

𝑛

𝑘=0

x[n] = 2n(n + 1)

So, x[2] = 22(2 + 1)

= 12

EE

1. [Ans. B]

Two discrete time signals x(n) and y (n) are said to be orthogonal, if

∑ x(n) y ∗ (n) = 0,∞n=0 whose * denotes complex conjugation………(1)

Or

∑ x(n) y(n) = 0,∞n=0 for real x(n) and y(n) ……….(2)

For the given x(n) , X(z) = ∑3n

2+nz2n∞

n=0

=1

2+ 1 z2 +……………

∴ x(n) =1

2δ(n) + 1δ(n + 2) +…………(3)

From the option given , according to the definition of Z.T:

y1(n) = (2

3)

n

u(n), y3(n) = 2−|n|

y4(n) = 1δ(n) + 3δ(n − 2) + 2δ(n − 4)

For Y2(z) = ∑ (5n − n)z−(2n+1)∞n=0

= 1z−1 + 4z−3 + 23z−5 +…………

∴ y2(n) = 1δ(n − 1) + 4δ(n − 3) + 23δ(n − 5)

It can be seen that equation (2) is satisfied only when y(n) = y2(n)

Signals and Systems

2. [Ans. D]

For the given X(z) =z

(z−a)2 with |z|>a

Let Y(z) = zn−1 X(z) =zn

(z−a)2

Residue of Y(z)= 1

1!

d

dz[(z − a)2Y(z)]|

z=a

=d

dz[zn]|

z=a

= n zn−1|z=a = n an−1 =1

anan

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

X(z) = z

(z−a)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 an−1u(n) →z

(z − a)2 , |z| > |a|

3. [Ans. A]

(1−1

2z−1). 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 − z−3

1 + z−3 + z−6

1 − z−3 1

1 − z−3

z−3

z−3 − z−6

z−6

Signals and Systems

z−6 − z−9

…………….

∴ x(z) = 1 + z−3 + z−6+. ….

∴ 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]z−n

∞

n=−∞

= ∑ 2n u[n]z−n

∞

n=−∞

= ∑ 2n. z−n

∞

n=0

= ∑ (2

z)

n

=1

1 − 2z−1

∞

n=0

ROC ⇒ |2z−1| < 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

4Z−1) > 0

Then, Z >1

4

4. [Ans. C]

Only this option gives poles inside unit circle.