Z Transform Inversion MCQ Quiz – Objective Question with Answer for Z Transform Inversion

11. What is the inverse z-transform of
X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\) if ROC is 0.5<|z|<1?

A. -2u(-n-1)+(0.5)nu(n)
B. -2u(-n-1)-(0.5)nu(n)
C. -2u(-n-1)+(0.5)nu(-n-1)
D. 2u(n)+(0.5)nu(-n-1)

Answer: B

The partial fraction expansion of the given X(z) is

\(X(z)= \frac{2z}{z-1}-\frac{z}{z-0.5}\)

In this case, ROC is 0.5<|z|<1 is a ring, which implies that the signal is two-sided. Thus one of the signals corresponds to a causal signal and the other corresponds to an anti causal signal.

Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply the inverse z-transform we get

x(n)= -2u(-n-1)-(0.5)nu(n).

 

12. What is the causal signal x(n) having the z-transform
X(z)=\(\frac{1}{(1+z^{-1})(1-z^{-1})^2}\)?

A. [1/4(-1)n+3/4-n/2]u(n)
B. [1/4(-1)n+3/4-n/2]u(-n-1)
C. [1/4+3/4(-1)n-n/2]u(n)
D. [1/4(-1)n+3/4+n/2]u(n)

Answer: D

The partial fraction expansion of X(z) is

\(X(z) = \frac{z}{4(z+1)} + \frac{3z}{4(z-1)} + \frac{z}2{(z-1)^2}\)

When we apply the inverse z-transform for the above equation, we get

x(n)=[1/4(-1)n+3/4+n/2]u(n).

 

13. The z-transform of a signal x(n) whose definition is given by \(X(z)=\sum_{n=0}^{\infty} x(n)z^{-n}\) is known as ________

A. Unilateral z-transform
B. Bilateral z-transform
C. Rational z-transform
D. None of the mentioned

Answer: A

The z-transform of the x(n) whose definition exists in the range n=-∞ to +∞ is known as a bilateral or two-sided z-transform. But in the given question the value of n=0 to +∞. So, such a z-transform is known as a Unilateral or one-sided z-transform.

 

14. For what kind of signals is one-sided z-transform unique?
A. All signals
B. Anti-causal signal
C. Causal signal
D. None of the mentioned

Answer: C

One-sided z-transform is unique only for causal signals because only these signals are zero for n<0.

 

15. What is the one sided z-transform X+(z) of the signal x(n)={1,2,5,7,0,1}?

A. z2+2z+5+7z-1+z-3
B. 5+7z+z3
C. z-2+2z-1+5+7z+z3
D. 5+7z-1+z-3

Answer: D

Since the one sided z-transform is valid only for n>=0, the z-transform of the given signal will be X+(z)= 5+7z-1+z-3.

 

16. What is the one-sided z-transform of x(n)=δ(n-k)?

A. z-k
B. zk
C. 0
D. 1

Answer: A

Since the signal x(n)= δ(n-k) is a causal signal i.e., it is defined for n>0 and x(n)=1 at z=k
So, from the definition of one-sided z-transform X+(z)=z-k.

 

17. What is the one-sided z-transform of x(n)=δ(n+k)?

A. z-k
B. 0
C. zk
D. 1

Answer: B

Since the signal x(n)=δ(n+k) is an anti causal signal i.e., it is defined for n<0 and x(n)=1 at z=-k. Since the one-sided z-transform is defined only for causal signal, in this case X+(z)=0.

 

18. If X+(z) is the one sided z-transform of x(n), then what is the one sided z-transform of x(n-k)?

A. z-k X+(z)

B. zk X+(z-1)

C. z-k \([X^+(z)+\sum_{n=1}^k x(-n)z^n]\); k>0

D. z-k \([X^+(z)+\sum_{n=0}^k x(-n)z^n]\); k>0

Answer: C

From the definition of one sided z-transform we have,

Z+{x(n-k)}=\(z^{-k}[\sum_{l=-k}^{-1} x(l) z^{-l}+\sum_{l=0}^{\infty} x(l)z^{-l}]\)

=\(z^{-k}[\sum_{l=-1}^{-k} x(l) z^{-l}+X^+ (z)]\)

By changing the index from l to n= -l, we obtain

Z+{x(n-k)}=\(z^{-k}[X^+(z)+\sum_{n=1}^k x(-n)z^n]\) ;k>0

 

19. If x(n)=an, then what is one sided z-transform of x(n-2)?

A. \(\frac{z^{-2}}{1-az^{-1}} + a^{-1}z^{-1} + a^{-2}\)

B. \(\frac{z^{-2}}{1-az^{-1}} – a^{-1}z^{-1} + a^{-2}\)

C. \(\frac{z^{-2}}{1-az^{-1}} + a^{-1}z^{-1} – a^{-2}\)

D. \(\frac{z^{-2}}{1+az^{-1}} + a^{-1}z^{-1} + a^{-2}\)

Answer: A

Given x(n)=an=>X+(z)=\(\frac{1}{1-az^{-1}}\)

We will apply the shifting property for k=2. Indeed we have

Z+{x(n-2)}=z-2[X+(z)+x(-1)z+x(-2)z2]

=z-2 X+(z)+x(-1)z-1+x(-2)

Since x(-1)=a-1 and x(-2)=a-2, we obtain

X1+(z)=\(\frac{z^{-2}}{1-az^{-1}}\) + a-1z-1 + a-2

 

20. If x(n)=an, then what is one sided z-transform of x(n+2)?

A. \(\frac{z^{-2}}{1-az^{-1}}\) + a-1z-1 + a-2

B. \(\frac{z^{-2}}{1-az^{-1}}\) – a-1z-1 + a-2

C. \(\frac{z^2}{1-az^{-1}}\) + a z + z2

D. \(\frac{z^2}{1+az^{-1}}\) – z2 – az

Answer: D

We will apply the time advance theorem with the value of k=2.We obtain,
Z+{x(n+2)}=z2 X+(z)-x(0)z2-x(1)z

=>X1+(z)=\(\frac{z^2}{1+az^{-1}}\) – z2 – az.

 

21. If X+(z) is the one sided z-transform of the signal x(n), then
\(\lim_{n \rightarrow \infty} x(n)=\lim_{z\rightarrow 1}(z-1) X^+(z)\) is called Final value theorem.

A. True
B. False

Answer: A

In the above theorem, we are calculating the value of x(n) at infinity, so it is called as final value theorem.

 

22. The impulse response of a relaxed LTI system is h(n)=anu(n), |a|<1. What is the value of the step response of the system as n→∞?

A. \(\frac{1}{1+a}\)

B. \(\frac{1}{1-a}\)

C. \(\frac{a}{1+a}\)

D. \(\frac{a}{1-a}\)

Answer: B

The step response of the system is y(n)=x(n)*h(n) where x(n)=u(n)
On applying z-transform on both sides, we get

Y(z)=\(\frac{1}{1-az^{-1}} \frac{1}{1-z^{-1}}=\frac{z^2}{(z-1)(z-A.}\) ROC |z|>|a|

Now

(z-1)Y(z)=\(\frac{z^2}{(z-A.}\) ROC |z|>|a|

Since |a|<1 the ROC of (z-1)Y(z) includes the unit circle. Consequently by applying the final value theorem

\(\lim_{n\rightarrow\infty}y(n)=\lim_{z\rightarrow 1}⁡ \frac{z^2}{z-a}=\frac{1}{1-a}\)

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