One Sided Z Transform MCQ Quiz – Objective Question with Answer for One Sided Z Transform

11. What is the step response of the system y(n)=ay(n-1)+x(n) -1<a<1, when the initial condition is y(-1)=1?

A. \(\frac{1}{1-a}\)(1+an+2)u(n)

B. \(\frac{1}{1+a}\)(1+an+2)u(n)

C. \(\frac{1}{1-a}\)(1-an+2)u(n)

D. \(\frac{1}{1+a}\)(1-an+2)u(n)

Answer: C

By taking the one sided z-transform of the given equation, we obtain
Y+(Z)=a[z-1Y+(z)+y(-1)]+X+(z)
Upon substitution for y(-1) and X+(z) and solving for Y+(z), we obtain the result

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

By performing the partial fraction expansion and inverse transforming the result, we have

y(n)=\(\frac{1}{(1-A.}(1-a^{n+2})u(n)\).

 

12. What is the unit step response of the system described by the difference equation?
y(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?

A. [1.099+1.088(0.9)n.\(cos(\frac{\pi}{3} n+5.2^o)\)]u(n)

B. [1.099+1.088(0.9)n.\(cos(\frac{\pi}{3} n-5.2^o)\)]u(n)

C. [1.099+1.088(0.9)n.\(cos(\frac{\pi}{3} n-5.2^o)\)]

D. None of the mentioned

Answer: B

The system function is H(z)=\(\frac{1}{1-0.9z^{-1}+0.81z^{-2}}\)

The system has two complex-conjugate poles at p1=0.9ejπ/3 and p2=0.9e -jπ/3

The z-transform of the unit step sequence is

X(z)=\(\frac{1}{1-z^{-1}}\)

Therefore,

Yzs(z) = \(\frac{1}{(1-0.9e^{jπ/3} z^{-1})(1-0.9e{-jπ/3} z^{-1})(1-z^{-1})}\)

=\(\frac{0.542-j0.049}{(1-0.9e^{jπ/3} z^{-1})} + \frac{0.542-j0.049}{(1-0.9e^{jπ/3} z^{-1})} + \frac{1.099}{1-z^{-1}}\)

and hence the zero state response is yzs(n)= [1.099+1.088(0.9)n.\

(cos(\frac{\pi}{3} n-5.2^o)\)]u(n)

Since the initial conditions are zero in this case, we can conclude that y(n)=yzs(n).

 

13. If all the poles of H(z) are outside the unit circle, then the system is said to be _____________

A. Only causal
B. Only BIBO stable
C. BIBO stable and causal
D. None of the mentioned

Answer: D

If all the poles of H(z) are outside a unit circle, it means that the system is neither causal nor BIBO stable.

 

14. If pk, k=1,2,…N are the poles of the system and |pk| < 1 for all k, then the natural response of such a system is called a Transient response.

A. True
B. False

Answer: A

If |pk| < 1 for all k, then ynr(n) decays to 0 as n approaches infinity. In such a case we refer to the natural response of the system as the transient response.

 

15. If all the poles have small magnitudes, then the rate of decay of signal is __________

A. Slow
B. Constant
C. Rapid
D. None of the mentioned

Answer: C

If the magnitudes of the poles of the response of any system are very small i.e., almost equal to zero, then the system decays very rapidly.

 

16. If one or more poles are located near the unit circle, then the rate of decay of the signal is _________

A. Slow
B. Constant
C. Rapid
D. None of the mentioned

Answer: A

If the magnitudes of the poles of the response of any system is almost equal to one, then the system decays very slowly or the transient will persist for a relatively long time.

 

17. What is the transient response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(πn/4)u(n) and the system is initially at rest?

A. (0.5)nu(n)
B. 0.5(6.3)nu(n)
C. 6.3(0.5)n
D. 6.3(0.5)nu(n)

Answer: D

The system function for the system is

H(z)=\(\frac{1}{1-0.5z^{-1}}\)

and therefore the system has a pole at z=0.5. The z-transform of the input signal is

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

=\(\frac{10(1-(\frac{1}{\sqrt{2}}) z^{-1})}{(1-0.5z^{-1})(1-e^{jπ/4} z^{-1})(1-e^{-jπ}/4 z^{-1})}\)

=\(\frac{6.3}{1-0.5z^{-1}} + \frac{6.78e^{-j28.7}}{(1-e^{jπ/4} z^{-1})} + \frac{6.78e^{j28.7}}{(1-e^{-jπ/4} z^{-1})}\)

The natural or transient response is

ynr(n)= 6.3(0.5)nu(n)

 

18. What is the steady-state response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(πn/4)u(n) and the system is initially at rest?

A. 13.56cos(πn/4 -28.7o)
B. 13.56cos(πn/4 +28.7o)u(n)
C. 13.56cos(πn/4 -28.7o)u(n)
D. None of the mentioned

Answer: C

The system function for the system is

H(z)=\(\frac{1}{1-0.5z^{-1}}\)
and therefore the system has a pole at z=0.5. The z-transform of the input signal is

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

=\(\frac{10(1-(\frac{1}{\sqrt{2}}) z^{-1})}{(1-0.5z^{-1})(1-e^{jπ/4} z^{-1})(1-e^{-jπ}/4 z^{-1})}\)

=\(\frac{6.3}{1-0.5z^{-1}} + \frac{6.78e^{-j28.7}}{(1-e^{jπ/4} z^{-1})} + \frac{6.78e^{j28.7}}{(1-e^{-jπ/4} z^{-1})}\)

The forced state response or steady-state response is
yfr(n)=13.56cos(πn/4 -28.70)u(n)

 

19. If the ROC of the system function is the exterior of a circle of radius r < ∞, including the point z = ∞, then the system is said to be ___________

A. Stable
B. Causal
C. Anti causal
D. None of the mentioned

Answer: B

A linear time-invariant system is said to be causal if and only if the ROC of the system function is the exterior of a circle of radius r < ∞, including the point z = ∞.

 

20. A linear time-invariant system is said to be BIBO stable if and only if the ROC of the system function _____________

A. Includes unit circle
B. Excludes unit circle
C. Is a unit circle
D. None of the mentioned

Answer: A

For an LTI system, if the ROC of the system function includes the unit circle, then the system is said to be BIBO stable.

 

21. If all the poles of H(z) are inside the unit circle, then the system is said to be ____________

A. Only causal
B. Only BIBO stable
C. BIBO stable and causal
D. None of the mentioned

Answer: C

If all the poles of H(z) are inside a unit circle, then it follows the condition that |z|>r < 1, which means that the system is both causal and BIBO stable.

 

22. A linear time invariant system is characterized by the system function H(z)=\(\frac{1}{1-0.5z^{-1}}+\frac{2}{1-3z^{-1}}\). What is the h(n) if the system is stable?

A. (0.5)nu(n)-2(3)nu(n)
B. (0.5)nu(-n-1)-2(3)nu(-n-1)
C. (0.5)nu(-n-1)-2(3)nu(n)
D. (0.5)nu(n)-2(3)nu(-n-1)

Answer: D

The system has poles at z=0.5 and at z=3.

Since the system is stable, its ROC must include unit circle and hence it is 0.5<|z|<3 . Consequently, h(n) is non causal and is given as
h(n)= (0.5)nu(n)-2(3)nu(-n-1).

 

23. A linear time-invariant system is characterized by the system function H(z)=\(\frac{1}{1-0.5z^{-1}}+\frac{2}{1-3z^{-1}}\). What is the ROC of H(z) if the system is causal?

A. |z|<3
B. |z|>3
C. |z|<0.5
D. |z|>0.5

Answer: B

The system has poles at z=0.5 and at z=3.
Since the system is causal, its ROC is |z|>0.5 and |z|>3. The common region is |z|>3. So, the ROC of the given H(z) is |z|>3.

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