Test: Z Domain System Analysis


10 Questions MCQ Test Digital Signal Processing | Test: Z Domain System Analysis


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Attempt Test: Z Domain System Analysis | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study Digital Signal Processing for Electrical Engineering (EE) Exam | Download free PDF with solutions
QUESTION: 1

 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? 

Solution:

Explanation: The system function is H(z)=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)=1/(1-z-1 )
Therefore,
Yzs(z) = 1/((1-0.9e^(jπ/3) z-1)(1-0.9e-jπ/3 z-1 )(1-z-1))
= (0.542-j0.049)/((1-0.9ejπ/3 z-1) ) + (0.542-j0.049)/((1-0.9e^(jπ/3) z-1 ) ) + 1.099/(1-z-1 )
and hence the zero state response is yzs(n)= [1.099+1.088(0.9)n.cos(πn/3-5.2o)]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)= yzs(n).

QUESTION: 2

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

Solution:

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

QUESTION: 3

 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 as Transient response. 

Solution:

Explanation: 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.

QUESTION: 4

 If all the poles have small magnitudes, then the rate of decay of signal is:

Solution:

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

QUESTION: 5

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

Solution:

Explanation: 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.

QUESTION: 6

 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:

Solution:

Explanation: 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 = ∞.

QUESTION: 7

A linear time invariant system is said to be BIBO stable if and only if the ROC of the system function:

Solution:

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

QUESTION: 8

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

Solution:

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

QUESTION: 9

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

Solution:

Explanation: 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).

QUESTION: 10

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

Solution:

Explanation: 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, ROC of given H(z) is |z|>3.

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