Test: Z Domain System Analysis
10 Questions MCQ Test Digital Signal Processing | Test: Z Domain System Analysis
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?
y(n)=0.9y(n-1)-0.81y(n-2)+x(n) under the initial conditions y(-1)=y(-2)=0?
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).
If all the poles of H(z) are outside the unit circle, then the system is said to be:
Explanation: If all the poles of H(z) are outside an unit circle, it means that the system is neither causal nor BIBO stable.
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.
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.
If all the poles have small magnitudes, then the rate of decay of signal is:
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.
If one or more poles are located near the unit circle , then the rate of decay of signal is:
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.
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:
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 = ∞.
A linear time invariant system is said to be BIBO stable if and only if the ROC of the system function:
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.
If all the poles of H(z) are inside the unit circle, then the system is said to be:
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.
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?
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).
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?
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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