Test: Z Domain System Analysis - Electronics and Communication Engineering (ECE) MCQ

Explanation: The system function is H(z)=1/(1-0.9z^-1+0.81z^-2 )
The system has two complex-conjugate poles at p1=0.9e^jπ/3 and p2=0.9e ^-jπ/3
The z-transform of the unit step sequence is
X(z)=1/(1-z^-1 )
Therefore,
Y_zs(z) = 1/((1-0.9e^(jπ/3) z^-1)(1-0.9e^-jπ/3 z^-1 )(1-z^-1))
= (0.542-j0.049)/((1-0.9e^jπ/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 y_zs(n)= [1.099+1.088(0.9)ⁿ.cos(πn/3-5.2^o)]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)= y_zs(n).

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

Explanation: If |p_k| < 1 for all k, then y_nr(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.

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.

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.

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 = ∞.

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.

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.

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).

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.