Test: Z Transform - Electronics and Communication Engineering (ECE) MCQ

Explanation: The z-transform of a real discrete time sequence x(n) is defined as a power of ‘z’ which is equal to , where ‘z’ is a complex variable.

Explanation: Since X(z) is a infinite power series, it is defined only at few values of z. The set of all values of z where X(z) converges to a finite value is called as Radius of Convergence(ROC).

Explanation: We know that, for a given signal x(n) the z-transform is defined as
Substitute the values of n from -2 to 3 and the corresponding signal values in the above formula
We get, X(z) = 2z² + 4z + 5 +7z^-1 + z^-3.

Explanation: We know that, the z-transform of a signal x(n) is 
Given x(n)= δ(n-k)=1 at n=k
=> X(z)=z^-k
From the above equation, X(z) is defined at all values of z except at z=0 for k>0.
So ROC is defined as Entire z-plane, except at z=0.

Explanation: For a given signal x(n), its z-transform 
 

Explanation:
Let x(n)= αⁿu(n)

Given x(n) = δ(n + 3)
We know that δ(n + 3) =

 

Explanation: We know that,

ROC of z-transform of aⁿu(n) is |z|>|a|. ROC of z-transform of bⁿu(-n-1) is |z|<|b|. By combining both the ROC's we get the ROC of z-transform of the signal x(n) as |a|<|z|<|b|

Explanation: Let us an example of anti causal sequence whose z-transform will be in the form X(z)=1+z+z² which has a finite value at all values of ‘z’ except at z=∞.So, ROC of an anti-causal sequence is entire z-plane except at z=∞.

Explanation: Let us plot the graph of z-transform of any two sided sequence which looks as follows.

From the above graph, we can state that the ROC of a two sided sequence will be of the form r2 < |z| < r1.

Explanation: The entire timing sequence is divided into two parts n=0 to ∞ and n=-∞ to 0.
Since the z-transform of the signal given in the questions contains both the parts, it is called as Bi-lateral z-transform.

Explanation: A discrete time LTI is BIBO stable, if and only if its impulse response h(n) is absolutely summable. That is,

Explanation: Since the value of z-transform tends to infinity, the ROC of the z-transform does not contain poles.

Explanation:

Given h(n)= aⁿ(n) (|a|<1) The z-transform of h(n) is H(z)=z/(z-a),ROC is |z|>|a| If |a|<1, then the ROC contains the unit circle. So, the system is BIBO stable

Explanation: The ROC of causal infinite sequence is of form |z|>r1 where r1 is largest magnitude of poles.