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

Explanation: For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called as ‘zeros’ of X(z).

Explanation: For a rational z-transform X(z) to be infinity, the denominator of X(z) is zero and the solutions of the denominator are called as ‘poles’ of X(z).

Explanation: If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z)=z -M+N.X'(z).
So, if N>M then z has a positive power. So, it has |N-M| zeros at origin.

Explanation: If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z)=z^-M+N.X'(z).
So, if N < M then z has a negative power. So, it has |N-M| poles at origin.

Explanation: From the given pole-zero plot, the z-transform of the signal has one zero at z=0 and one pole at z=a.
So, we obtain X(z)=z/(z-a)
By applying inverse z-transform for X(z), we get
x(n)= aⁿu(n).

Explanation: From the figure given, the z-transform of the signal has 8 zeros on circle of radius ‘a’ and 7 poles at origin.

Explanation: The z-transform of the given signal is X(z)= z/(z-a)
So, it has one pole at z=a and one zero at z=0.

Explanation: From the pole-zero plot, it is shown that r < 1, so the signal is a decaying signal.

Explanation: For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called as ‘zeros’ of X(z).

Explanation: We know that for an LTI system, y(n)=h(n)*x(n)
On applying z-transform on both sides we get, Y(z)=H(z).X(z)=>H(z)= ( Y(z))/(X(z) ).