Question

The impulse response of a relaxed LTI system is h(n)=a^nu(n), |a|<1. What is the value of the step response of the system as n→∞?

(a) \(\frac{1}{1+a}\)

(b) \(\frac{1}{1-a}\)

(c) \(\frac{a}{1+a}\)

(d) \(\frac{a}{1-a}\)

I have been asked this question in an internship interview.

My question is taken from One Sided Z Transform in division Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

The correct option is (b) \(\frac{1}{1-a}\)

To explain: The step response of the system is y(n)=x(n)*h(n) where x(n)=u(n)

On applying z-transform on both sides, we get

Y(z)=\(\frac{1}{1-az^{-1}} \frac{1}{1-z^{-1}}=\frac{z^2}{(z-1)(z-a)}\) ROC |z|>|a|

Now

(z-1)Y(z)=\(\frac{z^2}{(z-a)}\) ROC |z|>|a|

Since |a|<1 the ROC of (z-1)Y(z) includes the unit circle. Consequently by applying the final value theorem

\(\lim_{n\rightarrow\infty}y(n)=\lim_{z\rightarrow 1}⁡ \frac{z^2}{z-a}=\frac{1}{1-a}\)