Question

What is the step response of the system y(n)=ay(n-1)+x(n) -1<a<1, when the initial condition is y(-1)=1?

(a) \(\frac{1}{1-a}\)(1+a^n+2)u(n)

(b) \(\frac{1}{1+a}\)(1+a^n+2)u(n)

(c) \(\frac{1}{1-a}\)(1-a^n+2)u(n)

(d) \(\frac{1}{1+a}\)(1-a^n+2)u(n)

I had been asked this question in an online quiz.

This is a very interesting question from One Sided Z Transform in chapter Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

Correct choice is (c) \(\frac{1}{1-a}\)(1-a^n+2)u(n)

For explanation I would say: By taking the one sided z-transform of the given equation, we obtain

Y^+(Z)=a[z^-1Y^+(z)+y(-1)]+X^+(z)

Upon substitution for y(-1) and X^+(z) and solving for Y^+(z), we obtain the result

Y^+(z)=\(\frac{a}{1-az^{-1}} + \frac{1}{(1-az^{-1})(1-z^{-1})}\)

By performing the partial fraction expansion and inverse transforming the result, we have

y(n)=\(\frac{1}{(1-a)}(1-a^{n+2})u(n)\).