Question

Determine the impulse response for the cascade of two LTI systems having impulse responses h1(n)=\((\frac{1}{2})^2\) u(n) and h2(n)=\((\frac{1}{4})^2\) u(n).

(a) \((\frac{1}{2})^n[2-(\frac{1}{2})^n]\), n<0

(b) \((\frac{1}{2})^n[2-(\frac{1}{2})^n]\), n>0

(c) \((\frac{1}{2})^n[2+(\frac{1}{2})^n]\), n<0

(d) \((\frac{1}{2})^n[2+(\frac{1}{2})^n]\), n>0

I have been asked this question in an international level competition.

Question is from Analysis of Discrete time LTI Systems in section Discrete Time Signals and Systems of Digital Signal Processing

Answer 1

The correct choice is (b) \((\frac{1}{2})^n[2-(\frac{1}{2})^n]\), n>0

The explanation: Let h2(n) be shifted and folded.

so, h(k)=h1(n)*h2(n)=\(\sum_{k=-\infty}^{\infty} h_1 (k)h_2 (n-k)\)

For k<0, h1(n)= h2(n)=0 since the unit step function is defined only on the right hand side.

Therefore, h(k)=\((\frac{1}{2})^k (\frac{1}{4})^{n-k}\)

=>h(n)=\(\sum_{k=0}^n (\frac{1}{2})^k (\frac{1}{4})^{n-k}\)

=\((\frac{1}{4})^n \sum_{k=0}^n(2)^k\)

=\((\frac{1}{4})^n.(2^{n+1}-1)\)

=\((\frac{1}{2})^n[2-(\frac{1}{2})^n], n>0\)