Question

If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only he(n)?

(a) h(n)=2he(n)u(n)+he(0)δ(n), n ≥ 0

(b) h(n)=2he(n)u(n)+he(0)δ(n), n ≥ 1

(c) h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 1

(d) h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 0

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

This key question is from General Considerations for Design of Digital Filters topic in portion Digital Filters Design of Digital Signal Processing

Answer 1

Correct choice is (d) h(n)=2he(n)u(n)-he(0)δ(n), n ≥ 0

Easiest explanation: Given h(n) is causal and h(n)= he(n)+ho(n)

=>he(n)=1/2[h(n)+h(-n)]

Now, if h(n) is causal, it is possible to recover h(n) from its even part he(n) for 0≤n≤∞ or from its odd component ho(n) for 1≤n≤∞.

=>h(n)= 2he(n)u(n)-he(0)δ(n), n ≥ 0.