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 ho(n)?

(a) h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 0

(b) h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 1

(c) h(n)=2ho(n)u(n)-h(0)δ(n), n ≥ 1

(d) h(n)=2ho(n)u(n)-h(0)δ(n), n ≥ 0

I got this question in unit test.

My question is taken from General Considerations for Design of Digital Filters topic in division Digital Filters Design of Digital Signal Processing

Answer 1

The correct choice is (b) h(n)=2ho(n)u(n)+h(0)δ(n), n ≥ 1

For 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)= 2ho(n)u(n)+h(0)δ(n), n ≥ 1

since ho(n)=0 for n=0, we cannot recover h(0) from ho(n) and hence we must also know h(0).