1.

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 ≥ 0I 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»

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).



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