Question

What is the relation between H(k+α) and h(n)?

(a) H(k+α)=\(\sum_{n=0}^{M+1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M+1

(b) H(k+α)=\(\sum_{n=0}^{M-1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M-1

(c) H(k+α)=\(\sum_{n=0}^M h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M

(d) None of the mentioned

This question was posed to me in an interview for internship.

This is a very interesting question from Design of Linear Phase FIR Filters by Frequency Sampling Method topic in division Digital Filters Design of Digital Signal Processing

Answer 1

Correct option is (b) H(k+α)=\(\sum_{n=0}^{M-1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M-1

Easy explanation: We know that

ωk=\(\frac{2π}{M}\)(k+α) and H(ω)=\(\sum_{n=0}^{M-1} h(n)e^{-jωn}\)

Thus from substituting the first in the second equation, we get

H(k+α)=\(\sum_{n=0}^{M-1} h(n)e^{-j2π(k+α)n/M}\); k=0,1,2…M-1