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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 mentionedThis 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 |
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Answer» 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 |
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