What are the Fourier series coefficients for the signal x(n)=cosπn/3?(a) c1=c2=c3=c4=0,c1=c5=1/2(b) c0=c1=c2=c3=c4=c5=0(c) c0=c1=c2=c3=c4=c5=1/2(d) none of the mentionedThe question was asked during an interview for a job.My enquiry is from Frequency Analysis of Discrete Time Signal topic in division Frequency Analysis of Signals and Systems of Digital Signal Processing

What are the Fourier series coefficients for the signal x(n)=cosπn/3?

1.	What are the Fourier series coefficients for the signal x(n)=cosπn/3?(a) c1=c2=c3=c4=0,c1=c5=1/2(b) c0=c1=c2=c3=c4=c5=0(c) c0=c1=c2=c3=c4=c5=1/2(d) none of the mentionedThe question was asked during an interview for a job.My enquiry is from Frequency Analysis of Discrete Time Signal topic in division Frequency Analysis of Signals and Systems of Digital Signal Processing
Answer» The correct option is (a) c1=c2=c3=c4=0,c1=c5=1/2 Easiest EXPLANATION: In this CASE, f0=1/6 and hence x(n) is periodic with fundamental period N=6. Given SIGNAL is x(n)=cosπn/3=cos2πn/6=\(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{-j2πn/6}\) We know that -2π/6=2π-2π/6=10π/6=5(2π/6) Therefore, x(n)=\(\frac{1}{2} e^{j2πn/6}+\frac{1}{2} e^{j2π(5)n/6}\) Compare the above equation with x(n)=\(\sum_{k=0}^{N-1}c_k e^{j2πkn/N}\) So, we get c1=c2=c3=c4=0 and c1=c5=1/2.

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