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What is the expression for Fourier series coefficient ck in terms of the discrete signal x(n)?(a) \(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{j2πkn/N}\)(b) \(N\sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)(c) \(\frac{1}{N} \sum_{n=0}^{N+1}x(n)e^{-j2πkn/N}\)(d) \(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)The question was posed to me in an interview for job.The query is from Frequency Analysis of Discrete Time Signal topic in portion Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer»

Correct answer is (d) \(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)

To EXPLAIN: We know that, the Fourier series representation of a discrete signal x(n) is GIVEN as

x(n)=\(\sum_{n=0}^{N-1}c_k e^{j2πkn/N}\)

Now multiply both sides by the exponential e^-j2πln/N and summing the product from n=0 to n=N-1. Thus,

\(\sum_{n=0}^{N-1} x(n)e^{-j2πln/N}=\sum_{n=0}^{N-1}\sum_{k=0}^{N-1}c_k e^{j2π(k-l)n/N}\)

If we PERFORM summation over n first in the right hand side of above equation, we get

\(\sum_{n=0}^{N-1} e^{-j2πkn/N}\) = N, for k-l=0,±N,±2N…

= 0, otherwise

Therefore, the right hand side reduces to Nck

So, we obtain ck=\(\frac{1}{N} \sum_{n=0}^{N-1}x(n)e^{-j2πkn/N}\)


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