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Practice Problem

Compute the discrete Fourier transform of the discrete-time signal

$ x[n]= e^{-j \frac{1}{5} \pi n} $.

How does your answer related to the Fourier series coefficients of x[n]?

Share your answers below

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Answer 1

$ x[n]= e^{-j \frac{1}{5} \pi n}=cos(\frac{\pi n}{5})+jsin(\frac{\pi n}{5}) $.

period=10, therefor, by comparing with$ x[n]=e^{-j2\pi k_0 n/N} $.

we get $ N=10 $,$ k_0=1 $.

From DFT transfer pair, $ X[k]=10\delta[k-1] $. repeated with period 10.

Instructor's comment: Why do you need to write the exponential as sine and cosine in order to find the period? Can you find the period directly from the exponential? -pm

Answer 2

$ x[n]= e^{-j \frac{1}{5} \pi n} $.

$ period = {2*pi / (pi/5)} = 10 $.


$ x[n]=e^{-j2\pi k_0 n/N} $.

N= 10 , k0 = 1

Instructor's comment: How do you go from here to the answer below? Please justify. -pm

$ X[k]=10\delta[k-1] $.


Back to ECE438 Fall 2011 Prof. Boutin

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