Practice Question on "Digital Signal Processing"

Topic: Discrete Fourier Transform


Question

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] $.

Instructor's comment: Your answer is not a periodic signal, as it should be. -pm

Answer 3

$ x[n]= e^{-j \frac{1}{5} \pi n} = e^{-j \frac{2\pi}{10} n} $

By comparison with the IDT

$ \,x [n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{j \frac{2\pi}{N}kn} \, $

We see that x[n] can be sifted out of the sum on the right hand side of the definition.

In order for both exponents to match, k has to take on the value -1 (and -1 only).

This corresponds to X[k] being a delta function whose argument is [k+1].

Since the IDT the sum is multiplied by 1/(period=10), the gain of the delta must be 10 to make the overall RHS expression coefficient = 1.

These characteristics make the X[k] expression $ X[k] = 10\delta[k+1] $

Instructor's comment: Your answer is not a periodic signal, as it should be-pm

Answer 4

I believe some people might be using the wrong formula for comparing with either modulation or direct translation from exponential to delta function. Make sure you note where the negatives are or are NOT located. The closest solution I have seen to the answer I believe is completely correct so far is Answer 3. I am not sure if this is completely true but from what I understand in class the DFT is generally expressed as a one-sided transformation, meaning everything is expressed on the positive side of the Fourier domain. In order to get Answer 3 to the correct side all that needs to be done is shift the signal by a period N and it will be correct then.

$ X[k] = N\delta[k-(k_{o})] = N\delta[k-(N+k_{o}) $ so the actual answer should be $ X[k] = 10\delta[k-9] $ as I understand it. Or you could solve the problem like this if it is still unclear.

$ \begin{align} X [k] &= \sum_{n=0}^{N-1} x[n]e^{\frac{-j2\pi k n}{N}} \\ &= \sum_{n=0}^{N-1} e^{-j\frac{1}{5}\pi n}e^{\frac{-j2\pi k n}{N}} \text{ where N=10 because of periodicity of the signal} \\ &= \sum_{n=0}^{9} e^{j2\pi\frac{9}{10} n}e^{-j \frac{1}{10}2 \pi n k} \text{ shifting the signal by one period is like multiplying by 1} \\ &= \sum_{n=0}^{9} e^{-j\frac{1}{10}2\pi n (k-9)} \end{align} $

Then if you compare this equation to the IDFT formula and you get that

$ \begin{align} X [k] &= \sum_{n=0}^{N-1} x[n]e^{\frac{-j2\pi k n}{N}} \\ &= 10 \delta (k-9) \end{align} $

Instructor's comment: Your answer is not a periodic signal, as it should be. -pm

Answer 5

Answers 3 and 4 are the same because X[k] has a period of 10.

Instructor's comment: Good observation! -pm

Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva