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| + | ==Answer 3== | ||
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| + | <math>x[n]= e^{-j \frac{1}{5} \pi n} = e^{-j \frac{2\pi}{10} n}</math> | ||
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| + | By comparison with the IDT | ||
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| + | <math>\,x [n] = \frac{1}{N} \sum_{k=0}^{N-1} X[k] e^{j \frac{2\pi}{N}kn} \,</math> | ||
| + | |||
| + | 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 <math> X[k] = 10\delta[k+1]</math> | ||
| + | |||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 02:34, 6 October 2011
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]?
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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] $.
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] $
