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:<span style="color:green"> Instructor's comment: You need to learn to find the answer without using a table.  Now, I am not sure which table you used, but it must be wrong, since the anwer you obtained is not periodic with period <math>2\pi</math>.</span>
 
:<span style="color:green"> Instructor's comment: You need to learn to find the answer without using a table.  Now, I am not sure which table you used, but it must be wrong, since the anwer you obtained is not periodic with period <math>2\pi</math>.</span>
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:<span style="color:red"> TA's comment: You could rather say, X(ω) equals this from -π to π and it's 2π periodic. </span>
 
===Answer 2===
 
===Answer 2===
  

Revision as of 11:38, 20 September 2013


Practice Problem on Discrete-time Fourier transform computation

Compute the discrete-time Fourier transform of the following signal:

$ x[n]= \sin \left( \frac{2 \pi }{100} n \right) $

(Write enough intermediate steps to fully justify your answer.)


Share your answers below

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]=\sin \left( \frac{2 \pi}{100} \right) $


$ x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right) $

$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $

$ X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right) $


$ X_(\omega) = \frac{\pi}{j} \left( \delta \left({\omega - \frac{2 \pi}{100}}\right) - \delta \left({\omega + \frac{2 \pi}{100}}\right) \right) by DTFT table $

Instructor's comment: You need to learn to find the answer without using a table. Now, I am not sure which table you used, but it must be wrong, since the anwer you obtained is not periodic with period $ 2\pi $.
TA's comment: You could rather say, X(ω) equals this from -π to π and it's 2π periodic.

Answer 2

First, write the original function as: $ x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right) $


Then, for w = [-pi, pi] ( Instructor's comment: You need more justification here.)

$ X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right) $

$ X_(\omega) = \frac{100}{2j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right) $

which is really is:

$ X_(\omega) = rep_2pi \frac{50}{j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right) $

( Instructor's comment: You should make it clear which expressions are valid for all values of $ \omega $, and which expressions are only valid for $ \omega \in [-\pi, \pi ] $.

Answer 3

Xiang Zhang

We can separate (Instructor's comment: separate? Do you mean "write"?)the equation ( Instructor's comment: it's not an equation: it's a signal, or a function.) to the following function

$ x[n]=\frac{1}{2 j} \left( e^\frac{j 2 \pi n}{100} - e^\frac{- j 2 \pi n}{100} \right) $

Because based on Fourier transform equation,

$ X_(\omega) = \sum_{n = -\infty}^{\infty} x[n] e^{-j \omega n} $

Substitute in x[n]

$ X_(\omega) = \frac{1}{2 j} \left( \sum_{n = -\infty}^{\infty} e^{ \frac{j2 \pi n} {100} } e^{-j\omega n} - \sum_{n = -\infty}^{ \infty} e^{\frac{-j2 \pi n} {100} } e^{-j\omega n} \right) $

(Instructor's comment: Why write the equation above if you are going to use a FT pair from a table?)

From Discrete Fourier Transform pair,

$ x[n] = e^{-j\omega_0 n} $ DTFT to $ X_(\omega) = 2 \pi \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) $

(Instructor's comment: Careful above! The original signal was called x[n]; you can't reuse x[n] for a different signal.)

Hence, the function (Instructor's comment: Function? You mean "DTFT"?.) will be

$ X_(\omega) = \frac{\pi}j \left( \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) - \sum_{n = -\infty}^{ \infty} \delta \left( \omega+\omega_0 - 2\pi l \right) \right) $

(Instructor's comment: What is $ \omega_0 $?)

$ x[n]=\sin \left( \frac{2\pi}{100} n \right) $

(Instructor's comment: You don't need to re-write the signal.)


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