Difference between revisions of "Practice DTFT computation sine ECE438F13" - Rhea

Latest revision as of 11:36, 26 November 2013

Question

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.)

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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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