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Hence, the function will be | Hence, the function will be | ||
| − | <math> X_(\omega) = \frac{\pi}j \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) </math> | + | <math> 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) </math> |
<math>x[n]=\sin \left( \frac{2\pi}{100} n \right)</math> | <math>x[n]=\sin \left( \frac{2\pi}{100} n \right)</math> | ||
Revision as of 17:16, 12 September 2013
Contents
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.)
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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 $
Answer 2
Write it here.
Answer 3
We can separate the equation 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) $
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) $
Hence, the function 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) $
$ x[n]=\sin \left( \frac{2\pi}{100} n \right) $
