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| − | <math>x[n]=sin | + | <math>x[n]=\sin \left( 2pi/100 \right)</math> |
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<math>X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}</math> | <math>X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}</math> | ||
| − | <math>X_(\omega) = \sum_{n=-\infty}^{+\infty} e^{-j2 \pi | + | <math>X_(\omega) = \sum_{n=-\infty}^{+\infty} e^{-j2 \pi 100 n} e^{-j\omega n}</math> |
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 16:06, 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( 2pi/100 \right) $
$ x[n] = 1/(2*j)*(exp(j*2*pi/100*n)-exp(-j*2*pi/100*n)) $
$ X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n} $
$ X_(\omega) = \sum_{n=-\infty}^{+\infty} e^{-j2 \pi 100 n} e^{-j\omega n} $
Answer 2
Write it here.
