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===Answer 3=== | ===Answer 3=== | ||
| − | + | ||
| + | <math> | ||
| + | \mathcal{X}(\omega)=\mathcal{F}(x[n])=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} | ||
| + | =\sum_{n=-\infty}^{\infty}(u[n]-u[n-3])e^{-j\omega n} | ||
| + | =\sum_{n=-\infty}^{\infty}(\delta [n]+\delta [n-1]+\delta[n-2])e^{-j\omega n}</math> | ||
| + | |||
| + | <math> | ||
| + | =\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j\omega n}+\delta [n-1]e^{-j\omega n}+\delta[n-2]e^{-j\omega n}]) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | =\delta [0]e^{-j0\omega}+\delta [1]e^{-j\omega}+\delta[2]e^{-j2\omega}</math> | ||
| + | |||
| + | <math> | ||
| + | =e^{-j0\omega}\sum_{n=-\infty}^{\infty}\delta [n]+e^{-j\omega}\sum_{n=-\infty}^{\infty}\delta [n-1]+e^{-j2\omega}\sum_{n=-\infty}^{\infty}\delta [n-2]</math> | ||
| + | |||
| + | <math> | ||
| + | =1+e^{-j\omega}+e^{-j2\omega} | ||
| + | </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 16:44, 6 September 2011
Contents
Discrete-time Fourier transform computation
Compute the discrete-time Fourier transform of the following signal:
$ x[n]= u[n]-u[n-3] $
(Write enough intermediate steps to fully justify your answer.)
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
$ \mathcal{X}(\omega)=\mathcal{F}(x[n])=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} =\sum_{n=-\infty}^{\infty}(u[n]-u[n-3])e^{-j\omega n} =\sum_{n=-\infty}^{\infty}(\delta [n]+\delta [n-1]+\delta[n-2])e^{-j\omega n} $
$ =\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j\omega n}+\delta [n-1]e^{-j\omega n}+\delta[n-2]e^{-j\omega n}]) $
$ =\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j0\omega}+\delta [n-1]e^{-j\omega}+\delta[n-2]e^{-j2\omega}]) $
$ =e^{-j0\omega}\sum_{n=-\infty}^{\infty}\delta [n]+e^{-j\omega}\sum_{n=-\infty}^{\infty}\delta [n-1]+e^{-j2\omega}\sum_{n=-\infty}^{\infty}\delta [n-2] $
$ =1+e^{-j\omega}+e^{-j2\omega} $
- Instructor's comments: This is a bit long. Could you shorten your solution somehow? -pm
Answer 2
$ \mathcal{X}(\omega) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = \sum_{n=-\infty}^{\infty} (u[n] - u[n-3]) e^{-j\omega n} $
$ = \sum_{n=-\infty}^{\infty} u[n]e^{-j\omega n} - \sum_{n=-\infty}^{\infty}u[n-3]e^{-j\omega n} = \sum_{n=0}^{\infty}e^{-j\omega n} - \sum_{n=3}^{\infty}e^{-j\omega n} $
Let l = n-3
$ = \frac{1}{1-e^{-j\omega}} - \sum_{l=0}^{\infty}e^{-j\omega l}e^{-j\omega 3} = \frac{1}{1-e^{-j\omega}} - e^{-j\omega 3} \sum_{l=0}^{\infty}(e^{-j\omega})^{l} $
$ = \frac{1}{1-e^{-j\omega}} - e^{-j\omega 3}\frac{1}{1-e^{-j\omega}} $
Answer 3
$ \mathcal{X}(\omega)=\mathcal{F}(x[n])=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} =\sum_{n=-\infty}^{\infty}(u[n]-u[n-3])e^{-j\omega n} =\sum_{n=-\infty}^{\infty}(\delta [n]+\delta [n-1]+\delta[n-2])e^{-j\omega n} $
$ =\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j\omega n}+\delta [n-1]e^{-j\omega n}+\delta[n-2]e^{-j\omega n}]) $
$ =\delta [0]e^{-j0\omega}+\delta [1]e^{-j\omega}+\delta[2]e^{-j2\omega} $
$ =e^{-j0\omega}\sum_{n=-\infty}^{\infty}\delta [n]+e^{-j\omega}\sum_{n=-\infty}^{\infty}\delta [n-1]+e^{-j2\omega}\sum_{n=-\infty}^{\infty}\delta [n-2] $
$ =1+e^{-j\omega}+e^{-j2\omega} $
