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===Answer 1=== | ===Answer 1=== | ||
− | + | <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> | ||
+ | =\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j0\omega}+\delta [n-1]e^{-j\omega}+\delta[n-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> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. |
Revision as of 18:23, 3 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} $
Answer 2
Write it here.
Answer 3
write it here.