Revision as of 07:57, 4 September 2011 by Mboutin (Talk | contribs)

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.)


Share your answers below

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

Write it here.

Answer 3

write it here.


Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009