Revision as of 20:25, 6 September 2011

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

$ \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} $

$ =1+e^{-j\omega}+e^{-j2\omega} $

so

Answer 4

$ x[n] = u[n]-u[n-3] = \delta [n] + \delta [n-1] + \delta [n-2]; $

by Fourier's linearity,

$ \mathcal{X}(\omega)=\mathcal{F}(x[n])= \mathcal{F}(\delta [n]) + \mathcal{F}(\delta [n-1]) + \mathcal{F}(\delta [n-2]); $

from the table Discrete-time Fourier Transform Pairs and Properties

$ \mathcal{F}(\delta[n])= 1; $

$ \mathcal{F}(\delta[n-n0])= e^{-j\omega n0}; $

So one can conclude that

$ \mathcal{F}(x[n])= 1 + e^{-j\omega} + e^{-2j\omega}; $

Prove the properties and pairs if needed.

Answer 4

$ \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} $ $ =\delta(0)e^{-j\omega 0}+\delta(1-1)e^{-j\omega 1}+\delta(2-2)e^{-j\omega 2}=1+e^{-j\omega}+e^{-j2\omega} $

Answer 5

$ \begin{align} x[n] = u[n]-u[n-3] = \delta [n]+\delta [n-1]+\delta[n-2]\end{align} $

$ \begin{align} \mathcal{F}(x[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}]) \\=1*\delta [n]+e^{-j\omega}\delta [n-1]+e^{-j2\omega}\delta [n-2] \\=1+e^{-j\omega}+e^{-j2\omega} \end{align} $

Answer 6

$ x[n]= u[n]-u[n-3] $

can be written as

$ x[n] = \delta[n] + \delta [n-1] + \delta[n-2] $.

The first term fourier transforms to $ e^{j\omega 0} = 1 $ and the other terms are delayed versions of the first term.

Thus the complete fourier transform comes out to be

$ =1+e^{-j\omega}+e^{-j2\omega} $

Answer 7

$ \begin{align} \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=0}^{2}e^{-j\omega n}\\ &=1+e^{-j\omega}+e^{-j\omega 2} \end{align} $

Back to ECE438 Fall 2011 Prof. Boutin