Practice DTFT computation rect ECE438F11 - Rhea

Question

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.)

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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^{-j2\omega} \end{align} $

Answer 8

In discrete time, u[n] - u[n-3] is the same as

$ \delta [n] + \delta [n-1] + \delta [n-2] $

By the linearity property of Fourier Transforms, we can transform each term individually. The FT of x[n] is then:

$ \mathcal{X}(\omega) = 1+e^{-j\omega}+e^{-j\omega 2} $

Answer 9

$ \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^{-2j\omega } \end{align} $

Answer 10

$ \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}^{3}e^{-j\omega n}\\ &=1+e^{-j\omega}+e^{-2j\omega} +e^{-3j\omega} \end{align} $

TA's comments: The item $ e^{-3j\omega} $ is redundant because u[n]-u[n-3] has only three items that are non-zero. Be careful.

Answer 11

$ \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^{-2j\omega } \end{align} $

TA's comments: Answer 7,9,11 are great, and it's a simple problem with not much space for creativity in solution. but it doesn't contribute much by repeating the same thing.

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