Contents
Practice Problem on 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} $
$ =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} $
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
