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[[Category:ECE438Fall2011Boutin]] | [[Category:ECE438Fall2011Boutin]] | ||
[[Category:problem solving]] | [[Category:problem solving]] | ||
| − | = Discrete-time Fourier transform computation = | + | [[Category:discrete time Fourier transform]] |
| + | <center><font size= 4> | ||
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Discrete-time Fourier transform computation | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
Compute the discrete-time Fourier transform of the following signal: | Compute the discrete-time Fourier transform of the following signal: | ||
| Line 119: | Line 128: | ||
\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}]) | \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}]) | \\=\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}</math> | \end{align}</math> | ||
| − | <math> | + | ===Answer 6=== |
| − | + | <math> x[n]= u[n]-u[n-3] </math> | |
| − | <math> | + | can be written as |
| − | + | ||
| + | <math> x[n] = \delta[n] + \delta [n-1] + \delta[n-2] </math>. | ||
| + | |||
| + | The first term fourier transforms to <math> e^{j\omega 0} = 1 </math> and the other terms are delayed versions of the first term. | ||
| + | |||
| + | Thus the complete fourier transform comes out to be | ||
<math> | <math> | ||
| Line 131: | Line 147: | ||
</math> | </math> | ||
| + | ===Answer 7=== | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | ===Answer 8=== | ||
| + | In discrete time, u[n] - u[n-3] is the same as | ||
| + | |||
| + | <math> \delta [n] + \delta [n-1] + \delta [n-2] </math> | ||
| + | |||
| + | By the linearity property of Fourier Transforms, we can transform each term individually. The FT of x[n] is then: | ||
| + | |||
| + | <math> \mathcal{X}(\omega) = 1+e^{-j\omega}+e^{-j\omega 2} </math> | ||
| + | |||
| + | |||
| + | ===Answer 9=== | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
| + | ===Answer 10=== | ||
| + | <math>\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}</math> | ||
| + | |||
| + | <span style="color:red">TA's comments: The item <math>e^{-3j\omega}</math> is redundant because u[n]-u[n-3] has only three items that are non-zero. Be careful.</span> | ||
| + | ===Answer 11=== | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | ---- | ||
| + | <span style="color:red">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.</span> | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Latest revision as of 11:35, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Discrete-time Fourier transform computation
Contents
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.)
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} $
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.
