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[[Category:ECE438Fall2011Boutin]]
 
[[Category:ECE438Fall2011Boutin]]
 
[[Category:problem solving]]
 
[[Category:problem solving]]
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= Discrete-time Fourier transform computation =
+
[[Category:discrete time Fourier transform]]
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<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:
  
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----
 
----
 
===Answer 1===
 
===Answer 1===
−
Write it here.
+
<math>
 +
\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}</math>
 +
 
 +
<math>
 +
=\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j\omega n}+\delta [n-1]e^{-j\omega n}+\delta[n-2]e^{-j\omega n}])
 +
</math>
 +
 
 +
<math>
 +
=\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j0\omega}+\delta [n-1]e^{-j\omega}+\delta[n-2]e^{-j2\omega}])</math>
 +
 
 +
<math>
 +
=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]</math>
 +
 
 +
<math>
 +
=1+e^{-j\omega}+e^{-j2\omega}
 +
</math>
 +
:<span style="color:green">Instructor's comments: This is a bit long. Could you shorten your solution somehow? -pm </span>
 +
 
 
===Answer 2===
 
===Answer 2===
−
Write it here.
+
<math> \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}
 +
</math>
 +
 
 +
<math>                    = \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}
 +
</math>
 +
 
 +
Let l = n-3
 +
 
 +
<math>                    = \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}
 +
</math>
 +
 
 +
<math>                    = \frac{1}{1-e^{-j\omega}} - e^{-j\omega 3}\frac{1}{1-e^{-j\omega}}
 +
</math>
 +
 
 
===Answer 3===
 
===Answer 3===
−
write it here.
+
 
 +
<math>
 +
\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}</math>
 +
 
 +
<math>
 +
=\sum_{n=-\infty}^{\infty}(\delta [n]e^{-j\omega n}+\delta [n-1]e^{-j\omega n}+\delta[n-2]e^{-j\omega n}])
 +
</math>
 +
 
 +
<math>
 +
=\delta [0]e^{-j0\omega}+\delta [1]e^{-j\omega}+\delta[2]e^{-j2\omega}</math>
 +
 
 +
 
 +
<math>
 +
=1+e^{-j\omega}+e^{-j2\omega}
 +
</math>
 +
 
 +
so
 +
 
 +
===Answer 4===
 +
 
 +
<math>
 +
x[n] = u[n]-u[n-3] = \delta [n] + \delta [n-1] + \delta [n-2];
 +
</math>
 +
 
 +
by Fourier's linearity,
 +
 
 +
<math>
 +
\mathcal{X}(\omega)=\mathcal{F}(x[n])= \mathcal{F}(\delta [n]) + \mathcal{F}(\delta [n-1]) + \mathcal{F}(\delta [n-2]);
 +
</math>
 +
 
 +
from the table [[Table_DT_Fourier_Transforms|Discrete-time Fourier Transform Pairs and Properties]]
 +
 
 +
<math>
 +
\mathcal{F}(\delta[n])= 1;
 +
</math>
 +
 
 +
<math>
 +
\mathcal{F}(\delta[n-n0])= e^{-j\omega n0};
 +
</math>
 +
 
 +
So one can conclude that
 +
 
 +
<math>
 +
\mathcal{F}(x[n])= 1 + e^{-j\omega} + e^{-2j\omega};
 +
</math>
 +
 
 +
Prove the properties and pairs if needed.
 +
 
 +
===Answer 4===
 +
 
 +
<math>\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}</math>
 +
<math>
 +
=\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} </math>
 +
 
 +
 
 +
===Answer 5===
 +
 
 +
<math>\begin{align}
 +
x[n] = u[n]-u[n-3] = \delta [n]+\delta [n-1]+\delta[n-2]\end{align}</math>
 +
 
 +
<math>\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}</math>
 +
 
 +
===Answer 6===
 +
<math> x[n]= u[n]-u[n-3] </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>
 +
=1+e^{-j\omega}+e^{-j2\omega}
 +
</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

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.)

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^{-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.

Back to ECE438 Fall 2011 Prof. Boutin

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