Latest revision as of 11:33, 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:

$x[n]= \cos \left( \frac{2 \pi }{500} n \right)$

(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

$\mathcal{F}(x[n]) = \mathcal{F}(cos(\frac{2\pi}{500}n)) = \mathcal{F}(\frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}) =\frac{1}{2}( \mathcal{F}(e^{j\frac{2\pi}{500}n})+\mathcal{F}(e^{-j\frac{2\pi}{500}n}))$

$=\frac{1}{2}( \pi\sum_{l=-\infty}^{+\infty}\delta(w-\frac{2\pi}{500}-2\pi l) + \pi\sum_{l=- \infty}^{+\infty}\delta(w+\frac{2\pi}{500}-2\pi l) )$

Instructor's comment: You need to justify this step (i.e. the previous equality). -pm

$=\frac{\pi}{2} \sum_{l=-\infty}^{+\infty}[ \delta(w-\frac{2\pi}{500}-2\pi l)+\delta(w+\frac{2\pi}{500}-2\pi l) ]$

DTFT f(w) is periodic funtion, just need to include one period to be sufficient

Instructor's comment: Correct, but do not write your answer in such a way that it looks like the FT is zero outside of one period. -pm

Answer 2

$x[n] = \int_{-\pi}^{\pi} \mathcal{X} (w)e^{j\omega n} dw$

The input x[n] can can be written in the exponential form.

$x[n] = cos(\frac{2\pi}{500}n) = \frac{e^{j\frac{2\pi}{500}n} + e^{-j\frac{2\pi}{500}n}}{2}$

In order for the input x[n] to have such a value, (Please justify! -pm)

$\mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})$

Answer 3

$x[n] = \frac{2\pi}{500}n = \frac{e^{j\frac{2\pi}{500}n}}{2}+\frac{e^{-j\frac{2\pi}{500}n}}{2}$

Instructor's comment: Obviously, the previous line contains a mistake (missing a cosine). -pm)

$\mathcal{X} (\omega) = \pi (\delta(\omega - \frac{2\pi}{500}) + \delta(\omega + \frac{2\pi}{500}))$

Instructor's comment: You would get close to zero points for this answer, because a) it is not periodic, b) it is just "plugged in" without justification. -pm

Answer 4

$\begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) \\ = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align}$

$0 < \frac{2\pi}{500}n < \pi$

$-\pi < -\frac{2\pi}{500}n < \pi$

consider $-\pi < \omega < \pi$

$\begin{align}\mathcal{F}[x[n]] = 2\pi * \frac{1}{2} [\delta(\omega - \frac{2\pi}{500}n) + \delta(\omega + \frac{2\pi}{500}n)] \end{align}$

Instructor's comment: What about other values of omega? Also, be careful not to confuse the convolution symbol (*) with a multiplication. -pm

Answer 5

$\begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align}$

$F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})$ repp'ed every $2\pi$

Instructor's comment: And the justification for this last step is ??? -pm

Answer 6

$x[n]= \cos \left( \frac{2 \pi }{500} n \right)=\frac{1}{2}(e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n})$

$\mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]$

Instructor's comment: Please justify this last step! -pm

Answer 7

$\begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align}$

$F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})$

Note that the since we are dealing with a DT signal, it repeats every $2\pi$

Instructor's comment: How did you find out the FT of a complex exponential? Please justify. -pm

Answer 8

$\begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align}$

$\mathcal{X} (\omega) = F[x[n]] = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})$ (Why? -pm )

     $( \omega \in [-\pi,\pi])$  (This should be above the previous equation for clarity. -pm)

$\mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]$

Answer 9

$x[n]= \cos \left( \frac{2 \pi }{500} n \right)=\frac{1}{2}(e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n})$

$F[x[n]] = \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]$

Instructor's comment: You did not fully justify your answer. -pm

Answer 10

$\begin{align} x[n]= \cos \left( \frac{2 \pi }{500} n \right) = \frac{ e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2} \end{align}$

$\mathcal{X} (\omega) = F[x[n]] = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]$

Instructor's comment: Not enough justification for full credit. -pm

Note from the instructor: There is still not a single solution on this page that would deserve full credit. Please keep trying. -pm

Back to ECE438 Fall 2011 Prof. Boutin

@@ Line 1: / Line 1: @@
+[[Category:ECE438]]
 [[Category:ECE438Fall2011Boutin]]
 [[Category:problem solving]]
-= Discrete-time Fourier transform computation =
+[[Category:discrete-time Fourier transform]]
-Compute the discrete-time Fourier transform of the two following signal:
+<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:
 <math>
@@ Line 14: / Line 25: @@
 ----
 ===Answer 1===
-Write it here.
+<math>
+\mathcal{F}(x[n]) = \mathcal{F}(cos(\frac{2\pi}{500}n)) = \mathcal{F}(\frac{
+e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2})
+=\frac{1}{2}( \mathcal{F}(e^{j\frac{2\pi}{500}n})+\mathcal{F}(e^{-j\frac{2\pi}{500}n}))
+</math>
+<math>
+=\frac{1}{2}( \pi\sum_{l=-\infty}^{+\infty}\delta(w-\frac{2\pi}{500}-2\pi l) + \pi\sum_{l=-          \infty}^{+\infty}\delta(w+\frac{2\pi}{500}-2\pi l) )
+</math>
+:<span style="color:red">Instructor's comment: You need to justify this step (i.e. the previous equality). -pm</span>
+<math>
+=\frac{\pi}{2} \sum_{l=-\infty}^{+\infty}[ \delta(w-\frac{2\pi}{500}-2\pi l)+\delta(w+\frac{2\pi}{500}-2\pi l) ]
+</math>
+DTFT f(w) is periodic funtion, just need to include one period to be sufficient
+:<span style="color:green">Instructor's comment: Correct, but do not write your answer in such a way that it looks like the FT is zero outside of one period. -pm</span>
 ===Answer 2===
-Write it here.
+<math> x[n] = \int_{-\pi}^{\pi} \mathcal{X} (w)e^{j\omega n} dw
+</math>
+The input x[n] can can be written in the exponential form.
+<math> x[n] = cos(\frac{2\pi}{500}n)
+            = \frac{e^{j\frac{2\pi}{500}n} + e^{-j\frac{2\pi}{500}n}}{2}
+</math>
+In order for the input x[n] to have such a value, <span style="color:red">(Please justify! -pm)</span>
+<math> \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})
+</math>
 ===Answer 3===
-write it here.
+<math> x[n] = \frac{2\pi}{500}n
+            = \frac{e^{j\frac{2\pi}{500}n}}{2}+\frac{e^{-j\frac{2\pi}{500}n}}{2}
+</math>
+:<span style="color:red">Instructor's comment: Obviously, the previous line contains a mistake (missing a cosine). -pm)</span>
+<math> \mathcal{X} (\omega) = \pi (\delta(\omega - \frac{2\pi}{500}) + \delta(\omega + \frac{2\pi}{500}))
+</math>
+:<span style="color:red">Instructor's comment: You would get close to zero points for this answer, because a) it is not periodic, b) it is just "plugged in" without justification. -pm  </span>
+===Answer 4===
+<math>\begin{align}
+x[n]= \cos \left( \frac{2 \pi }{500} n  \right)
+\\ = \frac{
+e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}
+\end{align}</math>
+<math>0 < \frac{2\pi}{500}n < \pi </math>
+<math>-\pi < -\frac{2\pi}{500}n < \pi </math>
+consider <math> -\pi < \omega < \pi </math>
+<math>\begin{align}\mathcal{F}[x[n]] = 2\pi * \frac{1}{2} [\delta(\omega - \frac{2\pi}{500}n) + \delta(\omega + \frac{2\pi}{500}n)] \end{align}</math>
+:<span style="color:red">Instructor's comment: What about other values of omega? Also, be careful not to confuse the convolution symbol (*) with a multiplication.  -pm  </span>
+===Answer 5===
+<math>\begin{align}
+x[n]= \cos \left( \frac{2 \pi }{500} n  \right) = \frac{
+e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}
+\end{align}</math>
+<math> F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})
+</math> repp'ed every <math>2\pi</math>
+:<span style="color:red">Instructor's comment: And the justification for this last step is ???  -pm  </span>
+===Answer 6===
+<math>x[n]= \cos \left( \frac{2 \pi }{500} n  \right)=\frac{1}{2}(e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n})</math>
+<math> \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]</math>
+:<span style="color:red">Instructor's comment: Please justify this last step!  -pm  </span>
+===Answer 7===
+<math>\begin{align}
+x[n]= \cos \left( \frac{2 \pi }{500} n  \right) = \frac{
+e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}
+\end{align}</math>
+<math> F[x[n]] = \mathcal{X} (\omega) = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500}) </math>
+Note that the since we are dealing with a DT signal, it repeats every <math> 2\pi </math>
+:<span style="color:red">Instructor's comment: How did you find out the FT of a complex exponential? Please justify. -pm  </span>
+===Answer 8===
+<math>\begin{align}
+x[n]= \cos \left( \frac{2 \pi }{500} n  \right) = \frac{
+e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}
+\end{align}</math>
+<math> \mathcal{X} (\omega) = F[x[n]] = \pi \delta(\omega - \frac{2\pi}{500}) + \pi \delta(\omega + \frac{2\pi}{500})</math> <span style="color:red">(Why? -pm ) </span>
+     <math> ( \omega \in [-\pi,\pi])</math> <span style="color:red">(This should be above the previous equation for clarity. -pm) </span>
+<math> \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]</math>
+===Answer 9===
+<math>x[n]= \cos \left( \frac{2 \pi }{500} n  \right)=\frac{1}{2}(e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n})</math>
+<math>F[x[n]] = \mathcal{X} (\omega) = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]</math>
+:<span style="color:red">Instructor's comment: You did not fully justify your answer. -pm</span>
+===Answer 10===
+<math>\begin{align}
+x[n]= \cos \left( \frac{2 \pi }{500} n  \right) = \frac{
+e^{j\frac{2\pi}{500}n}+e^{-j\frac{2\pi}{500}n}}{2}
+\end{align}</math>
+<math>\mathcal{X} (\omega) = F[x[n]] = rep_{2\pi}\left[\pi \delta \left(\omega - \frac{2\pi}{500} \right) + \pi \delta \left(\omega + \frac{2\pi}{500} \right) \right]</math>
+:<span style="color:red">Instructor's comment: Not enough justification for full credit. -pm</span>
+----
+=<span style="color:red"> Note from the instructor: There is still not a single solution on this page that would deserve full credit. Please keep trying. -pm</span>=
 ----
 [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]

Difference between revisions of "Practice DTFT computation cosine ECE438F11" - Rhea

Latest revision as of 11:33, 26 November 2013

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Question

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Answer 1

Answer 2

Answer 3

Answer 4

Answer 5

Answer 6

Answer 7

Answer 8

Answer 9

Answer 10

Note from the instructor: There is still not a single solution on this page that would deserve full credit. Please keep trying. -pm

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