Line 1: Line 1:
−
[[Category:ECE]]
+
<p><br />
−
[[Category:ECE438]]
+
</p>
−
[[Category:2013_Fall_ECE_438_Boutin]]
+
<h1> <a href=":Category:Problem solving">Practice Problem</a> on Discrete-time Fourier transform computation </h1>
−
[[Category:problem solving]]
+
<p>Compute the discrete-time Fourier transform of the following signal:
−
 
+
</p><p><img _fckfakelement="true" _fck_mw_math="&#10;x[n]= \sin \left( \frac{2 \pi }{100} n  \right)&#10;" src="/rhea/images/math/b/f/0/bf0f97ec20b83c8416e3cd5d95395388.png" />
−
[[Category:discrete time Fourier transform]]
+
</p><p>(Write enough intermediate steps to fully justify your answer.)  
−
 
+
</p>
−
= [[:Category:Problem_solving|Practice Problem]] on Discrete-time Fourier transform computation =
+
<hr />
−
Compute the discrete-time Fourier transform of the following signal:
+
<h2>Share your answers below</h2>
−
 
+
<p>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!
−
<math>
+
</p><p><b>No need to write your name: we can find out who wrote what by checking the history of the page.</b>
−
x[n]= \sin \left( \frac{2 \pi }{100} n  \right)
+
</p>
−
</math>
+
<hr />
−
 
+
<h3>Answer 1</h3>
−
(Write enough intermediate steps to fully justify your answer.)  
+
<p><img _fckfakelement="true" _fck_mw_math="x[n]=\sin \left( \frac{2 \pi}{100} \right)" src="/rhea/images/math/2/2/3/2231dd05d43da978a3499a964d730f15.png" />
−
----
+
</p><p><br />
−
==Share your answers below==
+
<img _fckfakelement="true" _fck_mw_math="x[n] = \frac{1}{2j}  \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)" src="/rhea/images/math/8/3/6/83609c9889b65fcd96711aed18f6111c.png" />
−
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!
+
</p><p><img _fckfakelement="true" _fck_mw_math="X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}" src="/rhea/images/math/2/b/d/2bd6e9f4f8c6f4d0c8ba6b2119288a9f.png" />
−
 
+
</p><p><img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)" src="/rhea/images/math/7/f/9/7f9ab2bb6f04aadec1f5789b52ed7e47.png" />
−
'''No need to write your name: we can find out who wrote what by checking the history of the page.'''
+
</p><p><br />
−
----
+
<img _fckfakelement="true" _fck_mw_math="X_(\omega) =  \frac{\pi}{j} \left( \delta \left({\omega - \frac{2 \pi}{100}}\right) - \delta \left({\omega + \frac{2 \pi}{100}}\right) \right)  by  DTFT  table" src="/rhea/images/math/6/1/8/61814381fa1ee34cad6f25374d24430a.png" />
−
===Answer 1===
+
</p>
−
 
+
<h3>Answer 2</h3>
−
 
+
<p>First, write the original function as:
−
<math>x[n]=\sin \left( \frac{2 \pi}{100} \right)</math>
+
<img _fckfakelement="true" _fck_mw_math="x[n] = \frac{1}{2j}  \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)" src="/rhea/images/math/8/3/6/83609c9889b65fcd96711aed18f6111c.png" />
−
 
+
</p><p>Then, for w = [-pi, pi]
−
 
+
<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)" src="/rhea/images/math/7/f/9/7f9ab2bb6f04aadec1f5789b52ed7e47.png" />
−
<math>x[n] = \frac{1}{2j}  \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)</math>
+
</p><p><img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{100}{2j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)" src="/rhea/images/math/f/0/c/f0c3875a9621aabf1286b15e89b68e9f.png" />
−
 
+
</p><p>which is really is:
−
<math>X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}</math>
+
</p><p><img _fckfakelement="true" _fck_mw_math="X_(\omega) = rep_2pi \frac{50}{j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)" src="/rhea/images/math/3/7/b/37b3b44a22cf3b61840a7c8bb4486eec.png" />
−
 
+
</p><p><br />
−
<math>X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)</math>
+
</p>
−
 
+
<h3>Answer 3</h3>
−
 
+
<p>We can separate the equation to the following function
−
<math>X_(\omega) =  \frac{\pi}{j} \left( \delta \left({\omega - \frac{2 \pi}{100}}\right) - \delta \left({\omega + \frac{2 \pi}{100}}\right) \right)  by  DTFT  table</math>
+
</p><p><img _fckfakelement="true" _fck_mw_math="x[n]=\frac{1}{2 j} \left( e^\frac{j 2 \pi n}{100}  - e^\frac{- j 2 \pi n}{100}  \right)  " src="/rhea/images/math/0/5/b/05ba7655493b0c4e3b694ba6fac0539c.png" />
−
 
+
</p><p>Because based on Fourier transform equation,
−
===Answer 2===
+
</p><p><img _fckfakelement="true" _fck_mw_math="X_(\omega) = \sum_{n = -\infty}^{\infty} x[n] e^{-j \omega n}" src="/rhea/images/math/a/7/9/a79fed22e488f9ab5773eadadc46bcb0.png" />
−
 
+
</p><p>Substitute in x[n]
−
First, write the original function as:
+
</p><p><img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2 j} \left( \sum_{n = -\infty}^{\infty} e^{ \frac{j2 \pi n} {100} } e^{-j\omega n} - \sum_{n = -\infty}^{ \infty} e^{\frac{-j2 \pi n} {100} } e^{-j\omega n} \right)" src="/rhea/images/math/b/2/8/b283311397523aab2fbcfa322f3b759f.png" />
−
<math>x[n] = \frac{1}{2j}  \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)</math>
+
</p><p>From Discrete Fourier Transform pair,
−
 
+
</p><p><img _fckfakelement="true" _fck_mw_math=" x[n] = e^{-j\omega_0 n} " src="/rhea/images/math/1/1/9/119b4dfb4f6bcf5deb0663acaa69ca03.png" /> DTFT to <img _fckfakelement="true" _fck_mw_math=" X_(\omega) = 2 \pi \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right)  " src="/rhea/images/math/b/2/5/b25d82705497c4e62e1d068138dae962.png" />
−
Then, for w = [-pi, pi]
+
</p><p>Hence, the function will be
−
<math>X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)</math>
+
</p><p><img _fckfakelement="true" _fck_mw_math=" X_(\omega) =  \frac{\pi}j \left( \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) - \sum_{n = -\infty}^{ \infty} \delta \left( \omega+\omega_0 - 2\pi l \right) \right)  " src="/rhea/images/math/5/2/c/52cb905ce8f853f34e9fb0f152c902f0.png" />
−
 
+
</p><p><img _fckfakelement="true" _fck_mw_math="x[n]=\sin \left( \frac{2\pi}{100} n \right)" src="/rhea/images/math/b/f/0/bf0f97ec20b83c8416e3cd5d95395388.png" />
−
<math>X_(\omega) = \frac{100}{2j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)</math>
+
</p><p><br />
−
 
+
<a href="2013 Fall ECE 438 Boutin">Back to ECE438 Fall 2013</a>
−
which is really is:
+
</p><a _fcknotitle="true" href="Category:ECE">ECE</a> <a _fcknotitle="true" href="Category:ECE438">ECE438</a> <a _fcknotitle="true" href="Category:2013_Fall_ECE_438_Boutin">2013_Fall_ECE_438_Boutin</a> <a _fcknotitle="true" href="Category:Problem_solving">Problem_solving</a> <a _fcknotitle="true" href="Category:Discrete_time_Fourier_transform">Discrete_time_Fourier_transform</a>
−
 
+
−
<math>X_(\omega) = rep_2pi \frac{50}{j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)</math>
+
−
 
+
−
 
+
−
===Answer 3===
+
−
We can separate the equation to the following function
+
−
 
+
−
<math>x[n]=\frac{1}{2 j} \left( e^\frac{j 2 \pi n}{100}  - e^\frac{- j 2 \pi n}{100}  \right)  </math>
+
−
 
+
−
Because based on Fourier transform equation,
+
−
 
+
−
<math>X_(\omega) = \sum_{n = -\infty}^{\infty} x[n] e^{-j \omega n}</math>
+
−
 
+
−
Substitute in x[n]
+
−
 
+
−
<math>X_(\omega) = \frac{1}{2 j} \left( \sum_{n = -\infty}^{\infty} e^{ \frac{j2 \pi n} {100} } e^{-j\omega n} - \sum_{n = -\infty}^{ \infty} e^{\frac{-j2 \pi n} {100} } e^{-j\omega n} \right)</math>
+
−
 
+
−
From Discrete Fourier Transform pair,
+
−
 
+
−
<math> x[n] = e^{-j\omega_0 n} </math> DTFT to <math> X_(\omega) = 2 \pi \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right)  </math>
+
−
 
+
−
Hence, the function will be
+
−
 
+
−
<math> X_(\omega) =  \frac{\pi}j \left( \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) - \sum_{n = -\infty}^{ \infty} \delta \left( \omega+\omega_0 - 2\pi l \right) \right)  </math>
+
−
 
+
−
<math>x[n]=\sin \left( \frac{2\pi}{100} n \right)</math>
+
−
 
+
−
 
+
−
[[2013_Fall_ECE_438_Boutin|Back to ECE438 Fall 2013]]
+

Revision as of 18:07, 12 September 2013


<a href=":Category:Problem solving">Practice Problem</a> on Discrete-time Fourier transform computation

Compute the discrete-time Fourier transform of the following signal:

<img _fckfakelement="true" _fck_mw_math=" x[n]= \sin \left( \frac{2 \pi }{100} n \right) " src="/rhea/images/math/b/f/0/bf0f97ec20b83c8416e3cd5d95395388.png" />

(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!

No need to write your name: we can find out who wrote what by checking the history of the page.

Answer 1

<img _fckfakelement="true" _fck_mw_math="x[n]=\sin \left( \frac{2 \pi}{100} \right)" src="/rhea/images/math/2/2/3/2231dd05d43da978a3499a964d730f15.png" />


<img _fckfakelement="true" _fck_mw_math="x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)" src="/rhea/images/math/8/3/6/83609c9889b65fcd96711aed18f6111c.png" />

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \sum_{n=-\infty}^{+\infty} x[n] e^{-j\omega n}" src="/rhea/images/math/2/b/d/2bd6e9f4f8c6f4d0c8ba6b2119288a9f.png" />

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)" src="/rhea/images/math/7/f/9/7f9ab2bb6f04aadec1f5789b52ed7e47.png" />


<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{\pi}{j} \left( \delta \left({\omega - \frac{2 \pi}{100}}\right) - \delta \left({\omega + \frac{2 \pi}{100}}\right) \right) by DTFT table" src="/rhea/images/math/6/1/8/61814381fa1ee34cad6f25374d24430a.png" />

Answer 2

First, write the original function as: <img _fckfakelement="true" _fck_mw_math="x[n] = \frac{1}{2j} \left( e^{ \frac{j2 \pi}{100n}}-e^{- \frac{j2 \pi}{100n}} \right)" src="/rhea/images/math/8/3/6/83609c9889b65fcd96711aed18f6111c.png" />

Then, for w = [-pi, pi]

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2j} \left( \sum_{n=-\infty}^{+\infty} e^{ \frac{j2 \pi} {100} n} e^{-j\omega n} - \sum_{n=-\infty}^{+\infty} e^{\frac{-j2 \pi} {100} n} e^{-j\omega n} \right)" src="/rhea/images/math/7/f/9/7f9ab2bb6f04aadec1f5789b52ed7e47.png" />

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{100}{2j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)" src="/rhea/images/math/f/0/c/f0c3875a9621aabf1286b15e89b68e9f.png" />

which is really is:

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = rep_2pi \frac{50}{j} \left( \delta \left( \frac{100}{2pi}\omega - 1 \right) + \left( \frac{100}{2pi}\omega + 1 \right) \right)" src="/rhea/images/math/3/7/b/37b3b44a22cf3b61840a7c8bb4486eec.png" />


Answer 3

We can separate the equation to the following function

<img _fckfakelement="true" _fck_mw_math="x[n]=\frac{1}{2 j} \left( e^\frac{j 2 \pi n}{100} - e^\frac{- j 2 \pi n}{100} \right) " src="/rhea/images/math/0/5/b/05ba7655493b0c4e3b694ba6fac0539c.png" />

Because based on Fourier transform equation,

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \sum_{n = -\infty}^{\infty} x[n] e^{-j \omega n}" src="/rhea/images/math/a/7/9/a79fed22e488f9ab5773eadadc46bcb0.png" />

Substitute in x[n]

<img _fckfakelement="true" _fck_mw_math="X_(\omega) = \frac{1}{2 j} \left( \sum_{n = -\infty}^{\infty} e^{ \frac{j2 \pi n} {100} } e^{-j\omega n} - \sum_{n = -\infty}^{ \infty} e^{\frac{-j2 \pi n} {100} } e^{-j\omega n} \right)" src="/rhea/images/math/b/2/8/b283311397523aab2fbcfa322f3b759f.png" />

From Discrete Fourier Transform pair,

<img _fckfakelement="true" _fck_mw_math=" x[n] = e^{-j\omega_0 n} " src="/rhea/images/math/1/1/9/119b4dfb4f6bcf5deb0663acaa69ca03.png" /> DTFT to <img _fckfakelement="true" _fck_mw_math=" X_(\omega) = 2 \pi \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) " src="/rhea/images/math/b/2/5/b25d82705497c4e62e1d068138dae962.png" />

Hence, the function will be

<img _fckfakelement="true" _fck_mw_math=" X_(\omega) = \frac{\pi}j \left( \sum_{n = -\infty}^{ \infty} \delta \left( \omega-\omega_0 - 2\pi l \right) - \sum_{n = -\infty}^{ \infty} \delta \left( \omega+\omega_0 - 2\pi l \right) \right) " src="/rhea/images/math/5/2/c/52cb905ce8f853f34e9fb0f152c902f0.png" />

<img _fckfakelement="true" _fck_mw_math="x[n]=\sin \left( \frac{2\pi}{100} n \right)" src="/rhea/images/math/b/f/0/bf0f97ec20b83c8416e3cd5d95395388.png" />


<a href="2013 Fall ECE 438 Boutin">Back to ECE438 Fall 2013</a>

<a _fcknotitle="true" href="Category:ECE">ECE</a> <a _fcknotitle="true" href="Category:ECE438">ECE438</a> <a _fcknotitle="true" href="Category:2013_Fall_ECE_438_Boutin">2013_Fall_ECE_438_Boutin</a> <a _fcknotitle="true" href="Category:Problem_solving">Problem_solving</a> <a _fcknotitle="true" href="Category:Discrete_time_Fourier_transform">Discrete_time_Fourier_transform</a>
Retrieved from "https://www.projectrhea.org/rhea/index.php?title=Practice_DTFT_computation_sine_ECE438F13&oldid=58138"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva