(New page: ==Define a Periodic CT Signal and Compute its Fourier Series Coefficients== Let's start this process by defining our signal. For simplicities sake lets use the the signal <math> x(t) = 4...)
 
 
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[[Category:problem solving]]
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[[Category:ECE301]]
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[[Category:ECE]]
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[[Category:Fourier series]]
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[[Category:signals and systems]]
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== Example of Computation of Fourier series of a CT SIGNAL ==
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A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]]
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==Define a Periodic CT Signal and Compute its Fourier Series Coefficients==
 
==Define a Periodic CT Signal and Compute its Fourier Series Coefficients==
 
Let's start this process by defining our signal. For simplicities sake lets use the the signal  
 
Let's start this process by defining our signal. For simplicities sake lets use the the signal  
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The Fourier Series can be easily found by treating  
 
The Fourier Series can be easily found by treating  
  
<math> Asin(\omega_0t) = (A/ j2)*(e^{j\omega_0t} - e^{-j\omega_0t})  </math>
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<math> Asin(\omega_0t) = \frac{A*(e^{j\omega_0t} - e^{-j\omega_0t})}{2j} </math>
 
   
 
   
 
and  
 
and  
  
<math>  Acos(\omega_0t) = (A/2)*(e^{j\omega_0t} + e^{-j\omega_0t}) </math>
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<math>  Acos(\omega_0t) = \frac{A*(e^{j\omega_0t} + e^{-j\omega_0t})}{2} </math>
  
 
This alows us to to put x(t) in the form of
 
This alows us to to put x(t) in the form of
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which gives us
 
which gives us
  
<math> x(t) = (4/ j2)*(e^{j3t} - e^{-j3t}) + (8/2)*(e^{j7t} + e^{-j7t}) </math>
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<math> x(t) = \frac{4*(e^{j3t} - e^{-j3t})}{2j} + \frac{8*(e^{j7t} + e^{-j7t})}{2} </math>
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Simplifying and distributing
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<math> x(t) = \frac{2*e^{j3t} }{j}- \frac{2*e^{j3t} }{j} + 4e^{j7t} + 4e^{-j7t} </math>
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<math>\ a_{-3} = \frac{2}{j} </math>
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<math>\ a_{3}= \frac{-2}{j} </math>
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<math>\ a_{-7} = 4 </math>
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<math>\ a{_7} = 4 </math>
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all other <math> \ a_k = 0 </math>
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----
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[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 10:07, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Define a Periodic CT Signal and Compute its Fourier Series Coefficients

Let's start this process by defining our signal. For simplicities sake lets use the the signal

$ x(t) = 4sin(3t) + 8cos(7t) $

The Fourier Series can be easily found by treating

$ Asin(\omega_0t) = \frac{A*(e^{j\omega_0t} - e^{-j\omega_0t})}{2j} $

and

$ Acos(\omega_0t) = \frac{A*(e^{j\omega_0t} + e^{-j\omega_0t})}{2} $

This alows us to to put x(t) in the form of

$ x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t} $

which gives us

$ x(t) = \frac{4*(e^{j3t} - e^{-j3t})}{2j} + \frac{8*(e^{j7t} + e^{-j7t})}{2} $

Simplifying and distributing

$ x(t) = \frac{2*e^{j3t} }{j}- \frac{2*e^{j3t} }{j} + 4e^{j7t} + 4e^{-j7t} $

$ \ a_{-3} = \frac{2}{j} $

$ \ a_{3}= \frac{-2}{j} $

$ \ a_{-7} = 4 $

$ \ a{_7} = 4 $

all other $ \ a_k = 0 $


Back to Practice Problems on Signals and Systems

Alumni Liaison

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

Francisco Blanco-Silva