(New page: CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t) + \cos(2\omega_0 t+ \frac{\pi}{4})</math> <math>x(t) = 1+\frac {1}{2j} [e^(j\omega_0 t)-e^(-j\omega_0 t)]+\frac{1}{2}{e^[j(2\omega_0 t...)
 
 
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CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t) + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
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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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----
  
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<math>x(t) = 1+\frac {1}{2j} [e^(j\omega_0 t)-e^(-j\omega_0 t)]+\frac{1}{2}{e^[j(2\omega_0 t)+/frac {\pi}{4})]+e^[-j(2\omega_0 t)+/frac {\pi}{4})]}</math>
 
  
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<math>= \frac{e^{3j\pi t}}{2} + \frac{e^{-3j\pi t}}{2} + \frac{e^{4j\pi t}}{2j} - \frac{e^{-4j\pi t}}{2j} \,</math>
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CT Periodic Signal : <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
  
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I take <math>\omega_o \,</math> as <math>\pi \,</math> since both functions have a period based on it.
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<math>x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})})</math>
  
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The following is the coefficient of the signal:
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<math>x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})}</math>
  
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<math>a_3 = \frac{1}{2}\,</math>
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<math>x(t) = 1e^{0j\omega_0 t}+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j\frac {\pi}{4}}e^{2\omega_0 t}+\frac{1}{2}e^{-j\frac {\pi}{4}}e^{2\omega_0 t}</math>
  
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<math>a_{-3} = \frac{1}{2}\,</math>
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Hence we get,
  
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<math>a_{4} = \frac{1}{2j}\,</math>
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<math>a_0 = 1</math>
  
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<math>a_{-4} = -\frac{1}{2j}\,</math>
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<math>a_1 = \frac{1}{2j},</math>
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<math>a_{-1} = -\frac{1}{2j},</math>
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<math>a_2 = \frac{1}{2}e^{j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1+j),</math>
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<math>a_{-2} = \frac{1}{2}e^{-j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1-j),</math>
  
 
We can write the function in the following illiterations:
 
We can write the function in the following illiterations:
  
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<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math> where
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<math>a_0 = 1</math>
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<math>a_1 = \frac{1}{2j},</math>
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<math>a_{-1} = -\frac{1}{2j},</math>
  
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<math>a_3 = a_{-3} = \frac{1}{2}\,</math>
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<math>a_2 = \frac{\sqrt2}{4}(1+j),</math>
  
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<math>a_{4} = \frac{1}{2j} = -a_{-4}\,</math>
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<math>a_{-2} = \frac{\sqrt2}{4}(1-j),</math>
  
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<math>a_k = 0 , k \neq 3,-3,4,-4\,</math>
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<math>a_k = 0 , k \neq 0,1,-1,2,-2\,</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 09:54, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


CT Periodic Signal : $ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $

$ x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})}) $

$ x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})} $

$ x(t) = 1e^{0j\omega_0 t}+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j\frac {\pi}{4}}e^{2\omega_0 t}+\frac{1}{2}e^{-j\frac {\pi}{4}}e^{2\omega_0 t} $

Hence we get,

$ a_0 = 1 $

$ a_1 = \frac{1}{2j}, $

$ a_{-1} = -\frac{1}{2j}, $

$ a_2 = \frac{1}{2}e^{j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1+j), $

$ a_{-2} = \frac{1}{2}e^{-j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1-j), $

We can write the function in the following illiterations:

$ a_0 = 1 $

$ a_1 = \frac{1}{2j}, $

$ a_{-1} = -\frac{1}{2j}, $

$ a_2 = \frac{\sqrt2}{4}(1+j), $

$ a_{-2} = \frac{\sqrt2}{4}(1-j), $

$ a_k = 0 , k \neq 0,1,-1,2,-2\, $

Back to Practice Problems on Signals and Systems

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