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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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==The signal==
 
==The signal==
  
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<math>a_0=\frac{2}{\pi}\int_0^{2\pi}cos(2t)dt+\frac{3j}{2\pi}\int_0^{2\pi}sin(3t)dt</math>
 
<math>a_0=\frac{2}{\pi}\int_0^{2\pi}cos(2t)dt+\frac{3j}{2\pi}\int_0^{2\pi}sin(3t)dt</math>
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<math>a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi}-\frac{j}{2\pi}[cos(3t)]_0^{2\pi}</math>
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<math>a_0=\frac{1}{\pi}[sin(4\pi)-sin(0)]+\frac{j}{2\pi}[(cos(6\pi)-cos(0)]</math>
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<math>a_0=\frac{1}{\pi}[0]+\frac{j}{2\pi}[0]</math>
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<math>a_0=0\!</math>
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The exact same method can be used to compute any <math>a_k\!</math>. However, that would take quite a hefty sum of time, and I, unlike most double E students, have a life. Therefore, I will use complex exponential identities to do the rest of them. Here we go, home slice:
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<math>x(t) = 4cos(2t) + (3j)sin(3t)\!</math> turns into the following:
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<math>x(t) = \frac{4}{2}(e^{j2t} + e^{-j2t}) + \frac{3j}{2j}(e^{j3t} - e^{-j3t})</math>
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<math>x(t) = 2(e^{j2t} + e^{-j2t}) + \frac{3}{2}(e^{j3t} - e^{-j3t})</math>
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<math>x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t}</math>
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<math>a_2 = 2, a_{-2} = 2, a_3 = \frac{3}{2}, a_{-3} = -\frac{3}{2}\!</math><br><br>
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----
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=Comments=
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*Hey Virgil it's Joe.. I am not sure if its period is <math> 2\pi </math>. I would appreciate your explanation. (Jungu -Joe- Choi)
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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 09:55, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


The signal

The signal I chose to use is as follows:


$ x(t) = 4cos(2t) + (3j)sin(3t)\! $


The fundamental period, denoted as $ T\! $, of this signal is $ 2\pi\! $. The fundamental frequency, denoted $ \omega_0\! $, is defined as:


$ \omega_0 = \frac{T}{2\pi}\! $


The value of this is $ \frac{2\pi}{2\pi}\! $, which coincidently, by no planning of mine, turns out to be $ 1\! $.


Solution

We know that the equation for signal coefficients is as follows:


$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.


And the equation for fourier series of a function is as follows:


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


We first put our signal into the first equation, and we get this monster:


$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[4cos(2t) + (3j)sin(3t)]e^{0}dt $


We now solve.


$ a_0=\frac{2}{\pi}\int_0^{2\pi}cos(2t)dt+\frac{3j}{2\pi}\int_0^{2\pi}sin(3t)dt $


$ a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi}-\frac{j}{2\pi}[cos(3t)]_0^{2\pi} $


$ a_0=\frac{1}{\pi}[sin(4\pi)-sin(0)]+\frac{j}{2\pi}[(cos(6\pi)-cos(0)] $


$ a_0=\frac{1}{\pi}[0]+\frac{j}{2\pi}[0] $


$ a_0=0\! $


The exact same method can be used to compute any $ a_k\! $. However, that would take quite a hefty sum of time, and I, unlike most double E students, have a life. Therefore, I will use complex exponential identities to do the rest of them. Here we go, home slice:


$ x(t) = 4cos(2t) + (3j)sin(3t)\! $ turns into the following:


$ x(t) = \frac{4}{2}(e^{j2t} + e^{-j2t}) + \frac{3j}{2j}(e^{j3t} - e^{-j3t}) $


$ x(t) = 2(e^{j2t} + e^{-j2t}) + \frac{3}{2}(e^{j3t} - e^{-j3t}) $


$ x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t} $


$ a_2 = 2, a_{-2} = 2, a_3 = \frac{3}{2}, a_{-3} = -\frac{3}{2}\! $


Comments

  • Hey Virgil it's Joe.. I am not sure if its period is $ 2\pi $. I would appreciate your explanation. (Jungu -Joe- Choi)

Back to Practice Problems on Signals and Systems

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