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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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----
 
== Periodic CT Signal and Its Fourier Coefficients ==
 
== Periodic CT Signal and Its Fourier Coefficients ==
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A Fourier Series of a periodic CT signal is such that:<br>
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<math> x(t) = \sum_{n=-\infty}^\infty a_k * e^{j*k*w_0*t}</math>
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where
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<math> a_k = \frac{1}{T} \int_{0}^{T} x(t)* e</math><sup>(<math>-j*k*w_0*t</math>)</sup><math> \,\ dt</math>
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If our signal <math> x(t) </math> consists of only sine and cosine waves, we don't have to do all those complicated integrals in order to find the Fourier coefficients <math> a_k </math>.
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Take the signal <math> x(t) = 5cos(2t) - 4sin(5t) </math>.  The graph below proves that it is indeed periodic, with a period <math> T = 2\pi </math>.
  
 
[[Image:ECE301HW4p1_ECE301Fall2008mboutin.jpg]]
 
[[Image:ECE301HW4p1_ECE301Fall2008mboutin.jpg]]
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<math> \,\ sin(x) = \frac{1}{2j} * (e</math><sup>(jx)</sup> <math> \,\ - e</math><sup>(-jx)</sup><math> \,\ )</math>
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and
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<math> \,\ cos(x) = \frac{1}{2} * (e</math><sup>(jx)</sup> <math> \,\ + e</math><sup>(-jx)</sup><math> \,\ )</math>
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Therefore,
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<math> x(t) = 5 * \frac{1}{2} * (e</math><sup>(j2t)</sup> <math> \,\ + e</math><sup>(-j2t)</sup><math> \,\ ) - 4 * \frac{1}{2j} * (e</math><sup>(j5t)</sup> <math> \,\ - e</math><sup>(-j5t)</sup><math> \,\ )</math>
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<math> \,\ x(t) = \frac{5}{2} * e</math><sup>(j2t)</sup> <math> \,\ + \frac{5}{2} * e</math><sup>(-j2t)</sup><math> - \frac{4}{2j} * e</math><sup>(j5t)</sup> <math>+ \frac{4}{2j} * e</math><sup>(-j5t)</sup>
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The period <math> \,\ T = 2\pi </math> so if <math> \,\ w_0 = \frac{2\pi}{T} </math>, then <math> \,\ w_0 = 1 </math>.
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Our new equation can now be rewritten as:<br>
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<math> \,\ x(t) = \frac{5}{2} * e</math><sup>(<math>j*w_0*k_1*t</math>)</sup> <math> \,\ + \frac{5}{2} * e</math><sup>(<math>j*w_0*k_2*t</math>)</sup><math> - \frac{4}{2j} * e</math><sup>(<math>j*w_0*k_3*t</math>)</sup> <math>+ \frac{4}{2j} * e</math><sup>(<math>j*w_0*k_4*t</math>)</sup>
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<br><br>
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where <math>k_1,k_2,k_3,</math> and <math> k_4</math> correspond to { <math> \frac{5}{2},    \frac{5}{2} ,  -2 , \frac{2}{j}</math> }, respectively.
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Therefore, we can take the coefficients of the equation and the <math> k </math> values to determine the Fourier coefficients.
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<math>\mathbf{a_2} = \mathbf{\frac{5}{2}}</math><br><br>
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<math>\mathbf{a}</math><sub>-2</sub><math> = \mathbf{\frac{5}{2j}}</math><br><Br>
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<math>\mathbf{a_5} = \mathbf{-2}</math><br><br>
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<math>\mathbf{a}</math><sub>-5</sub><math> = \mathbf{\frac{2}{j}}</math><br><br>
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<math> a_x = 0 </math> elsewhere.
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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 09:58, 16 September 2013

Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


Periodic CT Signal and Its Fourier Coefficients

A Fourier Series of a periodic CT signal is such that:
$ x(t) = \sum_{n=-\infty}^\infty a_k * e^{j*k*w_0*t} $

where

$ a_k = \frac{1}{T} \int_{0}^{T} x(t)* e $($ -j*k*w_0*t $)$ \,\ dt $

If our signal $ x(t) $ consists of only sine and cosine waves, we don't have to do all those complicated integrals in order to find the Fourier coefficients $ a_k $.


Take the signal $ x(t) = 5cos(2t) - 4sin(5t) $. The graph below proves that it is indeed periodic, with a period $ T = 2\pi $.

ECE301HW4p1 ECE301Fall2008mboutin.jpg


$ \,\ sin(x) = \frac{1}{2j} * (e $(jx) $ \,\ - e $(-jx)$ \,\ ) $

and

$ \,\ cos(x) = \frac{1}{2} * (e $(jx) $ \,\ + e $(-jx)$ \,\ ) $

Therefore,

$ x(t) = 5 * \frac{1}{2} * (e $(j2t) $ \,\ + e $(-j2t)$ \,\ ) - 4 * \frac{1}{2j} * (e $(j5t) $ \,\ - e $(-j5t)$ \,\ ) $


$ \,\ x(t) = \frac{5}{2} * e $(j2t) $ \,\ + \frac{5}{2} * e $(-j2t)$ - \frac{4}{2j} * e $(j5t) $ + \frac{4}{2j} * e $(-j5t)


The period $ \,\ T = 2\pi $ so if $ \,\ w_0 = \frac{2\pi}{T} $, then $ \,\ w_0 = 1 $.

Our new equation can now be rewritten as:


$ \,\ x(t) = \frac{5}{2} * e $($ j*w_0*k_1*t $) $ \,\ + \frac{5}{2} * e $($ j*w_0*k_2*t $)$ - \frac{4}{2j} * e $($ j*w_0*k_3*t $) $ + \frac{4}{2j} * e $($ j*w_0*k_4*t $)

where $ k_1,k_2,k_3, $ and $ k_4 $ correspond to { $ \frac{5}{2}, \frac{5}{2} , -2 , \frac{2}{j} $ }, respectively.


Therefore, we can take the coefficients of the equation and the $ k $ values to determine the Fourier coefficients.

$ \mathbf{a_2} = \mathbf{\frac{5}{2}} $

$ \mathbf{a} $-2$ = \mathbf{\frac{5}{2j}} $

$ \mathbf{a_5} = \mathbf{-2} $

$ \mathbf{a} $-5$ = \mathbf{\frac{2}{j}} $

$ a_x = 0 $ elsewhere.


Back to Practice Problems on Signals and Systems

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