(CT Fourier Series)
 
(3 intermediate revisions by one other user not shown)
Line 1: Line 1:
 +
[[Category:problem solving]]
 +
[[Category:ECE301]]
 +
[[Category:ECE]]
 +
[[Category:Fourier series]]
 +
[[Category:signals and systems]]
 +
 +
== Example of Computation of Fourier series of a CT SIGNAL ==
 +
A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]]
 +
----
 
== CT Fourier Series ==
 
== CT Fourier Series ==
  
Line 8: Line 17:
  
 
<font size ="3">The fundamental frequency is <math>8\pi</math>.</font>
 
<font size ="3">The fundamental frequency is <math>8\pi</math>.</font>
 +
 +
<math>x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2})</math>
 +
 +
<math>x(t) = 1 + (1 + \frac{1}{2j})e^{j8\pi t} + (1 - \frac{1}{2j})e^{-j8\pi t} + (\frac{1}{2} e^{j(\frac{\pi}{4})}) e^{j16\pi t} + (\frac{1}{2} e^{-j(\frac{\pi}{4})}) e^{-j16\pi t}</math>
 +
 +
<font size ="3">The Fourier series coefficients are:</font>
 +
 +
<font size ="4">'''''<math>a_0 = 1</math>'''''</font>
 +
 +
<math>a_8 = 1 - \frac{1}{2}j</math>
 +
 +
<math>a_{-8} = 1 + \frac{1}{2}j</math>
 +
 +
<math>a_{16} = \frac{\sqrt{2}}{4}(1 + j)</math>
 +
 +
<math>a_{-16} = \frac{\sqrt{2}}{4}(1 + j)</math>
 +
 +
----
 +
[[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]]

Latest revision as of 11:03, 16 September 2013


Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"


CT Fourier Series

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_o t} $

Example

$ x(t) = 1 + sin(8\pi t) + 2cos(8\pi t) + cos(16\pi t + \frac{\pi}{4}) $

The fundamental frequency is $ 8\pi $.

$ x(t) = 1 + (\frac{e^{j8\pi t} - e^{-j8\pi t}}{2j}) + (e^{j8 \pi t} + e^{-j8 \pi t}) + (\frac{e^{j(16 \pi t + \frac{\pi}{4})} + e^{-j(16 \pi t + \frac{\pi}{4})}} {2}) $

$ x(t) = 1 + (1 + \frac{1}{2j})e^{j8\pi t} + (1 - \frac{1}{2j})e^{-j8\pi t} + (\frac{1}{2} e^{j(\frac{\pi}{4})}) e^{j16\pi t} + (\frac{1}{2} e^{-j(\frac{\pi}{4})}) e^{-j16\pi t} $

The Fourier series coefficients are:

$ a_0 = 1 $

$ a_8 = 1 - \frac{1}{2}j $

$ a_{-8} = 1 + \frac{1}{2}j $

$ a_{16} = \frac{\sqrt{2}}{4}(1 + j) $

$ a_{-16} = \frac{\sqrt{2}}{4}(1 + j) $


Back to Practice Problems on Signals and Systems

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett