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| + | [[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 signal == | == CT signal == | ||
| − | <math>x(t) = | + | |
| + | <math>x(t) = cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\,</math> | ||
== Coefficients == | == Coefficients == | ||
| Line 8: | Line 18: | ||
<math>4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math><br> | <math>4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math><br> | ||
| − | <math>x(t) = | + | <math>x(t) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math> |
<br> | <br> | ||
| − | <math>x(t) = | + | <math>x(t) = \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{5j2\pi t}{6}} + 2je^{\frac{-5j2\pi t}{6}}</math> |
| + | <br> | ||
| + | Then we can know the fundamental frequency is <math>\frac{\pi}{3}</math>. <br><br> | ||
| + | Also, we can get coefficients <math>a_2</math>,<math>a_{-2}</math>,<math>a_5</math>, | ||
| + | <math>a_{-5}</math>.<br><br> | ||
| + | <math>a_2 = a_{-2} = \frac{1}{2}, a_5 = -2j, a_{-5} = 2j, a_k = 0,</math>where k is not 2,-2,5,-5 | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:04, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT signal
$ x(t) = cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\, $
Coefficients
$ cos({\frac{2\pi t}{3}}) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} $
$ 4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $
$ x(t) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $
$ x(t) = \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{5j2\pi t}{6}} + 2je^{\frac{-5j2\pi t}{6}} $
Then we can know the fundamental frequency is $ \frac{\pi}{3} $.
Also, we can get coefficients $ a_2 $,$ a_{-2} $,$ a_5 $,
$ a_{-5} $.
$ a_2 = a_{-2} = \frac{1}{2}, a_5 = -2j, a_{-5} = 2j, a_k = 0, $where k is not 2,-2,5,-5
