(→CT Signal and its Fourier Series Coefficients) |
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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"]] | ||
| + | ---- | ||
| + | |||
Let the signal be | Let the signal be | ||
| − | <math>\ x(t) = \cos( | + | <math>\ x(t) = \cos(2t\pi /3) \sin(2t \pi /9) </math> |
| − | which has a fundamental frequency of | + | |
| + | which has a fundamental frequency of 9 | ||
Now computing its coefficients: | Now computing its coefficients: | ||
| − | <math>\ x(t)= (\frac{e^{ | + | <math>\ x(t)= (\frac{e^{j2t\pi /3} + e^{-j 2t\pi/ 3}}{2}) (\frac{e^{j 2t\pi /9} - e^{-j2t \pi /9}}{2j}) </math> |
| − | <math>\ x(t)= \frac{e^{ | + | <math>\ x(t)= \frac{e^{-j4t\pi /9} - e^{-j8t\pi /9} + e^{j8t\pi /9} - e^{j4t\pi /9}}{4j} </math> |
Now rearranging the terms: | Now rearranging the terms: | ||
| − | <math>\ x(t)= \frac{e^{ | + | <math>\ x(t)= \frac{-e^{j4t\pi /9} + e^{-j4t\pi /9} + e^{j8t\pi /9} - e^{j8t\pi /9}}{4j} </math> |
| + | |||
| + | <math>\ x(t)= \frac{-e^{j(2)2t\pi /9} + e^{-j(2)2t\pi /9} + e^{j(4)2t\pi /9} - e^{j(4)2t\pi /9}}{4j} </math> | ||
| + | |||
| + | Now, the Fourier coefficients can clearly be seen: | ||
| + | |||
| + | <math>\ a_{2}= \frac{-1}{4j} = \frac{j}{4}</math> | ||
| + | |||
| + | <math>a_{-2}= \frac{1}{4j} = \frac{-j}{4}</math> | ||
| + | |||
| + | <math>a_{4}= \frac{1}{4j} = \frac{-j}{4}</math> | ||
| + | |||
| + | <math>a_{-4}= \frac{-1}{4j} = \frac{j}{4}</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:59, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Let the signal be $ \ x(t) = \cos(2t\pi /3) \sin(2t \pi /9) $
which has a fundamental frequency of 9 Now computing its coefficients:
$ \ x(t)= (\frac{e^{j2t\pi /3} + e^{-j 2t\pi/ 3}}{2}) (\frac{e^{j 2t\pi /9} - e^{-j2t \pi /9}}{2j}) $
$ \ x(t)= \frac{e^{-j4t\pi /9} - e^{-j8t\pi /9} + e^{j8t\pi /9} - e^{j4t\pi /9}}{4j} $
Now rearranging the terms:
$ \ x(t)= \frac{-e^{j4t\pi /9} + e^{-j4t\pi /9} + e^{j8t\pi /9} - e^{j8t\pi /9}}{4j} $
$ \ x(t)= \frac{-e^{j(2)2t\pi /9} + e^{-j(2)2t\pi /9} + e^{j(4)2t\pi /9} - e^{j(4)2t\pi /9}}{4j} $
Now, the Fourier coefficients can clearly be seen:
$ \ a_{2}= \frac{-1}{4j} = \frac{j}{4} $
$ a_{-2}= \frac{1}{4j} = \frac{-j}{4} $
$ a_{4}= \frac{1}{4j} = \frac{-j}{4} $
$ a_{-4}= \frac{-1}{4j} = \frac{j}{4} $
