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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"]] | ||
+ | ---- | ||
+ | |||
== A periodic CT signal == | == A periodic CT signal == | ||
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− | Input CT signal: <math> x(t) = | + | Input CT signal: <math> x(t) = cos4t+sin2t</math> |
+ | |||
+ | <math>\,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j}</math> | ||
+ | |||
+ | |||
+ | <math>x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t}</math> | ||
+ | |||
+ | <math>a_4=\frac{1}{2}</math> | ||
+ | |||
+ | <math>a_{-4}=\frac{1}{2}</math> | ||
− | <math> | + | <math>a_2=\frac{1}{2j}</math> |
+ | <math>a_{-2}=\frac{-1}{2j}</math> | ||
− | <math> | + | otherwise <math>\,a_k</math> values are zero. |
+ | ---- | ||
+ | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] |
Latest revision as of 10:08, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
A periodic CT signal
Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
, where $ a_k $ is
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Input CT signal: $ x(t) = cos4t+sin2t $
$ \,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j} $
$ x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t} $
$ a_4=\frac{1}{2} $
$ a_{-4}=\frac{1}{2} $
$ a_2=\frac{1}{2j} $
$ a_{-2}=\frac{-1}{2j} $
otherwise $ \,a_k $ values are zero.