(New page: ==CT signal== The CT signal I will use is: <math>x(t) = 4cos(2t) + (3j)sin(3t)\!</math> <br>The fundamental period is 2*pi <br>we know that Wo=T/(2*pi), so: <br> Wo=2*pi) |
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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== | ||
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<br>we know that Wo=T/(2*pi), so: | <br>we know that Wo=T/(2*pi), so: | ||
<br> Wo=2*pi | <br> Wo=2*pi | ||
| + | ==Solution== | ||
| + | Equation to find signal coeffiecients is: | ||
| + | <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | ||
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| + | |||
| + | The Equation to find a fourier series is: | ||
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| + | |||
| + | <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | ||
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| + | |||
| + | By substituting signal into the first equation we get: | ||
| + | |||
| + | <math>a_0=\frac{1}{2\pi}\int_0^{2\pi}[4cos(2t) + (3j)sin(3t)]e^{0}dt</math> | ||
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| + | |||
| + | Solving this we get: | ||
| + | |||
| + | |||
| + | <math>a_0=\frac{2}{\pi}\int_0^{2\pi}cos(2t)dt+\frac{3j}{2\pi}\int_0^{2\pi}sin(3t)dt</math> | ||
| + | |||
| + | |||
| + | <math>a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi}-\frac{j}{2\pi}[cos(3t)]_0^{2\pi}</math> | ||
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| + | |||
| + | <math>a_0=\frac{1}{\pi}[sin(4\pi)-sin(0)]+\frac{j}{2\pi}[(cos(6\pi)-cos(0)]</math> | ||
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| + | <math>a_0=\frac{1}{\pi}[0]+\frac{j}{2\pi}[0]</math> | ||
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| + | |||
| + | <math>a_0=0\!</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:02, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT signal
The CT signal I will use is:
$ x(t) = 4cos(2t) + (3j)sin(3t)\! $
The fundamental period is 2*pi
we know that Wo=T/(2*pi), so:
Wo=2*pi
Solution
Equation to find signal coeffiecients is: $ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
The Equation to find a fourier series is:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
By substituting signal into the first equation we get:
$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[4cos(2t) + (3j)sin(3t)]e^{0}dt $
Solving this we get:
$ a_0=\frac{2}{\pi}\int_0^{2\pi}cos(2t)dt+\frac{3j}{2\pi}\int_0^{2\pi}sin(3t)dt $
$ a_0=\frac{1}{\pi}[sin(2t)]_0^{2\pi}-\frac{j}{2\pi}[cos(3t)]_0^{2\pi} $
$ a_0=\frac{1}{\pi}[sin(4\pi)-sin(0)]+\frac{j}{2\pi}[(cos(6\pi)-cos(0)] $
$ a_0=\frac{1}{\pi}[0]+\frac{j}{2\pi}[0] $
$ a_0=0\! $
