(New page: == Equations == Fourier series of x(t): <br> <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> Signal Coefficients: <br> <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</...) |
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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"]] | ||
| + | ---- | ||
== Equations == | == Equations == | ||
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<math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math> | <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math> | ||
| − | + | From Phil Cannon | |
| + | |||
| + | ==Input Signal== | ||
| + | |||
| + | <math>x(t)=cos(3*pi*t)cos(6*pi*t)\!</math> | ||
| + | <br> | ||
| + | <br> | ||
| + | <math>x(t)=[1/2*e^{j*2*pi*t}+1/2*e^{-j*2*pi*t}]*[1/2*e^{j*4*pi*t}+1/2*e^{-j*4*pi*t}]</math> | ||
| + | <br> | ||
| + | <math> =1/4*e^{j6pit}+1/4*e^{-j2pit}+1/4*e^{j2pit}+1/4*e^{-j6pit}</math> | ||
| + | <br> | ||
| + | <br> | ||
| + | The fundamental frequency is 2*pi | ||
| + | <br> | ||
| + | a3=1/4 | ||
| + | a-1=1/4 | ||
| + | a1=1/4 | ||
| + | a3=1/4 | ||
| + | all other ak=0 | ||
| + | |||
| + | ---- | ||
| + | [[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"
Equations
Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
Signal Coefficients:
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $
From Phil Cannon
Input Signal
$ x(t)=cos(3*pi*t)cos(6*pi*t)\! $
$ x(t)=[1/2*e^{j*2*pi*t}+1/2*e^{-j*2*pi*t}]*[1/2*e^{j*4*pi*t}+1/2*e^{-j*4*pi*t}] $
$ =1/4*e^{j6pit}+1/4*e^{-j2pit}+1/4*e^{j2pit}+1/4*e^{-j6pit} $
The fundamental frequency is 2*pi
a3=1/4
a-1=1/4
a1=1/4
a3=1/4
all other ak=0
