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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"]] | ||
| + | ---- | ||
| + | |||
==The signal== | ==The signal== | ||
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| − | <math>x(t) = 4cos( | + | <math>x(t) = 4cos(2t) + (3j)sin(3t)\!</math> |
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The value of this is <math>\frac{2\pi}{2\pi}\!</math>, which coincidently, by no planning of mine, turns out to be <math>1\!</math>. | The value of this is <math>\frac{2\pi}{2\pi}\!</math>, which coincidently, by no planning of mine, turns out to be <math>1\!</math>. | ||
| + | |||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | |||
| + | We know that the equation for signal coefficients is as follows: | ||
| + | |||
| + | |||
| + | <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | ||
| + | |||
| + | |||
| + | And the equation for fourier series of a function is as follows: | ||
| + | |||
| + | |||
| + | <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | ||
| + | |||
| + | |||
| + | We first put our signal into the first equation, and we get this monster: | ||
| + | |||
| + | |||
| + | <math>a_0=\frac{1}{2\pi}\int_0^{2\pi}[4cos(2t) + (3j)sin(3t)]e^{0}dt</math> | ||
| + | |||
| + | |||
| + | We now solve. | ||
| + | |||
| + | |||
| + | <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> | ||
| + | |||
| + | |||
| + | <math>a_0=\frac{1}{\pi}[sin(4\pi)-sin(0)]+\frac{j}{2\pi}[(cos(6\pi)-cos(0)]</math> | ||
| + | |||
| + | |||
| + | <math>a_0=\frac{1}{\pi}[0]+\frac{j}{2\pi}[0]</math> | ||
| + | |||
| + | |||
| + | <math>a_0=0\!</math> | ||
| + | |||
| + | |||
| + | The exact same method can be used to compute any <math>a_k\!</math>. However, that would take quite a hefty sum of time, and I, unlike most double E students, have a life. Therefore, I will use complex exponential identities to do the rest of them. Here we go, home slice: | ||
| + | |||
| + | |||
| + | <math>x(t) = 4cos(2t) + (3j)sin(3t)\!</math> turns into the following: | ||
| + | |||
| + | |||
| + | <math>x(t) = \frac{4}{2}(e^{j2t} + e^{-j2t}) + \frac{3j}{2j}(e^{j3t} - e^{-j3t})</math> | ||
| + | |||
| + | |||
| + | <math>x(t) = 2(e^{j2t} + e^{-j2t}) + \frac{3}{2}(e^{j3t} - e^{-j3t})</math> | ||
| + | |||
| + | |||
| + | <math>x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t}</math> | ||
| + | |||
| + | |||
| + | <math>a_2 = 2, a_{-2} = 2, a_3 = \frac{3}{2}, a_{-3} = -\frac{3}{2}\!</math><br><br> | ||
| + | ---- | ||
| + | =Comments= | ||
| + | *Hey Virgil it's Joe.. I am not sure if its period is <math> 2\pi </math>. I would appreciate your explanation. (Jungu -Joe- Choi) | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:55, 16 September 2013
Contents
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
The signal
The signal I chose to use is as follows:
$ x(t) = 4cos(2t) + (3j)sin(3t)\! $
The fundamental period, denoted as $ T\! $, of this signal is $ 2\pi\! $. The fundamental frequency, denoted $ \omega_0\! $, is defined as:
$ \omega_0 = \frac{T}{2\pi}\! $
The value of this is $ \frac{2\pi}{2\pi}\! $, which coincidently, by no planning of mine, turns out to be $ 1\! $.
Solution
We know that the equation for signal coefficients is as follows:
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
And the equation for fourier series of a function is as follows:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
We first put our signal into the first equation, and we get this monster:
$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[4cos(2t) + (3j)sin(3t)]e^{0}dt $
We now solve.
$ 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\! $
The exact same method can be used to compute any $ a_k\! $. However, that would take quite a hefty sum of time, and I, unlike most double E students, have a life. Therefore, I will use complex exponential identities to do the rest of them. Here we go, home slice:
$ x(t) = 4cos(2t) + (3j)sin(3t)\! $ turns into the following:
$ x(t) = \frac{4}{2}(e^{j2t} + e^{-j2t}) + \frac{3j}{2j}(e^{j3t} - e^{-j3t}) $
$ x(t) = 2(e^{j2t} + e^{-j2t}) + \frac{3}{2}(e^{j3t} - e^{-j3t}) $
$ x(t) = 2e^{j2t} + 2e^{-j2t} + \frac{3}{2}e^{j3t} - \frac{3}{2}e^{-j3t} $
$ a_2 = 2, a_{-2} = 2, a_3 = \frac{3}{2}, a_{-3} = -\frac{3}{2}\! $
Comments
- Hey Virgil it's Joe.. I am not sure if its period is $ 2\pi $. I would appreciate your explanation. (Jungu -Joe- Choi)
