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| − | + | [[Category:problem solving]] | |
| − | == CT SIGNAL == | + | [[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"]] | ||
| + | ---- | ||
I chose the signal: <math>f(t) = (3+j)cos(2t) + (10+j)sin(7t)\!</math> | I chose the signal: <math>f(t) = (3+j)cos(2t) + (10+j)sin(7t)\!</math> | ||
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We know that: | We know that: | ||
| − | <math>f(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> where <math>a_k=\frac{1}{T}\int_0^Tf(t)e^{-jk\ | + | <math>f(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> where <math>a_k=\frac{1}{T}\int_0^Tf(t)e^{-jk\omega_{0}t}dt</math>. |
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| − | <math>f(t) = | + | <math>f(t) = (3+j)\frac{e^{2jt}}{2} + (3+j)\frac{e^{-2jt}}{2} + (10+j)\frac{e^{7jt}}{2j} - (10+j)\frac{e^{-7jt}}{2j}\!</math> |
<br> | <br> | ||
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<math>a_{-7} = -\frac{10+j}{2j}\!</math> | <math>a_{-7} = -\frac{10+j}{2j}\!</math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:55, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
I chose the signal: $ f(t) = (3+j)cos(2t) + (10+j)sin(7t)\! $
FOURIER SERIES COEFFICIENTS
In order to find the fourier series coefficients, we must first understand the operations associated with taking the fourier transform of a signal. The fundamental period of the signal (above) is 2$ \pi\! $. We know that $ \omega_0\! $ = 2$ \pi / T\! $ (where T is the fundamental period). Therefore, the fundamental frequency is $ 1\! $.
We know that:
$ f(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $ where $ a_k=\frac{1}{T}\int_0^Tf(t)e^{-jk\omega_{0}t}dt $.
$ a_0=\frac{1}{2\pi}\int_0^{2\pi}[(3+j)cos(2t) + (10+j)sin(7t)]e^{0}dt $
$ a_0=\frac{1}{2\pi}[\frac{(3+j)sin(2t)}{2} + \frac{-(10+j)cos(7t)}{7}]_0^{2\pi} $
$ a_0=\frac{1}{2\pi}[\frac{-(10+j)}{7} - \frac{-(10+j)}{7}] $
$ a_0 = 0\! $
Now, we can use the same process to find $ a_k\! $. However, the most efficient way to solve for the coefficients is to use complex identities. First, we must consider the following complex identities.
$ sin(t) = \frac{e^{jt}-e^{-jt}}{2j}\! $
$ cos(t) = \frac{e^{jt}+e^{-jt}}{2}\! $
When we apply these identities (above) to our original function, we obtain the following equation:
$ f(t) = (3+j)cos(2t) + (10+j)sin(7t)\! $
$ f(t) = (3+j)\frac{e^{2jt}+e^{-2jt}}{2} + (10+j)\frac{e^{7jt}-e^{-7jt}}{2j}\! $
Now, we can proceed to multiply everything out as follows:
$ f(t) = (3+j)\frac{e^{2jt}}{2} + (3+j)\frac{e^{-2jt}}{2} + (10+j)\frac{e^{7jt}}{2j} - (10+j)\frac{e^{-7jt}}{2j}\! $
Therefore, we can conclude that the fourier series coefficients are:
$ a_{2} = \frac{3+j}{2}\! $
$ a_{-2} = \frac{3+j}{2}\! $
$ a_{7} = \frac{10+j}{2j}\! $
$ a_{-7} = -\frac{10+j}{2j}\! $
