(New page: ==Finding the Series== It will be helpful -- necessary even -- to find the fundamental period of the signal. In our case, the period of the overall signal is <math>2\pi</math>, so <math>\...) |
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| + | ==The Signal== | ||
| + | Consider the signal which is given by the following values and limits: | ||
| + | |||
| + | { -1 for (0 < t < 4) | ||
| + | { 1 for (4 < t < 8) | ||
| + | |||
| + | ==The Formulae== | ||
| + | Recall the Fourier Series formulae for the continuous time signal case: | ||
| + | |||
| + | <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | ||
| + | |||
| + | and | ||
| + | |||
| + | <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | ||
| + | |||
==Finding the Series== | ==Finding the Series== | ||
| − | + | First the fundamental period has to be found. The period of the overall signal is <math>8</math>, so <math>\omega_0</math> will be <math>\frac{2\pi}{8}=\pi/4</math>. | |
| − | + | To start, calculate the average of the function. Observe that the signal is -1 for 4 seconds and 1 for 4 seconds. Thus, the average, <math>a_0</math>, is 0. | |
| − | <math>a_0= | + | <math>a_0=0</math> |
Latest revision as of 13:32, 26 September 2008
The Signal
Consider the signal which is given by the following values and limits:
{ -1 for (0 < t < 4) { 1 for (4 < t < 8)
The Formulae
Recall the Fourier Series formulae for the continuous time signal case:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
and
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Finding the Series
First the fundamental period has to be found. The period of the overall signal is $ 8 $, so $ \omega_0 $ will be $ \frac{2\pi}{8}=\pi/4 $.
To start, calculate the average of the function. Observe that the signal is -1 for 4 seconds and 1 for 4 seconds. Thus, the average, $ a_0 $, is 0.
$ a_0=0 $
$ =\frac{7}{2\pi}\int_0^{2\pi}\sin(2t)dt + \frac{1+j}{2\pi}\int_0^{2\pi}\cos(3t)dt $
$ =\frac{-7}{4\pi}\cos(2t)|_0^{2\pi}+\frac{1+j}{6\pi}\sin(3t)|_0^{2\pi} $
$ =\frac{-7}{4\pi}(\cos(4\pi)-\cos(0))+\frac{1+j}{6\pi}(\sin(6\pi)-\sin(0))=0 $
After this point, integrating becomes quite tedious, so I'll revert to using complex exponential identities to continue the solution. Our signal then becomes
$ x(t)=\frac{7}{2j}(e^{2j}-e^{-2j})+\frac{1+j}{2}(e^{3j}+e^{-3j}) $
$ =\frac{7}{2j}(e^2e^j-e^{-2}e^j)+\frac{1+j}{2}(e^3e^j+e^{-3}e^j) $
Now the Fourier coefficients should be fairly obvious.
$ a_{-3}=a_3=\frac{1+j}{2} $
$ a_{-2}=\frac{-7}{2j} $
$ a_2=\frac{7}{2j} $
All other $ a_k=0 $.
