(→Periodic CT Signal and Its Fourier Coefficients) |
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<math> x(t) = \sum_{n=-\infty}^\infty a_k * e</math><sup>(<math>j*k*w_0*t</math>)</sup> | <math> x(t) = \sum_{n=-\infty}^\infty a_k * e</math><sup>(<math>j*k*w_0*t</math>)</sup> | ||
| + | where | ||
| − | Take the signal <math> x(t) = 5cos(2t) - 4sin(5t) </math>. The graph below proves that it is indeed periodic, with a period <math> T = \pi </math>. | + | <math> a_k = \frac{1}{T} \int_{0}^{T} x(t)* e</math><sup>(<math>-j*k*w_0*t</math>)</sup><math> \,\ dt</math> |
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| + | Take the signal <math> x(t) = 5cos(2t) - 4sin(5t) </math>. The graph below proves that it is indeed periodic, with a period <math> T = 2\pi </math>. | ||
[[Image:ECE301HW4p1_ECE301Fall2008mboutin.jpg]] | [[Image:ECE301HW4p1_ECE301Fall2008mboutin.jpg]] | ||
Revision as of 15:03, 25 September 2008
Periodic CT Signal and Its Fourier Coefficients
A Fourier Series of a periodic CT signal is such that:
$ x(t) = \sum_{n=-\infty}^\infty a_k * e $($ j*k*w_0*t $)
where
$ a_k = \frac{1}{T} \int_{0}^{T} x(t)* e $($ -j*k*w_0*t $)$ \,\ dt $
Take the signal $ x(t) = 5cos(2t) - 4sin(5t) $. The graph below proves that it is indeed periodic, with a period $ T = 2\pi $.
$ \,\ sin(x) = \frac{1}{2j} * (e $(jx) $ \,\ - e $(-jx)$ \,\ ) $
and
$ \,\ cos(x) = \frac{1}{2} * (e $(jx) $ \,\ + e $(-jx)$ \,\ ) $
Therefore,
$ x(t) = 5 * \frac{1}{2} * (e $(j2t) $ \,\ + e $(-j2t)$ \,\ ) - 4 * \frac{1}{5j} * (e $(j5t) $ \,\ - e $(-j5t)$ \,\ ) $
$ \,\ x(t) = \frac{5}{2} * e $(j2t) $ \,\ + \frac{5}{2} * e $(-j2t)$ - \frac{4}{5} * e $(j5t) $ + \frac{4}{5} * e $(-j5t)
The period $ \,\ T = 2\pi $ so if $ \,\ w_0 = \frac{2\pi}{T} $, then $ \,\ w_0 = 1 $.
