(New page: Let us take the periodic, CT signal: <math>3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math> ---- As we know, the Fourier Series for a CT signal is written as: <math>x(t) = \sum^{\infty}_{k = ...) |
|||
| Line 15: | Line 15: | ||
<math>x(t) = \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}</math> | <math>x(t) = \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t}</math> | ||
| + | |||
| + | Solving for a: | ||
| + | |||
| + | <math> a_1 = \frac{3}{2} </math> | ||
| + | |||
| + | <math> a_{-1} = \frac{3}{2} </math> | ||
| + | |||
| + | <math> a_2 = a_{-2} = 1 </math> | ||
| + | |||
| + | <math> a_k = 0</math> else | ||
Revision as of 09:18, 26 September 2008
Let us take the periodic, CT signal: $ 3cos(4\pi t) + e^{j\frac{2\pi}{5}t} $
As we know, the Fourier Series for a CT signal is written as:
$ x(t) = \sum^{\infty}_{k = -\infty}a_k e^{j k w_o t} $
Where $ a_k $ is: $ a_k = \frac{1}{T} \int^{T}_{0} x(t) e^{-j k w_o t} $
Our signal, x(t), can also be written as:
$ x(t) = \frac{3}{2}(e^{j 4 \pi t} + e^{-j 4 \pi t}) + e^{j \frac{2 \pi}{5} t} $
Solving for a:
$ a_1 = \frac{3}{2} $
$ a_{-1} = \frac{3}{2} $
$ a_2 = a_{-2} = 1 $
$ a_k = 0 $ else
