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<math>a_{-2} = \frac{\sqrt2}{4}(1-j),</math> | <math>a_{-2} = \frac{\sqrt2}{4}(1-j),</math> | ||
| − | <math>a_k = 0 , k \neq | + | <math>a_k = 0 , k \neq 0,1,-1,2,-2\,</math> |
Revision as of 04:07, 25 September 2008
CT Periodic Signal : $ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $
$ x(t) = 1+\frac {1}{2j} (e^{j\omega_0 t}-e^{-j\omega_0 t})+\frac{1}{2}(e^{j(2\omega_0 t+\frac {\pi}{4})}+e^{-j(2\omega_0 t+\frac {\pi}{4})}) $
$ x(t) = 1+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j(2\omega_0 t+\frac {\pi}{4})}+\frac {1}{2j}e^{-j(2\omega_0 t+\frac {\pi}{4})} $
$ x(t) = 1e^{0j\omega_0 t}+\frac {1}{2j} e^{j\omega_0 t}-\frac {1}{2j}e^{-j\omega_0 t}+\frac{1}{2}e^{j\frac {\pi}{4}}e^{2\omega_0 t}+\frac{1}{2}e^{-j\frac {\pi}{4}}e^{2\omega_0 t} $
Hence we get,
$ a_0 = 1 $
$ a_1 = \frac{1}{2j}, $
$ a_{-1} = -\frac{1}{2j}, $
$ a_2 = \frac{1}{2}e^{j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1+j), $
$ a_{-2} = \frac{1}{2}e^{-j\frac{\pi}{4}}=\frac{\sqrt2}{4}(1-j), $
We can write the function in the following illiterations:
$ a_0 = 1 $
$ a_1 = \frac{1}{2j}, $
$ a_{-1} = -\frac{1}{2j}, $
$ a_2 = \frac{\sqrt2}{4}(1+j), $
$ a_{-2} = \frac{\sqrt2}{4}(1-j), $
$ a_k = 0 , k \neq 0,1,-1,2,-2\, $
