Difference between revisions of "HW4.1 Xujun Huang ECE301Fall2008mboutin" - Rhea

Revision as of 04:06, 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 3,-3,4,-4\, $