(→Part B) |
|||
| Line 39: | Line 39: | ||
<math> = -\frac{1}{2}Ke^{-as}e^{j3\pi t}-\frac{1}{2}Ke^{-as}e^{-j3\pi t}</math><br> | <math> = -\frac{1}{2}Ke^{-as}e^{j3\pi t}-\frac{1}{2}Ke^{-as}e^{-j3\pi t}</math><br> | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
Revision as of 18:39, 26 September 2008
Part A
$ y(t) = K x(t-a) $
if $ x(t)=e^{jwt} $ was inputed to the system
$ y(t) = K e^{jw(t-a)} $
$ = K e^{-jwa}e^{jwt} $
eigen function is $ e^{-jwa} $
$ H(jw)=Ke^{-jwa} $
$ h(t)=K\delta (t-a) $
$ H(s)=\int_{-\infty}^{\infty}K\delta (\tau -a)e^{-s\tau}d\tau=Ke^{-as} $
Part B
I REFERRED TO RONY WIJAYA'S ANSWER
Signal defined in Question 1:
$ x(t) = cos(3\pi t+\pi) \! $
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}\, $
From Question 1:
$ = -\frac{1}{2}e^{j3\pi t}-\frac{1}{2}e^{-j3\pi t} $
With this expression we can conclude:
$ a_3 = -\frac{1}{2} $
$ a_{-3} = -\frac{1}{2} $
$ = -\frac{1}{2}Ke^{-as}e^{j3\pi t}-\frac{1}{2}Ke^{-as}e^{-j3\pi t} $
