Difference between revisions of "HW4.3 Seung Seob Lee ECE301Fall2008mboutin" - Rhea

Revision as of 17:40, 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} $

$ y(t) = -\frac{1}{2}Ke^{-as}e^{j3\pi t}-\frac{1}{2}Ke^{-as}e^{-j3\pi t} $