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

Revision as of 18:32, 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: $ x(t) = 3e^{j2\pi t}+3e^{-j2\pi t} + 4e^{j4\pi t}-4e^{-j4\pi t}\, $
With this expression we can conclude:
$ a_1 = 3\, $
$ a_{-1} = 3\, $
$ a_2 = 4\, $
$ a_{-2} = -4\, $

$ x(t) = 3H(s)e^{j2\pi t}+3H(s)e^{-j2\pi t} + 4H(s)e^{j4\pi t}-4H(s)e^{-j4\pi t}\, $

$ x(t) = 3j\omega_0e^{j2\pi t}+3j\omega_0e^{-j2\pi t} + 4j\omega_0e^{j4\pi t}-4j\omega_0e^{-j4\pi t}\, $

$ \omega_0\, $ value as the base frequency is 2

$ x(t) = 6je^{j2\pi t}+6je^{-j2\pi t} + 8je^{j4\pi t}-8je^{-j4\pi t}\, $