Difference between revisions of "HW4.3 Ben Horst ECE301Fall2008mboutin" - Rhea

Latest revision as of 16:34, 25 September 2008

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Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.5

System

y(t) = 3x(t) which is proven as an LTI system ( shown here)

Impulse Response

y($ \delta(t) $) = 3($ \delta(t) $)

=>impulse response = $ 3\delta(t) $

System Function

Find H(s):

H($ j\omega $) = $ \int_{-\infty}^{\infty}h(\tau)e^{-j\omega\tau}d\tau $, where $ j\omega $ is s.

H(s) = $ \int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau $

H(s) = $ \int_{-\infty}^{\infty}3\delta(\tau)e^{-s\tau}d\tau $

By the Sifting property, this is:

H(s) = $ 3e^0 $

thus,

H(s) = $ 3 $

Example Response

Input

From previous part of homework:

$ x(t) = 2\sin(6t) + 4\cos(3t) $

Info

From the previously computed math, we can determine all the coefficients: $ \ \ a_{-2} = 1; \ \ a_{-1} = 2; \ \ a_{0} = 0; \ \ a_1 = 2;\ \ a_{2} = -1 $

The fundamental period of the function is found from: $ e^{j\omega_0} $ where he period T = $ {2\pi \over \omega_o} $

Thus, the fundamental period = $ {2\pi \over 3} $

Response

Given that y(t) = $ \sum_{k=-\infty}^{\infty} a_kH(j\omega_0 k) e^{j\omega_0 k} $ from pg228 in Signals and Systems (Oppenheim & Willsky)

Thus:

y(t) = 1(3)$ e^{-2jt} $ + 2(3)$ e^{-1jt} $ + 0(3)$ e^{0jt} $ + 2(3)$ e^{1jt} $ - 1(3)$ e^{2jt} $

y(t) = 3$ e^{-2jt} $ + 6$ e^{-jt} $ + 6$ e^{jt} $ - 3$ e^{2jt} $