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<math>x(t) = 2\sin(6t) + 4\cos(3t)</math> | <math>x(t) = 2\sin(6t) + 4\cos(3t)</math> | ||
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| + | ===Info=== | ||
| + | From the [[HW4.1 Ben Horst _ECE301Fall2008mboutin| previously computed]] math, we can determine all the coefficients: | ||
| + | <math> \ \ a_{-2} = 1; \ \ a_{-1} = 2; \ \ a_{0} = 0; \ \ a_1 = 2;\ \ a_{2} = -1 </math> | ||
| + | |||
| + | The fundamental period of the function is found from: <math>e^{j\omega_0}</math> where he period T = <math>{2\pi \over \omega_o}</math> | ||
| + | |||
| + | Thus, the fundamental period = <math> {2\pi \over 3} </math> | ||
| + | |||
===Response=== | ===Response=== | ||
Revision as of 15:26, 25 September 2008
Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.4
Contents
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 $
H(s) = $ 3e^{-s}\int_{-\infty}^{\infty}\delta(\tau)e^{\tau}d\tau $
By the Sifting property, this is:
H(s) = $ 3e^{-s}e^0 $
thus,
H(s) = $ 3e^{-s} $
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} $
