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| − | [[Homework 4_ECE301Fall2008mboutin|<< Back to Homework | + | [[Homework 4_ECE301Fall2008mboutin|<< Back to Homework 4]] |
| − | Homework 4 Ben Horst: [[HW4. | + | Homework 4 Ben Horst: [[HW4.1 Ben Horst _ECE301Fall2008mboutin| 4.1]] :: [[HW4.3 Ben Horst _ECE301Fall2008mboutin| 4.3]] :: [[HW4.5 Ben Horst _ECE301Fall2008mboutin| 4.5]] |
---- | ---- | ||
| + | ==System== | ||
| + | y(t) = 3x(t) which is proven as an LTI system ([[HW2-C_Ben_Horst _ECE301Fall2008mboutin| shown here]]) | ||
| + | |||
| + | ==Impulse Response== | ||
| + | y(<math>\delta(t)</math>) = 3(<math>\delta(t)</math>) | ||
| + | |||
| + | =>impulse response = <math>3\delta(t)</math> | ||
| + | |||
| + | |||
| + | ==System Function== | ||
| + | Find H(s): | ||
| + | |||
| + | H(<math>j\omega</math>) = <math> \int_{-\infty}^{\infty}h(\tau)e^{-j\omega\tau}d\tau</math>, where <math>j\omega</math> is <em>s</em>. | ||
| + | |||
| + | |||
| + | H(s) = <math> \int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau</math> | ||
| + | |||
| + | |||
| + | H(s) = <math> \int_{-\infty}^{\infty}3\delta(\tau)e^{-s\tau}d\tau</math> | ||
| + | |||
| + | |||
| + | By the Sifting property, this is: | ||
| + | |||
| + | H(s) = <math>3e^0</math> | ||
| + | |||
| + | thus, | ||
| + | |||
| + | H(s) = <math>3</math> | ||
| + | |||
| + | ==Example Response== | ||
| + | ===Input=== | ||
| + | From previous part of homework: | ||
| + | |||
| + | <math>x(t) = 2\sin(6t) + 4\cos(3t)</math> | ||
| + | |||
| + | ===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=== | ||
| + | Given that y(t) = <math>\sum_{k=-\infty}^{\infty} a_kH(j\omega_0 k) e^{j\omega_0 k}</math> from pg228 in Signals and Systems (Oppenheim & Willsky) | ||
| + | |||
| + | |||
| + | Thus: | ||
| + | |||
| + | y(t) = 1(3)<math>e^{-2jt}</math> + 2(3)<math>e^{-1jt}</math> + 0(3)<math>e^{0jt}</math> + 2(3)<math>e^{1jt}</math> - 1(3)<math>e^{2jt}</math> | ||
| + | |||
| + | |||
| + | y(t) = 3<math>e^{-2jt}</math> + 6<math>e^{-jt}</math> + 6<math>e^{jt}</math> - 3<math>e^{2jt}</math> | ||
Latest revision as of 16:34, 25 September 2008
Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.5
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 $
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} $
