(started h(s)) |
(finished h(s)) |
||
| Line 9: | Line 9: | ||
y(<math>\delta(t)</math>) = 3(<math>\delta(t)</math>) | y(<math>\delta(t)</math>) = 3(<math>\delta(t)</math>) | ||
| − | =>impulse response = <math> | + | =>impulse response = <math>3\delta(t)</math> |
| Line 15: | Line 15: | ||
Find H(s): | Find H(s): | ||
| − | H( | + | 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> | ||
| + | |||
| + | |||
| + | H(s) = <math> 3e^{-s}\int_{-\infty}^{\infty}\delta(\tau)e^{\tau}d\tau</math> | ||
| + | |||
| + | |||
| + | By the Sifting property, this is: | ||
| + | |||
| + | H(s) = <math>3e^{-s}e^0</math> | ||
| + | |||
| + | thus, | ||
| + | |||
| + | H(s) = <math>3e^{-s}</math> | ||
==Example Response== | ==Example Response== | ||
| − | =Input= | + | ===Input=== |
| − | =Response= | + | ===Response=== |
Revision as of 15:10, 25 September 2008
Homework 4 Ben Horst: 4.1 :: 4.3 :: 4.4
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} $
