(→Dealing with Differentials) |
(→b) |
||
Line 51: | Line 51: | ||
====b==== | ====b==== | ||
+ | <math>H(j\omega) = \frac{1}{j\omega + 4}</math> |
Revision as of 13:57, 8 October 2008
Lecture 17 PDF requires Adobe Reader 7 or greater
Contents
[hide]Main Concepts
Fourier Transforms and the frequency response of a system.
- $ Y\big(\omega) = H(j \omega) X(\omega) $
The frequency response has a fundamental relationship to the unit step response through Fourier Transforms as follows
- $ H(j\omega\big) = \mathcal{H}(\omega) = \mathcal{F}\{ h(t) \} $
From this, the unit step response can be found
- $ h(t) = \mathcal{F}^{-1}\{ H(j\omega) \} $
Differential stuff....
Exercises
Tricky Fourier Transform
Compute the Fourier Transform of $ u\big(t-3) $
Dealing with Differentials
Given:
- $ \frac{d y(t) }{dx} + 4y(t) = x(t) $
- a) What is the frequency response of the system?
- b)What is the unit impulse response h(t) of the system?
a
First transform into the frequency domain
- $ \mathcal{F}\lbrace\frac{d y(t) }{dx} + 4y(t) = x(t)\rbrace $
Apply linearity
- $ \mathcal{F}\lbrace\frac{d y(t) }{dx}\rbrace + 4\mathcal{F}\lbrace y(t)\rbrace = \mathcal{F}\lbrace x(t)\rbrace $
Use the differentiation property to reduce the differential term
- $ j\omega\mathcal{F}\lbrace y(t)\rbrace + 4\mathcal{F}\lbrace y(t)\rbrace = \mathcal{F}\lbrace x(t)\rbrace $
Apply some arithmetic
- $ \mathcal{F}\lbrace y(t)\rbrace = \frac{1}{j\omega + 4}\mathcal{F}\lbrace x(t)\rbrace $
From the main concepts the frequency response is the portion in front of $ \mathcal{F}\lbrace x(t)\rbrace $
- $ H(j\omega) = \frac{1}{j\omega + 4} $
b
$ H(j\omega) = \frac{1}{j\omega + 4} $