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can be used to solve part b | can be used to solve part b | ||
| − | :: <math>h(t) = \mathcal{F}^{-1}\{ H(j\omega) \} = \mathcal{F}^{-1}\{ \frac{1}{4 + j\omega} \}</math> | + | :: <math>h(t) = \mathcal{F}^{-1}\{ H(j\omega) \} = \mathcal{F}^{-1}\Big\{ \frac{1}{4 + j\omega} \Big\}</math> |
:: <math> = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{j\omega t}}{4+j\omega}\,d\omega </math> | :: <math> = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{j\omega t}}{4+j\omega}\,d\omega </math> | ||
Revision as of 14:19, 8 October 2008
Lecture 17 PDF requires Adobe Reader 7 or greater
Contents
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
The fundamental relationship from the main concepts
can be used to solve part b
- $ h(t) = \mathcal{F}^{-1}\{ H(j\omega) \} = \mathcal{F}^{-1}\Big\{ \frac{1}{4 + j\omega} \Big\} $
- $ = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{j\omega t}}{4+j\omega}\,d\omega $
I don't know how to evaluate this integral, I'm not sure if it can be. Any thoughts?
Using the table
Because the function $ H\big(j\omega) $ has the form $ \frac{1}{a + j\omega}\,,a = 4 $ the inverse Fourier transform is of the type $ e^{-at}u\big(t) $
This results in the final answer of
- $ h(t) = e^{-4t}u\big(t) $
