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:<math> Y\big(\omega) = H(j \omega) X(\omega) </math> | :<math> Y\big(\omega) = H(j \omega) X(\omega) </math> | ||
| − | + | The frequency response has a fundamental relationship to the unit step response through Fourier Transforms as follows | |
| + | ::<math> H(j\omega\big) = \mathcal{H}(\omega) = \mathcal{F}\{ h(t) \}</math> | ||
| + | |||
| + | From this, the unit step response can be found | ||
| + | |||
| + | :: <math>h(t) = \mathcal{F}^{-1}\{ H(j\omega) \} </math> | ||
| + | |||
| + | Differential stuff.... | ||
==Exercises== | ==Exercises== | ||
| Line 42: | Line 49: | ||
:::<math>H(j\omega) = \frac{1}{j\omega + 4}</math> | :::<math>H(j\omega) = \frac{1}{j\omega + 4}</math> | ||
| + | |||
| + | ====b==== | ||
| + | The fundamental relationship from the main concepts | ||
| + | |||
| + | can be used to solve part b | ||
| + | :: <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> | ||
| + | |||
| + | I don't know how to evaluate this integral, I'm not sure if it can be. Any thoughts? | ||
| + | |||
| + | The integral as it stands cannot be evaluated. This is one of those problems where you have to find a function whose FT is <math>\frac{1}{j\omega +4}</math><br> | ||
| + | |||
| + | We need | ||
| + | : <math>\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt = \frac{1}{j\omega +4}</math> | ||
| + | |||
| + | Saying <math>h(t) = e^{-4t}u(t)\,</math> accomplishes this, and the table confirms it. | ||
| + | |||
| + | |||
| + | '''Using the table''' | ||
| + | |||
| + | Because the function <math>H\big(j\omega)</math> has the form <math> \frac{1}{a + j\omega}\,,a = 4 </math> the inverse Fourier transform is of the type <math>e^{-at}u\big(t) </math> | ||
| + | |||
| + | This results in the final answer of | ||
| + | |||
| + | :: <math>h(t) = e^{-4t}u\big(t)</math> | ||
Latest revision as of 16:07, 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?
The integral as it stands cannot be evaluated. This is one of those problems where you have to find a function whose FT is $ \frac{1}{j\omega +4} $
We need
- $ \int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt = \frac{1}{j\omega +4} $
Saying $ h(t) = e^{-4t}u(t)\, $ accomplishes this, and the table confirms it.
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) $
