Difference between revisions of "Lecture17 ECE301Fall2008mboutin" - Rhea

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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(j\omega) = \frac{1}{j\omega + 4} $