CT Fourier Transform
- CT Fourier Transform_ECE301Fall2008mboutin CT Fourier Transform
- CT Inverse Fourier Transform_ECE301Fall2008mboutin CT Inverse Fourier Transform
- CT Fourier Transform Properties
DT Fourier Transform
- DT Fourier Transform_ECE301Fall2008mboutin DT Fourier Transform
- DT Inverse Fourier Transform_ECE301Fall2008mboutin DT Inverse Fourier Transform
- DT Fourier Transform Properties
Frequency Response
Frequency response in CT and DT are very similar. They both have the form of $ \ Y(\omega) = H(\omega)X(\omega) $. This form is obtained through differential equations in CT and difference equations in DT.
Here is a CT example we did in class:
$ \frac{d}{dt}y(t) + 4y(t) = x(t) $ Taking the Fourier transform of both sides yields $ \ j\omega Y(\omega) + 4Y(\omega) = \mathcal{X}(\omega) $.
Solving for $ Y(\omega) $ gives us $ Y(\omega) = \frac{1}{j\omega +4}\mathcal{X}(\omega) $. From this $ H(\omega) $ can be observed as $ \frac{1}{j\omega +4} $.
By taking the Fourier inverse of $ \ H(\omega) $ the unit impulse response can be found.
$ \ h(t) $in this problem can be found from looking at the table of CT FT Pairs, so $ \ h(t) = e^{-4t}u(t) $
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