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*[[CT Fourier Transform_ECE301Fall2008mboutin]] {{:CT Fourier Transform}} | *[[CT Fourier Transform_ECE301Fall2008mboutin]] {{:CT Fourier Transform}} | ||
*[[CT Inverse Fourier Transform_ECE301Fall2008mboutin]] {{:CT Inverse Fourier Transform}} | *[[CT Inverse Fourier Transform_ECE301Fall2008mboutin]] {{:CT Inverse Fourier Transform}} | ||
| − | [[Table_of_Formulas_and_Properties_ECE301Fall2008mboutin#CT_Fourier_Transform_Properties]] | + | *[[Table_of_Formulas_and_Properties_ECE301Fall2008mboutin#CT_Fourier_Transform_Properties|CT Fourier Transform Properties]] |
== DT Fourier Transform == | == DT Fourier Transform == | ||
*[[DT Fourier Transform_ECE301Fall2008mboutin]] {{:DT Fourier Transform}} | *[[DT Fourier Transform_ECE301Fall2008mboutin]] {{:DT Fourier Transform}} | ||
*[[DT Inverse Fourier Transform_ECE301Fall2008mboutin]] {{:DT Inverse Fourier Transform}} | *[[DT Inverse Fourier Transform_ECE301Fall2008mboutin]] {{:DT Inverse Fourier Transform}} | ||
| + | *[[Table of Formulas and Properties_ECE301Fall2008mboutin#DT Fourier Transform Properties|DT Fourier Transform Properties]] | ||
| + | == Frequency Response == | ||
| + | Frequency response in CT and DT are very similar. They both have the form of <math>\ Y(\omega) = H(\omega)X(\omega)</math>. This form is obtained through differential equations in CT and difference equations in DT. | ||
| − | This entry contains | + | Here is a CT example we did in class: |
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| + | <math>\frac{d}{dt}y(t) + 4y(t) = x(t)</math> | ||
| + | Taking the Fourier transform of both sides yields <math>\ j\omega Y(\omega) + 4Y(\omega) = \mathcal{X}(\omega)</math>. | ||
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| + | Solving for <math>Y(\omega)</math> gives us <math>Y(\omega) = \frac{1}{j\omega +4}\mathcal{X}(\omega)</math>. From this <math>H(\omega)</math> can be observed as <math>\frac{1}{j\omega +4}</math>. | ||
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| + | By taking the Fourier inverse of <math>\ H(\omega)</math> the unit impulse response can be found. | ||
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| + | <math>\ h(t)</math>in this problem can be found from looking at the table of CT FT Pairs, so <math>\ h(t) = e^{-4t}u(t)</math> | ||
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| + | This entry contains some previously defined entries. They are placed all on this page in order save time navigation wise. | ||
Latest revision as of 16:12, 24 October 2008
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) $
This entry contains some previously defined entries. They are placed all on this page in order save time navigation wise.
