View source for Emily Blount: DT and CT frequency response for systems characterized by Linear, Constant coefficients, Difference Equations ECE301Fall2008mboutin - Rhea

==CT Frequency Response==

First the system:

<math>\sum_{k=0}^{N}a_k\frac{d^ky(t)}{dk^t}=\sum_{k=0}^{M}b_k\frac{d^kx(t)}{dt^k}</math>

Then the Fourier Transform:

<math>\sum_{k=0}^Na_k(j\omega)^kY(\omega)=\sum_{k=0}^Mb_k(j\omega)^kX(\omega)</math>

Then the frequency response H(jw):

<math>Y(\omega)=\frac{\sum_{k=0}^Mb_k(j\omega)^k}{\sum_{k=0}^Na_k(j\omega)^k}X(\omega)</math>

So:

<math>H(j\omega)=\frac{\sum_{k=0}^Mb_k(j\omega)^k}{\sum_{k=0}^Na_k(j\omega)^k}</math>


==DT Frequency Response==

First the system:

<math>\sum_{k=0}^Na_ky[n-k]=\sum_{k=0}^Mb_kx[n-k]</math>

Then the Fourier Transform:

<math>\sum_{k=0}^Na_ke^{-jk\omega}Y(\omega)=\sum_{k=0}^Mb_ke^{-jk\omega}X(\omega)</math>

So the frequency response is:

<math>H(e^{j\omega})=\frac{\sum_{k=0}^Mb_ke^{-jk\omega}}{\sum_{k=0}^Na_ke^{-jk\omega}}</math>