(New page: ==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...) |
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Latest revision as of 18:11, 24 October 2008
CT Frequency Response
First the system:
$ \sum_{k=0}^{N}a_k\frac{d^ky(t)}{dk^t}=\sum_{k=0}^{M}b_k\frac{d^kx(t)}{dt^k} $
Then the Fourier Transform:
$ \sum_{k=0}^Na_k(j\omega)^kY(\omega)=\sum_{k=0}^Mb_k(j\omega)^kX(\omega) $
Then the frequency response H(jw):
$ Y(\omega)=\frac{\sum_{k=0}^Mb_k(j\omega)^k}{\sum_{k=0}^Na_k(j\omega)^k}X(\omega) $
So:
$ H(j\omega)=\frac{\sum_{k=0}^Mb_k(j\omega)^k}{\sum_{k=0}^Na_k(j\omega)^k} $
DT Frequency Response
First the system:
$ \sum_{k=0}^Na_ky[n-k]=\sum_{k=0}^Mb_kx[n-k] $
Then the Fourier Transform:
$ \sum_{k=0}^Na_ke^{-jk\omega}Y(\omega)=\sum_{k=0}^Mb_ke^{-jk\omega}X(\omega) $
So the frequency response is:
$ H(e^{j\omega})=\frac{\sum_{k=0}^Mb_ke^{-jk\omega}}{\sum_{k=0}^Na_ke^{-jk\omega}} $
