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Amplitude Response

We can isolate the filter amplitude response $ G(\omega)$ by squaring and adding the above two equations:

a^2(\omega) + b^2(\omega) &=& G^2(\omega)\cos^2[\Theta(\omega)...
...ta(\omega)] + \sin^2[\Theta(\omega)]\right\}\\
&=& G^2(\omega).
This can then be simplified as follows:
G^2(\omega) &=& a^2(\omega) + b^2(\omega)\\
&=& [1 + \cos(\...
... \cos(\omega T) \\
&=&4 \cos^2\left(\frac{\omega T}{2}\right).
So we have made it to the amplitude response of the simple lowpass filter $ y(n) = x(n) + x(n - 1)$:

$\displaystyle G(\omega) = 2 \left\vert\cos\left(\frac{\omega T}{2}\right)\right\vert

Since $ \cos(\pi fT)$ is nonnegative for $ -f_s/2 \leq f \leq f_s/2$, it is unnecessary to take the absolute value as long as $ f$ is understood to lie in this range:

$\displaystyle \zbox {G(\omega) = 2 \cos(\pi f T)} \qquad \left\vert f\right\vert \leq \frac{f_s}{2} \protect$ (2.6)

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