Page 425 - Analog and Digital Filter Design

422 Analog and Digital Filter Design

Now, the required order is equal to 12. This is given by an equation that uses the
two variables k and L:
K(l/k) . K( ,/m)

= K(l/L). K(,/iqF)
Function K(x) is an elliptic integral of x, so four elliptic integrals are needed to
find the filter order. The elliptic integral itself takes the value to be integrated
as the starting point. It is a recursive equation. Here are the equations.

These equations look horrific, so an example of the elliptic integral algorithm’s
use is now given.

If k = 2, K(l/lc) = K(0.5) and the value to be integrated is Xo = 0.5. Xi =
w) 0.8660. This is then used to find X,.
=
Xi = (1 - 0.866)/(1+ 0.866) = 0.0718

XI’ = J(1- 0.0718l) = 0.9974. This is used to find Xz.
Xz = (1 - 0.9974)/(1+ 0.9974) = 1.3017 x
X2’ = J(1- 0.0013017’) = 0.999999152. This is used to find 1,.
X, = 4.2361 x

As X becomes small its effect diminishes, and when it is less than lo-’ it can be
ignored. Therefore the infinite limit to the product of (1 + X) is actually trun-
cated after a few iterations. In this case we can see that X, will be very small.

K(0.5)= ~/2(1+0.0718).(1+0.0013017).(1+4.2361x lo-’)= 1.68577.

Noise Bandwidth

Knowledge of a filter’s noise bandwidth can help in system design and
testing. Suppose you want to find the noise figure of an amplifier. You first
filter the amplifier’s output and measure the RMS (root mean square) output
voltage. You then divide by the amplifier’s gain, the filter bandwidth,