Page 414 - Analog and Digital Filter Design

4
Design Equations

and group delay is given by:

Unfortunately the attenuation formula is for a Bessel response with a one-second
delay. We must scale it using Table 2.1. Let's show how this works for n = 3.

real part = 15 - 6w'; imaginary part = l5jo - jd; a. = 15.

In the case of the third-order response, the scaling factor is 1.75. This factor
must be multiplied by the frequency ratio. For instance, to find the attenuation
at 5rad/s with a normalized response (lradls cutoff), the frequency ratio is 5
and, therefore, in both real and imaginary parts, w = 8.75.

ao
-20. L OG1o
real' -t imaginary'
= -2O.LOG(15/J191469+641876) 35.7dB
=
Compare this with the graph in Figure 2.6 showing attenuation versus frequency
and order. Polynomials for the Bessel response up to third order, Table A. 1 gives
some of the Bessel polynomial coefficients for higher orders.

The phase of a third-order response can be analyzed in a similar way to show
the near constant group delay.

Cp = tan-'(Imaginary H(jw)/Real H(jo))

real part = 15 - 6w': imaginary part = l5jw -jw3; = 15.
Cp = tad(jw.(l5 - d)/(15 - 60'))
=tan- [ ]
l j0(15-w')
(15 -60')
If w = 0.25, @ =tan-'[j0.25.(14.9375)/{14.625)] = 0.249999rad
If w= 0.5, 4 =tan-'[j0.5.(14.75)/(13.5)] = 0.499995rad
If w = 0.75, Cp = tan-'[j0.75.(14.4375)/( 11.625)] = 0.749922rad
If CL)= 1, @=tan-'[jl.{l4)/(9)] = 0.999459rad
If w= 1.5, @ = tan-l[j1.5.(12.75)/(1.5)] = 1.492525rad

Since the magnitude of the phase is almost the same as the frequency, the rate
of change is almost constant @e., = 1 rad/rad/s = 1 s). As the frequency increases
to w = 1.5 the error is only 0.007475, or 0.5%. In the normalized response. a