Using three VCVS sections in cascade, design a six-pole lowpass Bessel filter having half-power frequency f 0 =10kHz and minimum input impedance (magnitude) greater than 100kΩ.

Question

Using three VCVS sections in cascade, design a six-pole lowpass Bessel filter having half-power frequency [latex]f_0=10 kHz[/latex] and minimum input impedance (magnitude) greater than [latex] 100 k \Omega[/latex].

Accepted Answer

From Table 19.1, the values of the scale factor λ for the three sections are 1.607 , 1.692, and 1.908. This means, for example, that for the first section [latex] f_{R C}=1.607 f_0 [/latex], where [latex] f_0[/latex] is the specified overall half-power frequency. Thus the R C products for the three sections are:

Table 19.1 Design parameters for VCVS filters: N is the number of poles and [latex] \lambda [/latex] is the scaling factor for Bessel and Chebyshev sections [latex] ^6 [/latex]
N	Butterworth	Bessel		Chebyshev
	k	k	λ	k	λ
2	0.586	0.268	-1.738	0.842	1.231
4	0.152	0.084	1.432	0.582	0.597
	1.235	0.759	1.606	1.66	1.031
6	0.068	0.040	1.607	0.537	0.396
	0.586	0.364	1.692	1.448	0.768
	1.483	1.023	1.908	1.846	1.011
8	0.038	0.024	1.781	0.522	0.297
	0.337	0.213	1.835	1.379	0.599
	0.889	0.593	1.956	1.711	0.861
	1.610	1.184	2.192	1.913	1.006

[latex] \begin{aligned} (R C)_1 &=\frac{1}{2 \pi f_{R C 1}}=\frac{1}{2 \pi \lambda_1 f_0}=\frac{1}{2 \pi(1.607)(10 kHz )} \\ &=9.904 \mu s \\ (R C)_2 &=\frac{1}{2 \pi f_{R C 2}}=\frac{1}{2 \pi \lambda_2 f_0}=\frac{1}{2 \pi(1.692)(10 kHz )} \\ &=9.406 \mu s \\ (R C)_3 &=\frac{1}{2 \pi f_{R C 3}}=\frac{1}{2 \pi \lambda_3 f_0}=\frac{1}{2 \pi(1.908)(10 kHz )} \\ &=8.431 \mu s \end{aligned} [/latex]

We choose [latex] R=100 k \Omega [/latex] for all sections, for which value the input impedance (of the first section) will meet or exceed the specification [latex] \left|Z_{ ext {in }} ight| \geq 100 k \Omega[/latex]. The required values for the capacitances are then

[latex] \begin{aligned} &C_1=\frac{R C_1}{R}=\frac{9.904 \mu s }{100 k \Omega}=99.04 pF \\ &C_2=\frac{R C_2}{R}=\frac{9.406 \mu s }{100 k \Omega}=94.06 pF \\ &C_3=\frac{R C_3}{R}=\frac{8.431 \mu s }{100 k \Omega}=84.31 pF \end{aligned} [/latex]

[latex] { }^6[/latex] Adapted from Table 5.2 in Horowitz and Hill (ibid).

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