Question 19.5: Using three VCVS sections in cascade, design a six-pole lowp...

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 f_{R C}=1.607 f_0 , where f_0 is the specified overall half-power frequency. Thus the  R C  products for the three sections are:

 \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}

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

\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}

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