Question 15.3: Objective: Design a one-pole low-pass switched capacitor fil...

From Equation (15.32), the low-frequency gain is −(C_{1}/C_{2}), and the capacitance ratio must be (C_{1}/C_{2}) = 1. From Equation (15.32), the cutoff frequency is

T(jω) = − \frac{(1/f_{C}C_{2})}{(1/f_{C}C_{1})} \cdot \frac{1}{1  +  j \frac{(2π f )C_{F}}{f_{C}C_{2}}} = − \frac{C_{1}}{C_{2}} \cdot \frac{1}{1  +  j \frac{f}{f_{3  dB}}}         (15.32)

f_{3  dB} = \frac{f_{C}C_{2}}{2π C_{F}}
If we set the clock frequency to f_{C} = 10  kHz, then

\frac{C_{2}}{C_{F}} = \frac{2π f_{3  dB}}{f_{C}} = \frac{2π(10^{3})}{10  ×  10^{3}} = 0.628
Trade-offs: We can use standard-valued capacitors C_{1} = C_{2} = 75  pF. We would need C_{F} = C_{2}/0.628 = 75/0.628 = 119.4  pF. A standard-valued capacitor C_{F} = 120  pF can be used.
Comment: Since the low-frequency gain and cutoff frequency are both functions of capacitor ratios, the absolute capacitor values can be designed for compatibility with IC fabrication.