SC filter: 6 dB peak at 400 rad/s

Reading the plot. The response rises from 0 dB at 200 to 6 dB at 400 (one octave, i.e. 20 dB/decade): a zero at 200. At 400 the slope changes from +20 to −20 dB/decade, a change of −40 dB/decade, so a double pole at 400. It reaches 0 dB at 800 and stays flat: a zero at 800.

$$\begin{aligned} T(s) &= \frac{(1 + s/200)(1 + s/800)}{(1 + s/400)^2} = \frac{400^2}{200 \times 800}\cdot\frac{(s+200)(s+800)}{(s+400)^2} \\ &= \frac{(s+200)(s+800)}{(s+400)^2} \end{aligned}$$

(gain constant $= 160000/160000 = 1$.)

Building block (first-order SC section). An inverting op-amp stage with input admittance $Y_1 = sC_1 + 1/R_1$ and feedback admittance $Y_2 = sC_2 + 1/R_2$ gives

$$\begin{aligned} T(s) &= -\frac{Y_1}{Y_2} = -\frac{sC_1 + 1/R_1}{sC_2 + 1/R_2} = -\frac{C_1}{C_2}\cdot\frac{s + \frac{1}{R_1C_1}}{s + \frac{1}{R_2C_2}} \end{aligned}$$

Each resistor is replaced by a switched capacitor clocked at $f_c$ (two-phase, non-overlapping clock), for which $R = \frac{1}{f_c C_R}$. Then

$$T(s) = -\frac{sC_1 + f_cC_{R1}}{sC_2 + f_cC_{R2}}, \qquad \omega_z = \frac{f_cC_{R1}}{C_1}, \quad \omega_p = \frac{f_cC_{R2}}{C_2}$$

With $C_1 = C_2 = C$ the high-frequency gain is 1 and $C_{R1} = \frac{C\,\omega_z}{f_c}$, $C_{R2} = \frac{C\,\omega_p}{f_c}$. Two inverting stages in cascade give a positive overall gain.

Splitting into two first-order sections:

$$T(s) = \underbrace{\frac{s+200}{s+400}}_{T_1(s)} \times \underbrace{\frac{s+800}{s+400}}_{T_2(s)}$$

Choice of clock and capacitors. Highest break is 800 rad/s ($\approx 127$ Hz); take $f_c = 20$ kHz. Take $C_1 = C_2 = C = 100$ pF in both stages. With $C_R = C\omega/f_c$:

$$\begin{aligned} \text{Stage 1: } C_{R1} &= \frac{100\text{ pF}\times 200}{2\times 10^4} = 1\text{ pF}, & C_{R2} &= \frac{100\text{ pF}\times 400}{2\times 10^4} = 2\text{ pF} \\ \text{Stage 2: } C_{R1} &= \frac{100\text{ pF}\times 800}{2\times 10^4} = 4\text{ pF}, & C_{R2} &= \frac{100\text{ pF}\times 400}{2\times 10^4} = 2\text{ pF} \end{aligned}$$

Final circuit (two identical-form stages in cascade; each SC resistor is the toggle-switch circuit shown):

                    C2
            +-------||------+
            |               |
            +---[CR2-SC]----+
            |               |
       C1   |   |\          |
 Vi -+--||--+---|-\         |
     |      |   |  >--------+---> Vo
     +[CR1]-+ +-|+/
      (SC)    | |/
             GND

       phi1          phi2
 V1 ----o/o----+----o/o---- V2
               |
              ===  CR
               |
              GND

Clock: $f_c = 20$ kHz, two-phase non-overlapping. Equivalent resistors: 50 MΩ, 25 MΩ (stage 1) and 12.5 MΩ, 25 MΩ (stage 2).

Answer: $T(s) = \frac{(s+200)(s+800)}{(s+400)^2}$, two cascaded SC sections, $C = 100$ pF, switched capacitors 1 & 2 pF and 4 & 2 pF, $f_c = 20$ kHz. The design follows the asymptotes; the actual peak at $\omega = 400$ is 1.94 dB.