Inverting realization of 7(s+400)/(s+200)

Given $T(s) = 7\dfrac{s+400}{s+200}$: HF gain 7, DC gain 14, zero at −400, pole at −200. Only R, C and op-amps are used (no inductors).

Inverting configuration: $Z_1$ from the input to the inverting input, $Z_2$ in feedback:

$$T(s) = -\frac{Z_2}{Z_1} = -\frac{Y_1}{Y_2}$$

With $Y_1 = \dfrac{1}{R_1} + sC_1$ (parallel $R_1C_1$) and $Y_2 = \dfrac{1}{R_2} + sC_2$ (parallel $R_2C_2$):

$$T(s) = -\frac{C_1}{C_2}\cdot\frac{s + \frac{1}{R_1C_1}}{s + \frac{1}{R_2C_2}}$$

              +---R2---+
              |        |
              +---||---+
       R1     |   C2   |
  +--/\/\--+  |        |
Vin        +--+-|-\    |
  +---||---+    |  >---+--- Vo
       C1   GND-|+/

Matching:

$$\begin{aligned} \frac{C_1}{C_2} &= 7 \\ \frac{1}{R_1C_1} &= 400 \\ \frac{1}{R_2C_2} &= 200 \end{aligned}$$

Choose $C_2 = 0.1\ \mu$F, so $C_1 = 0.7\ \mu$F:

$$\begin{aligned} R_1 &= \frac{1}{400 \times 0.7\times10^{-6}} = 3.571\ \text{k}\Omega \\ R_2 &= \frac{1}{200 \times 0.1\times10^{-6}} = 50\ \text{k}\Omega \end{aligned}$$

Check: DC gain $= R_2/R_1 = 50/3.571 = 14 = 7 \times 400/200$. Correct.

Answer: $R_1 = 3.57\ \text{k}\Omega \parallel C_1 = 0.7\ \mu$F at the input, $R_2 = 50\ \text{k}\Omega \parallel C_2 = 0.1\ \mu$F in feedback. The circuit gives $-7(s+400)/(s+200)$; add a unity-gain inverting amplifier (two equal resistors) if the positive sign is required.