Non-inverting filter: zero at 8, pole at 4, gain 2

The required function (zero at $s=-8$, pole at $s=-4$) is

$$T(s) = K\frac{s+8}{s+4}, \qquad T(0) = K\cdot\frac{8}{4} = 2 \Rightarrow K = 1$$

So $T(s) = \dfrac{s+8}{s+4}$: DC gain 2, HF gain 1.

Non-inverting configuration: $Z_1$ from the inverting input to ground, $Z_2$ from output to inverting input:

$$T(s) = 1 + \frac{Z_2}{Z_1}$$

Use $Z_1 = R_1$ and $Z_2 = R_2 \parallel C_2$:

$$T(s) = 1 + \frac{R_2}{R_1(1+sR_2C_2)} = \frac{s + \frac{1}{R_2C_2}\left(1 + \frac{R_2}{R_1}\right)}{s + \frac{1}{R_2C_2}}$$

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

Matching:

$$\begin{aligned} \frac{1}{R_2C_2} &= 4 \\ 1 + \frac{R_2}{R_1} &= \frac{8}{4} = 2 \Rightarrow R_1 = R_2 \end{aligned}$$

Normalized: $C_2 = 1$ F, $R_1 = R_2 = 0.25\ \Omega$.

Practical values: choose $C_2 = 10\ \mu$F: $R_2 = \dfrac{1}{4 \times 10^{-5}} = 25\ \text{k}\Omega = R_1$.

Answer: $R_1 = R_2 = 25\ \text{k}\Omega$, $C_2 = 10\ \mu$F; DC gain 2, HF gain 1, zero at −8, pole at −4.