Active filter; non-inverting (s+8)/(s+2)

Active filter

An active filter is a filter built from resistors, capacitors and active devices (op-amps or transistors) instead of inductors. The active device provides gain and isolation between stages.

Design of $T(s) = (s+8)/(s+2)$

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

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

Here the zero (8) is larger than the pole (2), the high-frequency gain is 1 and the DC gain is $8/2 = 4$. Use $Z_1 = R_1$ and $Z_2 = R_2 \parallel C_2$:

$$\begin{aligned} T(s) &= 1 + \frac{R_2}{R_1(1 + sR_2C_2)} \\ &= \frac{s + \frac{R_1 + R_2}{R_1R_2C_2}}{s + \frac{1}{R_2C_2}} \end{aligned}$$

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

Matching coefficients:

$$\begin{aligned} \frac{1}{R_2C_2} &= 2 \\ \frac{1}{R_2C_2}\left(1 + \frac{R_2}{R_1}\right) &= 8 \Rightarrow 1 + \frac{R_2}{R_1} = 4 \Rightarrow R_2 = 3R_1 \end{aligned}$$

Normalized: $C_2 = 1$ F, $R_2 = 0.5\ \Omega$, $R_1 = 0.1667\ \Omega$.

Practical values (impedance scale only, since the poles are already in rad/s): choose $C_2 = 10\ \mu$F:

$$R_2 = \frac{1}{2 \times 10\times10^{-6}} = 50\ \text{k}\Omega, \qquad R_1 = \frac{R_2}{3} = 16.67\ \text{k}\Omega$$

Check: DC gain $1 + R_2/R_1 = 4 = 8/2$; HF gain 1.

Answer: $R_1 = 16.67\ \text{k}\Omega$, $R_2 = 50\ \text{k}\Omega$, $C_2 = 10\ \mu$F.