Active vs passive; realize 4(s+2)/(s+1)

Active vs passive filters

Realization of $T(s) = 4(s+2)/(s+1)$

Gains: HF gain = 4, DC gain = $4 \times 2/1 = 8$. The simple $R \parallel C$ feedback gives HF gain 1 only, so add a series resistor $R_a$ in the feedback path: $Z_1 = R_1$, $Z_2 = R_a + (R_b \parallel C)$.

$$\begin{aligned} T(s) &= 1 + \frac{1}{R_1}\left(R_a + \frac{R_b}{1 + sR_bC}\right) \\ &= \frac{R_1 + R_a}{R_1}\cdot\frac{s + \frac{1}{R_bC}\left(1 + \frac{R_b}{R_1 + R_a}\right)}{s + \frac{1}{R_bC}} \end{aligned}$$

         +--Ra--+--Rb--+
         |      |      |
         |      +--||--+
         |         C   |
   +-----+-|-\         |
   |       |  >--------+--- Vo
  R1 Vin---|+/
   |
  GND

Matching:

$$\begin{aligned} 1 + \frac{R_a}{R_1} &= 4 \Rightarrow R_a = 3R_1 \\ \frac{1}{R_bC} &= 1 \\ 1 + \frac{R_b}{R_1 + R_a} &= 2 \Rightarrow R_b = 4R_1 \end{aligned}$$

Normalized ($C = 1$ F): $R_b = 1\ \Omega$, $R_1 = 0.25\ \Omega$, $R_a = 0.75\ \Omega$.

Check: DC gain $= 1 + (R_a + R_b)/R_1 = 1 + 1.75/0.25 = 8$. Correct.

Practical values (impedance scale $k_m = 10^5$): $R_1 = 25\ \text{k}\Omega$, $R_a = 75\ \text{k}\Omega$, $R_b = 100\ \text{k}\Omega$, $C = 1/k_m = 10\ \mu$F.

Answer: $R_1 = 25\ \text{k}\Omega$, $R_a = 75\ \text{k}\Omega$, $R_b = 100\ \text{k}\Omega$, $C = 10\ \mu$F.