Cascade of first-order sections for 12/(s²+8s+12)

Factorize

$$T(s) = \frac{12}{s^2+8s+12} = \frac{12}{(s+2)(s+6)} = \underbrace{\left(\frac{-2}{s+2}\right)}_{T_1}\underbrace{\left(\frac{-6}{s+6}\right)}_{T_2}$$

Each factor is an inverting first-order low-pass section; the two minus signs cancel, giving the required positive sign. (Each section has dc gain $-1$, so the total dc gain is $12/12 = 1$.)

First-order inverting section

            +---[ R2 ]---+
            +---| C |----+
            |            |
Vin--R1-----+--(-)       |
                  OA >---+--- Vo
           GND--(+)

$$T(s) = -\frac{Z_f}{R_1} = -\frac{R_2/(1+sR_2C)}{R_1} = -\frac{1/(R_1C)}{s + 1/(R_2C)}$$

Section 1: $T_1 = -\dfrac{2}{s+2}$

$1/(R_2C) = 2$ and $1/(R_1C) = 2$. Normalized ($C = 1$ F): $R_1 = R_2 = 0.5\ \Omega$.

Section 2: $T_2 = -\dfrac{6}{s+6}$

$1/(R_2C) = 6$ and $1/(R_1C) = 6$. Normalized ($C = 1$ F): $R_1 = R_2 = 1/6 = 0.1667\ \Omega$.

Practical values

Impedance scale by $k_m = 10^6$ (choose $C = 1\ \mu$F in both sections):

Cascade

Vin-->[ Section 1: -2/(s+2) ]-->[ Section 2: -6/(s+6) ]--> Vo

Op-amp outputs have very low output impedance, so the sections do not load each other and the overall transfer function is the product:

$$T(s) = \frac{-2}{s+2}\cdot\frac{-6}{s+6} = \frac{12}{s^2+8s+12}$$

Answer: two inverting first-order sections in cascade: $R_1 = R_2 = 500$ k$\Omega$, $C = 1\ \mu$F (pole at 2 rad/s), and $R_1 = R_2 = 166.7$ k$\Omega$, $C = 1\ \mu$F (pole at 6 rad/s).