Non-inverting realization of (s+8)/(s+2)

Approach

A non-inverting op-amp amplifier with impedance $Z_1$ from the (-) input to ground and $Z_2$ in feedback has

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

Vin ---------(+)
                \
                 OA >----+---- Vo
                /        |
     +-------(-)         |
     |                   |
     +-------[ Z2 ]------+
     |
   [ Z1 ]
     |
    GND

Matching the given function

$$T(s) = \frac{s+8}{s+2} = 1 + \frac{6}{s+2} \;\Rightarrow\; \frac{Z_2}{Z_1} = \frac{6}{s+2}$$

Take $Z_1 = R_1$ and $Z_2 = R_2 \parallel C$:

$$Z_2 = \frac{R_2}{1 + sR_2C} = \frac{1/C}{s + 1/(R_2C)}, \qquad \frac{Z_2}{Z_1} = \frac{1/(R_1C)}{s + 1/(R_2C)}$$

Comparing:

$$\frac{1}{R_2C} = 2, \qquad \frac{1}{R_1C} = 6$$

Normalized values ($C = 1$ F): $R_2 = 0.5\ \Omega$, $R_1 = 1/6 = 0.1667\ \Omega$.

Check: $T(s) = 1 + \dfrac{6}{s+2} = \dfrac{s+8}{s+2}$. Zero at $s = -8$, pole at $s = -2$, dc gain $8/2 = 4$, high-frequency gain 1.

Practical values (impedance scaling)

Choose $C = 1\ \mu$F, i.e. $k_m = 10^6$ (frequencies unchanged):

$$R_2 = 0.5 \times 10^6 = 500\ \text{k}\Omega, \qquad R_1 = 0.1667\times10^6 = 166.7\ \text{k}\Omega, \qquad C = 1\ \mu\text{F}$$

Answer: non-inverting amplifier with $R_1 = 166.7$ k$\Omega$ to ground and $R_2 = 500$ k$\Omega$ in parallel with $C = 1\ \mu$F in feedback (normalized: $R_1 = 1/6\ \Omega$, $R_2 = 0.5\ \Omega$, $C = 1$ F). This form works because the zero is farther from the origin than the pole, so $|T| \ge 1$ at all frequencies.