Sallen-Key realization of 1/(s²+s+1) at 1.8 kHz

Sallen-Key low-pass circuit and transfer function

The Sallen-Key low-pass biquad uses one op-amp wired as a non-inverting amplifier of gain $K = 1 + R_B/R_A$, two resistors $R_1, R_2$ in series, a capacitor $C_1$ fed back from the output to node A, and a grounded capacitor $C_2$ at node B.

              C1 (feedback)
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         |
                         RA
                         |
                        GND

Derivation (ideal op-amp, so $V_o = K V_B$ and no current enters the + input):

Node B: the current through $R_2$ flows into $C_2$:

$$\frac{V_A - V_B}{R_2} = sC_2 V_B \;\Rightarrow\; V_A = V_B(1 + sR_2C_2)$$

Node A (KCL):

$$\frac{V_{in} - V_A}{R_1} = \frac{V_A - V_B}{R_2} + sC_1 (V_A - V_o)$$

Substituting $V_A$ and $V_B = V_o/K$ and simplifying:

$$T(s) = \frac{V_o}{V_{in}} = \frac{\dfrac{K}{R_1R_2C_1C_2}}{s^2 + s\left(\dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}\right) + \dfrac{1}{R_1R_2C_1C_2}}$$

Comparing with $T(s) = \dfrac{K\omega_o^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$:

$$\begin{aligned} \omega_o &= \frac{1}{\sqrt{R_1R_2C_1C_2}} \\ \frac{\omega_o}{Q} &= \frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2} \end{aligned}$$

DC gain $= K$.

Realization of $1/(s^2 + s + 1)$ with half-power frequency 1.8 kHz and all C = 10 nF

Comparing: $\omega_o = 1$, $\omega_o/Q = 1$, so $Q = 1$.

All capacitors are equal, so use the equal-element design ($R_1 = R_2 = 1$, $C_1 = C_2 = 1$):

$$K = 3 - \frac{1}{Q} = 3 - 1 = 2 \;\Rightarrow\; R_B = R_A$$

Half-power frequency. Because $Q = 1$ (not a Butterworth response), the $-3$ dB frequency is not $\omega_o$. For the normalized function,

$$|T(j\omega)|^2 = \frac{1}{(1-\omega^2)^2 + \omega^2} = \frac{1}{\omega^4 - \omega^2 + 1} = \frac12 \;\Rightarrow\; \omega^4 - \omega^2 - 1 = 0 \;\Rightarrow\; \omega^2 = \frac{1+\sqrt5}{2}$$

so the normalized half-power frequency is $\omega_h = 1.272$ rad/s. The 1.8 kHz half-power point therefore needs

$$\omega_o = \frac{2\pi \times 1800}{1.272} = \frac{11\,309.7}{1.272} = 8891\ \text{rad/s} \quad (f_o \approx 1.415\ \text{kHz})$$

Scaling ($C = 10$ nF):

$$\begin{aligned} k_f &= 8891\ \text{rad/s} \\ k_m &= \frac{1}{8891 \times 10\times10^{-9}} = 11\,247 \\ R_1 = R_2 &= 11.25\ \text{k}\Omega \end{aligned}$$

Take $R_A = R_B = 10\ \text{k}\Omega$.

The circuit gain is 2, but $T(s)$ has DC gain 1. To get unity gain, apply gain reduction with $\alpha = 1/K = 0.5$:

$$R_{1a} = \frac{R_1}{\alpha} = 22.49\ \text{k}\Omega, \qquad R_{1b} = \frac{R_1}{1-\alpha} = 22.49\ \text{k}\Omega$$

Answer: $R_2 = 11.25\ \text{k}\Omega$, $C_1 = C_2 = 10$ nF, $K = 2$, with $R_1$ split into two 22.5 kΩ resistors ($R_{1a} \parallel R_{1b} = 11.25\ \text{k}\Omega$). The realized function is $\dfrac{\omega_o^2}{s^2 + \omega_o s + \omega_o^2}$ with $\omega_o = 8891$ rad/s, whose half-power frequency is 1.8 kHz.

Note: if your course simply scales by $\omega_o = 2\pi(1800)$ (treating the normalized 1 rad/s as the half-power point), the values become $R_1 = R_2 = 8.84\ \text{k}\Omega$ and $R_{1a} = R_{1b} = 17.68\ \text{k}\Omega$. The true $-3$ dB point is then at $1.272 \times 1.8 \approx 2.29$ kHz.