Q and center frequency; realize −2000/(s²+500s+10⁶)

Quality factor and centre frequency of a low-pass biquad

A second-order low-pass biquad has

$$T(s) = \frac{K\omega_0^2}{s^2 + \dfrac{\omega_0}{Q}s + \omega_0^2}$$

• Centre (pole / natural) frequency $\omega_0$: the magnitude of the poles, $\omega_0 = |p| = \sqrt{\sigma^2 + \omega_d^2}$. It sets where the response turns over; at $\omega = \omega_0$ the phase is $-90^\circ$.
• Quality factor $Q$: $Q = \dfrac{\omega_0}{2\sigma}$, where $\sigma$ is the distance of the poles from the $j\omega$ axis. It measures how sharp the response is near $\omega_0$: $|T(j\omega_0)| = KQ$. A large $Q$ puts the poles close to the $j\omega$ axis and gives a high peak; $Q = 0.707$ gives the maximally flat (Butterworth) shape; $Q < 0.5$ gives real poles.

 |T|
  |        Q = 5 (sharp peak)
  |        /\
  |       /  \
 K|------/-.--\--- Q = 0.707 (flat)
  |       '. \ \
  |          '.\ \
  |             '-\----
  +-----------+--------> w
              w0

The pole angle from the negative real axis is $\psi = \cos^{-1}\!\big(1/2Q\big)$, so higher $Q$ means poles closer to the $j\omega$ axis.

Tow-Thomas circuit used

            R1 (sets Q)
          +--/\/\--+
          |   C1   |
          +---||---+
          |        |
Vin--R4---+--[OA1]-+--> V1 (band-pass)
          ^
          |  V1--R3-->[OA2 integrator, C2]--> V2 (LP)
          |  V2--r--->[OA3 inverter, r/r]---> V3=-V2
          |                                    |
          +------------------R2----------------+

All op-amp (+) inputs are grounded. OA1 is a lossy integrator ($R_1 \parallel C_1$ in feedback), OA2 an ideal integrator ($C_2$ in feedback), OA3 a unity-gain inverter.

Its inverting low-pass output is

$$\frac{V_3}{V_{in}} = \frac{-\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}$$

with $\omega_0 = 1/\sqrt{R_2R_3C_1C_2}$, $\omega_0/Q = 1/(R_1C_1)$, $K = R_2/R_4$. With $C_1 = C_2 = C$, $R_2 = R_3 = R$: $R = 1/(\omega_0C)$, $R_1 = QR$, $R_4 = R/K$.

Realization of $T(s) = \dfrac{-2000}{s^2 + 500s + 1{,}000{,}000}$

Step 1: Parameters.

$$\begin{aligned} \omega_0^2 &= 10^6 \;\Rightarrow\; \omega_0 = 1000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 500 \;\Rightarrow\; Q = \frac{1000}{500} = 2 \\ \text{dc gain} &= T(0) = \frac{-2000}{10^6} = -0.002, \quad \text{so } K = \frac{R_2}{R_4} = 0.002 \end{aligned}$$

The gain is negative, so the output is taken at $V_3$ (the inverter output), which gives $-\dfrac{1/(R_3R_4C_1C_2)}{s^2 + s/(R_1C_1) + 1/(R_2R_3C_1C_2)}$.

Step 2: Element values. Choose $C_1 = C_2 = C = 1\ \mu$F, $R_2 = R_3 = R$ (with $C = 0.1\ \mu$F, $R_4$ would be 5 M$\Omega$, which is too large, so $1\ \mu$F is chosen).

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{1000 \times 10^{-6}} = 1\ \text{k}\Omega \\ R_1 &= QR = 2 \times 1 = 2\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{1\ \text{k}\Omega}{0.002} = 500\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter)} \end{aligned}$$

Check: numerator $\dfrac{1}{R_3R_4C_1C_2} = \dfrac{1}{10^3 \times 5\times10^5 \times 10^{-12}} = 2000$; $\dfrac{1}{R_1C_1} = \dfrac{1}{2000\times10^{-6}} = 500$; $\dfrac{1}{R_2R_3C^2} = 10^6$. All coefficients match.

Answer: $C_1 = C_2 = 1\ \mu$F, $R_1 = 2$ k$\Omega$, $R_2 = R_3 = 1$ k$\Omega$, $R_4 = 500$ k$\Omega$, $r = 10$ k$\Omega$, output at $V_3$ (OA3).