Active Filters

Tow-Thomas low-pass biquad

The Tow-Thomas biquad is a two-integrator-loop circuit using three op-amps: a lossy integrator, an ideal integrator and an inverter in a feedback loop. It gives low-pass and band-pass outputs at the same time, has low sensitivity, and lets $\omega_0$, $Q$ and gain be tuned independently.

            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.

Derivation (ideal op-amps, so each inverting input is a virtual ground):

KCL at the inverting input of OA1:

$$\frac{V_{in}}{R_4} + \frac{V_3}{R_2} + V_1\left(\frac{1}{R_1} + sC_1\right) = 0$$

OA2 (integrator) and OA3 (inverter):

$$V_2 = -\frac{V_1}{sR_3C_2}, \qquad V_3 = -V_2 = \frac{V_1}{sR_3C_2} \;\Rightarrow\; V_1 = sR_3C_2V_3$$

Substituting $V_1$ into the KCL equation:

$$\begin{aligned} sR_3C_2V_3\left(\frac{1}{R_1} + sC_1\right) + \frac{V_3}{R_2} &= -\frac{V_{in}}{R_4} \\ V_3\left(s^2R_3C_1C_2 + \frac{sR_3C_2}{R_1} + \frac{1}{R_2}\right) &= -\frac{V_{in}}{R_4} \end{aligned}$$

Dividing by $R_3C_1C_2$:

$$\frac{V_3}{V_{in}} = \frac{-\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \qquad \frac{V_2}{V_{in}} = -\frac{V_3}{V_{in}}$$

Comparing with $T(s) = \dfrac{\pm K\omega_0^2}{s^2 + (\omega_0/Q)s + \omega_0^2}$:

$$\omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}}, \qquad \frac{\omega_0}{Q} = \frac{1}{R_1C_1}, \qquad K = \frac{R_2}{R_4}\ \text{(dc gain)}$$

$V_2$ is a non-inverting low-pass output and $V_3$ an inverting one; $V_1$ is a band-pass output.

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 450$, $\omega_d = 893.03$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 450^2 + 893.03^2 = 1{,}000{,}002.6 \;\Rightarrow\; \omega_0 \approx 1000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{1000}{900} = 1.111 \end{aligned}$$

Required transfer function (dc gain $K = 1.5$):

$$T(s) = \frac{1.5\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{1.5 \times 10^{6}}{s^2 + 900\,s + 10^{6}}$$

Design equations (choose $C_1 = C_2 = C$, $R_2 = R_3 = R$):

$$R = \frac{1}{\omega_0 C}, \qquad R_1 = QR = \frac{Q}{\omega_0 C}, \qquad R_4 = \frac{R_2}{K}, \qquad r = \text{any convenient value}$$

Step 2: Element values. Choose $C_1 = C_2 = C = 0.1\ \mu$F (a standard value that gives resistors in the k$\Omega$ range).

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{1000 \times 0.1\times10^{-6}} = 10\ \text{k}\Omega \\ R_1 &= QR = 1.111 \times 10\ \text{k}\Omega = 11.11\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{10\ \text{k}\Omega}{1.5} = 6.667\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 1.5.

Answer: $C = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).

In the equal-component Sallen-Key low-pass design the amplifier gain is fixed by the required $Q$: $K = 3 - 1/Q$. The filter's dc gain therefore equals $K$, which is often more than the gain wanted. The excess gain is removed without disturbing $\omega_0$ and $Q$ by replacing the input resistor $R_1$ with a voltage divider (gain reduction).

Sallen-Key low-pass circuit

                 C1
          +------||------------+
          |                    |
Vin--R1---+A--R2--+B---(+)     |
                  |     OA >---+--- Vo
                 C2  +-(-)     |
                  |  |         RB
                 GND +---------+
                     |
                     RA
                     |
                    GND

Amplifier gain $K = 1 + R_B/R_A$, so $V_B = V_o/K$.

Derivation. Node B: $\dfrac{V_A - V_B}{R_2} = sC_2V_B \Rightarrow V_A = V_B(1 + sR_2C_2)$.

Node A:

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

Eliminating $V_A$ and $V_B$:

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

so $\omega_0 = \dfrac{1}{\sqrt{R_1R_2C_1C_2}}$ and $\dfrac{\omega_0}{Q} = \dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}$. For the equal-component design ($R_1 = R_2 = R$, $C_1 = C_2 = C$): $\omega_0 = 1/RC$ and $\omega_0/Q = (3-K)/RC$, so $K = 3 - 1/Q$.

Compensation by input voltage divider

Suppose the required gain is $H < K$. Replace $R_1$ by two resistors $R_{1a}$ (series) and $R_{1b}$ (to ground):

Vin--R1a--+--> node A        Thevenin equivalent:
          |                  Vth = alpha*Vin
         R1b                 Rth = R1a || R1b = R1
          |
         GND

The Thevenin equivalent seen from node A is a source $\alpha V_{in}$ in series with $R_{1a}\parallel R_{1b}$, where

$$\alpha = \frac{R_{1b}}{R_{1a} + R_{1b}}, \qquad R_{1a}\parallel R_{1b} = \frac{R_{1a}R_{1b}}{R_{1a}+R_{1b}}$$

If we make $R_{1a}\parallel R_{1b} = R_1$, the network seen by node A is unchanged, so $\omega_0$ and $Q$ are unchanged, and the transfer function becomes

$$\frac{V_o}{V_{in}} = \alpha\,T(s) \;\Rightarrow\; \text{dc gain} = \alpha K = H \;\Rightarrow\; \alpha = \frac{H}{K}$$

Solving the two conditions:

$$R_{1a} = \frac{R_1}{\alpha}, \qquad R_{1b} = \frac{R_1}{1-\alpha}$$

Check: $R_{1a}\parallel R_{1b} = \dfrac{R_1^2/[\alpha(1-\alpha)]}{R_1/[\alpha(1-\alpha)]} = R_1$.

Example

For a 2nd order Butterworth filter ($Q = 0.7071$), $K = 3 - 1.414 = 1.586$. For unity overall gain ($H = 1$): $\alpha = 1/1.586 = 0.6306$, so $R_{1a} = R_1/0.6306 = 1.586R_1$ and $R_{1b} = R_1/0.3694 = 2.707R_1$. With $R_1 = 10$ k$\Omega$: $R_{1a} = 15.86$ k$\Omega$, $R_{1b} = 27.07$ k$\Omega$.

Notes

• Only the input resistor changes; $C_1$, $C_2$, $R_2$ and the gain resistors $R_A$, $R_B$ stay the same.
• This method can only reduce gain ($\alpha < 1$). If more gain is needed, gain enhancement (feeding $C_1$ from a tap of an output divider) or a separate amplifier stage is used.

In the equal-component Sallen-Key low-pass filter the amplifier gain is fixed by $Q$ ($K = 3 - 1/Q$), so the dc gain is fixed too. Gain enhancement raises the overall gain above $K$ while keeping $\omega_0$ and $Q$ unchanged.

Basic idea

The RC network only "sees" the voltage that is fed back through $C_1$ and the voltage at the (+) input. If the op-amp output is made larger by a factor $1/\beta$, but only $\beta V_o$ is fed back to $C_1$ (and set as $K$ times $V_B$), the RC network works exactly as before while the output is bigger.

Circuit

              C1
          +---||-------------------+
          |                        |
Vin--R1---+A--R2--+B--(+)          |
                  |     OA(mu)--+--|--- Vo
                 C2             |  |
                  |             Rc |
                 GND            +--+ Vo' = beta*Vo
                                |
                                Rd
                                |
                               GND
 OA(mu): non-inverting amp, mu = 1 + RB/RA

The op-amp output $V_o$ drives a divider $R_c$, $R_d$; capacitor $C_1$ is returned to the tap $V_o' = \beta V_o$ where $\beta = \dfrac{R_d}{R_c + R_d} < 1$ (the divider resistance $R_c\parallel R_d$ is made small compared with the impedance of $C_1$, or absorbed into the design).

Derivation

Let the non-inverting amplifier gain be $\mu = 1 + R_B/R_A$, so $V_o = \mu V_B$ and $V_o' = \beta\mu V_B$. The RC network now sees an "effective gain" $\beta\mu$ from node B to the point where $C_1$ returns. Writing the node equations exactly as for the normal circuit with $V_o$ replaced by $V_o'$:

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

For the required $Q$, set $\beta\mu = K = 3 - 1/Q$ (equal-component case). Then $\omega_0$ and $Q$ are the same as before, and the output is

$$\frac{V_o}{V_{in}} = \frac{1}{\beta}\cdot\frac{V_o'}{V_{in}} \;\Rightarrow\; \text{dc gain} = \frac{K}{\beta} > K$$

Design steps

1. Design the normal equal-component Sallen-Key for $\omega_0$ and $Q$; this gives $K = 3 - 1/Q$.
2. For a required dc gain $H > K$, choose $\beta = K/H$.
3. Set the op-amp gain $\mu = 1 + R_B/R_A = H$ and choose the divider $R_d/(R_c+R_d) = \beta$.

Example: Butterworth ($Q = 0.707$, $K = 1.586$) with required gain $H = 5$: $\beta = 1.586/5 = 0.317$, $\mu = 5$ ($R_B = 4R_A$), and the divider is $R_c = 2.15R_d$ (e.g. $R_d = 1$ k$\Omega$, $R_c = 2.15$ k$\Omega$).

For gain reduction (the opposite case) the input resistor $R_1$ is split into a voltage divider instead.

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). A non-inverting stage can only realize $|T| \ge 1$ forms like this one (zero farther from the origin than the pole).

Sallen-Key low-pass biquad

The Sallen-Key (positive-feedback, VCVS) biquad uses one op-amp as a non-inverting amplifier of gain $K$ with an RC network.

                 C1
          +------||------------+
          |                    |
Vin--R1---+A--R2--+B---(+)     |
                  |     OA >---+--- Vo
                 C2  +-(-)     |
                  |  |         RB
                 GND +---------+
                     |
                     RA
                     |
                    GND

Amplifier gain $K = 1 + R_B/R_A$, so $V_B = V_o/K$.

Derivation. Node B: $\dfrac{V_A - V_B}{R_2} = sC_2V_B \Rightarrow V_A = V_B(1 + sR_2C_2)$.

Node A:

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

Eliminating $V_A$ and $V_B$:

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

so $\omega_0 = \dfrac{1}{\sqrt{R_1R_2C_1C_2}}$ and $\dfrac{\omega_0}{Q} = \dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}$. For the equal-component design ($R_1 = R_2 = R$, $C_1 = C_2 = C$): $\omega_0 = 1/RC$ and $\omega_0/Q = (3-K)/RC$, so $K = 3 - 1/Q$.

Design for $T(s) = \dfrac{1}{s^2 + 0.765s + 1}$

Interpretation: "feedback resistors equal" is taken as the equal-resistor ($R_1 = R_2$) equal-capacitor design with $C = 0.01\ \mu$F.

Step 1: Parameters.

$$\omega_0^2 = 1 \Rightarrow \omega_0 = 1\ \text{rad/s}, \qquad \frac{\omega_0}{Q} = 0.765 \Rightarrow Q = \frac{1}{0.765} = 1.307$$

Step 2: Amplifier gain.

$$K = 3 - \frac{1}{Q} = 3 - 0.765 = 2.235, \qquad 1 + \frac{R_B}{R_A} = 2.235 \Rightarrow R_B = 1.235R_A$$

Choose $R_A = 10$ k$\Omega$, $R_B = 12.35$ k$\Omega$.

Step 3: Normalized design. $C_1 = C_2 = 1$ F, $R_1 = R_2 = 1\ \Omega$ gives $\omega_0 = 1$ and $\omega_0/Q = (3-K)/1 = 0.765$.

Step 4: Impedance scaling to $C = 0.01\ \mu$F: $k_m = \dfrac{1}{C\,\omega_0} = \dfrac{1}{10^{-8} \times 1} = 10^8$.

$$R_1 = R_2 = \frac{1}{\omega_0 C} = \frac{1}{1 \times 0.01\times10^{-6}} = 100\ \text{M}\Omega$$

Because the given $T(s)$ is normalized ($\omega_0 = 1$ rad/s), the resistors come out very large. For a practical cutoff, the same design is frequency scaled; e.g. for $\omega_0 = 10^4$ rad/s, $R_1 = R_2 = 1/(10^4 \times 10^{-8}) = 10$ k$\Omega$ with the same capacitors and the same $K$.

Step 5: Gain. The dc gain of this circuit is $K = 2.235$. The given $T(s)$ has unity dc gain, so the excess gain is removed by splitting $R_1$ into a divider: $\alpha = 1/2.235 = 0.4474$, $R_{1a} = R_1/\alpha = 2.235R_1$, $R_{1b} = R_1/(1-\alpha) = 1.810R_1$.

Answer: $C_1 = C_2 = 0.01\ \mu$F, $R_1 = R_2 = 100$ M$\Omega$ (for $\omega_0 = 1$ rad/s; 10 k$\Omega$ if scaled to $10^4$ rad/s), $K = 2.235$ ($R_A = 10$ k$\Omega$, $R_B = 12.35$ k$\Omega$); for unity gain, $R_1$ is replaced by $R_{1a} = 2.235R_1$ and $R_{1b} = 1.810R_1$.

RC-CR transformation converts an active RC low-pass filter into a high-pass filter of the same order and $\omega_0$ by replacing every resistor (of the frequency-determining RC network) by a capacitor and every capacitor by a resistor. In normalized form, $R_i \to C_i = 1/R_i$ and $C_j \to R_j = 1/C_j$. This is equivalent to the LP to HP transformation $s \to 1/s$, because the impedance $R$ becomes $1/(sC)$ and $1/(sC)$ becomes $R$. Resistors that set the amplifier gain ($R_A$, $R_B$) are not changed, since they do not depend on frequency.

Applying it to the Sallen-Key LPF

Sallen-Key LPF: $R_1$, $R_2$ in series path, $C_1$ feedback to output, $C_2$ to ground, gain $K$:

$$T_{LP}(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}$$

After RC-CR: the series elements become capacitors $C_1$, $C_2$; the feedback element and the grounded element become resistors $R_1$ (to output) and $R_2$ (to ground).

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

Analysing this circuit gives

$$T_{HP}(s) = \frac{Ks^2}{s^2 + s\left(\frac{1}{R_2C_1} + \frac{1}{R_2C_2} + \frac{1-K}{R_1C_1}\right) + \frac{1}{R_1R_2C_1C_2}}$$

For equal components ($R_1 = R_2 = R$, $C_1 = C_2 = C$): $\omega_0 = 1/RC$, $\omega_0/Q = (3-K)/RC$ (same as the LPF) and high-frequency gain $K$. So a normalized LPF with $C = 1$ F, $R = 1\ \Omega$ becomes an HPF with $R = 1\ \Omega$, $C = 1$ F at the same $\omega_0$ and $Q$.

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).

Tow-Thomas high-pass biquad

The basic Tow-Thomas loop (lossy integrator OA1, integrator OA2, inverter OA3) gives LP and BP outputs. A high-pass output is obtained by feeding the input to OA1 through a capacitor $C_f$ instead of the resistor $R_4$; the output is taken at OA1.

            R1
          +--/\/\--+
          |   C1   |
          +---||---+
          |        |
Vin--Cf---+--[OA1]-+--> V1 (high-pass output)
          ^
          |  V1--R3-->[OA2 integrator, C2]--> V2
          |  V2--r--->[OA3 inverter, r/r]---> V3=-V2
          |                                    |
          +------------------R2----------------+

Derivation. KCL at OA1's inverting input (virtual ground):

$$sC_fV_{in} + \frac{V_3}{R_2} + V_1\left(\frac{1}{R_1} + sC_1\right) = 0$$

From OA2 and OA3: $V_3 = -V_2 = \dfrac{V_1}{sR_3C_2}$. Substituting:

$$V_1\left(sC_1 + \frac{1}{R_1} + \frac{1}{sR_2R_3C_2}\right) = -sC_fV_{in}$$

Multiplying by $s/C_1$:

$$\frac{V_1}{V_{in}} = \frac{-\dfrac{C_f}{C_1}s^2}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}$$

This is a second-order high-pass function with $\omega_0 = 1/\sqrt{R_2R_3C_1C_2}$, $\omega_0/Q = 1/(R_1C_1)$ and high-frequency gain $-C_f/C_1$.

Low-pass Tow-Thomas used for the design

With the resistor $R_4$ at the input (figure above with $C_f$ replaced by $R_4$), the same analysis gives at $V_2$:

$$\frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \quad \omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}},\ \frac{\omega_0}{Q} = \frac{1}{R_1C_1},\ K = \frac{R_2}{R_4}$$

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 450$, $\omega_d = 893.03$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 450^2 + 893.03^2 = 1{,}000{,}002.6 \;\Rightarrow\; \omega_0 \approx 1000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{1000}{900} = 1.111 \end{aligned}$$

Required transfer function (dc gain $K = 1.5$):

$$T(s) = \frac{1.5\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{1.5 \times 10^{6}}{s^2 + 900\,s + 10^{6}}$$

Design equations (choose $C_1 = C_2 = C$, $R_2 = R_3 = R$):

$$R = \frac{1}{\omega_0 C}, \qquad R_1 = QR = \frac{Q}{\omega_0 C}, \qquad R_4 = \frac{R_2}{K}, \qquad r = \text{any convenient value}$$

Step 2: Element values. Choose $C_1 = C_2 = C = 0.1\ \mu$F (a standard value that gives resistors in the k$\Omega$ range).

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{1000 \times 0.1\times10^{-6}} = 10\ \text{k}\Omega \\ R_1 &= QR = 1.111 \times 10\ \text{k}\Omega = 11.11\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{10\ \text{k}\Omega}{1.5} = 6.667\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 1.5.

Answer: $C = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).

Tow-Thomas biquad circuit

The Tow-Thomas biquad is a two-integrator-loop circuit using three op-amps: a lossy integrator, an ideal integrator and an inverter in a feedback loop. It gives low-pass and band-pass outputs at the same time, has low sensitivity, and lets $\omega_0$, $Q$ and gain be tuned independently.

            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.

Derivation (ideal op-amps, so each inverting input is a virtual ground):

KCL at the inverting input of OA1:

$$\frac{V_{in}}{R_4} + \frac{V_3}{R_2} + V_1\left(\frac{1}{R_1} + sC_1\right) = 0$$

OA2 (integrator) and OA3 (inverter):

$$V_2 = -\frac{V_1}{sR_3C_2}, \qquad V_3 = -V_2 = \frac{V_1}{sR_3C_2} \;\Rightarrow\; V_1 = sR_3C_2V_3$$

Substituting $V_1$ into the KCL equation:

$$\begin{aligned} sR_3C_2V_3\left(\frac{1}{R_1} + sC_1\right) + \frac{V_3}{R_2} &= -\frac{V_{in}}{R_4} \\ V_3\left(s^2R_3C_1C_2 + \frac{sR_3C_2}{R_1} + \frac{1}{R_2}\right) &= -\frac{V_{in}}{R_4} \end{aligned}$$

Dividing by $R_3C_1C_2$:

$$\frac{V_3}{V_{in}} = \frac{-\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \qquad \frac{V_2}{V_{in}} = -\frac{V_3}{V_{in}}$$

Comparing with $T(s) = \dfrac{\pm K\omega_0^2}{s^2 + (\omega_0/Q)s + \omega_0^2}$:

$$\omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}}, \qquad \frac{\omega_0}{Q} = \frac{1}{R_1C_1}, \qquad K = \frac{R_2}{R_4}\ \text{(dc gain)}$$

$V_2$ is a non-inverting low-pass output and $V_3$ an inverting one; $V_1$ is a band-pass output.

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 450$, $\omega_d = 893.03$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 450^2 + 893.03^2 = 1{,}000{,}002.6 \;\Rightarrow\; \omega_0 \approx 1000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{1000}{900} = 1.111 \end{aligned}$$

Required transfer function (dc gain $K = 1.3$):

$$T(s) = \frac{1.3\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{1.3 \times 10^{6}}{s^2 + 900\,s + 10^{6}}$$

Design equations (choose $C_1 = C_2 = C$, $R_2 = R_3 = R$):

$$R = \frac{1}{\omega_0 C}, \qquad R_1 = QR = \frac{Q}{\omega_0 C}, \qquad R_4 = \frac{R_2}{K}, \qquad r = \text{any convenient value}$$

Step 2: Element values. Choose $C_1 = C_2 = C = 0.1\ \mu$F (a standard value that gives resistors in the k$\Omega$ range).

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{1000 \times 0.1\times10^{-6}} = 10\ \text{k}\Omega \\ R_1 &= QR = 1.111 \times 10\ \text{k}\Omega = 11.11\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{10\ \text{k}\Omega}{1.3} = 7.692\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

A 1000 rad/s (about 159 Hz) cutoff with $Q \approx 1.1$ passes the ECG band and removes higher-frequency noise and interference. Use 1% metal-film resistors (e.g. 7.68 k$\Omega$ for $R_4$, 11.1 k$\Omega$ for $R_1$) and low-noise op-amps.

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 7.692\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 1.3.

Answer: $C = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 7.692\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).

The Tow-Thomas biquad realizes $T(s) = \dfrac{K\omega_0^2}{s^2 + (\omega_0/Q)s + \omega_0^2}$ with three op-amps (lossy integrator, integrator, inverter).

            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 low-pass transfer function (output $V_2$) is

$$\frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \quad \omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}},\ \frac{\omega_0}{Q} = \frac{1}{R_1C_1},\ K = \frac{R_2}{R_4}$$

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 1000$, $\omega_d = 9949.87$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 1000^2 + 9949.87^2 = 99{,}999{,}913 \approx 10^8 \;\Rightarrow\; \omega_0 \approx 10{,}000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{10000}{2000} = 5 \end{aligned}$$

Required transfer function (dc gain $K = 1.5$):

$$T(s) = \frac{1.5\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{1.5 \times 10^{8}}{s^2 + 2000\,s + 10^{8}}$$

Design equations (choose $C_1 = C_2 = C$, $R_2 = R_3 = R$):

$$R = \frac{1}{\omega_0 C}, \qquad R_1 = QR = \frac{Q}{\omega_0 C}, \qquad R_4 = \frac{R_2}{K}, \qquad r = \text{any convenient value}$$

Step 2: Element values. Given $C_1 = C_2 = C = 0.01\ \mu$F.

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{10{,}000 \times 0.01\times10^{-6}} = 10\ \text{k}\Omega \\ R_1 &= QR = 5 \times 10\ \text{k}\Omega = 50\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{10\ \text{k}\Omega}{1.5} = 6.667\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.01\ \mu\text{F}$, $R_1 = 50\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 1.5.

Answer: $C = 0.01\ \mu\text{F}$, $R_1 = 50\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).