---
title: "Tow-Thomas 2nd order Butterworth LPF at 5 kHz"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:tow-thomas-butterworth-5khz"
source: https://www.bctengineer.com/q/semester_5/filter_design/tow-thomas-butterworth-5khz
site: BCT Engineering
---

# Tow-Thomas 2nd order Butterworth LPF at 5 kHz

Filter Design · Chapter 6: Active Filters

- Asked in: 2071 Shrawan, 4+4 marks
- Video: [BiQuad Filter Circuit Design and Transfer Function Analysis](https://www.youtube.com/watch?v=U3AR1XCkX8k) — Engineering Prof.
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/tow-thomas-butterworth-5khz
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

Draw the circuit diagram of Tow Thomas biquad filter and derive its lowpass transfer function. Design a second order Butterworth lowpass filter having half power frequency of 5 kHz using Tow Thomas biquad circuit. Your final circuit should have all capacitors of 0.001μF.

## Short answer

For a second-order Butterworth LPF with half-power frequency 5 kHz: ω0=2π×5000=31416 rad/s, Q=1/√2=0.7071. With C1=C2=0.001 μF and R2=R3=R: R=1/(ω0C)=31.83 kΩ, R1=QR=22.51 kΩ, R4=R2=31.83 kΩ for unity gain, r=10 kΩ. Output at V2.

## Answer

#### 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.

```text
            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: 2nd order Butterworth LPF, $f_c = 5$ kHz

**Step 1: Parameters.** For a 2nd order Butterworth filter the half-power frequency equals $\omega_0$ and $Q = 1/\sqrt2 = 0.7071$ (poles at $135^\circ$, from $B(s) = s^2 + \sqrt2 s + 1$).

$$
\omega_0 = 2\pi f_c = 2\pi \times 5000 = 31{,}415.9\ \text{rad/s}
$$

No gain is specified, so take dc gain $K = 1$:

$$
T(s) = \frac{\omega_0^2}{s^2 + \sqrt2\,\omega_0 s + \omega_0^2}
$$

**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** with the given $C_1 = C_2 = C = 0.001\ \mu$F:

$$
\begin{aligned}
R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{31{,}415.9 \times 10^{-9}} = 31.83\ \text{k}\Omega \\
R_1 &= QR = 0.7071 \times 31.83 = 22.51\ \text{k}\Omega \\
R_4 &= \frac{R_2}{K} = 31.83\ \text{k}\Omega \\
r &= 10\ \text{k}\Omega
\end{aligned}
$$

(Equivalently: normalized design $C = 1$ F, $R_2 = R_3 = R_4 = 1\ \Omega$, $R_1 = 0.7071\ \Omega$, then frequency scale by $k_f = 31{,}415.9$ and impedance scale by $k_m = 1/(k_f \times 10^{-9}) = 31{,}831$.)

**Answer:** $C_1 = C_2 = 0.001\ \mu$F, $R_1 = 22.51$ k$\Omega$, $R_2 = R_3 = R_4 = 31.83$ k$\Omega$, $r = 10$ k$\Omega$; output at $V_2$, half-power frequency 5 kHz, unity dc gain.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a second-order Butterworth low-pass, Q equals:
   A. 1
   B. 0.5
   C. 1.414
   D. 0.7071
   Answer: D. The denominator s^2 + sqrt2 s + 1 gives Q = 1/sqrt2. (Key point: Butterworth Q)

2. For second-order Butterworth, w0 for a 5 kHz half-power frequency is:
   A. 5000 rad/s
   B. 2π x 5000 = 31,416 rad/s
   C. 5000/2π rad/s
   D. 2 x 5000 rad/s
   Answer: B. For Butterworth the half-power frequency equals w0, and w0 = 2πf. (Key point: w0 from half-power)

3. With C = 0.001 µF, R = 1/(w0C) is about:
   A. 3.18 kΩ
   B. 318 kΩ
   C. 31.83 kΩ
   D. 22.5 kΩ
   Answer: C. 1/(31416 x 10^-9) = 31.83 kΩ. (Key point: R2 = R3)

4. Which gives R1 = QR?
   A. 31.83 kΩ
   B. 22.51 kΩ
   C. 45.0 kΩ
   D. 10 kΩ
   Answer: B. 0.7071 x 31.83 kΩ = 22.51 kΩ. (Key point: R1)

## Related in Filter Design

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