---
title: "Tow-Thomas band pass biquad transfer function"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:tow-thomas-bandpass-derivation"
source: https://www.bctengineer.com/q/semester_5/filter_design/tow-thomas-bandpass-derivation
site: BCT Engineering
---

# Tow-Thomas band pass biquad transfer function

Filter Design · Chapter 6: Active Filters

- Asked in: 2082 Bhadra, 5 marks
- Video: [Tow-Thomas Biquad Filters (2): Op-Amp RC Circuit](https://www.youtube.com/watch?v=OBJmKUgmBNc) — Semiconductor Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/tow-thomas-bandpass-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

Draw a neat circuit diagram of Tow-Thomas band-pass biquad and derive its transfer function.

## Short answer

The band-pass output at V1 gives T(s) = -s/(R4 C1) / [s² + s/(R1 C1) + 1/(R2 R3 C1 C2)]. Centre frequency ω0 = 1/√(R2 R3 C1 C2), bandwidth = 1/(R1 C1), peak gain = R1/R4. The circuit uses a lossy integrator (R1||C1 in feedback), an integrator (C2 feedback), and an inverter.

## Answer

#### Tow-Thomas band-pass biquad

The band-pass output of the Tow-Thomas biquad is taken at the output of the lossy integrator OA1.

```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.** KCL at the inverting input of OA1 (virtual ground):

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

OA2 integrates and OA3 inverts:

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

Substituting $V_3$:

$$
V_1\left(sC_1 + \frac{1}{R_1} + \frac{1}{sR_2R_3C_2}\right) = -\frac{V_{in}}{R_4}
$$

Multiplying both sides by $s/C_1$:

$$
\frac{V_1}{V_{in}} = \frac{-\dfrac{s}{R_4C_1}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}
$$

Comparing with the standard band-pass form $T(s) = \dfrac{-H(\omega_0/Q)s}{s^2 + (\omega_0/Q)s + \omega_0^2}$:

$$
\omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}}, \qquad \text{BW} = \frac{\omega_0}{Q} = \frac{1}{R_1C_1}, \qquad Q = \omega_0R_1C_1
$$

and the gain at the centre frequency $s = j\omega_0$ is

$$
|T(j\omega_0)| = \frac{1/(R_4C_1)}{1/(R_1C_1)} = \frac{R_1}{R_4}
$$

So $\omega_0$ is set by $R_2, R_3$, the bandwidth by $R_1$ and the peak gain by $R_4$, independently.

## Check yourself

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

1. At which node of the Tow-Thomas biquad is the band-pass output taken?
   A. Output of the inverter OA3
   B. Output of the ideal integrator OA2
   C. Output of the lossy integrator OA1
   D. The input node
   Answer: C. V1, the output of the lossy integrator OA1, is the band-pass output. (Key point: Where the band-pass output is)

2. What is the starting point of the derivation?
   A. KCL at the inverting input of OA1 (virtual ground)
   B. KVL around the capacitor C2 only
   C. Setting Vin = 0
   D. Assuming infinite Q
   Answer: A. Ideal op-amps give a virtual ground, so KCL at OA1 inverting node relates Vin, V3 and V1. (Key point: Starting equation)

3. From the derived T(s), the bandwidth w0/Q equals:
   A. 1/(R2C2)
   B. 1/(R1C1)
   C. 1/(R4C1)
   D. R1/R4
   Answer: B. The s coefficient in the denominator is 1/(R1C1). (Key point: Bandwidth result)

4. What is the magnitude of the gain at s = jw0?
   A. R1/R4
   B. R4/R1
   C. R2/R4
   D. 1
   Answer: A. |T(jw0)| = (1/(R4C1))/(1/(R1C1)) = R1/R4. (Key point: Centre gain)

## Related in Filter Design

- [Tow-Thomas band pass biquad design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-bandpass-design) — Filter Design, chapter 6
- [Tow-Thomas high pass derivation; LPF design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-highpass-derive-lpf-design) — Filter Design, chapter 6, asked 3×
- [Tow-Thomas LPF, poles -450 ± j893.03, gain 1.5](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-lpf-450-893) — Filter Design, chapter 6, asked 3×
- [Tow-Thomas LPF, poles -4000 ± j39799.4975](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-lpf-4000-39799) — Filter Design, chapter 6, asked 3×
- [Tow-Thomas LPF, poles -500 ± j2449.49](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-lpf-500-2449) — Filter Design, chapter 6, asked 2×
- [Tow-Thomas LPF for ECG (gain 1.3)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-ecg-lpf-design) — Filter Design, chapter 6

## Related in other subjects

- [Transfer function of RC low-pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Arc-lowpass-transfer-function) — Signals and Systems, chapter 5, asked 2×
- [Transfer function of DT low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-lowpass-transfer-function) — Signals and Systems, chapter 6