---
title: "Transfer function of op-amp integrator-type circuit"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-opamp-r1-r2-c"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-opamp-r1-r2-c
site: BCT Engineering
---

# Transfer function of op-amp integrator-type circuit

Control Systems · Chapter 2: Component Modeling

- Asked in: 2080 Baisakh, 4 marks
- Video: [Intro to Control - 2.4 Inverting OpAmp Transfer Function](https://www.youtube.com/watch?v=fC17rdE6uLI) — Engineering with Prof. Kim
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/tf-opamp-r1-r2-c
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find transfer function. [Figure: op-amp circuit; input ein applied through R1 to one op-amp input, with capacitor C connected from that input to the op-amp output; the other op-amp input is connected through R2 to the common (ground) line; output eout taken at the op-amp output]

## Short answer

The circuit is an inverting integrator. With virtual ground, input current E_in/R₁ equals feedback current −sC E_out, giving transfer function E_out(s)/E_in(s) = −1/(sR₁C).

## Answer

The circuit is an **inverting integrator** (the input, through $R_1$, and the capacitor both go to the inverting input). $R_2$ only balances the input bias current; for an ideal op-amp it carries no current, so the non-inverting input is at 0 V.

#### Assumptions (ideal op-amp)
- Infinite input impedance: no current into the op-amp terminals.
- Virtual ground: $V_- = V_+ = 0$ (no current in $R_2$, so no drop across it).

#### Derivation
Current through $R_1$:
$$
I_1(s) = \frac{E_{in}(s) - 0}{R_1}
$$
This whole current flows through $C$ (impedance $1/sC$):
$$
I_1(s) = \frac{0 - E_{out}(s)}{1/(sC)} = -sC\,E_{out}(s)
$$
Equating:
$$
\frac{E_{in}(s)}{R_1} = -sC\,E_{out}(s)
$$

#### Transfer function
$$
\frac{E_{out}(s)}{E_{in}(s)} = -\frac{1}{sR_1C}
$$
In time domain, $e_{out}(t) = -\dfrac{1}{R_1C}\displaystyle\int e_{in}\,dt$. So the circuit is an integrator with time constant $R_1C$ and a sign inversion; it gives a pole at $s = 0$ and is used to realise the integral (I) action of a controller. Usually $R_2$ is chosen equal to $R_1$ to reduce output offset.

## Check yourself

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

1. The circuit with R1 in series with the input and C in the feedback path is:
   A. A voltage follower
   B. An inverting differentiator
   C. An inverting integrator
   D. A non-inverting amplifier
   Answer: C. Feedback capacitor with input resistor forms an inverting integrator. (Key point: Circuit identification)

2. Why does R2 not affect the transfer function?
   A. R2 is in parallel with C
   B. No current flows into the op-amp inputs, so there is no drop across R2 and V+ = 0
   C. R2 is much larger than R1
   D. R2 is shorted by the capacitor
   Answer: B. Ideal infinite input impedance means no current in R2; it only balances bias current. (Key point: Virtual ground and R2)

3. What is Eout(s)/Ein(s)?
   A. -R1/(1 + sR1C)
   B. -sR1C
   C. 1/(sR1C)
   D. -1/(sR1C)
   Answer: D. Setting Ein/R1 = -sC Eout gives -1/(sR1C), an inverting integrator with a pole at s=0. (Key point: Transfer function)

## Related in Control Systems

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- [Transfer function of op-amp with Rs ∥ C feedback](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-rs-c-r1-r2) — Control Systems, chapter 2
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