---
title: "Transfer function of op-amp with Rs ∥ C feedback"
subject: "Control Systems"
chapter: 2
key: "semester_3/control_systems:tf-opamp-rs-c-r1-r2"
source: https://www.bctengineer.com/q/semester_3/control_systems/tf-opamp-rs-c-r1-r2
site: BCT Engineering
---

# Transfer function of op-amp with Rs ∥ C feedback

Control Systems · Chapter 2: Component Modeling

- Asked in: 2074 Asoj, 4 marks
- 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-rs-c-r1-r2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-2

## Question

Find transfer function of an op-amp model as below. [Figure: inverting amplifier; Vin through R1 to the inverting (−) input; feedback from output to the inverting input through Rs in parallel with capacitor C; non-inverting (+) input connected to ground through R2; output Vout]

## Short answer

Zf = Rs || (1/sC) = Rs/(1+sRsC), Z1 = R1. With virtual ground, Vout/Vin = -Zf/Z1 = -(Rs/R1)/(1+sRsC). This is an inverting first-order low-pass filter with DC gain -Rs/R1 and pole at -1/(RsC). R2 provides bias compensation and does not affect the transfer function.

## Answer

This is an **inverting amplifier** with a feedback impedance $Z_f = R_s \parallel C$; $R_2$ on the non-inverting input only compensates bias current.

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

#### Impedances
$$
Z_1 = R_1, \qquad Z_f = R_s \parallel \frac{1}{sC} = \frac{R_s\cdot\frac{1}{sC}}{R_s + \frac{1}{sC}} = \frac{R_s}{1 + sR_sC}
$$

#### Derivation
Current through $R_1$ equals current through $Z_f$:
$$
\frac{V_{in}(s) - 0}{R_1} = \frac{0 - V_{out}(s)}{Z_f}
$$
$$
\frac{V_{out}(s)}{V_{in}(s)} = -\frac{Z_f}{Z_1}
$$

#### Transfer function
$$
\frac{V_{out}(s)}{V_{in}(s)} = -\frac{R_s}{R_1}\cdot\frac{1}{1 + sR_sC}
$$
This is a first-order lag (inverting low-pass) circuit:
- DC gain $= -R_s/R_1$
- pole at $s = -\dfrac{1}{R_sC}$, time constant $\tau = R_sC$

For example, with $R_1 = 10\ \text{k}\Omega$, $R_s = 100\ \text{k}\Omega$, $C = 1\ \mu\text{F}$, the gain is $-10/(1 + 0.1s)$. If $R_s \to \infty$ the circuit becomes a pure integrator $-1/(sR_1C)$.

## Check yourself

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

1. Zf for Rs in parallel with C is:
   A. Rs/(1 + sRsC)
   B. sRsC
   C. Rs + 1/(sC)
   D. Rs C
   Answer: A. Rs || 1/(sC) = Rs/(1 + sRsC). (Key point: Feedback impedance)

2. Vout/Vin equals:
   A. -R1/(Rs(1 + sRsC))
   B. -(Rs/R1) / (1 + sRsC)
   C. (Rs/R1)(1 + sRsC)
   D. (1 + Rs/R1)
   Answer: B. Gain is -Zf/Z1 for an inverting amplifier. (Key point: Transfer function)

3. The pole and DC gain are:
   A. Pole at -R1C
   B. Pole at -1/(R1C), gain Rs
   C. Pole at 0, gain 1
   D. Pole at s = -1/(RsC), DC gain -Rs/R1
   Answer: D. At s=0 gain is -Rs/R1; denominator zero at s = -1/(RsC), a lag. (Key point: Behaviour)

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