---
title: "Thermocouple time constant and frequency response"
subject: "Instrumentation and Measurement"
chapter: 4
key: "semester_4/instrumentation_and_measurement:thermocouple-first-order-response-numerical"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/thermocouple-first-order-response-numerical
site: BCT Engineering
---

# Thermocouple time constant and frequency response

Instrumentation and Measurement · Chapter 4: Dynamic Response of Measurement System

- Asked in: year not given, 8 marks
- Video: [Time Constant and Response of Thermocouple](https://www.youtube.com/watch?v=Gr0drXXI8xU) — Rahul Bawane
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/thermocouple-first-order-response-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-4

## Question

A thermocouple of time constant 2 s, initially at 25 $^\circ$C, is suddenly plunged into a bath at 200 $^\circ$C. Treat it as a first-order system of unit static sensitivity. Find (a) the time taken to indicate 190 $^\circ$C, (b) the indication after 5 s, (c) the time to reach 99% of the step, and (d) the amplitude ratio and phase lag if the bath temperature varies sinusoidally at 0.1 Hz.

## Short answer

Using T(t) = 200 - 175e^(-t/2). (a) Time to 190°C is 5.72 s. (b) Reading at 5 s is 185.6°C. (c) 99% step time is 9.21 s. (d) At 0.1 Hz, ωτ = 1.2566, amplitude ratio 0.623, phase lag 51.5°.

## Answer

For a first-order system with unit sensitivity, a step from $T_i = 25\ ^\circ$C to $T_\infty = 200\ ^\circ$C gives

$$
T(t) = T_\infty - (T_\infty - T_i)e^{-t/\tau} = 200 - 175\,e^{-t/2}
$$

#### (a) Time to indicate 190 $^\circ$C
$$
190 = 200 - 175e^{-t/2} \Rightarrow e^{-t/2} = \frac{10}{175} \Rightarrow t = -2\ln(0.05714) = 5.72\ \text{s}
$$

#### (b) Reading after 5 s
$$
T = 200 - 175e^{-5/2} = 200 - 175(0.0821) = 185.6\ ^\circ\text{C}
$$

#### (c) Time for 99% of the step
$$
e^{-t/\tau} = 0.01 \Rightarrow t = \tau\ln 100 = 2(4.605) = 9.21\ \text{s}
$$

#### (d) Sinusoidal bath temperature at 0.1 Hz
$\omega = 2\pi(0.1) = 0.6283$ rad/s, so $\omega\tau = 1.2566$.

$$
M = \frac{1}{\sqrt{1 + (\omega\tau)^2}} = \frac{1}{\sqrt{1 + 1.579}} = 0.623
$$

$$
\phi = -\tan^{-1}(\omega\tau) = -\tan^{-1}(1.2566) = -51.5^\circ
$$

The thermocouple shows only 62.3% of the true amplitude and lags by $51.5^\circ$ (time lag $= 51.5/360 \times 10 = 1.43$ s).

**Answer: (a) 5.72 s; (b) 185.6 $^\circ$C; (c) 9.21 s; (d) amplitude ratio 0.623, phase lag $51.5^\circ$.**

## Check yourself

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

1. For a step from 25 to 200 °C with tau = 2 s, the indication is
   A. 200 - 175 e^(-t/2)
   B. 25 + 175 e^(-t/2)
   C. 200 e^(-t/2)
   D. 200 - 25 e^(-t/2)
   Answer: A. T = T_inf - (T_inf - T_i)e^(-t/tau). (Key point: Response formula)

2. What does the thermocouple read after 5 s?
   A. 25 °C
   B. 185.6 °C
   C. 175 °C
   D. 199 °C
   Answer: B. 200 - 175 e^-2.5 = 200 - 175(0.0821) = 185.6 °C. (Key point: Reading after 5 s)

3. Time to reach 99% of the step is
   A. 4.6 s
   B. 9.21 s
   C. 5.72 s
   D. 10 s
   Answer: B. t = tau ln 100 = 2(4.605) = 9.21 s. (Key point: 99% time)

4. At 0.1 Hz, ωτ = 1.2566. The amplitude ratio and phase lag are
   A. 0.623 and 51.5°
   B. 0.623 and 38.5°
   C. 0.79 and 51.5°
   D. 1.0 and 0°
   Answer: A. M = 1/sqrt(1+1.579) = 0.623 and φ = -tan^-1(1.2566) = -51.5°. (Key point: Frequency response)

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