---
title: "RMS of sinusoid using log-antilog amplifiers"
subject: "Advanced Electronics"
chapter: 6
key: "semester_4/advanced_electronics:rms-sinusoid-log-antilog"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/rms-sinusoid-log-antilog
site: BCT Engineering
---

# RMS of sinusoid using log-antilog amplifiers

Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications

- Asked in: 2069 Chaitra, 7 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/rms-sinusoid-log-antilog
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6

## Question

How can you find the RMS value of a sinusoidal signal using log and antilog amplifiers? Explain with necessary derivations and circuit diagrams.

## Short answer

The circuit squares the input by taking 2·ln|Vi|, subtracts ln(Vo), antilogs to get Vi²/Vo, which is averaged by a low-pass filter. Feedback forces Vo = √(avg(Vi²)), giving true RMS. For a sinusoid Vi = Vm sin ωt, Vo = Vm/√2 = 0.707·Vm, e.g. 7.07 V for Vm = 10 V.

## Answer

The RMS value is Vrms = √( average of Vi² ). A log-antilog circuit computes this **implicitly**: it forms Vi²/Vo, averages it and feeds the result back as Vo. This avoids a separate squarer whose output range would be very large.

#### Principle

```text
Vo = avg( Vi² / Vo )
Vo is DC, so Vo = avg(Vi²)/Vo
Vo² = avg(Vi²)
Vo  = √avg(Vi²) = Vrms
```

#### Circuit (block form)

```text
Vi -->[ABSOLUTE ]--|Vi|-->[LOG]--V1--x2--+
      [VALUE CKT]                         |
                                     [SUBTRACT]--V3
         +-->[LOG]--V2------------->(−)   |
         |                           [ANTILOG]--V4
         |                                |
         |                         [LOW-PASS RC]
         |                         [ (averager) ]
         |                                |
         +--------------------------------+--> Vo
```

#### Derivation

1. Precision full-wave rectifier gives |Vi| (log amp needs a positive input; |Vi|² = Vi²).
2. Log amp and gain of 2:
```text
V1 = 2V_T ln(|Vi|/Vref) = V_T ln(Vi²/Vref²)
```
3. Log of the output (feedback):
```text
V2 = V_T ln(Vo/Vref)
```
4. Subtractor:
```text
V3 = V1 − V2 = V_T ln( Vi² / (Vref·Vo) )
```
5. Antilog:
```text
V4 = Vref·e^(V3/V_T) = Vi² / Vo
```
6. Low-pass filter (time constant much larger than signal period) takes the average:
```text
Vo = avg(Vi²/Vo) = avg(Vi²)/Vo
Vo = √avg(Vi²)
```

#### For a sinusoid Vi = Vm sin ωt

```text
avg(Vi²) = (1/T)∫₀ᵀ Vm² sin²ωt dt
         = (Vm²/T)∫₀ᵀ (1 − cos2ωt)/2 dt
         = Vm²/2
Vo = √(Vm²/2) = Vm/√2 = 0.707·Vm
```

Example: for Vi = 10 sin ωt V, the circuit gives Vo = 7.07 V DC.

Advantages: it is a **true RMS** converter (correct for any waveform, not only sine), and its dynamic range is wide because the internal signal Vi²/Vo stays close to Vi in size.

## Check yourself

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

1. What equation does the implicit RMS circuit satisfy at steady state?
   A. Vo = avg(Vi^2)
   B. Vo = avg(Vi^2)/Vo, so Vo = sqrt(avg Vi^2)
   C. Vo = avg(Vi)
   D. Vo = peak(Vi)
   Answer: B. Vo is DC, so Vo^2 = avg(Vi^2). (Key point: Implicit principle)

2. What does the antilog stage give?
   A. Vi^2/Vo
   B. Vi Vo
   C. Vi/Vo^2
   D. sqrt(Vi) Vo
   Answer: A. V3 = VT ln(Vi^2/(Vref Vo)), so V4 = Vi^2/Vo. (Key point: Antilog output)

3. What is the role of the low-pass RC?
   A. Squares the signal
   B. Rectifies the signal
   C. Removes Vref
   D. Averages Vi^2/Vo; its time constant must be much longer than the signal period
   Answer: D. Averaging produces a steady DC output. (Key point: LPF role)

4. For Vi = 10 sin(wt) V, what is Vo?
   A. 10 V
   B. 5 V
   C. 14.14 V
   D. 7.07 V
   Answer: D. 10/sqrt(2) = 7.07 V. (Key point: Sine example)

## Related in Advanced Electronics

- [Short note: RMS of sine wave using log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ashort-note-rms-log-antilog) — Advanced Electronics, chapter 6, asked 5×
- [Short note: Applications of log-antilog amplifiers](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ashort-note-log-antilog-applications) — Advanced Electronics, chapter 6, asked 2×
- [Design Vo = (Vx/Vy)^α·Vz with log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adesign-vx-vy-power-alpha) — Advanced Electronics, chapter 6, asked 2×
- [Multifunction converter using log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amultifunction-converter) — Advanced Electronics, chapter 6
- [Log-antilog circuit: Vo in terms of Vi and VR](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-antilog-circuit-vo-vi) — Advanced Electronics, chapter 5, asked 3×
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