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

# Short note: RMS of sine wave using log-antilog

Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications

- Asked in: 2074 Asoj, 3 marks; 2074 Chaitra, 3 marks; 2073 Shrawan, 3 marks; 2073 Chaitra, 5 marks; 2072 Chaitra, 3 marks (asked 5 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/short-note-rms-log-antilog
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6

## Question

Write a short note on generation of RMS output of sinusoidal wave (process of RMS detection of sine wave) using log antilog application.

## Short answer

A precision rectifier, log amp, and a gain of –2 produce v² after antilog conversion. Averaging gives the mean square, and a subsequent log amp with gain –1/2 followed by an antilog yields the square root, giving true RMS output (Vrms = 0.707 Vm for a sine wave).

## Answer

The RMS value of a signal is the square root of the mean of its square:

```text
Vrms = √( (1/T) ∫₀^T v² dt )
```

For a sine wave v = Vm sin ωt, Vrms = Vm/√2 = 0.707 Vm. Log and antilog amplifiers can compute this directly. Squaring and square-rooting become multiplication of a log by 2 and by ½.

**Block diagram (explicit method):**
```text
 v(t) -> [Precision rectifier |v|]
          -> [LOG] -> [x(-2)] -> [ANTILOG] = v^2
          -> [LPF / averager]              = mean(v^2)
          -> [LOG] -> [x(-1/2)] -> [ANTILOG]
          -> Vrms
```

**Steps:**
1. A precision full-wave rectifier gives |v|, because a log amp accepts only one polarity.
2. The log amp gives −K ln|v|. An inverting amplifier of gain −2 makes it 2K ln|v|. The antilog amp then gives **v²**.
3. A low-pass filter or averager, with a time constant much longer than the signal period, gives the **mean of v²**. For a sine wave this is Vm²/2.
4. A second log amp, a gain of −½ and an antilog amp take the **square root**:

```text
Vo = √( mean(v²) ) = √(Vm²/2) = Vm/√2 = 0.707 Vm
```

**Implicit method (fewer stages):** the output is fed back into the log–antilog block to form Vo = mean(v²/Vo). At steady state Vo² = mean(v²), so Vo = Vrms directly. This gives a wider dynamic range.

The result is a **true-RMS** reading, correct for any waveform and not only sine waves. It is used in true-RMS voltmeters and power measurement. Example IC: AD536.

## Check yourself

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

1. Why is a precision full-wave rectifier used first?
   A. A log amp accepts only one polarity of input
   B. To reduce the amplitude
   C. To filter out noise
   D. To increase gain
   Answer: A. ln of a negative number is undefined, so |v| is needed. (Key point: Why rectifier)

2. How is v^2 obtained from log and antilog blocks?
   A. Log, then divide by 2, then antilog
   B. Antilog then log
   C. Add the log to itself after averaging
   D. Log, multiply by 2 (gain -2), then antilog
   Answer: D. 2 ln|v| in antilog gives v^2. (Key point: Squaring)

3. What is the role of the low-pass filter or averager?
   A. Removes the square root
   B. Rectifies the signal
   C. Gives the mean of v^2, with time constant much longer than the signal period
   D. Gives the peak value
   Answer: C. The mean of v^2 is needed before the square root. (Key point: Averager)

4. What is the output for a sine wave of peak Vm?
   A. Vm sqrt(2)
   B. Vm/sqrt(2) = 0.707 Vm
   C. Vm/2
   D. 2Vm/pi
   Answer: B. Mean(v^2) = Vm^2/2, and its root is Vm/sqrt(2). (Key point: Final result)

## Related in Advanced Electronics

- [RMS of sinusoid using log-antilog amplifiers](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Arms-sinusoid-log-antilog) — Advanced Electronics, chapter 6
- [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×
- [Need for log-antilog circuits; analog multiplier](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Awhy-log-antilog-multiplier) — Advanced Electronics, chapter 6, asked 2×
- [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×
- [Multiplier and divider using log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amultiplier-divider-log-antilog) — Advanced Electronics, chapter 6
- [Multifunction converter using log-antilog](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amultifunction-converter) — Advanced Electronics, chapter 6

## Related in other subjects

- [RMS and average of half-wave rectified sine](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Arms-average-half-wave) — Fundamental of Electrical and Electronics Engineering, chapter 2
- [Generation of sinusoidal emf](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asinusoidal-emf-generation-angular-velocity) — Basic Electrical Engineering, chapter 5
- [QAM: need, generation and detection](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aqam-need-generation-detection) — Communication Systems, chapter 7
- [Average and RMS of rectified wave](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Aaverage-rms-rectified-wave-10v) — Basic Electrical Engineering, chapter 5