---
title: "Distance by phase comparison method"
subject: "Surveying I"
chapter: 2
key: "semester_3/surveying_i:phase-comparison-distance"
source: https://www.bctengineer.com/q/semester_3/surveying_i/phase-comparison-distance
site: BCT Engineering
---

# Distance by phase comparison method

Surveying I · Chapter 2: Distance Measurements

- Asked in: 2068 Chaitra, 6 marks
- Video: [ELECTRONIC DISTANCE MEASUREMENT (EDM-PART-3)( Phase comparison method) in Hindi.](https://www.youtube.com/watch?v=mxk3xsTcYAU) — Ekta Dwivedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/phase-comparison-distance
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-2

## Question

Explain briefly how a distance can be measured by the method of phase comparison.

## Short answer

A modulated continuous wave is sent to a reflector; the phase difference φ between transmitted and received signals is measured. The distance is D = ½(N + φ/2π)λ, where N is the integer number of wavelengths, resolved by using multiple modulation frequencies.

## Answer

In the phase-comparison method of EDM, a continuous carrier wave is modulated with a known frequency, sent to a reflector and compared with the returned wave. The phase difference gives the distance.

#### Principle
Let the modulation frequency be $f$ and velocity $v$, so wavelength $\lambda = v/f$. The wave travels from the instrument A to the reflector B and back, a total of $2D$. Hence

$$
2D = N\,\lambda + \Delta\lambda, \qquad \Delta\lambda = \frac{\phi}{2\pi}\,\lambda
$$

$$
D = \frac{1}{2}\left(N + \frac{\phi}{2\pi}\right)\lambda
$$

where $N$ is the integer number of complete wavelengths in the double path and $\phi$ the phase difference (in radians) measured by the instrument.

```text
 Sent wave      /\  /\  /\  /\ ...
 Received wave   /\  /\  /\ ...
                  |<-->| phase shift (phi)
```

#### Steps
1. The transmitted wave and the received wave are compared in a phase detector, and $\phi$ (the fractional part $\Delta\lambda$) is obtained.
2. The unknown integer $N$ cannot be found from one frequency. Hence several modulation frequencies are used, e.g. a low frequency (long wavelength, e.g. 1500 m) for the coarse distance and a higher frequency (short wavelength, e.g. 10 m) for fine reading. The readings of different wavelengths remove the ambiguity.
3. The microprocessor combines the results, applies atmospheric corrections and displays the distance.

#### Example
For $\lambda = 20$ m, $N = 10$ and $\phi = 90^\circ$ ($\tfrac{1}{4}$ cycle):

$$
D = \frac{1}{2}(10 + 0.25)\times 20 = 102.5\ \text{m}
$$

## Check yourself

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

1. What is compared in the phase-comparison method?
   A. Sent and received waves, giving phase difference phi
   B. Two tapes
   C. Two prisms
   D. Temperatures
   Answer: A. The phase detector compares transmitted and returned modulated waves. (Key point: Principle)

2. The distance in terms of N, phi and lambda is
   A. D = (N + phi/2pi) lambda
   B. D = 1/2 (N + phi/2pi) lambda
   C. D = 2 (N + phi/2pi) lambda
   D. D = N lambda
   Answer: B. Double path 2D = N lambda + (phi/2pi) lambda, so D is half. (Key point: Formula)

3. Why are several frequencies used?
   A. To resolve the unknown whole number N: coarse long wavelength and fine short wavelength
   B. To save battery
   C. To avoid atmosphere
   D. For redundancy of the prism
   Answer: A. One frequency gives only the fractional part. (Key point: Ambiguity)

4. lambda = 20 m, N = 10, phi = 90 degrees: what is D?
   A. 102.5 m
   B. 205 m
   C. 100 m
   D. 51.25 m
   Answer: A. D = 1/2 (10 + 0.25) x 20 = 102.5 m. (Key point: Worked example)

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