---
title: "Short note: Stereo encoder"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:short-note-stereo-encoder"
source: https://www.bctengineer.com/q/semester_6/communication_systems/short-note-stereo-encoder
site: BCT Engineering
---

# Short note: Stereo encoder

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2080 Chaitra (CS I), 5 marks; 2073 Magh (CS I), 5 marks; 2071 Magh (CS I), 4 marks (asked 3 times)
- Video: [CT#28 FM stereo Multiplexing in communication system || EC Academy](https://www.youtube.com/watch?v=1Jh3tK-tDN4) — EC Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/short-note-stereo-encoder
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

Write a short note on stereo encoder.

## Short answer

A stereo encoder forms L+R and L-R signals, low-pass filters and pre-emphasizes them, modulates L-R onto a 38 kHz subcarrier via DSB-SC, adds a 19 kHz pilot, and sums them to produce composite baseband m(t)=(L+R)+(L-R)cos(2π·38000t)+K cos(2π·19000t) that frequency-modulates the carrier. Total baseband bandwidth is 53 kHz.

## Answer

A **stereo encoder** (stereo multiplexer) combines the left (L) and right (R) audio channels into one composite baseband signal for FM stereo broadcasting, in a way that mono receivers can still use.

**Block diagram**

```text
 L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
     |    ^                            ^ ^  to FM
 R --|----+                            | |  mod.
     +-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
          ^ (R)                 ^ 38k    |
                  [19k osc]->[x2]        |
                      |                  |
                      +--19 kHz pilot----+
```

**Working**

1. A matrix circuit (adder and subtractor) forms $L+R$ (mono-compatible sum) and $L-R$ (difference).
2. Both are low-pass filtered to 15 kHz and pre-emphasized.
3. $L-R$ is DSB-SC modulated in a balanced modulator onto a 38 kHz sub-carrier, giving sidebands from 23 to 53 kHz.
4. The 38 kHz sub-carrier is made by doubling a 19 kHz crystal oscillator. A small 19 kHz **pilot** (about 10% deviation) is added so the receiver can regenerate the 38 kHz carrier in phase.
5. The composite signal is

$$
m(t) = (L+R) + (L-R)\cos(2\pi\,38000\,t) + K\cos(2\pi\,19000\,t)
$$

**Spectrum of composite signal**

```text
 |  L+R      pilot    L-R DSB-SC
 |______       |    ____    ____
 |      |      |   |    |  |    |
 +------+------+---+----+--+----+---> f (kHz)
 0     15     19  23   38      53
```

The total baseband bandwidth is 53 kHz, which then frequency-modulates the carrier. A mono receiver uses only 0–15 kHz ($L+R$); a stereo receiver recovers $L-R$ as well and forms $L$ and $R$ by adding and subtracting.

## Check yourself

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

1. Why can a mono receiver still use a stereo broadcast?
   A. It decodes the 38 kHz subcarrier
   B. The L+R sum is sent in the normal 0–15 kHz baseband
   C. Only the L channel is sent in baseband
   D. The pilot carries the audio
   Answer: B. A mono receiver uses only the 0–15 kHz L+R part and ignores the rest. (Key point: Mono compatibility)

2. How is the L−R signal transmitted?
   A. DSB-SC on a 38 kHz subcarrier, occupying 23–53 kHz
   B. Directly in baseband 0–15 kHz
   C. As AM with carrier on 19 kHz
   D. As SSB at 15 kHz
   Answer: A. A balanced modulator puts L−R on a suppressed 38 kHz subcarrier, giving sidebands 23–53 kHz. (Key point: Role of L−R)

3. What is the 19 kHz pilot for?
   A. To carry the L+R audio
   B. To widen the bandwidth to 75 kHz
   C. To let the receiver regenerate the 38 kHz subcarrier in phase
   D. To act as a noise reference for de-emphasis
   Answer: C. The pilot, at half 38 kHz, is doubled at the receiver to give the exact frequency and phase of the subcarrier. (Key point: Pilot tone)

4. Which is the composite signal?
   A. (L+R) + (L−R)cos(2π·19k t) + K cos(2π·38k t)
   B. (L−R) + (L+R)cos(2π·38k t)
   C. (L+R) + (L−R)cos(2π·38k t) + K cos(2π·19k t)
   D. (L+R)(L−R) + K
   Answer: C. Sum in baseband, difference DSB-SC at 38 kHz and a small 19 kHz pilot are added. (Key point: Composite signal)

## Related in Communication Systems

- [Stereo FM encoder and decoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astereo-fm-encoder-decoder) — Communication Systems, chapter 3, asked 2×
- [Pre-emphasis in FM broadcasting; stereo encoder](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apre-emphasis-stereo-encoder) — Communication Systems, chapter 3, asked 2×
- [Stereo FM encoder and decoder block diagram](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Astereo-fm-encoder-decoder-blocks) — Communication Systems, chapter 3
- [PSD–autocorrelation relation; stereo FM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apsd-autocorrelation-stereo-fm) — Communication Systems, chapter 3
