---
title: "Short note: M-ary baseband communication system"
subject: "Communication Systems"
chapter: 6
key: "semester_6/communication_systems:short-note-mary-baseband"
source: https://www.bctengineer.com/q/semester_6/communication_systems/short-note-mary-baseband
site: BCT Engineering
---

# Short note: M-ary baseband communication system

Communication Systems · Chapter 6: Baseband Digital Data Transmission

- Asked in: 2079 Bhadra (CS II), 5 marks; 2075 Chaitra (CS II), 5 marks; 2072 Chaitra (CS II), 2.5 marks (asked 3 times)
- Video: [DC#51 M ary PAM transmission and Eye pattern || EC Academy](https://www.youtube.com/watch?v=UuF_eHAu9c8) — 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-mary-baseband
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6

## Question

Write a short note on M-ary baseband communication system.

## Short answer

M-ary baseband groups log2 M bits into one symbol, sending one of M amplitudes. This lowers symbol rate and required bandwidth by a factor of log2 M, but levels are closer, demanding higher power to maintain error rate, trading power for bandwidth efficiency.

## Answer

In an **M-ary baseband system**, the transmitter does not send one bit per pulse. It groups $k = \log_2 M$ bits into one symbol and sends one of $M$ distinct pulse amplitudes (M-ary PAM) for each symbol. Binary transmission is the special case $M = 2$.

#### Working
- The bit stream is split into groups of $k$ bits, e.g. $M = 4$ gives 2 bits per symbol.
- Each group is mapped (usually with Gray code) to one of the levels $\pm 1, \pm 3, \dots, \pm (M-1)$ times $A$.
- The receiver samples each symbol and compares it with $M - 1$ thresholds to decide the level, then maps the level back to bits.

```text
 4-ary PAM (Gray coded)
 +3A  --  10
 +1A  --  11
 -1A  --  01
 -3A  --  00
 bits 10 01 11 00 -> levels +3A -1A +1A -3A
```

#### Key relations
$$
R_s = \frac{R_b}{\log_2 M}, \qquad B_{min} = \frac{R_s}{2} = \frac{R_b}{2 \log_2 M}
$$

So for the same bit rate, the needed bandwidth falls by a factor of $\log_2 M$. The bandwidth efficiency rises to $2\log_2 M$ b/s/Hz (Nyquist).

#### Advantages
- Higher data rate in a band-limited channel (e.g. telephone lines, 2B1Q in ISDN).
- Lower symbol rate, so less ISI and slower symbol timing.

#### Disadvantages
- For the same peak power, the levels are closer together, so the noise margin is smaller and the error probability rises. Keeping the same error rate needs more transmit power (roughly 6 dB more for each doubling of $M$ at large $M$).
- The receiver needs more decision thresholds and accurate gain control.

M-ary signalling therefore trades **power for bandwidth**.

## Check yourself

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

1. In an M-ary baseband system with M = 8, how many bits does each symbol carry?
   A. 8
   B. 4
   C. 2
   D. 3
   Answer: D. k = log2 M = log2 8 = 3 bits per symbol. (Key point: Bits per symbol)

2. For a given bit rate Rb, how does the minimum bandwidth change with M-ary signalling?
   A. B = Rb/(2 log2 M), reduced by log2 M
   B. B = Rb x log2 M, increased
   C. B = Rb/2, unchanged
   D. B = M x Rb/2
   Answer: A. Rs = Rb/log2 M and Bmin = Rs/2, so bandwidth falls by a factor log2 M. (Key point: Symbol rate and bandwidth)

3. How many decision thresholds does an M-ary PAM receiver need?
   A. M + 1
   B. M
   C. M - 1
   D. log2 M
   Answer: C. To distinguish M levels, M - 1 thresholds are needed. (Key point: Receiver decision)

4. What is the main cost of using M-ary instead of binary signalling?
   A. Larger bandwidth
   B. Higher symbol rate and more ISI
   C. Smaller noise margin, so more power is needed for the same error rate
   D. No clock recovery possible
   Answer: C. For the same peak power the levels lie closer together, so the error probability rises; M-ary trades power for bandwidth. (Key point: Trade-off)

## Related in Communication Systems

- [Binary baseband error probability; comparison with M-ary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-prob-binary-vs-mary) — Communication Systems, chapter 9, asked 3×
- [Error probability in binary baseband system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aerror-probability-binary-baseband) — Communication Systems, chapter 9, asked 2×
- [Error probability of M-ary system](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amary-error-probability) — Communication Systems, chapter 9, asked 2×
- [ISI in baseband systems; ideal and practical solutions](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aisi-ideal-practical-solutions) — Communication Systems, chapter 6, asked 2×
- [FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-10khz-100mhz-75khz) — Communication Systems, chapter 3
- [Error probability of binary ASK extended to M-ary](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aask-error-prob-mary) — Communication Systems, chapter 9

## Related in other subjects

- [M-ary QAM; 16-QAM vs 16-PSK](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amary-qam-vs-psk) — Wireless Communication, chapter 4
- [Block diagram of digital communication system](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Ablock-diagram-digital-comm) — Data Communication, chapter 1, asked 8×
- [Communication system and block diagram](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Acomm-system-block-diagram) — Basic Electronics Engineering, chapter 5, asked 7×
- [Elements of optical communication system](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aoptical-communication-elements) — Propagation and Antenna, chapter 6, asked 3×