---
title: "Quadrature amplitude modulation (QAM)"
subject: "Data Communication"
chapter: 5
key: "semester_5/data_communication:qam-explanation"
source: https://www.bctengineer.com/q/semester_5/data_communication/qam-explanation
site: BCT Engineering
---

# Quadrature amplitude modulation (QAM)

Data Communication · Chapter 5: Data Encoding and Modulation

- Asked in: 2076 Chaitra, 4 marks; 2076 Asoj, 6 marks; 2075 Asoj, 6+2 marks; 2071 Shrawan, 8 marks (asked 4 times)
- Video: [L 30 | Quadrature Amplitude Modulation QAM | Analog Comunication | Principles of Communication |GATE](https://www.youtube.com/watch?v=Y3gzVnZ7SFI) — Dopamine
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/qam-explanation
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-5

## Question

Explain Quadrature Amplitude Modulation (QAM) with transmitter and receiver block diagram and a constellation diagram (e.g. 16-QAM or 32-QAM).

## Short answer

QAM combines amplitude and phase modulation by modulating two quadrature carriers (cos and sin) with independent data streams. At the receiver, coherent demodulation recovers I and Q signals. 16-QAM has 16 constellation points, each carrying 4 bits, doubling spectral efficiency.

## Answer

**Quadrature Amplitude Modulation (QAM)** combines amplitude modulation and phase modulation. Two independent message streams amplitude-modulate two carriers of the same frequency that are 90° apart (in quadrature): cos(2πfct) (in-phase, I) and sin(2πfct) (quadrature, Q). Both are summed and sent on the same channel, so twice the data is sent in the same bandwidth.

s(t) = I(t) cos(2πfct) - Q(t) sin(2πfct)

#### Transmitter
```text
 bits   +-----+ I bits +-----+ I(t) +---+
 -----> | S/P |------->| DAC |----->| x |--+
        |     |        +-----+      +---+  v
        |     | Q bits +-----+ Q(t) +---+ (+)--> s(t)
        |     |------->| DAC |----->| x |--^
        +-----+        +-----+      +---+
       top mixer: cos(wct); bottom mixer: -sin(wct)
```
1. The serial bit stream is split into groups of n bits (4 for 16-QAM).
2. Half of the bits go to the I branch and half to the Q branch; each half is converted to a multilevel PAM signal (for 16-QAM: 2 bits map to four levels ±1, ±3).
3. Each is multiplied by its carrier; the products are added.

#### Receiver
```text
         cos(wct)
            |   +-----+
 s(t) --+-->x-->| LPF |-> I --+
        |       +-----+       |   +---------+
        |                     +-->| decision|--> bits
        |       +-----+       |   | + P/S   |
        +-->x-->| LPF |-> Q --+   +---------+
            |   +-----+
        -sin(wct)
```
Coherent demodulation multiplies s(t) by the two carriers and low-pass filters; the I and Q signals are recovered and decoded into bits.

#### Constellation of 16-QAM (4 bits per symbol)
```text
  Q
  3 |  o    o  |  o    o
  1 |  o    o  |  o    o
 ---+-----------+----------- I
 -1 |  o    o  |  o    o
 -3 |  o    o  |  o    o
     -3  -1    1    3
```
There are 16 points (amplitude and phase combinations). For 32-QAM there are 32 points. With 2ⁿ points, each symbol carries n bits, so the bit rate = n × baud rate. Example: at 2400 baud, 16-QAM gives 9600 bps. More points make the system more sensitive to noise (points are closer together).

## Check yourself

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

1. QAM combines:
   A. Frequency and amplitude modulation
   B. Amplitude and phase modulation using two carriers 90° apart
   C. Two carriers of different frequency
   D. Phase modulation only with one carrier
   Answer: B. I and Q carriers in quadrature. (Key point: Concept)

2. Why does QAM send twice the data in the same bandwidth?
   A. It uses two channels
   B. Two independent streams on quadrature carriers share one channel
   C. It doubles the carrier frequency
   D. It halves the baud rate only
   Answer: B. Quadrature carriers don't interfere. (Key point: Benefit)

3. Bits per symbol in 16-QAM:
   A. 2
   B. 8
   C. 4
   D. 16
   Answer: C. 16 = 2^4. (Key point: 16-QAM)

4. At 2400 baud with 16-QAM bit rate is:
   A. 2400 bps
   B. 4800 bps
   C. 19200 bps
   D. 9600 bps
   Answer: D. 4 x 2400. (Key point: Bit rate)

## Related in Data Communication

- [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×

## Related in other subjects

- [QAM modulation, demodulation and constellation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aqam-mod-demod-constellation) — Communication Systems, chapter 7
- [SSB figure of merit and 32-QAM constellation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Assb-figure-of-merit-32qam) — Communication Systems, chapter 9
- [Spread spectrum and DSSS transmitter/receiver](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aspread-spectrum-dsss-tx-rx) — Wireless Communication, chapter 4
- [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