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Quadrature amplitude modulation (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.

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

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

 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

         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)

  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).

✓ Answer checked by Claude Sonnet 5.5

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