- 2081 Bhadra (CS II) · 5 marks
QAM modulation, demodulation and constellation
Short answer
QAM modulates two quadrature carriers with multi-level amplitudes, achieving high spectral efficiency. Modulator: bits split and mapped to levels, then multiply quadrature carriers and sum. Demodulator: two correlators with local quadrature carriers and slicers recover levels. 16-QAM constellation is a 4×4 Gray-coded grid, each symbol carrying 4 bits.
Explain the working of Quadrature Amplitude Modulation (QAM) with appropriate modulation, demodulation block, and constellation diagram.
Quadrature Amplitude Modulation (QAM) is a digital modulation in which each symbol is sent by changing both the amplitude and phase of the carrier. It is built from two amplitude-modulated carriers of the same frequency in quadrature, so an -point QAM carries bits per symbol.
For 16-QAM, .
Modulator
+->[2-bit to 4-level]-a_k->(X)<-phi1
bits -->[S/P] |
4 bits/sym | (+)--> s(t)
+->[2-bit to 4-level]-b_k->(X)<-phi2
, .
- Serial-to-parallel converter sends 2 bits to the I branch and 2 bits to the Q branch.
- Each 2-bit group is mapped (Gray code) to a level: 00 to -3, 01 to -1, 11 to +1, 10 to +3.
- The levels amplitude-modulate the quadrature carriers and the outputs are summed.
Demodulator
+->(X)->[int 0..T]->[4-level slicer]--+
| ^ phi1 |
r(t) ----+ [P/S]-> bits
| v phi2 |
+->(X)->[int 0..T]->[4-level slicer]--+
Each correlator recovers one coordinate ( and plus noise). The slicer compares it with thresholds , picks the nearest level, and the level-to-bits converter plus P/S give the data. Carrier and symbol timing recovery are needed (coherent detection).
16-QAM constellation (Gray coded)
Q (phi2)
0010 0110 | 1110 1010 +3
|
0011 0111 | 1111 1011 +1
--------------+--------------> I (phi1)
0001 0101 | 1101 1001 -1
|
0000 0100 | 1100 1000 -3
-3 -1 +1 +3
First two bits select the I level, last two the Q level; neighbouring points differ by one bit, so most symbol errors cause only one bit error. The minimum distance between points is .
Merits
- 4 bits/symbol: bandwidth is 1/4 of BPSK for the same bit rate.
- Better error performance than 16-PSK at the same average power.
- Not constant envelope, so linear amplifiers are required.
✓ Answer checked by Claude Opus 5.5 (high effort)
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