- 2078 Chaitra · 5 marks
QAM modulation and demodulation
Short answer
QAM varies amplitude and phase using I and Q carriers, sending multiple bits per symbol. The modulator splits bits into multilevel I and Q streams multiplied by carriers. The coherent demodulator correlates and makes multilevel decisions. QAM has larger symbol distances than M-PSK for same average power.
Explain QAM modulation and demodulation with its required diagram.
Quadrature Amplitude Modulation (QAM) is an M-ary scheme in which both the amplitude and the phase of the carrier change. Two independent amplitude-modulated carriers, (in-phase, I) and (quadrature, Q), are added so that each symbol carries bits.
where are integer levels, e.g. for 16-QAM, and is the energy of the lowest-amplitude signal.
Modulator
+-> [2-to-L level] -> a_k -->(X)<- cos wc t
binary | |
data -->[S/P] (+)--> QAM
| |
+-> [2-to-L level] -> b_k -->(X)<- -sin wc t
- The serial data is split into two streams (half the bits each).
- Each stream is grouped into bits and converted to one of levels.
- The levels multiply the I and Q carriers and the products are added.
Demodulator (coherent)
+->(X)->[int 0..T]->[L-level decision]-+
| ^ cos wc t |
r(t) ----+ [P/S]-> data
| v -sin wc t |
+->(X)->[int 0..T]->[L-level decision]-+
- Two correlators (multiply and integrate) recover the I and Q amplitudes; the orthogonality of cos and sin removes cross-talk.
- Each output is compared with thresholds to decide the level.
- Level-to-bits conversion and a parallel-to-serial converter give the data back.
A carrier recovery circuit is needed because detection is coherent.
16-QAM constellation
Q
* * | * * +3
|
* * | * * +1
-----------+-----------> I
* * | * * -1
|
* * | * * -3
-3 -1 +1 +3
16 points, 4 bits per symbol; with Gray coding, neighbours differ by one bit. For the same , QAM has the same bandwidth efficiency as M-ary PSK ( bits per symbol), but because its points are spread over the whole plane rather than on one circle, they are farther apart for the same average power, so QAM gives a lower error probability.
✓ Answer checked by Claude Opus 5.5 (high effort)
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