Digital modulation
Introduced| English | Meaning |
|---|---|
| constellation | A set of signal-symbol points represented in a signal space such as the in-phase/quadrature plane. |
How many bits per hertz?
- The link's bits must ride a carrier. A modem chooses how many bits each symbol carries.
- More bits per symbol means more data per hertz, but noise knocks the symbols off more easily. This lesson prices that trade.
Constellations
- A constellation 星座图 plots each symbol as a point: ASK changes amplitude, PSK changes phase, and QAM changes both.
- $M$ points carry $\log_{2}M$ bits per symbol. Noise moves the received point; an error happens when it lands nearer a different point.
How many bits does each symbol of 64-QAM carry?
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$\log_{2}64 = 6$.
Match each scheme to what it changes.
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QAM uses both dimensions, which is why it packs points most efficiently.
Equal distance, unequal energy
- Compare constellations at the same minimum distance $d$ between points, since that sets the error rate.
- QPSK's four points on a circle need $E = d^{2}/2$. 16-QAM's 4×4 grid averages $E = 2.5d^{2}$: five times as much, or 7.0 dB.
At equal minimum distance, 16-QAM needs 5 times QPSK's symbol energy. How many dB is that? Give 1 decimal place.
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$10\log_{10}5 = 6.99$ dB.
Worked example: the energy ledger
- Per symbol: QPSK $0.5d^{2}$ for 2 bits; 16-QAM $2.5d^{2}$ for 4 bits.
- Per bit: $0.25d^{2}$ against $0.625d^{2}$, a factor of 2.5 or 4.0 dB. That matches the 4 dB gap in published error-rate curves.
Per bit, 16-QAM needs 0.625d² and QPSK 0.25d². How many dB more does 16-QAM need? Give 1 decimal place.
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The ratio is 2.5, and $10\log_{10}2.5 = 3.98$ dB.
The error-rate curve
- In white noise, BPSK's bit error rate is $Q\!\left(\sqrt{2E_{b}/N_{0}}\right)$, where $Q$ is the tail of a normal distribution.
- $E_{b}/N_{0}$, the energy per bit over the noise density, is the fair way to compare schemes.
E_b/N_0 is 5 dB. What is √(2E_b/N_0), the argument of Q for BPSK? Give 2 decimal places.
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5 dB is a ratio of 3.16, and $\sqrt{2 \times 3.16} = 2.51$.
Worked example: the cliff
- At $E_{b}/N_{0} = 5$ dB (3.16): $Q(2.51) \approx 6 \times 10^{-3}$. At 9 dB (7.94): $Q(3.98) \approx 3 \times 10^{-5}$.
- At 12 dB: about $10^{-8}$. A few dB move the error rate by orders of magnitude, so link budgets count every dB.
BPSK's E_b/N_0 rises from 5 dB to 9 dB. What happens to the bit error rate?
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The curve is a cliff, so 4 dB more energy per bit buys more than two orders of magnitude.
Three common slips
- Comparing at equal average energy. Compare at equal minimum distance, or per bit.
- Ignoring the normalisation. Check whether a curve is per bit or per symbol.
- "More points, more speed, free." Bits grow as $\log_{2}M$, but energy grows much faster.
Going from QPSK to 16-QAM doubles the bits per symbol at no cost in energy.
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At the same error rate it needs about 4 dB more energy per bit.
Recap
- $M$ points carry $\log_{2}M$ bits.
- At equal $d$: QPSK $d^{2}/2$, 16-QAM $2.5d^{2}$ (7 dB per symbol, 4 dB per bit).
- BPSK: $Q(\sqrt{2E_{b}/N_{0}})$, a steep cliff.