More bits per symbol did not create free range
If 16-QAM maps four bits while QPSK maps two, did we double throughput and range for free?
The recurring illustrative sensor case needs an uncoded mapped rate of 20 kbit/s. A rushed review selects 16-QAM because “four bits per symbol” sounds like twice QPSK's throughput and therefore extra range. The first half can be true only if symbol rate is held fixed. The range claim does not follow: under equal average symbol energy, 16 points must occupy the same energy budget, so nearest points become much closer.
Think about itUnder equal average symbol energy and symbol rate, what happens to minimum distance when QPSK becomes square 16-QAM?
It falls from √2 ≈ 1.414 for unit-average QPSK to 2/√10 ≈ 0.632 for unit-average 16-QAM. The raw mapped rate doubles, but the detector receives less geometric margin and the transmitter must support multiple ideal magnitudes.
Evidence Internal coordinate sums independently checked by regression tests; no channel, range, or measured performance is represented.
Bits, symbols, labels, and rates
A bit sequence exists at D0 or D1. A mapper groups k bits, selects one member of an M-ary alphabet where k = log2M, and emits an ideal D2 symbol label during each interval Ts. At R3, the detector turns one observed decision-time coordinate back into a point index and then its label. A point label is bookkeeping; the transmitted waveform still needs pulse shaping, sampling, conversion, and RF hardware.
20 kbit/s uncoded QPSK → 10 ksymbol/s
QPSK has M = 4 and k = 2, so 20,000 bit/s ÷ 2 bit/symbol = 10,000 symbol/s. Adding rate-1/2 coding for the same payload doubles coded mapper bits before any framing, pilots, or preamble overhead.
D3 might hold eight complex samples per symbol so a digital filter can represent the transition. R3 still makes one idealized symbol decision per interval. Sample rate and symbol rate are related by an implementation choice, not synonyms.
Common constellation families
BPSK, QPSK, M-PSK, and square M-QAM are different D2 geometries and receiver contracts. OQPSK delays one binary branch so I and Q do not switch simultaneously; its four ideal decision locations match QPSK, while its transition path and shaped-envelope behavior differ. A static constellation does not specify those paths, occupied spectrum, PAPR, eye opening, carrier recovery, or an RF waveform.
| Family | Signal-space geometry | Mapped bits/symbol | Receiver knowledge | What the geometry says |
|---|---|---|---|---|
| BPSK | Two antipodal points | 1 | Coherent phase reference | Largest spacing among the listed unit-energy planar candidates |
| QPSK | Four equal-radius points | 2 | Coherent phase and timing | OQPSK changes transition timing, not the four ideal decision states |
| M-PSK | M phases on one ideal radius | log₂M | Increasingly precise phase knowledge | Angular spacing shrinks as M increases |
| Square M-QAM | Grid with multiple radii | log₂M | Coherent phase, timing, and gain | Efficient mapped rate; smaller spacing and stronger linearity demand |
| Orthogonal 2-FSK | Two orthogonal correlator dimensions | 1 | Coherent or noncoherent detector | Not faithfully represented as one ordinary instantaneous I/Q point |
Orthogonal 2-FSK can be represented by two basis-correlator outputs, such as (√Es, 0) and (0, √Es). Drawing it as two ordinary single-time complex-envelope points without naming those dimensions would conceal the receiver and tone-orthogonality assumptions.
Think about itCan a static ideal constellation reveal occupied bandwidth, sample rate, pulse-shaped trajectories, PAPR, or timing margin?
No. It gives ideal decision-time coordinates and labels. Pulse shape and symbol timing add trajectories, spectrum, sample requirements, peak behavior, and an eye diagram in later stages.
Normalize before comparing
The portfolio default is E[|ak|²] = 1 at D2. Unit-average QPSK uses (±1 ± j)/√2. Square 16-QAM starts with I,Q ∈ {±1, ±3} and divides every coordinate by √10 because the average two-dimensional energy is 10. This makes an equal-Esgeometry comparison honest.
| Unit-average candidate | Average Es | Peak |a| | dmin | Peak/average energy |
|---|---|---|---|---|
| BPSK | 1 | 1 | 2 | 1 |
| QPSK | 1 | 1 | √2 = 1.4142 | 1 |
| 8-PSK | 1 | 1 | 2 sin(π/8) = 0.7654 | 1 |
| 16-QAM | 1 | √1.8 = 1.3416 | 2/√10 = 0.6325 | 1.8 |
Unit-minimum-distance normalization instead forces dmin = 1. That is useful for comparing energy required per unit spacing, but average energies then differ: QPSK has Es = 0.5 and 16-QAM has Es = 2.5. Mixing those normalizations can reverse apparent SNR, error, and PA conclusions.
Go deeperWhy a unit-average constellation still says nothing about watts
The coordinates are dimensionless numerical symbols at D2. Converting them to average power requires a pulse-energy convention, symbol rate, scaling through D3/A0/A1, gains and losses, and a declared physical reference plane. Es = 1 is not “one watt.”
Noise clouds and nearest-point decisions
The lab uses rk = sk + nk at R3. The noise is circular complex AWGN: independent zero-mean I and Q components each have variance N0/2 under the stated unit-symbol-energy convention. With equal priors and equal AWGN variance, maximum likelihood reduces to the minimum squared Euclidean distance.
| Candidate | Nearest-point region description |
|---|---|
| BPSK | Decide the negative-I point for I < 0; positive-I point for I ≥ 0. |
| QPSK | Decide by I and Q signs; the axes divide four quadrants. |
| 8-PSK | Decide the closest ideal phase; radial boundaries lie halfway between adjacent angles. |
| Unit-average 16-QAM | Slice I and Q independently at −2/√10, 0, and +2/√10. |
An error occurs when the observation crosses one of those regions. Noise does not “move toward a bit”; bits are labels attached after a geometric decision. Nor are scatter points snapshots of the GHz carrier: they are decision-time coordinates after the receiver has projected the waveform into its I/Q basis.
Gray labels change bit consequences
Labeling changes no coordinate and no nearest-point region. It changes the Hamming distance between a transmitted label and a mistaken one. Gray labeling arranges common nearest neighbors to differ by one bit; a reference binary ordering can place two-bit differences across a nearest boundary.
- Gray neighbor00 → 01One symbol error, one bit error.
- Gray diagonal00 → 11One symbol error, two bit errors—but a longer geometric jump.
- Binary boundary01 → 10One nearest-neighbor symbol error can flip two bits.
That is a high-SNR Gray-neighbor approximation for applicable constellations, not an identity. Diagonal or multi-boundary errors, non-Gray labels, unequal probabilities, coding, and a different detector change the bit consequences.
Think about itIf you relabel a fixed noisy 16-QAM run without moving any point, which metrics must remain unchanged?
Coordinates, detected point indices, symbol-error count, and SER remain unchanged. The bit-error count and BER can change because the point-index labels have changed.
Es/N0, Eb/N0, and SNR bookkeeping
Under an uncoded k-bit mapper, Es = kEb. Therefore Es/N0 is 10 log10(k) dB above Eb/N0. For QPSK k = 2, so the offset is exactly 3.0102999566 dB. Code rate, framing, pilots, spreading, overhead, and the chosen definition of “bit” must be written before extending this conversion.
8.00 dB Eb/N0 → 11.0103 dB Es/N0
With unit Es, Eb = 0.5. At Eb/N0 = 6.309573 linear, N0 = 0.0792447 and each I/Q noise component has variance 0.0396223, so σI,Q = 0.199054 normalized coordinate units.
“8 dB SNR” is incomplete without signal/noise bandwidth, averaging, reference plane, waveform power convention, coding/overhead boundary, and detector assumptions. Eb/N0is useful precisely because those bookkeeping choices must be made explicit.
Analytical reference versus finite simulation
For coherent BPSK and Gray QPSK in flat AWGN with perfect carrier and timing knowledge, the exact uncoded bit-error reference is Pb = Q(√(2Eb/N0)). At 8.00 dB, the linear ratio is 6.309573, the Q-function argument is 3.55234, and Pb ≈ 1.909 × 10−4. In 100,000 independent bits the expected count is about 19.1, not a guaranteed observation.
A 512-symbol scatter is excellent for seeing geometry and poor for validating a 10−4tail. The default deterministic run below happens to observe zero errors in 1,024 bits. It therefore displays a count and finite-run caveat rather than claiming BER = 0.
Constellation Decision Lab
Select the lowest-complexity mapping that can meet a 20 kbit/s raw-rate target and a 10−3 coherent-AWGN reference target at a fixed 10 ksymbol/s. Then stress the uncorrected detector with noise and phase error. This is calculated and deterministically simulated data, not an instrument capture.
Default QPSK case loaded.
- Raw mapped rate
- 20 kbit/s Rs = 10 ksymbol/s
- Minimum distance
- 1.414 Unit-average coordinates
- Empirical SER
- 0 0 / 512
- Empirical BER
- 0 0 / 1,024
- Analytical BER
- 0.0001909 Exact Gray-QPSK reference
Dashed lines are ideal nearest-point boundaries. Correct samples are circles; decision errors are crosses. Ideal symbols use four repeated shapes plus direct bit labels. No residual phase rotation; the detector axes match the transmitter reference.
Es/N0 = Eb/N0 + 10 log10(k) = 11.01 dB; σI,Q = √(N0/2) = 0.19905 coordinate units.Es = 1.0000, Eb = 0.5000, k = 2; energy is in normalized coordinate units, not watts.
Uncoded coherent detection in flat circular complex AWGN with perfect timing and carrier knowledge.
SER: 0 observed in 512 trials; not proof of zero. A rule-of-three 95% upper bound is about 5.86e-3. BER: 0 observed in 1,024 trials; not proof of zero. A rule-of-three 95% upper bound is about 2.93e-3.
Ideal coordinate table
| Index | Bits | I | Q |
|---|---|---|---|
| 0 | 00 | 0.707 | 0.707 |
| 1 | 01 | -0.707 | 0.707 |
| 2 | 11 | -0.707 | -0.707 |
| 3 | 10 | 0.707 | -0.707 |
Representative received symbols
Focus a row to locate its observation on the plot. Error rows are selected first.
Same-normalization candidate comparison
| Gray candidate | Bits/symbol | Raw rate | Es | dmin | Peak radius | Exact AWGN BER at 8.00 dB |
|---|---|---|---|---|---|---|
| BPSK | 1 | 10 kbit/s | 1.000 | 2.000 | 1.000 | 0.0001909 |
| QPSK | 2 | 20 kbit/s | 1.000 | 1.414 | 1.000 | 0.0001909 |
| 8-PSK | 3 | 30 kbit/s | 1.000 | 0.765 | 1.000 | Not implemented |
| 16-QAM | 4 | 40 kbit/s | 1.000 | 0.632 | 1.342 | Not implemented |
Model contract. PRBS-9 bits use x9 + x5 + 1 from state 0x1FF. Noise uses a pinned xorshift32 seed and Box–Muller draw order I then Q. The kth symbol receives θ[k] = θ0 + kΔθ/(N−1), AWGN is added after rotation, and an unrotated nearest-Euclidean detector resolves exact ties to the lowest point index. Changing only the label map preserves coordinates, samples, decisions, and SER; BER can change.
Structured impairments are not AWGN
Circular AWGN produces roughly isotropic clouds. Real error vectors can have structure: common phase rotates every point, frequency offset accumulates rotation through a burst, I/Q gain or quadrature error skews the geometry and creates an image, DC offset translates the cloud, compression bends magnitudes, and timing error or ISI makes observations depend on neighboring symbols. BER alone cannot uniquely identify which mechanism caused an error.
The same wrong point index can result from noise, reference rotation, timing history, or compression. Preserve symbol time order and compare multiple synchronized views before selecting a mitigation.
Update mapping and decision record
Version 5 records a baseline, not a universal winner. Under the supplied teaching constraints—20 kbit/s raw mapped rate, 10 ksymbol/s, 8 dB Eb/N0, coherent flat AWGN reference, and a preference for low complexity—Gray QPSK is the lowest-order listed candidate that meets both the raw-rate and analytical 10−3reference target.
| Candidate | Rate to carry 20 kbit/s | Receiver / PA implication | Decision |
|---|---|---|---|
| BPSK | 20 ksymbol/s | Coherent; one ideal radius | Reject at the fixed 10 ksymbol/s constraint: only 10 kbit/s. |
| Gray QPSK | 10 ksymbol/s | Coherent; one ideal radius at D2 | Baseline: closes raw rate and 8 dB AWGN reference with the fewest listed states. |
| Gray 16-QAM | 5 ksymbol/s | Coherent gain/phase; multiple radii and tighter unit-energy spacing | Reject for baseline: adds margin and linearity cost without a stated rate need. |
| Orthogonal 2-FSK | 20 ksymbol/s | May allow noncoherent detection; tone separation spends frequency support | Does not close at fixed 10 ksymbol/s; retain only if PA/reference benefits justify a changed symbol-rate/bandwidth plan. |
- V5 · mapping
- Gray QPSK; point order 00, 01, 11, 10 counterclockwise from quadrant I. Mapping labels are explicit rather than implied by “QPSK.”
- V5 · normalization
- Unit average symbol energy at D2: points (±1 ± j)/√2, Es = 1, dmin = √2, peak radius = 1.
- V5 · rate boundary
- Illustrative uncoded 20.0 kbit/s mapper rate at 10.0 ksymbol/s. Payload/net rate after coding, framing, pilots, and retransmission is not claimed.
- V5 · energy bookkeeping
- At 8.00 dB Eb/N0, Es/N0 = 11.0103 dB for k = 2. Units are normalized coordinates at D2/R3.
- V5 · decision rule
- At R3, ideal coherent nearest-Euclidean detection with equal priors; QPSK quadrants are split by I = 0 and Q = 0.
- V5 · error evidence
- Exact reference BER ≈ 1.909 × 10−4. The default seeded 512-symbol run observes 0/512 symbol and 0/1,024 bit errors with a finite-tail caveat.
- V5 · rejected candidate
- 16-QAM dmin = 0.6325 at equal Es, versus QPSK 1.4142; its extra mapped rate is not required and its PA/receiver demands remain unclosed.
- V5 · explicit deferrals
- Pulse shaping and bandwidth, timing/carrier loops, coding/framing, fading/channel, PA waveform behavior, EVM conformance, standard mode, and measured BER.
A constellation defines desired decision-time geometry. It does not define the continuous pulse, finite bandwidth, receive filter, or timing margin needed to make those samples; Pulse Shaping, ISI & Decisions will own that next step.
Artifact Waveform decision record v5 · mapping, normalization, rate/energy bookkeeping, decision rule, error evidence, rejected alternative, and named omissions.
Check your understanding
Answer each question in your own words, then reveal the model answer.
01A 20 kbit/s uncoded mapper uses QPSK. What symbol rate is required, and what changes if rate-1/2 coding is added?
Model answerUncoded QPSK maps two bits per symbol, so 20 kbit/s requires 10 ksymbol/s. If 20 kbit/s is the payload rate and rate-1/2 coding is added, the coded mapper input becomes at least 40 kbit/s before framing or pilots, so the QPSK symbol rate becomes at least 20 ksymbol/s.
02Write the unit-average QPSK coordinates and minimum distance.
Model answerThe four points are (±1/√2, ±1/√2), so every point has energy one. Nearest horizontal or vertical neighbors differ by 2/√2, giving dmin = √2.
03At Eb/N0 = 8 dB, what Es/N0 applies to uncoded QPSK?
Model answerQPSK has k = log2(4) = 2 bits per symbol, so Es/N0 = Eb/N0 + 10 log10(2) = 8 + 3.0103 = 11.0103 dB. This conversion assumes the declared uncoded mapping and consistent energy reference.
04Why can Gray and binary labeling have the same SER but different BER?
Model answerThe point coordinates and nearest-point detector are unchanged, so each received coordinate produces the same symbol decision and SER. The bit strings assigned to those point indices differ, so the Hamming distance of a mistaken decision—and therefore BER—can differ.
05A 512-symbol run reports no bit errors. What may you conclude?
Model answerOnly that this deterministic finite run observed zero errors. It does not prove BER is zero or validate a low tail. For Gray QPSK the run contains 1,024 bits, and the rule-of-three 95% upper bound is roughly 3/1,024 ≈ 2.9 × 10⁻³.
06Why is Gray QPSK a defensible baseline for the recurring case, and what remains unresolved?
Model answerAt 10 ksymbol/s it supplies the required uncoded 20 kbit/s and is the lowest-order listed candidate that does so; at 8 dB its coherent-AWGN reference BER is about 1.91 × 10⁻⁴. The choice still needs pulse shaping, synchronization, coding/framing, channel/fading, PA behavior, bandwidth, and measurement evidence.
Sources and further study
The mathematical claims are paraphrased from established digital-communications references and checked against independent coordinate sums and numerical evaluation. Bibliographic and document versions were rechecked on 5 September 2026. All plots are calculated or qualitative teaching graphics; none is measured evidence.
Detection, geometry, and error probability
- P02-S2 · Informative: J. G. Proakis and M. Salehi, Digital Communications, 5th ed., Chapters 4–5.
- P02-S9 · Informative: B. Sklar, Digital Communications: Fundamentals and Applications, 2nd ed., Chapters 1–4.
- P02-S3 · Informative/open: MIT OpenCourseWare, Unified Engineering: Signals and Systems, for signal-space foundations.
Impairment orientation and checked fixtures
- P02-S8 · Informative: Rohde & Schwarz, Understanding Error Vector Magnitude, version 01.00, October 2022, for qualitative constellation-impairment orientation—not conformance limits.
- Derived: unit-average coordinate sums, minimum distances, xorshift32/Box–Muller samples, nearest-point decisions, and the 8 dB Q-function fixture are regression tested under constellation-lab/1.0.
- Review boundary: an RF review should independently verify N0/2 per-component variance, Q-function convention, reference plane, and whether real receiver synchronization justifies the analytical comparison.