Learning tool · bits → waves → room → receiver

Signals,
up close.

A constellation is a picture of choices. The antenna sends a changing voltage; the room adds delayed copies; the receiver tries to recover the choices. Follow those transformations by changing one thing at a time.

In plain words Choose a symbol, build its wave, disturb it and let the receiver try to recover it. Then move a receiver through a room and watch two signal paths add together.

Wi-Fi, generation by generation The estimator from zero

English explanations from first principles, with equations and inspectable values. These are synthetic teaching models, not fleet captures. Reviewed 21 September 2026. The standards lab explains where each mechanism fits in Wi-Fi 4–7.

A point is a recipe for a wave

Start with 16-QAM. Four coded bits choose one of sixteen points. Its horizontal coordinate, I, weights a cosine; its vertical coordinate, Q, weights a sine shifted by a quarter cycle. Adding those components makes one waveform with a particular amplitude and phase.

Try this: keep Q fixed and move I. Then change the sign of Q. Look at the point, the bit label and the wave together. Higher QAM orders provide more choices, not a faster-moving radio wave.

Fig. 1 · from a label to a waveformInteractive teaching model

Worked 16-QAM examples: I and Q levels are −3, −1, 1, 3 divided by √10
I	Q	Amplitude	Phase
0.316	0.316	0.447	45°
0.949	0.316	1.000	18.435°
0.316	−0.316	0.447	−45°

s(t) = I cos(2πft) − Q sin(2πft). Unit average constellation energy does not make every point equally strong. This panel uses illustrative Gray labels and a slowed carrier; it is not the IEEE bit-mapping table.

Changing I or Q changes the waveform. The cursor connects an instant on the summed wave to its two components; the selected point remains constant throughout this symbol.References: QAM normalization and mapping · I/Q modulation

Why do adjacent labels differ by one bit?

The panel applies a Gray code separately to the I and Q positions, then concatenates the two labels. A horizontal or vertical neighbor differs by one bit, so a nearest-neighbor symbol mistake can change only one bit. Noise can also cause larger jumps. Real Wi-Fi adds coding, interleaving and a specified bit mapping; a displayed symbol mistake is not a decoded packet failure.

The receiver sees a cloud of estimates

A channel can shrink and rotate a constellation. Noise spreads each received estimate around its expected position. The receiver first compensates for the estimated channel, then interprets the remaining uncertainty. A hard decision chooses the nearest point; practical decoders can use soft information too.

Try this: select “Rotate the points” and compare the two panels. Then choose “Deep fade.” Correcting the rotation is easy when the channel is known exactly; dividing by a small gain also enlarges the noise.

Fig. 2 · before and after equalizationInteractive teaching model

Exact noiseless examples for the symbol x = 1 + j0
Channel h	Received y = hx	Corrected y/h
1	1 + j0	1 + j0
0.5	0.5 + j0	1 + j0
j	0 + j1	1 + j0

With noise, y/h = x + n/h. A gain of 0.25 multiplies noise amplitude by four after correction. The interactive view uses 512 synthetic samples and reports their exact error counts, not a measured BER or confidence interval.

Channel correction removes a known gain and phase, but it does not erase noise. The same seeded draws are reused while you change a control, so the visual comparison has one cause at a time.References: EVM definition · QAM decisions · Constellation impairments

EVM, SNR and errors answer different questions

Error vector magnitude (EVM) measures the distance from estimates to their intended symbols, normalized here by the batch's mean reference-symbol energy. The noise control sets Es/N₀ before channel gain. Neither number alone says how many packets a device will deliver: coding, channel estimation, synchronization and the channel's variation also matter. At the same average constellation energy, denser QAM puts decision boundaries closer together.

Wi-Fi adds many tones into one waveform

The QAM point becomes a complex coefficient Xₖ on one subcarrier. OFDM adds those subcarriers in time. An FFT at the receiver separates them again when frequency and timing align. Each subcarrier completes an integer number of cycles over the useful symbol interval.

Try this: turn one tone down to zero, then restore it and change its phase. Finally move the receiver's frequency offset away from zero and watch energy appear in other FFT bins.

Fig. 3 · four tones, one sum, then an FFTInteractive teaching model

Illustrative four-tone model using the HE/EHT useful-symbol duration
Quantity	Value
Useful interval T	12.8 µs
Tone separation 1/T	78.125 kHz
Active bins k	−2, −1, 1, 2
0.8 µs prefix / total interval	5.88%
Half-bin offset	39.0625 kHz

With ideal timing and zero frequency offset, the DFT recovers each coefficient and zero on unused bins. A sufficient prefix helps tolerate channel delay spread; it cannot by itself remove carrier-frequency error. This panel has no multipath and makes no CP performance prediction.

Changing one coefficient changes the whole time-domain waveform. A frequency offset disrupts the separation at the receiver. The cyclic prefix is a copy of the symbol tail, not an extra information-bearing symbol.References: OFDM and cyclic prefix · Synchronization · Foundational corpus

Where is the actual radio carrier?

The OFDM graph shows complex baseband: a lower-frequency mathematical description of the signal. A transmitter converts that signal to a real waveform around its RF carrier; the receiver downconverts it. The 78.125 kHz spacing is between tones inside a channel, whereas 5.2 GHz is an example of the channel's location in the spectrum. The time axis here is in microseconds; the first panel's two-cycle carrier was deliberately slowed.

The room changes what arrives

Place an AP and a receiver in a room. A direct signal arrives first and a wall reflection takes a longer route. At the receiver they add as waves, with a phase difference set by the excess distance and frequency. The combination can reinforce one frequency and weaken another.

Try this: move the receiver a few centimetres, change the carrier, then set the reflected amplitude to zero. Keep the axes fixed as you compare the frequency response. This is a deployment sketch you can operate, not a survey of our lab.

Fig. 4 · one room, two signal pathsInteractive teaching model

Authored geometry: 10 × 6 m room, AP (1, 1.5), receiver (7, 3.5), reflection from y = 0
Quantity	Value
Direct path √40	6.325 m
Reflected path √61	7.810 m
Excess distance	1.486 m
Excess delay at c = 299,792,458 m/s	4.956 ns
No echo: normalized \|H(f)\|	1 at every frequency

H(f) = 1 + a exp(−j2πfΔτ), with the direct path normalized and its common delay removed. Echo amplitude is an assumption; wall loss, reflection phase, people, antennas, diffraction and hardware effects are not modeled.

Moving the receiver changes the relative path delay and the resulting channel. Wider frequency coverage reveals more of its frequency dependence; it does not create extra physical paths.References: Multipath channels · Foundational corpus

The thesis connection: the next section puts people into a room like this one. QAM concerns the sent symbols X. CSI estimates the channel H that transformed them, using known training signals. Payload decisions and channel estimates serve different jobs. For actual device output, continue to One record; for the research implications, return to the standards lab's sensing section.

People change which copies arrive

The receiver never sees one clean wave. It sees the direct copy, one copy off each wall and one copy off each person, and it can only measure their sum. A body does two things to that sum: it weakens every copy that passes through it, and it adds a new copy of its own. Either change moves the sum, and the CSI value changes.

Try this: drag one person onto the line between the AP and the STA and watch the direct arrow fade. Then add people one at a time, press “Everyone walks” and compare the spread of the centre tone with one person and with eight.

Fig. 5 · a room with walls and peopleInteractive teaching model

Authored geometry and assumed constants
Quantity	Value
Room	4 × 3 m, flat, four walls
AP / STA	(0.6, 1.4) m / (3.4, 1.6) m
Direct path	2.807 m
Carrier and wavelength	2.437 GHz, 12.30 cm
Wall bounce	amplitude × −0.5
Body	radius 0.2 m, a crossing path keeps 0.25 of its amplitude, scatters like a 1 m² target
People	0 to 8

H(f) = Σᵢ gᵢ · 0.25^nᵢ · exp(−j2πf Lᵢ / c), where Lᵢ is the length of path i, gᵢ its free-space gain (1/L direct, −0.5/L for a wall image, √(σ/4π)/(d₁d₂) for a body) and nᵢ the number of bodies it crosses. One bounce per path, no body-to-body echoes, no diffraction, antennas or hardware offsets.

Every path is an arrow whose length is its strength and whose angle is set by its length in wavelengths. The STA measures only the sum. Moving a body by a few centimetres turns its own arrow and can swing the sum between strong and weak.References: Multipath channels · Foundational corpus

Why each extra person is harder to see: a new person adds an arrow whose angle is unrelated to the others, and dims the paths the body stands in. In this model the spread over time rises with the count, but more slowly than the count itself: eight walkers do not give eight times the spread of one. The average level can fall or rise, depending on whether bodies block more power than their echoes add. Try it: note the spread with one walker, then with four and with eight. The constants here are teaching assumptions, not a fitted body model, and the model does not predict counting accuracy. For what the fleet actually measures with people present, continue to How a level is made.

Try it

Predict before moving a control

Can two 16-QAM points have equal amplitude but different phases?
Reveal the reasoning
Yes. For example, (1,1)/√10 and (1,−1)/√10 have the same radius but angles of +45° and −45°. Find them in the first panel.
Will ideal channel correction make every noisy symbol correct?
Reveal the reasoning
No. Division removes the known channel factor; n/h remains. Compare a weak channel with a unit-gain channel using the same noise seed.
Does a longer cyclic prefix fix the frequency-offset leakage shown here?
Reveal the reasoning
No. This receiver discards the prefix before its FFT. The useful interval and frequency mismatch stay the same; compare the recovered-bin table.
Why does the room response become flat when the echo is disabled?
Reveal the reasoning
Only the normalized direct path remains: H(f) = 1. This model removes its common phase and delay, so position-dependent interference requires the second path.
Does a second person standing next to the first one double the change in the channel?
Reveal the reasoning
No. The second body adds its own arrow with its own angle, and it can also shadow the first body's path. The two arrows can cancel as easily as add. Place two people side by side in section 5 and read the change against the empty room.

Read on

Sources and assumptions

These references support the mechanisms and definitions. Symbol labels, noise seeds, tone choices and room geometry are authored for these lessons. No panel uses live measurements or establishes a Wi-Fi sensing accuracy result.

MathWorks: QAM modulation, symbol ordering and average-power normalization.
Analog Devices AN-1039: I/Q modulation and constellation impairments.
MathWorks: QAM hard and soft demodulation.
MathWorks: EVM and normalization choices.
MathWorks: OFDM modulation and the cyclic prefix.
MathWorks: OFDM transmitter, receiver and synchronization.
MathWorks: multipath channels and frequency selectivity.
Alan Bensky, Short-range Wireless Communication (2019), chapter 4, “Communication protocols and modulation,” OFDM passages in the research corpus.

Continue with a real channel record, or return to Wi-Fi generations and shared airtime.

Signals,up close.