The previous lesson gave you three mechanisms — reflection, diffraction, scattering — and ended by pointing at the consequence it could not fit. If every one of those mechanisms delivers a copy of your signal by a different route, the receiver does not get one signal. It gets a handful, or a hundred, each delayed by a different amount, each attenuated differently, each arriving with its own phase. They add. Sometimes they add up and sometimes they add to nearly nothing, and the difference between those two outcomes can be forty decibels and half a footstep apart. This lesson is about the statistics of that sum: what the received level does over time, over frequency and over position, and which four numbers an engineer needs to predict it.
Write down what the receiver actually sees. The transmitter sends one waveform; the channel hands over a sum of scaled, delayed copies of it. Each copy has picked up an amplitude ai, a delay τi and, because delay at a carrier frequency is phase, a phase rotation θi. Nothing here is new physics — it is exactly the superposition you met in M2-L3, applied to a set of copies rather than to two tuning forks:
Take the smallest interesting case: two copies of equal amplitude. If they arrive in phase, the field doubles, and doubling a field quadruples a power, so the received power is 10 log₁₀(4) = +6.0 dB relative to one copy alone (M2-L4 for the decibel, and note that this is a 20 log₁₀ amplitude ratio expressed as a power ratio). If they arrive in anti-phase, the field is 1 − 1 = 0 and the received power is, in principle, zero — an infinitely deep null. Real nulls are finite because the amplitudes are never exactly equal, but 30 dB nulls are entirely ordinary.
What flips one into the other is a change of half a wavelength in the differential path length, because half a wavelength is 180° of carrier phase. At 2 GHz the wavelength is λ = 3×10⁸/2×10⁹ = 0.15 m, so λ/2 is 7.5 cm. That single number explains a great deal of everyday experience: a handset moved a few centimetres on a desk changes its reception, because a few centimetres of movement can change a reflected path by the several centimetres needed to swing that phasor from addition to subtraction. The fading pattern is a standing-wave field in the room, and you are walking through it.
Fading is not attenuation. The mechanisms of M8-L2 remove energy from the direct path once and for all; fading redistributes what arrives, moment by moment and channel by channel, around a mean that path loss already set. So the two are separate lines in a link budget: path loss sets the average, and fading forces you to hold a margin above the average so the deep moments still decode.
Suppose there is no line of sight at all — a handset indoors, a car in a street of tall buildings — so that every arriving copy is a scattered one and none of them is much bigger than the others. Then the in-phase and quadrature parts of the sum are each the total of many small independent contributions, the central limit theorem applies to both, and the envelope of the sum turns out to follow the Rayleigh distribution. Its cumulative distribution is unusually friendly:
Substitute and evaluate. For a fade 10 dB deep, 10−1 = 0.1, so P = 1 − e−0.1 = 1 − 0.9048 = 0.0952, i.e. 9.5% of the time. For 20 dB, 10−2 = 0.01 and P = 1 − e−0.01 = 1 − 0.99005 = 0.00995, i.e. 1.0%. For 30 dB, 10−3 = 0.001 and P = 1 − e−0.001 = 0.0009995, i.e. 0.10%:
| Fade depth below mean power | 10−x/10 | P = 1 − exp(−10−x/10) | In words |
|---|---|---|---|
| 3 dB | 0.5012 | 0.394 | 39% of the time — a Rayleigh channel is below its own mean most of the time |
| 10 dB | 0.1 | 0.0952 | 9.5%, about one moment in ten |
| 20 dB | 0.01 | 0.00995 | 1.0%, about one moment in a hundred |
| 30 dB | 0.001 | 0.0009995 | 0.10%, about one moment in a thousand |
Look at the middle column and the pattern falls out: for small arguments 1 − e−ε ≈ ε, so beyond about 10 dB the probability of a fade is simply 10−x/10 — each extra 10 dB of depth is ten times rarer. That is the rule of thumb engineers actually quote, and now you know both where it comes from and where it stops being exact: at 3 dB the approximation would predict 50% and the true answer is 39%.
Notice also what the 3 dB row is telling you. A Rayleigh channel spends most of its life below its own mean power, because the mean is dragged upward by rare constructive peaks. Designing to the average received power is therefore not a conservative choice but an optimistic one, and that asymmetry is the entire reason fade margins exist.
Now let the receiver see the transmitter. One component — the direct ray, or a single strong specular reflection — is now much larger than any individual scatterer, and the phasor sum is a big fixed vector with a small random one wobbling on the end of it. The envelope follows the Rician distribution, and the one parameter that matters is the ratio between the two, the K-factor:
The two limits are the two lessons either side of this paragraph. Let the scattered power go to zero and K → ∞: there is one path, nothing fluctuates, and you are back in the free space of M8-L1. Let the dominant component go to zero and K = 0, which is −∞ dB: the Rician distribution reduces exactly to the Rayleigh distribution of the previous section. Rayleigh is not a different model; it is Rician with the interesting term deleted. Between the limits, K tells you how much of the received power is trustworthy:
| K | Dominant path carries K/(K+1) | P(fade > 10 dB) | Typical of |
|---|---|---|---|
| −∞ dB (K = 0) | 0% | 9.5% | Rayleigh: dense urban NLOS, indoors around a corner |
| 0 dB (K = 1) | 50% | 7.3% | A weak or partly obstructed direct path |
| 5 dB (K = 3.16) | 76% | 2.5% | Indoor line of sight, same room, small office |
| 10 dB (K = 10) | 90.9% | 0.07% | Good indoor or short outdoor line of sight |
| 15–20 dB and up | 97–99% | negligible | Open rural line of sight, a fixed microwave link on a mast |
Read the third column across and the design consequence is stark. Going from Rayleigh to a merely mediocre K = 5 dB cuts the frequency of 10 dB fades from 9.5% to 2.5% — almost a factor of four — and reaching K = 10 dB cuts it by more than a hundred. This is why a fixed link is aimed at a clear line of sight and bolted down, and why the same radio that needs 20 dB of margin on a mobile handset needs almost none on a rooftop dish. The K-factor, not the transmit power, is what buys reliability.
So far the copies only differed in phase. They also differ in arrival time, and that opens a second, independent way for the channel to hurt you. Collect the arrival delays and their powers and you can summarise the whole spread with one number: the rms delay spread στ, the power-weighted standard deviation of the delays. It is small indoors, where the extra paths are metres long, and large outdoors, where they are hundreds of metres long.
Why does it matter? Because two frequency components separated by enough will experience different multipath sums — a delay τ rotates a component at f by 2πfτ, so a change in f changes the relative phases and therefore the whole sum. The frequency separation over which the channel stays roughly the same is the coherence bandwidth Bc, and it is inversely proportional to delay spread:
An indoor office channel typically shows στ ≈ 50 ns. Then 5στ = 250 ns = 2.5×10−7 s, and Bc = 1/(2.5×10−7) = 4×10⁶ Hz = 4 MHz. An urban macrocell channel typically shows στ ≈ 1 µs. Then 5στ = 5 µs = 5×10−6 s, and Bc = 1/(5×10−6) = 2×10⁵ Hz = 200 kHz. Twenty times the delay spread, one twentieth of the coherence bandwidth, exactly as the reciprocal demands.
Two conventions in one course, on purpose. M3-L4 quoted an HF coherence bandwidth of about 160 Hz from a 1 ms delay spread, using 1/(2πστ). This lesson uses 1/(5στ), which for the same 1 ms would give 200 Hz. The two are not in conflict and neither is wrong: 5 versus 2π is a ratio of 1.257, so the conventions disagree by about 20% — nothing, against delay spreads that vary by factors of twenty between environments. Quote which form you used and the number is unambiguous; quote a bare “coherence bandwidth” and it is not.
Now compare your signal’s bandwidth with Bc and the channel sorts itself into two regimes:
In the time domain those notches are the same fact seen differently: if the delay spread is a significant fraction of a symbol period, each symbol’s echoes land on top of its successors. That is intersymbol interference, the problem M6-L2 introduced and solved with Nyquist filtering and raised-cosine pulses — except that M6-L2 assumed the only source of ISI was the pulse shape, which the transmitter controls. Multipath ISI is inflicted by the environment, and no choice of transmit filter removes it. The arithmetic is brutal and quick: in that urban channel with στ = 1 µs, a 5 Msymbol/s link has a symbol period of 1/5×10⁶ = 200 ns, so the echoes of one symbol smear across 1 µs / 200 ns = 5 symbol periods.
There is a way out, and it is not a better filter. If a wide signal suffers because it is wider than Bc, split it into many narrow subcarriers that are each narrower than Bc, and every one of them sees flat fading that a single complex number can undo. That is the reasoning behind OFDM, which is Module 10’s subject and not this lesson’s — but you now know precisely which problem it was invented to solve, and the two numbers, 4 MHz and 200 kHz, that set its subcarrier spacing.
Delay spread told you how the channel varies across frequency. Motion tells you how it varies across time. Move the receiver, or the transmitter, or merely the van that one of your reflections bounces off, and every path length changes at its own rate, so every phasor rotates at its own rate. A path whose length is shortening at speed v has its frequency raised by the Doppler shift:
Take a 2 GHz carrier, where λ = 0.15 m. At 100 km/h the speed is 100 000/3600 = 27.8 m/s, so fd = 27.8 × 2×10⁹/3×10⁸ = 27.8 × 6.667 = 185 Hz (equivalently v/λ = 27.8/0.15 = 185 Hz, which is the easier way to do it in your head). The coherence time is then Tc = 0.423/185 = 0.00228 s = 2.3 ms. Now walk instead: 5 km/h is 1.39 m/s, fd = 1.39 × 6.667 = 9.3 Hz, and Tc = 0.423/9.3 = 0.0457 s = 46 ms. Twenty times the speed, one twentieth of the coherence time — the same reciprocal shape as delay spread and coherence bandwidth, because it is the same mathematics with time and frequency exchanged.
Both numbers have a physical reading that is easier to trust than the formula. At 2 GHz the fading pattern in space repeats roughly every half wavelength, 7.5 cm, so a pedestrian at 1.39 m/s crosses about 18 fades a second and a car at 27.8 m/s crosses about 370. That is why reception on foot degrades gently and reception in a moving car breaks up in bursts.
With Tc in hand the fast/slow distinction is a single comparison against the symbol period Ts. If Ts > Tc, the channel changes during a symbol: that is fast fading, and it is genuinely nasty, because no receiver can estimate a channel that will not hold still long enough to be measured. If Ts < Tc the channel is effectively constant across each symbol and only drifts between them: slow fading, which is what nearly every practical system is designed to live in.
Put the driving number in. With Tc = 2.28 ms, fast fading would need Ts > 2.28 ms, i.e. a symbol rate below 1/0.00228 = 440 symbols per second. A 1 Msymbol/s link has Ts = 1 µs, which is 2280 times shorter than Tc, so it is firmly in slow fading even at motorway speed. Fast fading is a problem for very low rate links, for very high carrier frequencies where fd scales up with fc, and for aircraft and satellites — not for a phone in a car.
One more distinction matters more than students expect, because the two things are often both called “fading”:
They stack: the log-normal shadowing sets the mean that the Rayleigh or Rician envelope then fluctuates around. Shadowing statistics, its standard deviation in decibels, and the log-distance path-loss model it attaches to are M8-L4’s job, so this lesson deliberately stops at naming it.
Fading cannot be removed, only outmanoeuvred, and every technique for doing so exploits the fact that a deep fade is a coincidence — a particular sum, at a particular place, at a particular frequency, at a particular moment. Change any of those and the coincidence is unlikely to repeat. That single idea generates the whole toolbox:
And when none of that is enough, you pay for the rest in power. A link budget must include a fade margin: decibels held in reserve, above the level that would just work on average, so that the link survives the fades it is statistically certain to meet. The Rayleigh table above is how that number is chosen — if you can tolerate an outage 1% of the time, you need about 20 dB of margin on a Rayleigh channel, and if you can tolerate 0.1%, you need about 30 dB. What that margin does to bit error rate is Module 9’s subject, and how it is entered into a budget is M9-L4.