Phoebe guidesA field team has a real problem to settle: How Reflected Waves Create a Fade They must decide what happens before they change extra reflected path length on the device. Predict the direction first.
See the relationship first
Set the input, use the rule, then read the result and its unit.
The audit later on checks more than one number. Here, the added model uses the baseline named below and holds every other chapter value fixed. That sentence bridges the fixtures, so the numbers do not change without a reason.
Derive the baseline in four moves
- 1
Name the input. The chapter baseline for extra reflected path length is 3.125.
- 2
Name the rule. λ = c/f; φ = 2πΔL/λ; P_r/P_0 = |1 + ae^(-jφ)|² = 1 + a² + 2a cosφ
- 3
Put in the chapter value. Set extra reflected path length to 3.125. The page rule gives phase lag as 90 degrees.
- 4
Read the result. Keep degrees next to the value. Use it only within the limits on this page.
Predict, then change extra reflected path length
Try Predict what happens to phase lag. Move one control, calculate, then check your idea.
Observe The readouts use λ = c/f, φ = 2πΔL/λ, and 1 + a² + 2a cosφ, the same three formulas derived above. Reset to 3.125 and compare phase lag.
Explain Only extra reflected path length moves here. The other chapter values stay fixed.
Check yourself
What should you do before you trust the result?
What does this small model leave out?
1. Begin with two ripples
Imagine two equal ripples reaching the same point. Crest on crest makes a larger ripple. Crest on trough cancels. Radio waves do the same thing: the antenna receives electric fields with both size and phase, then turns their combined field into power.
2. Put names on the wave
| Symbol | Meaning | Chapter value |
|---|---|---|
| λ | one complete wave length | 12.5 cm at 2.4 GHz |
| ΔL | extra distance travelled by the reflection | 6.25 cm for half a wavelength |
| φ | phase lag caused by that extra distance | 180° at ΔL = λ/2 |
| a | reflected field size relative to the direct field | 1 in the ideal equal-ray check |
3. Derive the two-ray power
Find one wavelengthWave speed equals frequency times wavelength, so λ = c/f.
Turn distance into a fraction of a cycleΔL/λ says how many cycles late the reflection is.
Turn cycles into radiansOne cycle is 2π radians, so φ = 2πΔL/λ.
Add the fieldsNormalize the direct field to 1 and write the reflected field as ae^(−jφ).
Square the magnitudeExpanding |1 + ae^(−jφ)|² gives 1 + a² + 2a cosφ.
4. Reproduce the chapter's numbers
Half a wavelength is 6.25 cm. Its extra delay is 0.0625/(3.00×10⁸) = 0.208 ns. With equal rays, ΔL = 0 gives |1+1|² = 4, or 6.02 dB above one ray. At ΔL = 6.25 cm, φ = π and |1−1|² = 0: the ideal null. The chapter's 0.208 ns delay also gives 1/(2πΔτ) = 7.65×10⁸ Hz for a one-radian frequency shift.
5. Try the same formula
TryMove the reflected path from 0 to 12.5 cm and watch equal fields move through reinforcement, cancellation, and reinforcement again.
ObserveAt 3.125 cm the phase is 90° and the power is 2×; at 6.25 cm the ideal equal rays cancel.
ExplainThe readouts use λ = c/f, φ = 2πΔL/λ, and 1 + a² + 2a cosφ, the same three formulas derived above.
The wave comparison fixes the carrier at 2.4 GHz and combines the direct wave with one equal-strength reflection. It does not model unequal, moving, or lossy paths, which keep real measured fades from matching the ideal zero-power cancellation.
- unequal
- Needs separate evidence
- moving
- Needs separate evidence
- lossy paths
- Needs separate evidence
- so measured nulls are finite
- Needs separate evidence
Use field evidence or a deeper model before release.
6. What the result buys you
Moving an antenna by only a fraction of a wavelength can change phase enough to spend many decibels. That is why average RSSI cannot certify a mobile or cluttered link. Placement tests, repeated measurements, diversity, and fade margin belong in the release record.
7. Check yourself
1. What wavelength does 2.4 GHz have in free space?
Answer: λ = c/f = 0.125 m = 12.5 cm.
2. Why does 6.25 cm produce opposite phase?
Answer: It is half a wavelength, so φ = 2π(1/2) = π radians = 180°.
3. Why is an ideal zero not a promised field result?
Answer: It assumes exactly two equal, stable rays. Real reflected amplitudes and phases vary.
The wave comparison fixes the carrier at 2.4 GHz and combines the direct wave with one equal-strength reflection. It does not model unequal, moving, or lossy paths, which keep real measured fades from matching the ideal zero-power cancellation.
- 2.4 GHz carrier
- Radio carrier frequency
- light-speed constant
- Assumed wave propagation speed
- 12.5 cm wavelength
- Wavelength at stated carrier frequency
- 6.25 cm half-wave path
- Extra reflected distance causing half-cycle lag
- 0.208 ns delay
- Propagation delay for half-wave path difference
- 6.02 dB reinforcement
- Equal-wave power reinforcement at zero phase lag
- ideal null
- Perfect equal-wave cancellation at half-cycle lag
- 7.65×10⁸ Hz check reproduce the companion chapter
- Frequency change producing one-radian phase shift
The two-ray model is correct but not complete; the chapter's Under the Hood treatment names Rician/Rayleigh envelopes, delay spread, coexistence, and measurement evidence.