Phoebe guidesA field team faces an unresolved physical question: Why 6 dB Nearly Doubles Range They must answer it before changing gateway antenna gain on the real device. Predict the direction first.
See the relationship before changing it
The figure reads from left to right. The blue card is gateway antenna gain. The middle card applies this page's relationship. The green card is free-space ceiling. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline for gateway antenna gain is 6.
- 2
Name the relationship. FSPL_max = P_t + G_node + G_gateway - sensitivity; d = 10^[(FSPL_max + 147.55 - 20log₁₀f)/20]
- 3
Substitute the chapter fixture. Set gateway antenna gain to 6. The page ledger gives free-space ceiling as 869 km.
- 4
Read the result. Keep km beside the value. Use it only inside the technical boundary on this page.
Predict, then change gateway antenna gain
Try Predict the direction of free-space ceiling. Move one control, calculate, then check your prediction.
Observe The outputs use the closing link budget and the chapter's free-space loss equation, then undo 20 log₁₀d exactly as derived above. Reset the control to 6 and compare free-space ceiling.
Explain Only gateway antenna gain moves here. The other chapter fixtures remain fixed.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. Antenna gain reshapes power
An antenna does not create extra transmitter power. Gain concentrates the available power in some directions and gives up coverage in others. In the preferred direction, that concentration enters the link budget just like extra decibels of transmit power.
2. Understand the logarithm
Free-space path loss includes 20 log₁₀d. The logarithm turns multiplication of distance into addition of decibels. To undo it, divide the dB change by 20 and raise 10.
Power or effective aperture uses 10 log₁₀, so the corresponding power factor is 10^(gain/10).
3. Derive the free-space ceiling
Form effective radiated powerEIRP = P_t + G_node.
Set the closing boundaryAt the free-space ceiling, received power equals sensitivity.
Make path loss the subjectFSPL_max = P_t + G_node + G_gateway − sensitivity.
Insert free-space lossFSPL = 20log₁₀d + 20log₁₀f − 147.55.
Undo the logarithmIsolate 20log₁₀d, divide by 20, then raise 10.
4. Reproduce the LoRaWAN example
With 14 dBm transmit power, 0 dBi node gain, and −130 dBm sensitivity, a 0 dBi gateway allows 14 + 0 − (−130) = 144 dB of free-space loss. At 868 MHz that solves to about 435 km. A 6 dBi gateway raises the allowance to 150 dB and the ceiling to about 869 km. The range factor is 10^(6/20) = 1.995×; the power/aperture factor is 10^(6/10) = 3.98×, matching the chapter's 0.00951 m² to 0.0378 m² example.
5. Try the same formula
TryMove gateway gain from 0 to 6 dBi and watch allowed path loss, free-space ceiling, and the range multiplier change together.
ObserveEach added dB raises the loss allowance by one dB, but distance grows by the logarithmic factor 10^(G/20).
ExplainThe outputs use the closing link budget and the chapter's free-space loss equation, then undo 20 log₁₀d exactly as derived above.
The widget shows a mathematical free-space ceiling with no fade reserve.
- It is not a coverage prediction
- Needs separate evidence
Use field evidence or a deeper model before release.
6. What the result buys you
Gain can buy back some budget without changing the radio, but it changes the antenna pattern. A real design subtracts cable loss, obstruction, fading, interference, regulatory limits, and a release margin before estimating useful range, then validates it in the field.
7. Check yourself
1. What maximum path loss follows from 14 dBm and −130 dBm with 0 dBi gain?
Answer: 14 − (−130) = 144 dB.
2. What range factor does 6 dB imply in free space?
Answer: 10^(6/20) = 1.995×, almost double.
3. Why is 869 km not promised coverage?
Answer: It omits terrain, curvature, foliage, fading, interference, pattern direction, losses, and reserve.
These are the chapter inputs, worked results, and named teaching assumptions.
- 14 dBm
- Radio power level
- 25 mW
- Power or power-loss value
- 868 MHz
- Frequency, sample rate, or event rate
- −130 dBm
- Radio power level
- 0/6 dBi
- Chapter input or worked result
- 144/150 dB
- Gain, loss, margin, or level ratio
- 435/869 km
- Distance, wavelength, or size
- 1.995×
- Percentage, ratio, or gain
- 3.98×
- Percentage, ratio, or gain
- 0.00951/0.0378 m²
- Distance, wavelength, or size
The derivation is correct, but terrain, Earth curvature, foliage, antenna pattern, legal EIRP, cable loss, fading, and margin make real range much shorter; follow the chapter's deeper selection and validation process.