Math Bridge: RSSI and Free-space Distance

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Math BridgeSensorsStruggle-friendly runway

How −65 dBm becomes 17.7 metres

One careful inversion of the chapter's free-space equation, including the code error that produced 169 m.

Phoebe, the physics guidePhoebe guides
The one targetSolve the MHz path-loss equation for distance.
The chapter case2,400 MHz and RSSI = −65 dBm.
What it buys youSpot a 9.55× unit-and-logarithm error.

A field team faces an unresolved physical question: How -65 dBm becomes 17.7 metres They must answer it before changing rssi in dbm 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 rssi in dbm. The middle card applies this page's relationship. The green card is correct ideal distance. 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.

RSSI in dBm changes correct ideal distance An input card leads through the page relationship to the correct ideal distance result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The readouts use the same logarithm and inverse-power step derived above.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for rssi in dbm is -65.

  2. 2

    Name the relationship. d=10^[(27.55-20log₁₀f_MHz-RSSI)/20]

  3. 3

    Substitute the chapter fixture. Set rssi in dbm to -65. The page ledger gives correct ideal distance as 17.7 m.

  4. 4

    Read the result. Keep m beside the value. Use it only inside the technical boundary on this page.

Predict, then change rssi in dbm

Try Predict the direction of correct ideal distance. Move one control, calculate, then check your prediction.

-65
Chapter baseline
Correct ideal distance

Observe The readouts use the same logarithm and inverse-power step derived above. Reset the control to -65 and compare correct ideal distance.

Explain Only rssi in dbm moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only rssi in dbm moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only rssi in dbm. It replays the chapter fixture; it is not a field prediction.

1. Read decibels as a ratio scale

A logarithm compresses a multiplication ratio into addition. The term 20 log₁₀d therefore says that multiplying distance by ten adds 20 dB of ideal free-space loss.

2. Keep the frequency unit attached

With distance in metres and frequency in hertz, FSPL = 20log₁₀d + 20log₁₀f − 147.55. Writing f in megahertz adds 120 inside the frequency term, rebasing the constant to −27.55.

FSPL = 20log₁₀d + 20log₁₀f_MHz − 27.55

3. Make distance the subject

1

Use the quiz referenceFrom a 0 dBm ideal reference, FSPL = −RSSI.

2

Move known terms20log₁₀d = 27.55 − 20log₁₀f_MHz − RSSI.

3

Divide by 20log₁₀d = (27.55 − 20log₁₀f_MHz − RSSI)/20.

4

Undo log base tend = 10^[(27.55 − 20log₁₀f_MHz − RSSI)/20].

4. Try the corrected quiz

d=10^[(27.55−20log₁₀f_MHz−RSSI)/20]

TryMove RSSI and stop at the chapter's −65 dBm.

20 log₁₀(2,400)
Correct ideal distance
Using broken 48 dB term
Overestimate

ObserveAt −65 dBm, the proper 67.6 dB frequency term gives 17.7 m; the broken 48 dB term gives about 169 m.

ExplainThe readouts use the same logarithm and inverse-power step derived above.

Technical boundaries.

Free space assumes one unobstructed path and a 0 dBm reference.

Walls, antennas, body shadowing, AGC error, and multipath can dominate a real RSSI reading
Needs separate evidence

Use field evidence or a deeper model before release.

5. Work the chapter numbers

20log₁₀(2,400) = 67.6 dB. Then d = 10^[(27.55−67.6+65)/20] = 17.7 m. Substituting 48 gives about 169 m; 169/17.7 = 9.55.

6. Separate model error from sensor evidence

The corrected arithmetic removes a software error. It does not turn RSSI into a calibrated ruler. The chapter's channel-state-information detector deliberately treats movement-driven channel variation as its evidence.

7. Check yourself

Why is the MHz constant 27.55?
Answer: Converting Hz to MHz shifts the frequency logarithm by 120 dB, changing 147.55 to 27.55.
What was wrong with 20×2.4?
Answer: The formula needs 20log₁₀(2,400), not multiplication by the GHz number.
Is 17.7 m a measured range?
Answer: No. It is the ideal free-space result for the quiz's assumptions.
Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

2,400 MHz
Frequency, sample rate, or event rate
−65 dBm
Radio power level
67.6 dB
Gain, loss, margin, or level ratio
17.7 m
Distance, wavelength, or size
169 m
Distance, wavelength, or size
9.55×
Percentage, ratio, or gain

The page corrects the equation but does not claim RSSI is a field-accurate distance sensor.

Return to the chapter with the maths unpacked →