Math Bridge: The 8.38 dB Frequency Penalty

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Why does 2.4 GHz lose 8.38 dB against 915 MHz?

Derive free-space path loss from spherical spreading and receive aperture, then reproduce the chapter comparison.

Pete, the guidePete guides
The one targetDerive the FSPL shape and unit constant.
The chapter case100 m: 915 MHz = 71.7 dB; 2.4 GHz = 80.0 dB.
What it buys youUse FSPL as a reviewable baseline, never a range promise.

A field team faces an unresolved physical question: Why does 2.4 GHz lose 8.38 dB against 915 MHz? They must answer it before changing frequency 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 frequency. The middle card applies this page's relationship. The green card is wavelength. 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.

Frequency changes wavelength An input card leads through the page relationship to the wavelength result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. Frequency enters twice in the power ratio, so the dB term is 20log10(f). Space is not absorbing the extra energy; the ideal receive aperture is smaller.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for frequency is 2400.

  2. 2

    Name the relationship. FSPLdB=20log10(dkm)+20log10(fMHz)+K; K=20log10(4π)+180-20log10(c)

  3. 3

    Substitute the chapter fixture. Set frequency to 2400. The page ledger gives wavelength as 0.125 m.

  4. 4

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

Predict, then change frequency

Try Predict the direction of wavelength. Move one control, calculate, then check your prediction.

2400
Chapter baseline
Wavelength

Observe Frequency enters twice in the power ratio, so the dB term is 20log10(f). Space is not absorbing the extra energy; the ideal receive aperture is smaller. Reset the control to 2400 and compare wavelength.

Explain Only frequency 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 frequency moves. Field effects named in the page's technical boundary stay fixed.

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

1. Power spreads before an antenna catches it

A transmitter’s power spreads over a sphere whose area is 4πd². An ideal receive antenna catches only an effective aperture λ²/(4π). Dividing transmitted by received power joins those two physical ideas.

Pete: Keep the units and the assumptions beside every number.

2. Name every algebra move

1

Spread the powerS=Pt/(4πd²).

2

Catch one apertureAe=λ²/(4π), with λ=c/f.

3

Form the ratioFSPL=Pt/Pr=(4πdf/c)², then use 10 log10 on power.

3. The logarithm separates the ledger

FSPLdB=20log10(dkm)+20log10(fMHz)+K; K=20log10(4π)+180−20log10(c)

The +180 converts kilometres and megahertz into metres and hertz. With c=3.00×10⁸ m/s, K is 32.4 dB; the conventional rounded constant is 32.45 dB.

4. Try one controlled change

FSPLdB=20log10(dkm)+20log10(fMHz)+K; K=20log10(4π)+180−20log10(c)

TryMove the carrier from 915 MHz toward 2.4 GHz while distance stays at 100 m.

Frequency
Wavelength
Unit constant
FSPL at 100 m
915 MHz baseline
Frequency penalty
Aperture ratio

ObserveAt 2.4 GHz the calculator gives about 80.0 dB, against about 71.7 dB at 915 MHz: an 8.38 dB penalty.

ExplainFrequency enters twice in the power ratio, so the dB term is 20log10(f). Space is not absorbing the extra energy; the ideal receive aperture is smaller.

Technical boundaries.

FSPL is an ideal free-space baseline, not a complete release model.

isotropic antennas
Required antenna model
clear line of sight
Required path condition
material, Fresnel, fading, polarization, cable, or interference loss
Excluded loss terms

Real link budgets restore those terms and need field evidence.

5. Reproduce the chapter values

Using c=3.00×10⁸ m/s gives K=32.4. Then 20log10(0.1)+20log10(915)+32.4=71.7 dB and the same line at 2400 MHz gives 80.0 dB. Their difference is 20log10(2400/915)=8.38 dB.

6. Carry the evidence forward

Record formula units, distance, carrier, antenna gains, receiver mode, deployment losses, reserve policy, and representative packet/RSSI checks. A clean baseline becomes useful only when its omissions are visible.

7. Check yourself

Why is the distance coefficient 20?
Answer: Power loss contains d²; 10log10(d²)=20log10(d).
Does free space absorb more at 2.4 GHz?
Answer: No. The ideal aperture term changes with wavelength; the model contains no material absorption.
Does 80.0 dB prove a 100 m link works?
Answer: No. It is one baseline loss term, not a complete link budget or field result.
Honesty boundary.

These are the worked values and named assumptions for this bridge.

100 m
Comparison path distance
915 MHz
Reference carrier
71.7 dB
Reference FSPL
2.4 GHz
Changed carrier
80.0 dB
Changed FSPL
8.38 dB
Frequency penalty
32.4 dB
Rounded unit constant

FSPL assumes isotropic antennas, clear line of sight, and no material, Fresnel, fading, polarization, cable, or interference loss. Real link budgets restore those terms and need field evidence.

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