Math Bridge: Z-Wave Frequency Advantage

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Math BridgeZ-WaveSub-GHz link

How much link advantage comes from the lower carrier?

Compare wavelength, free-space loss, and ideal range before any wall or antenna claim is added.

Eddie, the electronics guideEddie guides
The one targetTurn one carrier-frequency choice into a bounded link comparison.
The chapter case868/908 MHz Z-Wave against the crowded 2.4 GHz band.
What it buys youAn honest frequency-only hypothesis for the chapter's home mesh.

A field team has a real problem to settle: How much link advantage comes from the lower carrier? They must decide what happens before they change sub-ghz carrier on the device. Predict the direction first.

See the relationship first

The figure reads from left to right. The blue card is sub-ghz carrier. The middle card uses this page's rule. The green card is sub-ghz wavelength. Follow the arrows: 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.

Sub-GHz carrier changes sub-ghz wavelength An input card leads through the page rule to the sub-ghz wavelength result. SET INPUT ONE CONTROL USE RULE predict calculate check units READ RESULT
Follow the arrows. One frequency ratio appears as a linear distance ratio and as a logarithmic dB difference.

Derive the baseline in four moves

  1. 1

    Name the input. The chapter baseline for sub-ghz carrier is 868.

  2. 2

    Name the rule. λ868 = 3x10⁸ / 868x10⁶ = 0.3456 m; λ2400 = 0.1250 m ΔFSPL = 20 log10(2400/868) = 8.83 dB dsame-budget ratio = 2400/868 = 2.765

  3. 3

    Put in the chapter value. Set sub-ghz carrier to 868. The page rule gives sub-ghz wavelength as 0.346 m.

  4. 4

    Read the result. Keep m next to the value. Use it only within the limits on this page.

Predict, then change sub-ghz carrier

Try Predict what happens to sub-ghz wavelength. Move one control, calculate, then check your idea.

868
Chapter baseline
Sub-GHz wavelength

Observe One frequency ratio appears as a linear distance ratio and as a logarithmic dB difference. Reset to 868 and compare sub-ghz wavelength.

Explain Only sub-ghz carrier moves here. The other chapter values stay fixed.

Check yourself

What should you do before you trust the result?
Answer: Predict its direction, use the shown rule, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only sub-ghz carrier moves. Field effects named in the page limits stay fixed.

Honesty note. This teaching model changes only sub-ghz carrier. It replays the chapter baseline; it is not a field prediction.

1. Start with the physical story

Lower frequency means longer wavelength. At the same distance, power, and antenna gains, it also means less free-space path loss.

Eddie: The frequency advantage is real, but “travels through walls better” still needs material and installation evidence.

2. Name every algebra move

1

Find wavelengthsDivide wave speed by each carrier frequency.

2

Form a ratioDivide 2.4 GHz by the Z-Wave carrier.

3

Convert to dBApply 20 log10 to the frequency ratio.

4

Hold budget fixedUse the inverse-frequency ratio for ideal equal-loss distance.

3. Reproduce the chapter case

λ868 = 3×10⁸ / 868×10⁶ = 0.3456 m; λ2400 = 0.1250 m
ΔFSPL = 20 log10(2400/868) = 8.83 dB
dsame-budget ratio = 2400/868 = 2.765

A 30 m ideal sub-GHz reference maps to 10.85 m at 2.4 GHz under this frequency-only comparison.

4. Try one real input

TryMove the sub-GHz carrier while 2.4 GHz and the reference link budget stay fixed.

Sub-GHz carrier
Sub-GHz wavelength
2.4 GHz wavelength
Wavelength ratio
Free-space advantage
Equal-budget range ratio
2.4 GHz comparison range
Mains repeater share

ObserveAt 868 MHz, the wavelength and ideal equal-budget range are 2.765 times the 2.4 GHz values, equal to an 8.83 dB free-space advantage.

ExplainOne frequency ratio appears as a linear distance ratio and as a logarithmic dB difference.

Technical boundaries.

This isolates carrier frequency; it is not a house-coverage model.

Free space
The comparison holds power, antenna gains, sensitivity, and allowable loss equal.
Building
Walls, bodies, furniture, diffraction, absorption, fading, and floor geometry need measurement.
Mesh
Thirteen repeaters among 35 devices is the chapter example, not a repeater-sizing formula.

Correct, not complete: the ratio does not approve a floor plan or route design.

5. Use the result in the lab

Keep antenna height and power comparable, then map received power and route health at both candidate bands through the actual building.

6. Record the evidence state

Keep carrier plan, transmit power, antenna gains, sensitivity, room and wall path, device state, RSSI, retries, route changes, and dead zones.

7. Check yourself

Does 8.83 dB prove a signal crosses every wall?
Answer: No. It is the frequency-only free-space difference.
Why is the ideal range ratio 2.765?
Answer: At fixed allowable path loss, distance scales inversely with frequency.
Does the 13/35 ratio size a mesh?
Answer: No. It only restates the chapter's example population.
Honesty boundary.

The bridge turns “sub-GHz helps” into one explicit frequency-only claim.

Computed
Wavelengths, dB advantage, ideal range ratio, and comparison distance are reproducible.
Specified
Power, antennas, sensitivity, region, reserve, and route policy come from the selected system.
Observed
Building loss, RSSI, retries, dead zones, and route health determine coverage.

Correct, not complete: survey the real building before releasing the mesh.

Return to Z-Wave Overview with the frequency advantage bounded →