Math Bridge: Band Reach and Capacity Trade

← Back to IoT Frequency Bands and Licensing
Math BridgeFrequency bandsStruggle-friendly runway

What reach toll buys the wider high-band channel?

Keep the ideal frequency-loss comparison beside, but separate from, Shannon's bandwidth capacity bound.

Eddie, the electronics guideEddie guides
The one targetCompare a frequency toll and an ideal bandwidth reward without claiming one cancels the other.
The chapter case868 MHz versus 5 GHz; 20 versus 80 MHz at 20 dB SNR.
What it buys youAn auditable band trade-off before field testing.

A field team faces an unresolved physical question: What reach toll buys the wider high-band channel? They must answer it before changing selected 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 selected 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.

Selected 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 changes wavelength and ideal path loss; bandwidth multiplies capacity only after an SNR has been supplied.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for selected frequency is 5000.

  2. 2

    Name the relationship. ΔFSPL868→5000 = 20log10(5000/868) = 15.21 dB power ratio = (5000/868)^2 = 33.18 η = log2(1 + 100) = 6.658 bit/s/Hz C20 = 133.2 Mbit/s; C80 = 532.7 Mbit/s

  3. 3

    Substitute the chapter fixture. Set selected frequency to 5000. The page ledger gives wavelength as 0.060 m.

  4. 4

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

Predict, then change selected frequency

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

5000
Chapter baseline
Wavelength

Observe Frequency changes wavelength and ideal path loss; bandwidth multiplies capacity only after an SNR has been supplied. Reset the control to 5000 and compare wavelength.

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

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

1. Start with the physical story

Higher frequency costs ideal same-distance received power. Wider bandwidth can carry more information at the same signal-to-noise ratio. They describe different parts of a design.

Eddie: A capacity ceiling is not a promise that the link closes, and a low loss toll is not a promise of enough spectrum.

2. Name every algebra move

1

Compare frequencyForm f2/f1 and use 20log10 for the dB toll.

2

Compare powerSquare the frequency ratio.

3

Convert SNRTurn 20 dB into a linear ratio with 10^(20/10) = 100.

4

Find spectral efficiencyCompute log2(1 + SNR).

5

Apply bandwidthMultiply spectral efficiency by channel bandwidth.

3. Reproduce the chapter case

ΔFSPL868→5000 = 20log10(5000/868) = 15.21 dB
power ratio = (5000/868)^2 = 33.18
η = log2(1 + 100) = 6.658 bit/s/Hz
C20 = 133.2 Mbit/s; C80 = 532.7 Mbit/s

The 80 MHz bound is four times the 20 MHz bound only because SNR is held equal. Real SNR, protocol overhead, coexistence, and permitted channels can change.

4. Try one real input

TryMove the selected band frequency. Watch wavelength and ideal path toll move while the fixed 80 MHz capacity case stays separate.

Selected frequency
Wavelength
Quarter wave
Frequency ratio
Ideal loss toll
Power ratio
SNR ratio
Spectral efficiency
80 MHz bound
20 MHz bound
Bandwidth ratio

ObserveAt 5 GHz the ideal loss toll is 15.21 dB, while the 80 MHz ideal bound remains 532.7 Mbit/s at the fixed 20 dB SNR.

ExplainFrequency changes wavelength and ideal path loss; bandwidth multiplies capacity only after an SNR has been supplied.

Technical boundaries.

Both sides are ideal comparisons, not an installed service forecast.

Loss
The frequency toll excludes materials, antennas, power rules, and fading.
Capacity
Shannon is an upper bound and excludes protocol overhead, sharing, coding limits, and application traffic.
Regulation
Usable bandwidth and power depend on region, licence, band, and device.

Correct, not complete: this ledger does not choose a band or promise throughput.

5. Use the result in the decision

Write the reach toll and capacity opportunity as separate hypotheses. Verify allowed spectrum, final hardware, service margin, interference, duty cycle, payload, and operations.

6. Record the evidence state

Record region, licence, centre frequency, bandwidth, EIRP, antenna, measured SNR and service, coexistence, device support, and the retest trigger.

7. Check yourself

Why is 15.21 dB also about a 33-to-1 power ratio?
Answer: The frequency ratio is squared for power, matching 10^(15.21/10).
Why is 80 MHz four times 20 MHz in this example?
Answer: Shannon capacity is linear in bandwidth when the same SNR is held fixed.
Does the 533 Mbit/s result promise application throughput?
Answer: No. It is an ideal upper bound before protocol and deployment losses.
Honesty boundary.

This ledger places two ideal limits side by side.

Reach
The frequency-only toll is not a full link budget.
Rate
Shannon's bound is not delivered throughput.
Choice
Regulation, hardware, service, and evidence decide the band.

Correct, not complete: this ledger does not choose a band or promise throughput.

Return to IoT frequency bands with the maths unpacked →