Math Bridge: Bluetooth Channel Capacity

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Math BridgeBluetooth and BLEStruggle-friendly runway

Why is 2 Mbit/s sensible inside a 13.3 Mbit/s ceiling?

Convert SNR from decibels, apply Shannon’s capacity formula, and compare an upper bound with practical low-power radio rates.

Radio Remi, the guideRadio Remi guides
The one targetCompare Bluetooth rates with a channel ceiling.
The chapter case1 MHz Classic and 2 MHz BLE channels at 20 dB SNR.
What it buys youStop treating unused theoretical capacity as design failure.

A technician must decide whether quarter-wave antenna length is safe before changing bluetooth carrier frequency on the real device. The result is unresolved until the rule and units are checked. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is bluetooth carrier frequency. The middle card applies this page's rule. The green card is quarter-wave antenna length. 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 model keeps those stated values fixed and changes only bluetooth carrier frequency, so the numeric fixture does not switch without explanation.

Bluetooth carrier frequency changes quarter-wave antenna length An input card leads through the rule quarter wave = 75,000 / frequency in MHz to the quarter-wave antenna length result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. Higher carrier frequency shortens the ideal quarter-wave antenna.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 2440 MHz.

  2. 2

    Name the relationship. quarter wave = 75,000 / frequency in MHz

  3. 3

    Substitute with units. 75,000 / 2,440 = 30.74 mm

  4. 4

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

Predict, then change bluetooth carrier frequency

Try Predict the direction of quarter wave = 75,000 / frequency in MHz. Test another bluetooth carrier frequency, then compare quarter-wave antenna length.

2440 MHz
Chapter baseline
Quarter-wave antenna length

Observe Higher carrier frequency shortens the ideal quarter-wave antenna. Reset bluetooth carrier frequency to 2440 and compare quarter-wave antenna length.

Explain Higher carrier frequency shortens the ideal quarter-wave antenna.

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 bluetooth carrier frequency moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only bluetooth carrier frequency. It is not a field prediction.

1. A decibel SNR must become a ratio

Shannon’s formula uses a plain power ratio, not decibels. At 20 dB, SNRlinear=10^(20/10)=100. Then one more term, +1, enters the logarithm.

Radio Remi: Keep the theoretical ceiling separate from payload throughput and energy cost.

2. Name the algebra moves

1

Undo decibelsSNRlinear=10^(SNRdB/10).

2

Apply bandwidthC=B log2(1+SNRlinear).

3

Compare a real rateuse=Rchapter/C×100%.

3. Frequency sets other trade-offs

λ=c/f; ΔFSPL=20log10(fhigh/flow)

The 2.4 GHz band gives about a 12.3 cm wavelength. Compared at equal distance, 5 GHz adds 6.38 dB loss, while 900 MHz would save 8.52 dB.

4. Try one controlled change

C=B log2(1+10^(SNRdB/10))

TryChange only the link SNR. Classic and BLE channel widths and the chapter’s 3 and 2 Mbit/s rates stay fixed.

Linear SNR
Classic ceiling (Mbit/s)
BLE ceiling (Mbit/s)
3 Mbit/s / ceiling
2 Mbit/s / ceiling
2.44 GHz wavelength
5 GHz loss penalty
900 MHz loss saving

ObserveAt 20 dB, the ceilings are 6.66 and 13.3 Mbit/s. The chapter rates use about 45.1% and 15.0% of those ideal ceilings.

ExplainMore SNR raises the mathematical ceiling logarithmically. Bluetooth can still choose a lower, robust modulation rate to reduce receiver and transmitter complexity and power.

Technical boundaries.

Shannon capacity is an ideal information-theory bound, not an advertised application rate.

Channel model
Assumes bandwidth-limited additive noise rather than every Bluetooth impairment
Practical rate
Protocol overhead, coding, modulation, interference, and implementation reduce throughput
Same-distance loss
Frequency-only comparison holds distance and antennas fixed

Use measured packet error, throughput, and current for a product trade-off.

5. Reproduce the chapter comparison

At 20 dB, SNR=100 and log2(101)=6.658. Multiplying by 1 MHz gives 6.66 Mbit/s; 3/6.658=45.1%. A 2 MHz channel gives 13.3 Mbit/s; 2/13.316=15.0%. These ratios compare chapter headline rates with ceilings, not payload efficiencies.

6. Carry the evidence forward

Record occupied bandwidth, PHY mode, coding, measured SNR distribution, interference, packet error, retransmissions, goodput, current, receiver complexity, antenna, distance, and environment.

7. Check yourself

Why can’t 20 dB be inserted directly as SNR=20?
Answer: The formula needs a linear power ratio; 20 dB means 10²=100.
Does 13.3 Mbit/s mean BLE should deliver that payload rate?
Answer: No. It is an ideal ceiling before protocol, coding, interference, and implementation costs.
Why might a radio deliberately use only 15% of the ceiling?
Answer: Simpler robust modulation and hardware can save power and tolerate real channel variation.
Honesty boundary.

The page compares the chapter’s headline PHY rates with one ideal ceiling.

6.66 and 13.3 Mbit/s
Shannon bounds at exactly 20 dB SNR
45.1% and 15.0%
Headline-rate-to-bound ratios, not measured efficiency
12.3 cm
Wavelength near the chapter’s 2.44 GHz centre example

The result explains headroom; it does not rank complete Bluetooth implementations.

Return to Bluetooth Assessment with the maths unpacked →