Math Bridge: Bandwidth and UWB Range Resolution

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Math BridgeUWBTime of flight

Why does 499.2 MHz turn into a roughly 30 cm ranging floor?

Follow inverse bandwidth into the chapter's two-way distance equation.

Eddie, the electronics guideEddie guides
The one targetCompute the timing and range screen set by channel bandwidth.
The chapter case499.2 MHz UWB channel 5, a 20 MHz comparison, and a 0.25 m robot boundary.
What it buys youA hard floor that clock averaging and extra exchanges cannot manufacture away.

A field team has a real problem to settle: Why does 499.2 MHz turn into a roughly 30 cm ranging floor? They must decide what happens before they change bandwidth on the device. Predict the direction first.

See the relationship first

The figure reads from left to right. The blue card is bandwidth. The middle card uses this page's rule. The green card is timing floor. 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.

Bandwidth changes timing floor An input card leads through the page rule to the timing floor result. SET INPUT ONE CONTROL USE RULE predict calculate check units READ RESULT
Follow the arrows. Bandwidth buys timing detail. Exchanges can reduce random noise, but they cannot recover a pulse edge the channel filtered out.

Derive the baseline in four moves

  1. 1

    Name the input. The chapter baseline for bandwidth is 499.2.

  2. 2

    Name the rule. Δt = 1/(499.2x10⁶) = 2.003 ns d = (3x10⁸ x 2.003x10⁻⁹)/2 = 0.3005 m At 20 MHz: Δt = 50 ns and d = 7.50 m

  3. 3

    Put in the chapter value. Set bandwidth to 499.2. The page rule gives timing floor as 2.0 ns.

  4. 4

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

Predict, then change bandwidth

Try Predict what happens to timing floor. Move one control, calculate, then check your idea.

499.2
Chapter baseline
Timing floor

Observe Bandwidth buys timing detail. Exchanges can reduce random noise, but they cannot recover a pulse edge the channel filtered out. Reset to 499.2 and compare timing floor.

Explain Only bandwidth 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 bandwidth moves. Field effects named in the page limits stay fixed.

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

1. Start with the physical story

A receiver cannot mark an edge more sharply than the channel preserves it. More bandwidth admits a shorter pulse, so the arrival-time window and the corresponding two-way distance window shrink together.

Eddie: Name the physical limit first; the algebra then has one honest job.

2. Name every algebra move

1

Invert bandwidthDivide one second by bandwidth to find the timing floor.

2

Convert unitsMultiply seconds by one billion to report nanoseconds.

3

Carry time into distanceMultiply by wave speed and divide by two for a round trip.

4

Compare screensDivide bandwidths and compare the range floor with the robot boundary.

3. Reproduce the chapter case

Δt = 1/(499.2×10⁶) = 2.003 ns
d = (3×10⁸ × 2.003×10⁻⁹)/2 = 0.3005 m
At 20 MHz: Δt = 50 ns and d = 7.50 m

The 0.3005 m screen reproduces the chapter's roughly 0.30 m result; implementation can be worse, never magically sharper than missing bandwidth.

4. Try one real input

TryMove occupied bandwidth and watch the same inverse relation drive timing, pulse length, and range resolution.

Bandwidth
Timing floor
One-way pulse length
Two-way range screen
Resolution gain vs 20 MHz
Share of 0.25 m boundary
20 MHz range screen

ObserveAt 499.2 MHz, the timing floor is 2.00 ns and the two-way range screen is 0.300 m, 24.96x sharper than the 20 MHz comparison.

ExplainBandwidth buys timing detail. Exchanges can reduce random noise, but they cannot recover a pulse edge the channel filtered out.

Technical boundaries.

This is a bounded formula screen, not a deployment approval.

Approximation
1/B is a physics-scale screen, not a complete correlator error model.
Propagation
Multipath and non-line-of-sight bias can dominate the nominal floor.
Safety
A 0.300 m screen does not certify a 0.25 m exclusion boundary.

Correct, not complete: use the measured state named above before release.

5. Use the result in the lab

Compare the screen with the required boundary, then measure bias and spread across line-of-sight and obstructed positions.

6. Record the evidence state

Keep channel, bandwidth, preamble and correlator settings, antenna delay calibration, geometry, reference distance, bias, and spread.

7. Check yourself

Does more exchange averaging beat the 1/B floor?
Answer: It can reduce random variation, but it cannot restore timing detail absent from the channel.
Why divide by two in the distance equation?
Answer: The named time is a round-trip interval, while range is one-way distance.
Does 499.2 MHz guarantee 30 cm accuracy?
Answer: No. It supplies a resolution screen; calibration, clock, multipath, and first-path detection add error.
Honesty boundary.

The bridge keeps calculation, chosen inputs, and field evidence separate.

Computed
Timing floor, pulse length, two-way range screen, and comparison ratios.
Specified
Occupied bandwidth, wave speed, reference bandwidth, and robot boundary.
Observed
Bias, repeatability, first-path quality, and errors under real geometry and obstruction.

Correct, not complete: this page does not certify hardware, coverage, safety, capacity, or compliance.

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