Math Bridge: Antenna Gain and Beamwidth

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Math BridgeOne learning thread

Why antenna gain narrows coverage

Follow a fixed transmit-power budget from a sphere to the chapter’s 6 dBi rack-aisle patch, including EIRP, on-axis density, and beamwidth.

Phoebe guides this bridge
One targetConnect gain to its coverage cost.
Chapter caseTwenty sensors, 17 dBm radio, 20 dBm cap.
What it buys youTest both battery and coverage gates.

See the relationship before changing it

The figure reads from left to right. The blue card is antenna gain. The middle card applies this page's rule. The green card is linear antenna gain. 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 antenna gain, so the numeric fixture does not switch without explanation.

Antenna gain changes linear antenna gain An input card leads through the rule linear gain = 10^(gain in dBi / 10) to the linear antenna gain result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. More directional gain concentrates ideal radiated power into less solid angle.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 6 dBi.

  2. 2

    Name the relationship. linear gain = 10^(gain in dBi / 10)

  3. 3

    Substitute with units. 10^(6 / 10) = 3.98 times

  4. 4

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

Predict, then change antenna gain

Try Predict the direction of linear gain = 10^(gain in dBi / 10). Test another antenna gain, then compare linear antenna gain.

6 dBi
Chapter baseline
Linear antenna gain

Observe More directional gain concentrates ideal radiated power into less solid angle. Reset antenna gain to 6 and compare linear antenna gain.

Explain More directional gain concentrates ideal radiated power into less solid angle.

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 antenna gain moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only antenna gain. It is not a field prediction.

1. Begin with power spread over area

An isotropic reference spreads transmit power Pt equally over a sphere. At distance d, that sphere has area 4πd².

Siso = Pt / (4πd²)

Phoebe: A directional antenna does not create energy. It moves energy away from some directions and into others.

2. Turn concentration into gain

A full sphere contains 4π steradians. If the useful lobe occupies solid angle Ω, the concentration ratio is:

G = 4π/Ω    and    Sdir = G Siso
1

Compare areasFull-sphere angle ÷ lobe angle gives the dimensionless gain G.

2

Multiply density by gainInside the lobe, the same Pt produces G times the isotropic density.

3

Convert ratio to decibelsG(dBi) = 10log10(G).

3. Respect the 20 dBm EIRP cap

EIRP(dBm) = Pt(dBm) + G(dBi)
1

Baseline17 dBm + 0 dBi = 17 dBm EIRP, leaving 3 dB below the chapter’s cap.

2

Make conducted power the subjectPt = EIRP − G.

3

Substitute the patch gainPt = 20 − 6 = 14 dBm.

4

Find the power ratio10−3/10 = 0.501× the conducted power.

4. Find the on-axis payoff

1

Find the EIRP change20 − 17 = 3 dB.

2

Convert decibels to ratio103/10 = 2.00×.

The on-axis power density doubles even though conducted power is roughly halved. The chapter calls this a 2 dB range win; the exact range change still depends on propagation and receiver thresholds.

5. Calculate the beamwidth bill

The chapter uses the Kraus estimate G ≈ 41253/(θEθH) with angles in degrees.

1

Convert 6 dBi to linear gainG = 106/10 = 100.6 = 3.98.

2

Cross-multiplyθEθH = 41253/G.

3

Substitute and divide41253/3.98 = 10362 deg².

4

Assume a symmetric beamθ² = 10362.

5

Take the square rootθ = √10362 ≈ 102°, or about ±51° from boresight.

6. Check yourself

1. With 6 dBi gain and a 20 dBm cap, what conducted power is allowed?

14 dBm, because 20 − 6 = 14.

2. What does a 3 dB on-axis rise mean as a ratio?

10^(3/10) = 2.00× power density.

3. What coverage warning follows from the 102° estimate?

Sensors more than about 51° off boresight may sit beyond the half-power lobe, so placement must be measured.

7. Honesty boundary

These are the chapter inputs, worked results, and named teaching assumptions.

The isotropic radiator is a reference
Current or responsivity value
not a physical antenna
Current or responsivity value
The Kraus relation is an estimate for a simple main lobe
Current or responsivity value
sidelobes
Chapter input or worked result
efficiency
Chapter input or worked result
polarization
Chapter input or worked result
reflections
Time, interval, or service-life value
cable loss
Chapter input or worked result
mounting
Chapter input or worked result
the exact regulatory domain matter
Current or responsivity value
Conducted RF power is not automatically proportional to whole-device battery current
Charge or energy value

Go deeper in Compare Finalists and Validate, then use a site survey and production antenna measurements.