Calculate broadcast range, transmit power, or frequency with a simplified free-space propagation model using any two known values.

Broadcast Distance Calculator

Enter any 2 values to calculate the missing variable

Assumes free-space propagation, isotropic antennas (0 dBi), and a reference received power of 1 µW (−30 dBm). Real-world coverage can differ greatly due to antenna height, terrain, clutter, and receiver sensitivity.

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Broadcast Distance Formula

The calculator uses a simplified free-space form of the Friis transmission equation. It assumes isotropic antennas, 0 dBi antenna gain, no cable loss, no obstruction loss, and a reference received power of 1 µW, which is -30 dBm.

To calculate distance:

D = (c) / (4π f)√((P) / (Pᵣ))

To calculate frequency:

f = (c) / (4π D)√((P) / (Pᵣ))

To calculate transmit power:

P = Pᵣ((4π fD) / (c))²
  • D = broadcast distance in meters
  • f = frequency in hertz
  • P = transmit power in watts
  • Pr = reference received power, fixed at 1 µW or 0.000001 W
  • c = speed of light, 299,792,458 m/s
  • π = pi, approximately 3.14159

If you enter frequency and power, the calculator solves for distance. If you enter power and distance, it solves for the frequency that would produce that free-space result. If you enter frequency and distance, it solves for the required transmit power. Unit selections are converted to hertz, watts, and meters before the formula is applied, then converted back to the unit you selected.

Typical Broadcast Frequency Ranges

These ranges help you choose a reasonable frequency input. The calculator can accept kHz, MHz, or GHz, but broadcast services normally fall into specific bands.

Service or band Typical frequency range Common input unit
AM radio 530 to 1700 kHz kHz
FM radio 87.5 to 108 MHz MHz
VHF television 54 to 216 MHz MHz
UHF television 470 to 608 MHz in many current allocations MHz
Microwave links 1 GHz and higher GHz

How Power and Frequency Affect the Result

Change Effect in this free-space model Reason
Double transmit power Distance increases by about 1.414 times Distance follows the square root of power
Quadruple transmit power Distance doubles √4 = 2
Double frequency Distance is cut in half Distance is inversely proportional to frequency
Double distance Required power becomes 4 times higher Power follows distance squared

Example

Example 1: Calculate broadcast distance

Suppose the frequency is 100 MHz and the transmit power is 1,000 W.

D = (299792458) / (4π(100000000))√((1000) / (0.000001))

The result is about 7,544 m, or 7.544 km. This is the simplified free-space distance to the 1 µW reference received power level.

Example 2: Calculate required transmit power

Suppose the frequency is 100 MHz and the target distance is 10 km.

P = 0.000001((4π(100000000)(10000)) / (299792458))²

The result is about 1,756 W, or 1.756 kW.

FAQ

Is this the same as real broadcast coverage?

No. This is a free-space estimate using a fixed received power reference of 1 µW. Real broadcast coverage depends on antenna height, antenna pattern, terrain, buildings, trees, ground conductivity, receiver sensitivity, interference, polarization, and regulatory limits. The result is best treated as an idealized physics estimate, not a coverage guarantee.

Why does a lower frequency give a longer distance?

In this formula, distance is inversely proportional to frequency. If power stays the same, lowering the frequency increases the calculated free-space distance. This is one reason lower-frequency signals often appear to travel farther, although real propagation also depends on many environmental factors.

Why does increasing power not increase distance by the same amount?

Distance depends on the square root of transmit power. For example, doubling power only increases the calculated distance by about 41%. To double the calculated distance, you need four times the transmit power under the same assumptions.