How could I determine if 2 antennas could communicate knowing the distance between them, the power/gain of transmitting/receiving, and the sensitivity level of the receiver?

  • \$\begingroup\$ Google RF link budget calculator. \$\endgroup\$ Commented Aug 30, 2019 at 18:34
  • 1
    \$\begingroup\$ Unless you are talking about spacecraft, you are missing an issue so serious as to mostly dwarf those you are considering: horizon (and any potential mechanisms for propagation beyond). Also in pretty much any case, interference sources. Practically speaking, start by looking at what is usually used for applications like yours - not just power levels, but modulation types, receiver performance, antenna type and siting (towers) etc. \$\endgroup\$ Commented Aug 30, 2019 at 18:52

2 Answers 2


We consider communicating across the town between buildings (1,000 meters) and across the country between high hills?? (1,000,000 meters).

The bigger the antenna, the better. Lets assume Citizens Band, near 30MHz with wavelength of 10 meters, and quarterwave verticals of approximately 100 inches. This size antenna (which needs a "ground counterpoise" of some wires underneath) can be strung up the wall of your living room; should you lack a cathedral ceiling, simply add some CLCL variable-capacitors fixed-inductors to have a fine resonance at your carrier; this is called antenna-tuner or matching.

Assume 10 bits per second; this allows very low power or very long range. And assume simple ON/OFF modulation, or BPSK, or perhaps differential QPSK which is self-synchronizing (there is no ambiguity). You could go for dense constellations, but the symbol rate becomes very low and phasenoise and carrier acquisition become big problems.

Assume thermal floor at room temperature is the famous -174 dBm/rootHertz which comes from K*T (Boltzmann constant and Kelvin temperature) being exactly 4.00e-21 watts/Hertz at 290 Kelvin.

Assume SignalNoiseRatio of 10dB. Depending on single or double-sided bandwidth, you will have approximately 0.1% Bit Error Rate BER), or 10^-3 rate. 10dB allows plenty of margin to bring Bit Error Detection and Correction into your system; this reduces lost-packets and the need to re-transmit packets.

Assume (front end losses plus Noise Figure) to total 4dB. This is 1/2.5 loss.

We make no allowance for multipathing, foliage, or interference, or timing-recovery. Nor for long-distance ionospheric reflections.

This is the signal strength we need:

-174 thermal floor at 1 cycle-per-second (1 Hertz) bandwidth, with

  • 10dB for 10Hz bandwidth

  • 10dB SignalNoiseRatio (this ignores the wonders of TurboCodes, etc)

  • 4dB losses and Noise Figure of that first transistor

and all this becomes

-174 + 24 = -150dBm. Given -120dBm across 50 ohms is 0.632 microVolts Peak Peak, we are 30 dB (31.6X) weaker, at 20 nanoVolts Peak Peak. Just a number.

Remember the -150dBm energy floor.

Assume the antennas have gain=1 (compared to an ideal isotropic antenna). So we don't worry about antennas any more.

Now, since the energy spreads out in X and in Y as the RF energy propagates, we have range*range reduction in available energy at the receiver. There also is a hemispheric bit of math, involving 4*PI.

The result is the PATHLOSS: 22dB + 10*log10[ (range/wavelength)^2 ]

Lets consider two situations:

A) across town 1,000 meters

B) across the country 1,000,000 meters

Across_town: range/wavelength = 1,000/10 = 100

with PATHLOSS (the spreading out of energy in 3_D) of 22+40 == 62dB. Which is a little more than 1Million:1 energy spreading.

Given we need -150 from the across-town receiver antenna, we need -150+62 or -88dBm into the Transmitter antenna in that highrise across the park. How big is this? since 0dBm/50_Ohm is 0.632vPP, and -88dB is near 1/31,000 we need 0.632/31,000 = 20 microVolts PeakPeak into the transmitter antenna.

For a robust link, you should consider pseudo-random spreading at the transmitter, and then a synchronized de-spread and low-pass-filter at 10Hz, in your receiver frontend. These systems may have regulations applying. I don't know.

Now for ACROSS THE COUNTRY. give the across-town math for 1,000 meters, we easily extend this to 1,000,000 meters.

The PATHLOSS increases by 1,000 ^2 or one Million X power, or 1 Thousand X voltage. Thus you need 20 milliVolts Peak Peak, at Citizens Band.


So what are we learning here? With physically large antennas for long wavelengths and low carrier frequencies, and using low low symbol/bit/baud rates, you can use very low power.

However, you must design the system to be able to track (search for carrier) transmitter frequency wander, caused by cheap XTAL oscillators.

You must tolerate enormous blockers either on top of your signal, or at some small offset frequency. In either case, you need receiver front-end amplifiers with high overload specs, such as the IP2 and IP3 etc. This requires high currents in the front-end LNAs (or transformer feedback to linearize). Also to handle blockers on top of your signal (either shared-channel or otherwise), you should examine code-spreading at TX, and then correlations to de-spread in the Receiver. A code-spread ratio of 1,024:1 should give you a 60dB better tolerance for energy on top of your channel. You must still have a very high purity (very low distortion) front end in the receiver. By the way, vacuum tubes may inherently provide the needed performance (except for the thermal noise floor?). With transistors, expect to operate at high currents and use discrete resistors to linearize the operating point.

PS there may be errors herein.


There are two aspects:

1) Power transfer of the transmission

Friis transmission equation - Wikipedia provides a calculation.

\$P_{received} = \frac{Gain_{transmis \, antenna} * Gain_{recieve \,anatenna} *\lambda^2}{(4*\pi*R)^2} *P_{transmit}\$

Along with the sensitivity of the receiver and the transmitted power you can calculate if the transmission will work.

2) The Friies formula implies several conditions for a transmission.

  • The antennas are in line of sight.
  • The antennas have to resonate at the same frequency.
  • The polarization of both antennas has to match.
  • The antennas are well matched.
  • No additional free-space attenuation

If these conditions are not fulfilled there are additional losses.

  • \$\begingroup\$ Even in a case where this applies (ie, line of site) it still does not yield a useful answer, as you must consider the desired signal vs. noise and interference... \$\endgroup\$ Commented Aug 30, 2019 at 18:54

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