Legend has it that Stephen Hawking was warned that he would lose half his readers for every equation in “A Brief History of Time.”  His editor persuaded him to remove them.  Hawking removed all but $E=MC^2$ because Einstein’s mass-energy relation was essential to the rest of the book.  In the same way, we cannot understand radio communications without the communication link equation (Friis’ formula).  Hawking’s editor assumed Hawking’s readers would be airport bookstore pop science readers: the kind who like to sound smart at cocktail parties.  AP readers are far smarter. Therefore, in spite of the equation(s), I am confident I will lose none of my readers.  Expressed logarithmically, I will keep 3 dB more readers than Stephen Hawking.  We should see the following exercise as part of a greater goal of learning as much math as possible.  Your kids will make more money the more math they learnSo will you.

The following explanation of Friis’ transmission formula is adapted chapter 2 of Stutzman and Thiele’s excellent book “Antenna Theory and Design (2nd)”.  The formula is expressed in terms of transmitted power, transmit and receive antenna gains and wavelength, though more terms such as match and cross-polarization losses can be added.  This formula is used to calculate received signal power in communication links.

$P_{r} = P_{t,W or mW}\frac{G_{t,linear}G_{r,linear} \lambda^2 }{(4\pi R)^2} \hfill(1)$

$P_{t}$ is the power transmitted by your friend to you.  On a data sheet, this can be expressed as a linear value in watts (W) or milliwatts (W) or as a logarithm (dBm): 10 x log10 (power in milliwatts)

$G_{t}$ is the gain of your friend’s antenna.  Though the above formula is linear, this is usually given as a logarithmic value of dBi on a datasheet: 10 x log (linear gain of the antenna/the gain of an isotropic antenna)

$G_{r}$ is the gain of your antenna

$\lambda^2$ is the square of the wavelength in meters.  The smaller the wavelength, the greater the frequency.

$R$ is the distance in meters between you and your friend.

$(4\pi R)^2$ is the surface area of a sphere with radius R.  This is spreading loss of the transmission.  The further away you are from your friend, the more the power spreads out as on the surface of a sphere and the less power there is to receive.

To perform the calculations, take the base-10 logarithm of both sides of the equation.  Gain values on antenna datasheets are already in dB.  Power values on radio data sheets may be expressed in watts, milliwatts, or dBm.  If they’re expressed in linear units, convert to dBm as explained above.

$P_{r,dBm} = P_{t,dBm}+G_{t,dBi}+G_{r,dBi}+20log(\lambda)-20log(4\pi R) \hfill (2)$

Let’s use (2), the Log form of Friis’ formula to compute an example.

My friend is using his VX-6E to transmit 10 watts at 146.40 MHz in the 2 meter band.  We are 1 km apart.  My friend is using a field-expedient half-wave wire dipole the he learned how to make in NC Scout’s RTO class.  I bought a 2m slim-jim to use with my portable antenna mast in suburbia.  How much power will I receive from my friend?  Will it be enough?

# The top of the equation

$\bf P_{t}$ is 10 watts = $10x10^3$ milliwatts.  Convert to dBm as follows: $10LOG_{10}\frac{10x10^3mW}{1mW}$= +40 dBm

$\bf G_{t}$ is the gain of my friend’s field-expedient wire dipole is 2.51 dBi

$\bf G_{r}$ is the gain of my slim-jim receive antenna given by the datasheet is 6 dBi.

$\bf \lambda^2$ The wavelength is  $\frac{3x10^8 m/s}{146.40x10^6Hz}$ = 2.05 meters (good).   The log form of the wavelength squared is $10LOG_{10}((2.05 m)^2) = 20 LOG_{10}(2.05)$= 6.24 dB

# The bottom of the equation

$\bf (4\pi R)^2$ is $20xLOG_{10}(4x3.14x1x10^3 m)$ = 82 dB

# The result:

Add the top and subtract the bottom:

+40 dBm + 2.51 + 6 + 6.24 – 82 = -27.5 dBm.  In milliwatts, this is $10^\frac{-27.5}{10}$= .00177 milliwattsThis is the power you receive from your friend over the communication link.  Is it enough?  Is my radio sensitive-enough to receive it?

The sensitivity of my Baofeng UV-5R is .2 microvolts @ 12 dB SINAD.  We’ll explore receiver sensitivity in another article.  For now, use equation 9 and assume 50 ohms to convert .2 microvolts to dBm.  I compute a sensitivity of -121 dBm, which is less than -27.5 dBm so I still receive the transmission!

Note that $\bf P_{t} + G_{t}$ (the transmitter power plus the transmitter antenna gain) is the effective radiated power (ERP) of my friend’s radio.  It’s almost always cheaper and easier to improve the antenna gains in a communication link than increase the transmitter power.  Next time (hopefully), we will discuss antenna gain.

Bonus: Re-calculate the power received using the jungle-foliage attenuation for 100 MHz.  Do you still receive the signal? (This is easy, don’t over-think it).