For a school assignment I need to calculate a link budget between two half wave dipole antennas, but I am stuck at calculating the polarisation losses in the antennas. the antennas are ideal and matched. The antennas are 6 KM apart, the signal strengt Ps = 10 W, frequency f= 100 MHz, angle between the antennas is 50 degree and the max Gain of the antennas is 2,13dB. enter image description here

I have to calculate the link budget between S and E3. What the teacher did the following to get the Gains with polarisation loss:

Gs3 = 10log (10^2,13 * (sin^3(90+50))^2 ) = -1,707dBi

G3s = 10log(sin^2(90-50) ) = - 1,707dBi

The problem is that I am not getting these numbers when I use these formulas and different now I am seriously confused about how I am supposed to calculate this.

I do know that the original formula for polarisation loss in dB is: 20 log( Cos^2(a)). But even when I use this formula I am not getting the same numbers as what the teacher gave us.

  • $\begingroup$ Those calculations by your teacher make no sense. How is it they have the same value? And what is Gs3 and G3s? And they calculations don't even add up: by my reckoning the first is equal to 0.403, and the 2nd is -0.384. $\endgroup$ Nov 14, 2016 at 13:36
  • $\begingroup$ I think Gs3 is supposed to be Gain of S to E3 and G3s is Gain of E3 to s. I have played around and made a few corrections to the formula and now I am getting really close to the answer the teacher gave us, but I am still unsure if its correct. new formula: 10Log(10^(2,13/10)* (sin(90+50))^2 = -1,7086 10Log(10^(2,13/10)* (sin(90-50))^2 = -1,7086 $\endgroup$
    – MarkerDave
    Nov 14, 2016 at 14:31
  • 2
    $\begingroup$ The gain between the inputs/outputs of any antenna are identical. Further, the received power will be identical if the transmit power is equal. Lastly, there will not be any polarization mismatch between two ideal dipoles in the same 3D plane (as pictured here). The only loss would because by not being directed at each other, which is a function of the angle and line of sight distance. $\endgroup$
    – Andrew W.
    Nov 17, 2016 at 22:41

1 Answer 1


I think I can decide the intent of the formula. It's wrong, as you've written it.
I assume the drawing is a flat 2D picture of three antennas in space, not a perspective view of three antennas on the earth's surface.

In this case I can see what you're trying to do:

Gain of a dipole is 2.13 dBi.

The radiation pattern of a dipole can be approximated by $\sin^2(\theta)$.

So you could say $10\log_{10}(10^{2.13/10} \sin^2(90-\alpha))$.
Or better, $2.13 + 20\log_{10}(\sin(90-\alpha))$.
Equals $2.13 - 3.8 = - 1.73$.

This is simply the gain of dipoleE3 in the direction of dipoleS.
Wikipedia has a good page with this picture:
enter image description here

There is no polarisation loss as they're in the same vertical plane. Your formula for that does look correct, but for dipoles not in the same horizontal plane, to find the angle a you might need some tricky trigonometry.


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