This is maybe a very basic question, which is driving me insane, because I can't remember the theory and neither I can seem to find any source about the topic.

I have an antenna, whose SPR (sector power ratio) is defined as the ratio between the co-polar field amplitude irradiated in a defined sector (let's say from -60° to +60° in respect to antenna's boresight), and the co-polar field amplitude irradiated outside of it plus the cr-polar field amplitude irradiated anywhere.

All the formulas I am finding simply reduce the problem to:

Sigma(Ampl Co in Sector) / [Sigma(Ampl Co out Sector) + Sigma(Ampl Cr everywhere)]

where Sigma is the sum of the amplitudes by each angular unit.

Now, the doubt that was raised to me was that since the two polarizations (+45° and -45°) lie with 90° angular difference from each other, they would lie on different vectorial planes too, therefore the sum shouldn't be the "normal" one.

Shouldn't the sum in this case be the quadratic sum? So basically:

Sigma(P Co in Sector) / Sqrt{[Sigma(P Co out Sector)]^2 + [Sigma(P Cr everywhere)]^2}

Is the following formula correct?

10*log{Sigma(P Co in Sector) / [Sigma(P Co out Sector)^2 + Sigma(P Cr everywhere)^2]}
  • \$\begingroup\$ 360 sphere/ cone angle of radiated power = antenna gain max. ( assuming impedance matched) \$\endgroup\$ Sep 13 '16 at 7:17
  • \$\begingroup\$ This is not quite what I was asking :/ \$\endgroup\$ Sep 13 '16 at 8:26

Powers add directly; signals like voltage or current have to add as \$\sqrt{A^2+B^2}\$ if they are un-harmonically related RMS values.

Electrical power is proportional to \$V^2\$ hence a normal "add" is appropriate. In other words, power is a scalar quantity and not a vector quantity.

  • \$\begingroup\$ Power is still a scalar quantity and does not add vectorially. \$\endgroup\$
    – Andy aka
    Sep 14 '16 at 10:48
  • \$\begingroup\$ @Noldor130884 - you keep accepting then unaccepting this answer - what is troubling you? \$\endgroup\$
    – Andy aka
    Sep 15 '16 at 10:33
  • \$\begingroup\$ sorry I saw this comment only now. Well, basically I'm not sure this really answers my question... \$\endgroup\$ Jun 13 '17 at 6:00

I didn't think about the properties of the log... Of course a log(sqrt( ... ) ) is equal to 0.5 * log( ... ) therefore the formula was correct


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