physical logic of axial ratio measurement

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Hello,there is an intresting youtube manual regarding measuring axial ratio in anechoic chamber shown bellow.
in the video i think they have a circularily polarzied antenna which is being tested by linear polarised antenna.
They rotate the linear antenna ,thus changing the polarization of the reference antenna.
So if we calculate the polarization loss factor of linear with respect to circularly polarised.
by the formula bellow it remains the same(PLF).

In CST we have CO-polarization and Cross polarization,so given the manual in the link bellow
our R_C is Amplitude in RH/ Amplitude in LH
In the results shown bellow R_C=R_H/L_H=7.8-(-11)=18dB
So out AXIAL ratio will be 0dB almost.
The same thing will happen if we use linear polarization case.
its CO polarization will be much higher then the cross polarization and by this formula We will get AXIAL ratio of 0dB Even in linear.
Where did i go wrong using this formula?
Thanks.














 

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yefj said:
The same thing will happen if we use linear polarization case.
its CO polarization will be much higher then the cross polarization and by this formula We will get AXIAL ratio of 0dB Even in linear.
Click to expand...
No. Linear polarization corresponds to R_C = 1, respectively axial ratio = inf.
 

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Hello FVM,another problem is a mathematical one, suppose we have E_0=h^+v^ without the complex number in "v"
if L=v^ stays the same then we get PLF 0.5=-3dB the same as E_0=h^+v^
|(h^+v^) dot v^|=1
|(h^+jv^) dot v^|=1
so 45 linear is the same as circular polarised?
Where is a problem in my mathematical logic?
Thanks.


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Hi there,

The value that CST gives you for the Axial Ratio is correct. Co-Pol - Cross pol = 19.23dB (at Theta =0 degrees).
Cross-pol Discrimination (XPD)=Co-Pol - Cross Pol = 20log ((AR+1)/(AR-1))=19.23dB.
AR is the Voltage Axial Ratio. If you solve the above equation, you'll find that AR=1.24 (linear). 20log(AR)=~1.906 dB, which is what CST shows you.

If your antenna is circularly polarised, it will be able to receive a linearly polarised signal from a transmitter, but the received power will be 3dB lower.
 
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