I want to measure the axial ratio (Not in simulation software but with the fabricated model). Kindly suggest me the steps.
One clarification to Marcelo's method. A scalar only approach like he describes will work, but you have to start with a rotational search on the polarization axis to find the minimum signal (the minor axis of the ellipse) and then the cross-pol orientation will be the major axis. The reason for looking for the minimum is that the null is generally narrower in angle than the peak and thus you can very accurately find the null of a linear antenna, while if you find the apparent peak, 90 degrees away may not be in the middle of the deepest part of the null. Note of course that for a "clean" elliptical polarization, you should just be able to find the minimum and maximum signals. They SHOULD be at 90 degrees from each other. Once you find the major and minor elliptical polarization components, the ratio of those two will give you the AR.
Any other random starting orientation other than the max/min won't be correct. For example, if you start with the dipole at 45 degrees to the major/minor axes, the cross pol will also be at 45 degrees from the major/minor axes and you'll measure circular polarization, whether it's linear (major axis only), circular (major = minor) or elliptical. With vector data you don't need to figure out the polarization orientation since you can measure the phase relationship between the two components you measure and calculate the necessary rotations to get the major and minor axes.
This is a straightforward output of a passive vector antenna pattern measurement, but it's unclear what equipment you have access to. You can determine axial ratio by measuring the magnitude and phase of two orthogonal polarizations. Normally this would be two linear polarizations (e.g. vertical and horizontal or theta and phi) measured using a vector network analyzer (VNA). From that you can determine the elliptical components and use the ratio of major to minor axis to determine the axial ratio. Let me know if you need more info.
Dear Kiran
As Michael pointed out, it depends on your available instrumentation. But basically you will measure the power received by a dipole (for instance), tuned to the same frequency you're interested in. Write it down and measure again but this time the dipole is rotated 90 degrees (orthogonal) to the previous position. THe relation of power between both measurements will give you the AR - if both measurements are the same you'll have a circular polarized wave (AR=1 or 0 dB).
One clarification to Marcelo's method. A scalar only approach like he describes will work, but you have to start with a rotational search on the polarization axis to find the minimum signal (the minor axis of the ellipse) and then the cross-pol orientation will be the major axis. The reason for looking for the minimum is that the null is generally narrower in angle than the peak and thus you can very accurately find the null of a linear antenna, while if you find the apparent peak, 90 degrees away may not be in the middle of the deepest part of the null. Note of course that for a "clean" elliptical polarization, you should just be able to find the minimum and maximum signals. They SHOULD be at 90 degrees from each other. Once you find the major and minor elliptical polarization components, the ratio of those two will give you the AR.
Any other random starting orientation other than the max/min won't be correct. For example, if you start with the dipole at 45 degrees to the major/minor axes, the cross pol will also be at 45 degrees from the major/minor axes and you'll measure circular polarization, whether it's linear (major axis only), circular (major = minor) or elliptical. With vector data you don't need to figure out the polarization orientation since you can measure the phase relationship between the two components you measure and calculate the necessary rotations to get the major and minor axes.
It should be noted that dipole mentioned above may receive cross polarized component
of e. m. wave emitted by your antenna. It is due to the dipole's nonideal geometry. In such instance, the dipole can be replaced for example with a rectangular horn.
From th other hand the horn may be much more easy-to-handle in the high-frequency band.
- Badges
- Science topic