How phase shift changes when we rotate linearly polarized patch antenna?

Georgy.Moshkin

Full Member level 4
I have some idea, but can't find simple answer to one of my questions. It is probably relates radar polarimetry.
Setup is very simple: single transmitter, single receiver.
Transmitter configuration: single patch antenna TX-1, vertical polarization
Receiver configuration: two patch antennas RX-1, RX-2 , vertical polarization.
Distance between transmitter and receiver is much larger than wavelength (>100 wavelengths).
Relative arrangement of transmitter and receiver is symmetrical. Obviously in this case signals received by RX-1 and RX-2 are identical, having same amplitude and phase. All RX signals are sampled exactly at patch feeding points.

Here is my question: If we start slowly rotate RX-2 antenna relative to it's own center of symmetry, what would be relative phase shift between RX-1 and RX-2?
I think that phase shift will vary from 0 to 180 deg. as we rotate RX-2 from 0 to 180 deg. Certainly amplitude will drop rapidly near 90 deg because polarization does not match, but I am mostly interested in relative phase shift of received signals when RX-2 is rotated by few degrees.

Pierre_B

Newbie level 6
Hi,

If we are considering :
• You are in farfield
• A vertical array of 2 elements (RX-1, RX-2)
• A phase shift of the second element only
• Same power feeding for both elements (RX-1, RX-2)
• Maximum directivity at 0°
If you increase the phase shift until you are reaching 180°, you are going to have a Null at 0°. This means that your receiver is not going to receive something in the 0° plan

If you are shifting by a few degrees you are going to have a tilt and of course your maximum directivity is not going to be at 0°. So you can still put your receiver higher to compensate that.

Please find attached a short PDF with several examples. You can see from the picture that you are close to 3dB losses at 90°. So for few degrees you are going to loose like -0.1 dB if you are considering you main lobe at 0° only.

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Georgy.Moshkin

points: 2

Georgy.Moshkin

Full Member level 4
I am interested in relative phase shift between RX1 and RX2. Obviously there will be a phase shift of 180 degrees if we flip it vertically. What about all other angles.

Pierre_B

Newbie level 6
Hi,

My answer was about the relative phase shift between RX1 and RX2 for several cases.
Have you look at the PDF file ? You have several cases like a relative phase shift of 22.5° between your dipoles and you have the effect on the pattern.

Georgy.Moshkin

points: 2

Georgy.Moshkin

Full Member level 4
Thanks for your answer, and the file is great, but did you physically rotate the antenna?
I'll try to rephrase. Let's eliminate antenna array to avoid possible confusion. Also remove any transmitter, we just have a perfect plane wave in space with a known electromagnetic field parameters in each point, for example Ex(t) Ey(t) Ez(t) . Consider we have single patch antenna and this plane wave is constantly coming and exciting currents at it's edges.

At some moment of time T1 we will observe received sinusoidal signal at feeding point with a certain phase φ.
Now, what if this patch was physically rotated by angle α around it's center, how would signal phase φ change depending on physical rotation angle α at moment of time T1? So there are two angles, one for signal, and one for physical rotation of patch around it's center. And I hope that φ is dependent on α, and how linear is it.

I've found some hints in circularly polarized antenna designs using sequential rotating feeding, but they deal with discrete rotation angles 0°, 90°, 180° and 270°, and polarization is circular.
I know for sure that for a simple linearly polarized patch antenna physical rotation α=180° will result in signal phase shift φ=180° (phase shift relative to the case when patch physical rotation α=0° )

Again, thanks for your attention.

Last edited:

Pierre_B

Newbie level 6
Hi,

You are welcome for the answer.

For me, if you have an antenna with a vertical polarization as a reference with α=0° as the rotation of the patch and φ = 0 ° as the phase shift then :
• If you rotate the patch then this is going to apply a phase shift compare to your reference of 0°. Moving α means you are moving φ of the same amount compare to your reference. If you are not taking any reference moving α doesn't mean you have an electrical phase shift.
In your case if you are taking α=90° (for example) then you are passing from a vertical polarization to an horizontal polarisation. This means you are going (in theory) receive no signal compare to your reference.

Here are some results of the attenuation that you are going to have compare to your reference :
• α=φ=45° -> Losses = -3dB
• α=φ=25° -> Losses = -0.85dB
• α=φ=20° -> Losses = -0.54dB
• α=φ=15° -> Losses = -0.30dB
• α=φ=10° -> Losses = -0.13dB
• α=φ=-> Losses = -0.033dB
• α=φ=2.5° -> Losses = -0.0082dB
• α=φ=-> Losses = -0.0013dB
I'm hopping this is going to help you