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

[Georgy.Moshkin]

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]

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.

 

[Georgy.Moshkin]

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]

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]

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 E_x(t) E_y(t) E_z(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.

 

[Pierre_B]

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=5° -> Losses = -0.033dB
• α=φ=2.5° -> Losses = -0.0082dB
• α=φ=1° -> Losses = -0.0013dB

I'm hopping this is going to help you

 

[Georgy.Moshkin]

I value your effort to explain things, but your answer (or my question?) is probably misleading. I performed some EM simulation and it seems that signal phase shift φ is not linearly dependent on physical rotation angle α. Note that angle α has nothing to do with radiation pattern, gain, etc. Angle α is antenna physical rotation in substrate plane. Currently I am observing that phase φ is almost constant and only changes rapidly when α approaches 90°. It is probably have something to do with antenna symmetry.

Consider following example: let's think about slotted waveguide array antenna. Radiating slot is very similar to edge of patch antenna.
If α=φ was true, then we could not use half wavelength spacing for slot arrays with rotation based conductance weighting. In small arrays slot angles are often close to α=45°. If additional phase shift exists and equal to φ=45°, then we surely need compensate it by moving slot along waveguide by 45° phase length. As we know it is never the case, and slot spacing is always 180° of waveguide phase length. Also if α=φ was true, then there already was some cost effective microwave angle encoders based on pair of patch antennas on the market, but we do not see any.

 

[Pierre_B]

Hi,

thanks for your answer, let's wait if someone as another idea

I found that α=φ if you are comparing your dipole before and after the rotation (you have a reference and then your are moving the angle and you compare it : physical rotation and phase shift before and after) but if you are taking only a simple dipole of course α is not equal to φ (the rotation of the dipole doesn't move the phase of it).

What i know from it is the slot are often 45° (slant +45 / -45) for the network operator polarization and you need to respect the distance between your element to avoid the grating lobe.

When you have some elements on a reflector plate with the same polarization you can turn some a little bit to optimize your pattern. Of course EM simulation or real farfield measurement can easily verifiy that.

Let's wait if someone else can help you !

Cheers,

 

[Georgy.Moshkin]

Currently the only simple way I've found is changing size (height) of patch antenna. This works well only with wideband patch, because increasing height of classical patch is shifting resonant frequency by significant amount. Rotation works to some extent, but usable phase shift is limited due to increased losses near 90° physical angle. There are many possible solutions, and I am trying to find one with minimal losses and maximum phase tuning range, and small element size and constant outer dimensions. Interestingly, I've found few papers using rotated elements for reflectarrays to obtain required phase shift for each element. Somehow It works, but it is unclear if feeding antenna have circular or linear polarization, and if array performs polarization conversion.

 

[volker@muehlhaus]

Currently I am observing that phase φ is almost constant and only changes rapidly when α approaches 90°.

This is what I would expect. The diagonal polarization can be represented by two orthogonal components, and only one component is received because it matches your receive antenna. That one scales in amplitude (!) when you rotate the antenna, and changes sign when rotation crosses the 90°. But that's it, I would expect a cos(angle) dependency in amplitude and that's what gives you the opposite sign = phase jump.