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E-plane and H-plane identification in microstrip planar array

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kae_jolie

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I am confused about which plane is the E-plane and which plane is the H-plane in a microstrip planar array.

There are two scenarios I find in the literature:

Scenario 1: E-plane is x-z plane where Φ= 0° and H-plane is the y-z plane where Φ= 90°

Scenario 2: E-plane is y-z plane where Φ= 90° and H-plane is the x-z plane where Φ= 0°


Could someone explain to me how I can tell for sure which is E-plane and which is H-plane? Is it related to the orientation of the length and width of the patch to the x- or y- axis? So if array is aligned along the x-axis, Scenario 1 is valid and if array is aligned along y-axis, Scenario 2 is valid? By array alignment along x-axis, I mean the patch width of each element of the array is aligned or in parallel with the x-axis.

Thanks.
 

Thanks, vfone. This diagram makes sense for an antenna oriented along the z-axis. In this case, the H-plane is on x-y plane and E-plane could be either the x-z or y-z plane. In the case of a planar microstrip antenna which lies on the x-y plane, the E-plane and H-plane is different and could be confused as I explained in my previous post.

Thanks.
 

H- and E-field planes depend on the antenne orientation and are defined based on the main beam direction. The E-plane is the plane that has the direction of propagation and the E-field in common. As E and H are perpendicular to eachother, the H-plane is perpendicular to the E-plane.

If the you feed a half wave resonating horizontally oriented patch from left or right side, the current in the patch is along (or parallel with) the x-axis. So there is an oscillating E- field between the left and right side. The upwards going radiated field (positive Z direction) has its E-field along the x-axis, so the E-plane is the XZ-plane.

The H-field is oriented along (or parallel with) the y-axis (H-field is perpendicular to current flow), hence the H-field plane for the main beam direction is the YZ plane.

If you would feed from positive or negative Y direction, the patch current oscillates along the y-axis. Then the E-field plane is the YZ plane and the H-field plane is the XZ plane.

When you have a vertically polarized antenna that radiates towards positive x-direction (with zero elevation), the E-plane is the XZ plane. As the H-field is in the horizontal plane (XY plane), the H-plane is the XY plane. Note the in this case an elevation angle of 0 degrees and pointing in positive X direction, means theta = 90 and phi=0
 
ok. still somewhat confused. in the case of this array configuration at
https://www.google.com/search?q=mic...jp%2Fmwel2012%2Fresearch%2Fmsa-e.html;686;310

which is the E-plane and which is the H-plane? Note that the current direction is not the same on all patches since they are going in opposite direction.

Thanks.

- - - Updated - - -

WimRFP, you mentioned that E-plane is the plane that has the direction of propagation.....in the case of a dipole, for example, located on the z-axis, the E-plane is the x-z plane but the direction of propagation is in the x-y plane, correct?
 

In your link, (patch array with inset feed), the current oscillates along the x-axis (for all patches). The E-field polarization of this array is aligned with the x-axis. Assuming that it radiates upwards, the E-plane is the XZ plane. The H-field is in Y-direction, hence H-plane is the YZ plane.

So if you want to draw the E-plane pattern, you need to set phi=0 and you plot gain versus theta angle. For the H-plane pattern, you set phi=90 and you plot gain versus theta angle.

It seems that the transmission line length between opposite patches is different, very likely this is to get in phase currents in all patches, may be in combination with some detuning of patches to get the required phase (but this is not related to your question).

the E plane is that plane that has the propagation direction (that is a vector) AND E-field vector in common.

For the dipole, the maximum radiation is in the XY-plane (that is theta = 90, phi = 0...360). Of course it also radiates at positive and negative elevation, but with less intensity. There are propagation vectors that leave from the origin and go to all directions in the XY plane, so the E-plane can be XZ, YZ, or any other plane that has the Z-axis in common.

If you want to plot the H-plane gain, you set theta=90 and phi goes from 0...360. You know you find something around 2.1 dB for all phi).

For antennas that have their main beam gain not on the x, y, or z axis, you can just follow the rule that the E-plane has the propagation direction and the E-field vector in common. The H-plane rotates 90 degrees around the propagation direction.
 
WimRFP, you mentioned "the current oscillates along the x-axis" Why is that? Is it because current oscillates along the wide (W) side of the microstrip patch? If so, why is that? Why not the other side (L)?
How about a square patch? Which axis does the current oscillate along? Both sides of a square patch are equal.

Thanks.
 

If you have a square or round half wave resonating patch (without any slots or other features on it), the position of the feed determines how the standing wave develops.

Assuming that you look from positive z-direction towards the XY plane:

When you feed from left (negative x) or right (positive x) direction, the voltage maximums are at the left and right side of the patch, like in a half wave dipole that is on the x-axis. The charge to enable the high voltage on the left AND right side causes the patch current to oscillate from left to right. This is the same as in the half wave dipole. The mechanical equivalent is a square sized cup filled with water. When you move it from left to right only (that is along the x-axis), you can see the standing wave and most of the horizontal movement of the water is in the middle, like in the half wave resonating dipole.

When you excite from the top (positive y) or bottom (negative y), the current oscillates in Y direction and the high voltage edges are the upper and low edge. It is like moving the cup filled with water along the y-axis.

So whether the E-plane is the XZ or YZ plane depends on whether your feed is on the x-axis or y-axis. Somewhat OT, when you feed from both axis, but 90 degrees out of phase, you have circular polarization.

When the patch shape isn't symmetrical (for example by adding a slot under 45 degrees, you may get current oscillating in y-direction also when feeding from x-direction. This is used to get circular polarization from a single feed.

Just see the half wave resonating patch as a very wide electrically half wave dipole above a ground plane.
 
what if I feed a rectangular patch antenna with a coax probe located right at the center of the patch? Which axis would the feed be considered at? X or Y? Which plane would be considered the E- or H- plane? From my understanding, the E-plane will be the wide side of the rectangular patch and the H-plane will be the short side of the rectangular patch? Is this correct?

Then, this brings up another question: what if I feed a square patch antenna with a coax probe located right at the center of the patch? Which is the E- and which is the H-plane? I think at this matter it does not matter. Either plane is E-plane or H-plane. Is this correct?

Thanks.
 
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If you use a probe feed right or left of the center (center of patch in origin), the feed is on the x-axis, hence the current oscillates along the x-axis (that is from left to right and vise versa). The energy that goes in positive Z direction has its E-field parallel with the x-axis. So the E-plane is the XZ plane and the H-plane is the YZ plane.

as long as you don't get a higher mode resonance also a (almost) square patch would have its E-plane in the XZ direction. If you have an EM simulator, try to get a vector current density animation, then you will see that in the square patch the current oscillates along (parallel with) the X-axis (when feeding left or right from the center). There will be low current running in y-direction.
 
could you elaborate on "If you use a probe feed right or left of the center"? What do you mean? The probe feed is exactly at the center of the patch and fed from the bottom. How will it be left or right of the center? I am missing something.
 

If your probe feed is in the center of the patch, you don't excite a half wave resonance in the patch. With "center" I mean the crossing of the two center lines (or diagonal lines) that you can draw for a rectangular patch. You can see the center as the center of mass or gravity.

Assuming that the center is in the origin of the XY plane, "right of the center" is a position on the positive X-axis.
 

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