Patch and dipole antenna

  • Thread starter legyptien21
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  • #1
legyptien21
Hi

Ok I have a contradiction in my mind:

Patch antenna is called a leaky wave resonant cavity because of the magnetic slots which radiates. We all agree on the fact that they are standing waves with clearly identified max and min as a characteristic of a specific mode inside this cavity.

1) My question is if we have standing waves, in my mind we would have a max reflection (=1) on the magnetic walls so how comes we have leakage if we have full reflection on the wall ??

2) If we have NOT full reflection on the wall so it means there is no standing wave in the cavity right ? the max and mins are moving ?

Thanks
 

Answers and Replies

  • #2
davenn
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Patch antenna is called a leaky wave resonant cavity because of the magnetic slots which radiates. We all agree on the fact that they are standing waves with clearly identified max and min as a characteristic of a specific mode inside this cavity.
That's not what I am familiar as a patch antenna

what you described is a slotted waveguide

8+8WaveGuideDiag.gif


these are patch antennas .....

11310432_f260.jpg


Antenne_patch_2_4_GHz.jpg



so what type of antenna are you really talking about ?

Dave
 
  • #3
legyptien21
Thanks for your answer. I'm definately talking about the 2 last pics, the patch antenna !!! I try to understand why your thing that I'm describing a slotted waveguide. Maybe because I wrote magnetic slot ? I was talking about the fringing field which can be "remplaced" by 2 magnetic current. The patch is equivalent to 2 magnetic slots....

when I talk about reflection on walls, I talk about the 4 magnetic walls which are delimiting the patch cavity....
 
  • #4
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What is a magnetic wall?
 
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  • #6
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It's my understanding that a patch antenna works by having the electric field fringe between the patch and the ground plane. The magnetic field is caused by currents moving down the edge of the patch and the resulting image current moving in the ground plane. The "end" of the antenna where the standing wave typically forms depends on where the feed point is and the shape of the patch.
 
  • #7
legyptien21
look what I don't understand is that an incident wave is brought by the transmission line then a reflection occurs inside the cavity (between the metallic patch and the ground plane) at the end of the cavity (limit of the metallic patch). This reflection isn't a full reflection for me since there is a leakage at the end. so it's not a standing wave inside the cavity ? I'm even confused on the phenomena of a partially reflected wave at a load which interfere with itself. It shouldnt be a standing wave if there is NOT a reflection coefficient of 1....to me...

let me know what you think

thanks
 
  • #8
Baluncore
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It shouldnt be a standing wave if there is NOT a reflection coefficient of 1....to me...
Any reflected wave will generate standing waves with the incident wave. The position of those nodes and nulls will be determined by the nature of the reflector.

Once a wave is resonant in a structure, with a Q of say 100, radiation of energy through a slot or fringing field need only be 1% of the circulating energy per cycle for the full power available to be radiated.
 
  • #9
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As a rough rule of thumb, elements less than 1/10th of a wavelength can be lumped together. Since the fringe is less than that, it can be considered a sudden shift to 377Ω; i.e. an open. This will cause a reflection coefficient (∫˜) ~= 1. (Of course nothing is perfect, and this presumption less than most.) As Baluncore pointed out, we don't need ∫˜ to be 1 to get standing waves.

Remember, EM waves add linearly (at least in most dielectrics). So a ∫˜ of 1/10th will give a standing wave of 1/10th the amplitude (compared with ∫˜= 1). The rest of the wave goes on its merry way, but there's now a new wave with 1/10th the energy just standing there...
 
  • #10
legyptien21
The position of those nodes and nulls will be determined by the nature of the reflector.
You anticipated my question. Suppose we are in a high permittivity dielectric cavity which is in contact with air. You said a standing wave appear whatever the reflection coefficient at the interface, fine but it means that the position of the node and nulls will not "move" along the cavity ? Look I'm gonna tell you what disturb me: the amplitude of the standing wave is :

Vi*(1+alpha) with alpha the coefficient of reflection and Vi the incident wave amplitude. If it's a full reflection, it's the simple case for me, alpha = 1 and the amplitude of the association of the 2 waves is 2 *Vi. Suppose we have no reflection, it should be a progressive wave with alpha = 0 , the position of the max or min is moving at C (speed of light). At any infinitesimal reflection we will have a standing wave ??!!!! How comes we have such a discontinuity in speed ??!! Another question is that what is the min value of a standing wave in a NON full reflection coefficient ?

thanks

Ps: Sorry lots of questions but it has to be extremely clear....
 
  • #11
legyptien21
The rest of the wave goes on its merry way, but there's now a new wave with 1/10th the energy just standing there...
Interesting ! This means to me that the whole wave obtained after reflection is an addition of the standing wave that you described and a progressive wave. If I had the ability to see the wave (the whole one) with my eyes, the node and the nulls will note be standing, they will be moving at the speed of C right ? Because this speed is introduce by the progressive wave... You see where I wanna go right ? I wanna talk about the discontinuity...
 
  • #12
Baluncore
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1) My question is if we have standing waves, in my mind we would have a max reflection (=1) on the magnetic walls so how comes we have leakage if we have full reflection on the wall ??
The leakage of signal is not through the open or conductive reflector but from the part of the resonant cavity exposed through a slot or aperture in the cavity wall.
A resonant dipole has full reflection at both ends and maintains a standing wave. Yet it still transmits a 2.15 dBi signal broadside.
 
  • #13
legyptien21
The leakage of signal is not through the open or conductive reflector but from the part of the resonant cavity exposed through a slot or aperture in the cavity wall.
A resonant dipole has full reflection at both ends and maintains a standing wave. Yet it still transmits a 2.15 dBi signal broadside.
That's fine, I'm following you. I thought we have a standing wave only when there is a full reflection. I agree with you we have a standing wave with any reflection and the amplitude of this standing wave is as small as the reflection is. This is solved ok.

Now what about the whole wave please ? the whole one is NOT a standing wave right ? Their should be a mathematical way to calculate the speed of the full wave. I'm sure it goes at C because of it's progressive part... I'm sure I missed something because of the speed discontinuity...
 
  • #14
legyptien21
Ok I have an answer for this apparent discontinuity, I will give my opinion later if someone is interested. Actually the partially reflected case if the same as a progressive wave modulated (with addition not multiplication like in an amplitude modulation) by a standing wave. The amplitude of the standing wave and the one of the progressive wave depends on the reflection coefficient. I think we agree on that...

Now I have only one question. what the value of the minimum of the whole wave ?
 
  • #15
Baluncore
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Now I have only one question. what the value of the minimum of the whole wave ?
If only the world was so simple. What "whole wave" in what "environment" ?
 
  • #16
legyptien21
A resonant dipole has full reflection at both ends and maintains a standing wave. Yet it still transmits a 2.15 dBi signal broadside.
Yes but this radiation is not the result of a leakage because since there is a full reflection there couldn't be a leakage, you agree ? Is this radiation is the result of leakage for you ?

I will talk about whole wave later....
 
  • #17
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A standing wave is not stationary in the sense of propagation. If it were a single wave like a soliton, it would move right along. The standing effect is typically (but not exclusively) caused when two waves moving in opposite directions interfere with each other. Since an incident wave train and its reflected wave wave train (with ∫˜=1) are just such a pair, they form a standing wave. But both wave trains are propagating at full speed.

At ∫˜<1, there will be a small standing wave equal to the reflected wave plus as much of the incident wave as needed to match that. This will be swamped by the incident wave for ∫˜<<1, but will still exist.

Other types of standing waves exist, but aren't really what we are talking about.

What is a "whole wave"? Waves add linearly, so a standing wave will typically be twice the amplitude of the incident wave. But with clever (or perhaps not so clever) design a waveguide resonator could bounce the wave back and forth six ways from Sunday until the voltage breaks down the dielectric and lets the magic smoke out.
 
  • #18
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Since PECs don't exist, nothing has a ∫˜of 1. But it has a nominal value of 1 (Claimed anyway. It sounds reasonable, but it's not my understanding of what goes on.). Since it is a resonator, the voltage keeps building until the small leakage power (out of the antenna) matches the incoming power. At least that's my understanding of the claim. (I'd buy it but haven't done the math myself.)
 
  • #19
legyptien21
A standing wave is not stationary in the sense of propagation. If it were a single wave like a soliton, it would move right along. The standing effect is typically (but not exclusively) caused when two waves moving in opposite directions interfere with each other. Since an incident wave train and its reflected wave wave train (with ∫˜=1) are just such a pair, they form a standing wave. But both wave trains are propagating at full speed.
By your last sentence you demonstrate that both waves are propagating and we all agree on that. However how can you say that the addition of these 2 waves which are giving standing wave will not propagate ? your link say the opposite of that no ?
 
  • #20
Baluncore
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Now I have only one question. what the value of the minimum of the whole wave ?
I will talk about whole wave later....
No. You talked about it before. Your term “whole wave” has no obvious meaning.

If you will not clarify your statements and you will not answer questions then your misuse of the terminology will continue and you will only waste the time of respondents. The more you write, the more nonsensical your thinking on this subject appears to be. That needs to change.
 
  • #21
legyptien21
At ∫˜<1, there will be a small standing wave equal to the reflected wave plus as much of the incident wave as needed to match that. This will be swamped by the incident wave for ∫˜<<1, but will still exist.
No. You talked about it before. Your term “whole wave” has no obvious meaning.

If you will not clarify your statements and you will not answer questions then your misuse of the terminology will continue and you will only waste the time of respondents. The more you write, the more nonsensical your thinking on this subject appears to be. That needs to change.
Ok Whole wave means the addition of the stationnary wave and the propagating one (which happen when there is partial reflection)
Look it's not that I don't wanna clarify it's that I try to solve one question after the other one. If you can look at my message 16 and if I can understand your point of view on it, after that we can talk about the whole wave which is a not clear term I agree...
 
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  • #22
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By your last sentence you demonstrate that both waves are propagating and we all agree on that. However how can you say that the addition of these 2 waves which are giving standing wave will not propagate ? your link say the opposite of that no ?
Because they are propagating in two different directions.

The addition is done linearly. It is done point by point in a 3+time space. So each point sees two time varying waves and adds them. In some times and places they add. In some times and places they subtract. In addition there are a few points where they always cancel.

For ∫˜=1 they are the same amplitude, so that place forms a zero point. 1/4th wavelength later (away from the reflective surface) they double which causes a rise and fall of twice the amplitude of the incident wave.
 
  • #23
legyptien21
By your last sentence you demonstrate that both waves are propagating and we all agree on that. However how can you say that the addition of these 2 waves which are giving standing wave will not propagate ? your link say the opposite of that no ?
ok now it's my mistake, "will propagate" because you stated that the stationary will propagate : your sentence was "A standing wave is not stationary in the sense of propagation. I"
 
  • #24
legyptien21
ok one last question, after that I won t bother...

If we have a transmission line missmatched at one of its end, we will have a partially reflected wave. How is the distribution of current along the transmission line ? My main question is the maximum and minimum of current will move along this line or not ?

I have an opinion but I don`t wanna influence you so don`t read the following unless you know the answer.

To me since there is a partially reflected wave, the max and min should move. We do have a standing wave but we have to add to it a progressive wave which is bringing the movement to the distribution.

Thanks for your time...
 
  • #25
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This is not a competition. I was trying to help you understand my thinking. Perhaps I'm way off base and you are right. That's fine with me.

In my experience reality doesn't really care what I think.
 

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