Why do patch antennas have leakage if they have standing waves?

[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

 

[davenn]

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

these are patch antennas ...

so what type of antenna are you really talking about ?

Dave

 

[legyptien21]

Thanks for your answer. I'm definitely 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...

 

[Jeff Rosenbury]

What is a magnetic wall?

 

[legyptien21]

What is a magnetic wall?

https://books.google.ca/books?id=_e...esc=y#v=onepage&q=patch magnetic wall&f=false

You see the boundary condition implied by a PEC (pefect electric conductor) ? A PMC (magnetic conductor) is the one which implies the opposite condition. For example, if you have an interface of air and a VERY VERY high dielectric constant matter, this matter will be as good PMC as its epsilon is high...

 

[Jeff Rosenbury]

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.

 

[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 shouldn't be a standing wave if there is NOT a reflection coefficient of 1...to me...

let me know what you think

thanks

 

[Baluncore]

It shouldn't 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.

 

[Jeff Rosenbury]

As a rough rule of thumb, elements less than 1/10^th 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 1/10^th will give a standing wave of 1/10^th the amplitude (compared with ∫˜= 1). The rest of the wave goes on its merry way, but there's now a new wave with 1/10^th the energy just standing there...

 

[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 going to 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...

 

[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 want to go right ? I want to talk about the discontinuity...

 

[Baluncore]

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.

 

[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...

 

[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 ?

 

[Baluncore]

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" ?

 

[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...

 

[Jeff Rosenbury]

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 let's the magic smoke out.

 

[Jeff Rosenbury]

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.)

 

[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 ?

 

[Baluncore]

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.

 

[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 want to 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...

 

[Jeff Rosenbury]

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. 1/4^th wavelength later (away from the reflective surface) they double which causes a rise and fall of twice the amplitude of the incident wave.

 

[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"

 

[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 want to 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...

 

[Jeff Rosenbury]

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.

 

[Baluncore]

There are two quite independent waves propagating on the TL. One is in the forward direction from the source, the other is the reflected component traveling back towards the source. The sum of those two propagating waves will vary along the line and will form a standing wave pattern. That standing wave can be seen by measuring the total voltage across, or current flowing in the line, which will show nodes or nulls in fixed positions along the line.

The propagating waves cannot see each other and do not influence each other, they are traveling waves, propagating in opposite directions on the same line.

 

[legyptien21]

This is not a competition.

! I never said or thought it s a competition. I do try to understand you. That s why I wanted to know (and asked) why did you say "A standing wave is not stationary in the sense of propagation.". I m not saying you are wrong, I just wanted to understand what do you mean by that.

sorry if I offended you...

 

[legyptien21]

The sum of those two propagating waves will vary along the line and will form a standing wave pattern. That standing wave can be seen by measuring the total voltage across, or current flowing in the line, which will show nodes or nulls in fixed positions along the line.

excellent ! We are touching here something which drives me crazy for a long time. We were saying in all what we said until now that in the general case of interference between the incident and reflected wave, we will have a standing wave PLUS a propagating wave. If we measure the total voltage as you say, it will give the standing PLUS (addition mathematically) a propagating wave.

I`m stuck with that for a long time, I have to understand that. To me the nodes or nulls won t be fixe in the general case (partially reflection).

 

[Baluncore]

in the general case of interference between the incident and reflected wave, we will have a standing wave PLUS a propagating wave.

No. There is an incident wave and a reflected wave, both of which are propagating, but in different directions. The sum of those is a fixed standing wave pattern. There is no "PLUS a propagating wave" as you put it.

 

[legyptien21]

Vi*sin (wt - kx ) + Vr*sin (wt + kx ) = Vr*(sin (wt - kx) + sin (wt + kx)) + C*sin (wt + kx) with Vi = Vr + C

Vr*(sin (wt - kx) + sin (wt + kx)) represent a standing wave if we use the Simpson formula...

Do we agree ? if not why ?

 

[Baluncore]

Do we agree ? if not why ?

No.
Where do the equations come from ?
What do the symbols mean ?

 

[legyptien21]

No.
Where do the equations come from ?
What do the symbols mean ?

the equations come fro my mind. Vi and Vr are the amplitude of the incident and reflected wave. K: wave number. w=2*pi*f with f the frequency. C is the amplitude of a wave which is part of the incident wave since we decide Vi = Vr + C. I hope it s ok now.

 

[Baluncore]

C is the amplitude of a wave which is part of the incident wave since we decide Vi = Vr + C.

So you are saying that the part of the incident wave that is not reflected, but which passes through the impedance mismatch, is reflected as C.
That is nonsensical.
If Vi = Vr + C, then C is not reflected and so is no longer propagating on the line. It is lost from the line.

 

[legyptien21]

So you are saying that the part of the incident wave that is not reflected,

I never said such a thing nowhere, sorry. I did a simple derivation, and I don t think I did a mistake mathematically. Then I interpretated the results as a superposition of standing wave and propagating wave. You may not agree with this physical interpretation but at least we should agree on the mathematical derivation... Do we ?

thanks

 

[Baluncore]

No.

Then I interpretated the results as a superposition of standing wave and propagating wave.

You cannot have a superposition of a standing wave and a propagating wave.