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Legyptien AntennaTheory.com Newbie
Joined: 27 May 2015 Posts: 4

Posted: Wed May 27, 2015 6:24 pm Post subject: patch and dipole antenna 


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[/url] 

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helix Antenna Theory Regular
Joined: 29 Jan 2015 Posts: 64

Posted: Sun May 31, 2015 4:11 am Post subject: 


you should avoid using the term "leaky wave" as it has an other, quite specific, meaning in antenna engineering... (see http://en.wikipedia.org/wiki/Leaky_wave_antenna)
reflection at a discontinuity in a transmission line does not have to be binary "all or nothing", you can have partial reflection (0 < gamma < 1), and therefore varying degrees of standing waves on a transmission line. ...this is the concept of VSWR.
understanding how a slotted line works should help you with these concepts, the minima and maxima are not "moving around": http://en.wikipedia.org/wiki/Slotted_line
so the reflection at the edges of the patch is indeed only partial reflection, allowing some RF power to flow into free space, while also creating a standing wave on the microstrip patch.
a more complex way to think about patch antennas utilizes cavity modes. a rectangular patch represents a rectangular cavity, but for conceptual and mathematical simplicity, the side walls can be modeled as perfect magnetic conductor (PMC) since they approximate an open circuit boundary condition (no current flow => Htan = 0). in reality, they are imperfect open circuit boundaries (which explains how power can flow through the side wall surfaces). ...the cavity model is handy when you encounter a circular patch, for example.
the classic paper by richards, lo and harrison (1981) explains how the cavity model is corrected for imperfect reflection at the open edges. but i'm sure you can find this information elsewhere. 

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Legyptien AntennaTheory.com Newbie
Joined: 27 May 2015 Posts: 4

Posted: Fri Jun 12, 2015 2:41 pm Post subject: 


Thanks for your answer.
If we consider a PMC (no power flow) instead of the reality so does it mean that the antenna does radiate ? I know it radiates because of the fringing fields which "leave" the metallic patch and "reach" the dielectric. In other words I would like to know if we can relate the imperfection of the wall with the radiation.
To me if we assume a PMC so it means that there is no power flow so no radiation. No power flow means no fringing field according to you ?
thanks 

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Legyptien AntennaTheory.com Newbie
Joined: 27 May 2015 Posts: 4

Posted: Fri Jun 12, 2015 3:21 pm Post subject: 


Oh I was going to forget an other question. As far as I remember the higher is the dielectric permittivity (on which the metallic patch is), the better is the PMC approximiation. However everyone is using the formula obtained from the cavity model for all kind of permittivity especially if we want to have a wideband antenna, we use a 2.2 relative permittivity...
Do you agree that the missmatch between the reality and the PMC approximation is increasing when the permitivity of the dielectric is decreasing (happen when we want to increase the bandwith). 

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helix Antenna Theory Regular
Joined: 29 Jan 2015 Posts: 64

Posted: Sat Jun 13, 2015 2:57 am Post subject: 


to answer your first question: yes, in a model with PMC side walls, there is no radiation. it is just a mental model, or way to think about how something works. various models have various levels of accuracy. usually, the simple models are easy to understand but are less accurate quantitatively. the impedance, resonant frequency and Q derived from a cavity model with PMC side walls will differ from that of the real antenna (measured or modeled on a computer). the paper i mentioned before shows how to accurately correct a cavity model to account for incomplete reflection (i.e., radiation) from the side walls.
i'm not sure i know the answer to your second question, at least in terms of the "perfectness" of the PMC. in the classic transmission line model of a rectangular patch antenna, the admittance of the edge Y = G + j*B. B is the susceptance and is actually proportional to the effective dielectric constant of the microstrip forming the cavity. for air, the effective dk is 1. the susceptance goes UP from there, i suppose until it's very large. the radiation conductance G is, to first order, constant with dk. so from a transmission line termination standpoint, the greater the dk, the greater the microstrip tline reflection coefficient, and the more "perfect" the reflection at the side wall will be.
perhaps related to this is that the patch is getting physically smaller as the dk goes up (while the frequency of operation is staying the same), and due to electrically small antenna theory (chu theory) the Q must go up. 

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Legyptien AntennaTheory.com Newbie
Joined: 27 May 2015 Posts: 4

Posted: Sat Jun 13, 2015 9:59 pm Post subject: 


Thanks for your answer. I'm sure about the fact that the approximation of the PMC is as true as the relative permittivity of the dielectric become high. So to be honest I'm shocked that we are taking an assumption (PMC) that suppose "deny" the first purpose of an antenna which is to radiate. Anyway...
I have an other question very basic on the dipole. If we are taking a half wavelength dipole and we look on the voltage and current distribution on the dipole. If we consider the dipole length either 0.5*Lambda or 0.47*Lambda, we will always have a max of current in the center of the dipole and a minimum of voltage in the center of the dipole. By minimum I mean 0 since there is only standing wave (full reflection) : it's not a matter of approximation of PMC here...
Thanks for your time... 

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helix Antenna Theory Regular
Joined: 29 Jan 2015 Posts: 64

Posted: Tue Jun 16, 2015 3:05 am Post subject: 


w.r.t. PMC walls on patches  again, it's just a mental model, and does a firstorder job of telling you what the fields inside the cavity look like and what the radiation pattern would look like. it's definitely not useless, and the PMC cavity model can be corrected to account for radiation (see the paper i cited in my first response).
w.r.t dipoles  ok this is a good question, but i think you may be relying too much on quasistatic circuit analogies. for one thing, there really is no such thing as a (welldefined) voltage potential in this problem... the geometry is a half wavelength long, and the curl of E is not zero (otherwise there could be no B field), and so you cannot really define a static voltage. to be honest, i'm not exactly sure why they draw the pictures of "voltage" as they do in text books. i would take those diagrams with a grain of salt... they may be a voltage corresponding to the purely staticlike term of the dipole field, but honestly i'm not sure and i've never seen a quantitative explanation of that "voltage" quantity accompanying the diagram.
the current distribution you describe appeals to me much more, so let's talk about that. basically, where you feed the dipole at the center, you're impressing a voltage there. and at resonance, the impressed voltage and resulting current are in phase, meaning you get real power flow (into radiation).
in some sense, you ARE getting "perfect" reflection of the current in the wire at the ends of the dipole (i = 0) and this is basically what gives rise to the current distribution/standing wave BUT, the wave propagation along the wire is lossy... power flows into free space. so imagine what you'd have if you took a length of coax that had 3dB of copper+dielectric loss and put a perfect open or short at the end: you'd measure 6dB return loss. the dipole in a way is like this, but the loss is radiation loss (this is the whole point of the device). 

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