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Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Oh no! The age old question of "How does the electricity get out of the wires into the air".
I found this nearly impossible to understand, and I'm not even sure that I understand it now. It helped me a lot to realise that the electricity was never in the wires in the first place. Electric current may flow in conductors, but electric power always travels in the space between conductors, no matter if it's DC or light.
If you visualise it this way, then an antenna is just a clever launching ramp or funnel-type structure, that coaxes the RF power out of the gap between the feedlines and lets it spread out smoothly into space. Microwave antennas show this most clearly, with their horns and waveguides.
As for the 377 ohm thing: You can't prod a DMM into free space to check it. But if you took some conducting sheet with a resistance of 377 ohms per square, and pressed the cut end of a piece of 50 ohm coax onto it, you would read 50 ohms between core and screen on an ohm meter.
If you put this same sheet across the end of a microwave waveguide, it would be a perfectly matching dummy load.
But, an open-ended waveguide or piece of coax is a very poor match that reflects most of the power, even though the end is "covered" with free space. You have to go and think about that for a while...
Registered Member #29
Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
As for the 377 ohm thing: You can't prod a DMM into free space to check it. But if you took some conducting sheet with a resistance of 377 ohms per square, and pressed the cut end of a piece of 50 ohm coax onto it, you would read 50 ohms between core and screen on an ohm meter.
If the coax is air dielectric, of course.. Otherwise you will need to correct with a factor 1/sqrt(eps_r). eps_r is the relative permittivity of the dielectric.
If you put this same sheet across the end of a microwave waveguide, it would be a perfectly matching dummy load.
This is not exactly true.. The wave impedance of a hollow waveguide is not the same as free space... It depends on the waveguide mode and frequency.... That said, if you use a material which matches the WG wave impedance, it will make a perfect dummy load (in fact, this is one way of simulating open structures...ao-called absorbing boundary conditions).
Banned on 3/17/2009. Registered Member #487
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Actually Tom, the higher the frequency, the more easily RF is influenced by surrounding objects.
I knew that much (and not much more) I just meat or thought they had an easier time squeezing through smaller areas like for example the screen on your microwave door has to be kept small because of wavelength of the microwaves being shorter than lets say a screen around a tesla coil.
I remember at my old job we had a receiver that ran at 800Mhz instead of our usual 27 and it was MUCH more sensitive to everything. Some of the coils in it were only .1 diameter with 1 or two turns.
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
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Wow guys... I didn't expect that much replies, I'l try to cover all of them up...
I actually ran in trouble yesterday because I couldn't stop thinking about this instead of studying grammar, I ended up awake until 3AM.
Chris Russell wrote ... In order for an antenna to radiate, the basic requirement is this: alternating current must flow a distance through a conductor. The more current and the longer the conductor, the more the antenna radiates. Let's look at your halfwave dipole example a little more closely. Imagine what happens along the length of your dipole antenna, if a 10V sine wave is applied at the feed point. When one end is at 10V, the other end of the dipole is at 0V. They, of course, switch off rapidly, setting up a standing wave. The ends of the antenna are swinging up and down in potential, while at the feedpoint, there is almost no change in voltage at all. These large voltage potentials cause RF current to flow through the antenna. Current flows from one half of the antenna to the other, reaching maximum current right at the feedpoint. This current does the job of radiating. The total amount of current that actually flows is determined by your antenna's feedpoint impedance. In this case, roughly 73 ohms, purely resistive.
Ok, this is apparently going to be a huge paradigm shift for me, much bigger than I imagined.
I figured (or at least thought to have) this out yesterday after drawing it on paper. I was left stunned. I'll try to get straight how I understood this:
Time needed for any information, including change in voltage, to travel the length of one dipole pole is equal to 1/4 of the period of the input signal. That is because total length of dipole is passed by c during one half-period.
So, this means that voltage at the end of dipole will be zero when middle node is at peak, and vice versa.
Now the situation is... that I actually have a potential gradient across the wire due to limited speed of c...?
And this potential gradient causes a current to flow, through the pole, with resistance encountered by the flow being radiation resistance?
I can't deny that this actually makes lots of sense, but the implication is something that after all this time I find very bizzare:
If I take a battery, and two pieces of wire with completely negligible capacitance between them, and connect them quickly to battery poles... ... for one brief moment, I'll have massive current flowing out of battery into an uncompleted circuit, and turning a small amount of energy into EM radiation?
I guess I don't need to say how strange that looks.
Or in case I'm completely wrong I'd like to be verified before going further.
WaveRider wrote ...
Antennas are not popular study topics because truly understanding them requires dedicated work... Building simple ones by trial and error can be satisfying...but the understanding requires almost religious devotion... The reference to quantum mechanics indicates that the understanding is just not there.
Waverider, understanding is not closely there, it's farther away for me than you might imagine.
I'm pretty much failing to see what's wrong with QM explanation if it fits everything else, as it should. Any EM radiation is said to be composed of photons and nearfield fields of virtual photons... scenically at least... But I don't want to get into anything like that for now.
Also, notions of voltage on radiators in antennas have little meaning since the electric scalar potential cannot be defined over lengths that correspond to a wavelength.... Currents, on the other hand can be uniquely defined and all wire antenna models solve for currents...
Yes.. but the biggest question is what makes the current to flow into an virtually open circuit at all? I wonder how would you comment on Chris's post.
Far-field transports energy and consists of both electric and magnetic field at right angles in phase quadrature (wikipedia for Poynting vector). They fall of at r^2, so actually have the chracteristic of a fixed an=mount of power being reduced in intensity by being spread out on the surface of an expanding sphere surface.
Yes. What I'm trying to figure out here, is how *true*, farfield radiation works.
Steve Conner wrote ... Oh no! The age old question of "How does the electricity get out of the wires into the air".
I found this nearly impossible to understand, and I'm not even sure that I understand it now. It helped me a lot to realize that the electricity was never in the wires in the first place. Electric current may flow in conductors, but electric power always travels in the space between conductors, no matter if it's DC or light.
If you visualize it this way, then an antenna is just a clever launching ramp or funnel-type structure, that coaxes the RF power out of the gap between the feedlines and lets it spread out smoothly into space. Microwave antennas show this most clearly, with their horns and waveguides.
Steve, I understand completely how you feel. Just after reading Chris's post, my reaction was O______O. Because nowhere on the internet I saw anything like that, or at least at language understandable to me.
What do you mean by saying that power is always in space between conductors? Looks like purely poetic meaning, how should I apply that to my questions?
As for the 377 ohm thing: You can't prod a DMM into free space to check it. But if you took some conducting sheet with a resistance of 377 ohms per square, and pressed the cut end of a piece of 50 ohm coax onto it, you would read 50 ohms between core and screen on an ohm meter.
What do you mean by 377 ohms per 'square'? meter? Cubic meter? Or infinite fudge of 377ohms/m^3 covering everything around the coax cable?
Now...how would I measure that? With DC?
What role would coax play then being ''50 ohm''? As far as I figured out, that is characteristic impedance sqrt of the cable, so what would it do at all?
And I don't see why would I then have 1/4 resistance too, so I'm not quite understanding what are you trying to say?
Characteristic impedance isn't anything closely like radiation resistance which is just like ohmic resistance. (?).
Does an antenna have a characteristic impedance too?
Why is a 1/2 wave dipole told to be ''resonant''?
I really don't know a better way to say that, , but does a halfwave dipole need to be ''pumped up'' over a number of cycles before it's impedance drops to ''working'' 50 ohms or like so?
I mean, just like a series RLC circuit needs to be pumped up over many cycles until current if finally limited mostly by R, and during first half-cycle it has impedance of sqrtL/C? DO you get what I mean?
You guys really made my day now... I don't care for hours I used up for this but I should really go studying now.
... not Russel! Registered Member #1
Joined: Thu Jan 26 2006, 12:18AM
Location: Tempe, Arizona
Posts: 1052
Time needed for any information, including change in voltage, to travel the length of one dipole pole is equal to 1/4 of the period of the input signal. That is because total length of dipole is passed by c during one half-period.
So, this means that voltage at the end of dipole will be zero when middle node is at peak, and vice versa.
Now the situation is... that I actually have a potential gradient across the wire due to limited speed of c...?
And this potential gradient causes a current to flow, through the pole, with resistance encountered by the flow being radiation resistance?
I would say that sums up how a halfwave dipole works quite well. Excellent work!
Yes.. but the biggest question is what makes the current to flow into an virtually open circuit at all? I wonder how would you comment on Chris's post.
Electric field gradients along the dipole make current flow, as you said above. The far ends of the dipole are at opposite potentials, causing current to flow. The ends of the dipole act as the plates of a very large capacitor, reaching a high potential because of the propagation delay. Currents rush back and forth to attempt to equalize the potential, encountering radiation resistance as they do, which conveniently happens to look and act just like ohmic resistance.
If I take a battery, and two pieces of wire with completely negligible capacitance between them, and connect them quickly to battery poles... ... for one brief moment, I'll have massive current flowing out of battery into an uncompleted circuit, and turning a small amount of energy into EM radiation?
For a moment, yes, current will flow, limited by the impedance of the antenna. Keep in mind that even if each leg of the dipole were 1km long, the current flow would only last three millionths of a second. If you could switch the polarity of the battery 300,000 times per second, then you could get a decent amount of current flowing in the antenna. Alternately, if you then removed the battery and shorted both legs of the dipole together, current would suge back and forth until it was dissipated as heat or radiation -- though the antenna would have to be charged to extremely high voltage to store any energy at all. Spark gap transmitters used to operate on a similar principle: charge each leg of the antenna to very high, opposite voltages, until a spark gap between both legs shorts them together. Current rushes one way and then changes direction over and over as the antenna "rings down." Doing this many times per second created a very rough, ugly, and broadband signal, but it did work.
What do you mean by 377 ohms per 'square'? meter? Cubic meter? Or infinite fudge of 377ohms/m^3 covering everything around the coax cable?
Now...how would I measure that? With DC?
It simply defines the relationship between electric and magnetic fields (E and H) in a vacuum. It doesn't have an area, and you'll have a heck of a time measuring it directly. If you can imagine two parallel conductors of infinite length, over any given length there is going to be a ratio of capacitance to inductance (governed by mu and epsilon), resulting in a characteristic impedance. In order for electromagnetic waves to propagate through space, thus acting as a transmission line of sorts, the impedance of free space has to have a real value. If it were 0 ohms, a changing magnetic field would not give rise to an electric field, since a potential cannot exist across a short. If it were infinite, no current could flow, and thus no magnetic fields could arise. This ties in to the whole "electric power always travels in the space between conductors" thing from earlier. We're not getting electricity out of the wires and into the air, all we're doing is trying to couple our electricity to the great big transmission line that surrounds us all.
What role would coax play then being ''50 ohm''? As far as I figured out, that is characteristic impedance sqrt of the cable, so what would it do at all?
And I don't see why would I then have 1/4 resistance too, so I'm not quite understanding what are you trying to say?
Characteristic impedance isn't anything closely like radiation resistance which is just like ohmic resistance. (?).
Does an antenna have a characteristic impedance too?
Why is a 1/2 wave dipole told to be ''resonant''?
A dipole is said to be "resonant" when the inductive and capacitive reactances are equal, thus cancelling eachother out. This leaves behind only the resistance of the dipole, which is composed of the ohmic resistance of the wire and the radiation resistance of the antenna. For a halfwave dipole, the ohmic resistance of the wire is usually negligible, and the radiation resistance is roughly 73 ohms. That is the impedance of a halfwave dipole in free space. Characteristic impedances are for transmission lines.
The basic idea of characteristic impedance is to allow you to bring the purely resistive load of the antenna right to the transmitter without introducing any reactive components. A 50 ohm coax cable terminated with a 50 ohm resistor will present a 50 ohm resistive load to the other end of the cable regardless of the length of the cable or the frequency used, if you ignore losses for the moment. Using the correct characteristic impedance of coax allows maximum power transfer between source and load. Most amateur radio transmitters are designed to drive a 50 ohm load -- note that they don't care if it's 50 ohms of radiation resistance or 50 ohms of carbon resistors -- and will therefore need a 50 ohm cable to obtain an ideal match. You could use 75 ohm cable in a pinch, but there would be an impedance mismatch, which would lead to some power being reflected back to the transmitter.
If you're wondering why 50 ohms, that's a somewhat more complicated question. Long story short, higher voltages would appear on 75 ohm cable, thus reducing the power handling capability of the cable for a given dielectic strength. Lower impedance would mean more current in the cable, and thus higher losses, so it's something of a compromise between power handling and loss.
I really don't know a better way to say that, , but does a halfwave dipole need to be ''pumped up'' over a number of cycles before it's impedance drops to ''working'' 50 ohms or like so?
I mean, just like a series RLC circuit needs to be pumped up over many cycles until current if finally limited mostly by R, and during first half-cycle it has impedance of sqrtL/C? DO you get what I mean?
You're thinking of Q. A dipole does have a Q factor that causes it to "ring" a bit, meaning that some of the cycles overlap, but the radiation resistance is so high that this is rarely a factor. In the case of a shortened dipole with loading coils, the Q can get fairly high. The result of this is that the bandwidth of the antenna suffers.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
If I take a battery, and two pieces of wire with completely negligible capacitance between them, and connect them quickly to battery poles... ... for one brief moment, I'll have massive current flowing out of battery into an uncompleted circuit, and turning a small amount of energy into EM radiation?
If they are open wires, then the caapcitiance will not be negligable. Everything's a transmission line, with the antenna wires being no exception. The input impedance, given by sqrt(L/C) (each per unit length) will typically be a few 10s of ohms, or a few 100s depending on the geomtery. As you connect the battery (it doesn't matter whather the battery is a theoretical zero output Z, finite like 50ohms, or a mess of its own internal transmission lines, but the simpler cases are easier to compute) then a current will flow into to chracterisitic impedance of the transmission line. So a 10v battery would send a 100mA pulse into a 100ohm line. The pulse of current with a voltage present will radiate some EM field.
It's only after the energy has travelled along the line, "found out" whether there's a load, or open circuit, or short circuit at the end, and so been reflected or not in the appropriate phase, that the battery current eventually settles down to its long term value. In this case, a 100ohm load means no reflection. An o/c means reflection of an anti-phase current, which drops the current to zero.
A short circuit means reflection of in-phase current which doubles the battery output. If the battery is itself zero Z out, then there's a further reflection and the output current increases further. The current builds a t rate dependant on the length and impedance of the line, or in otherwords, its inductance.
I've described this as if it's a single bit of 100ohm line, but of course that's not how an antenna looks. If you "play wavefront" and walk down the wire, looking around you to see what the wavefront sees, then a dipole will start lowish impedance at the feedpoint where the conductors are close together and have high capacitance, then have a gradually increasing impedance towards the ends. This only chages the detail of the solution, not the form. You can regard each change in impedance of the line as generating its own reflection. As you segment the line into more and more smaller sections, or better still represent the impedance analytically as a function of distance, then it approaches the behaviour of a real dipole.
What do you mean by 377 ohms per 'square'? meter? Cubic meter? Or infinite fudge of 377ohms/m^3 covering everything around the coax cable?
"Per square" means just that, it's just ohms, with no length dimension(s). Imagine a pieice of "resistance card", which is a thin film of conductive crud, cured onto cardboard. Cut a 1" square, fit electrodes to the top and bottom edges, and measure the resisitance. This is the resistance per square. Now cut a 2" square and do the same. It's the same resisitance, and you can verify that two 1" squares in parallel, in series with another two 1" squares, will give the same result. Actually, ressitance card exists because, before computers existed, it was the preferred method for solving 2D electromagnetic field problems (less messy than a 3D field tank, and could be done with DC). You would draw a cross section of your transmission line onto ressitance card with silver ink, put a voltage on the terminals, and measure the input current or plot the voltages across the space.
Conductive plastic component bags ought to make a good source of suitable material, however the surface contact can be patchy, and the resistivity is not particularly uniform across a sheet due to variations in the thickness and mix.
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Hi guys - I'm sorry, but I didn't want to reply yesterday as I was too tired and afraid that I won't make a good reply.
After Chris's post, you guys may have thought ''wtf, how didn't you know that''? I didn't. Not even close. I had no clue how could a current flow into an 'open circuit' as it appeared.
Still there's way too much things I don't know, but I feel I've at least bitten it.
I'll try to make my replies shortest in order for them to be most understandable:
Chris Russell wrote ... Electric field gradients along the dipole make current flow, as you said above. The far ends of the dipole are at opposite potentials, causing current to flow. The ends of the dipole act as the plates of a very large capacitor, reaching a high potential because of the propagation delay. Currents rush back and forth to attempt to equalize the potential, encountering radiation resistance as they do, which conveniently happens to look and act just like ohmic resistance.
OK, this is the newt thing I need to get over.
Everything seems to be implying that there is ''some Other (capital O) kind of resonance'' happening there which has nothing to do with L and C. To this day I'm trying to figure out what that means.
Chris Russell wrote ... The ends of the dipole act as the plates of a very large capacitor,
Are you talking about real C, which charges up, or dipole can store the energy in some other means? Let's say my dipole is nanoscopically tiny superconducting wire with hardly any capacitance at all.
Chris Russell wrote ... reaching a high potential because of the propagation delay.
Do you mean I can get much higher voltage at the ends of the antenna than on feedpoint?
Is it some kind of resonant rise?
Chris Russell wrote ... Currents rush back and forth to attempt to equalize the potential, encountering radiation resistance as they do, which conveniently happens to look and act just like ohmic resistance.
Will the current run into linearly lower resistance if wire is shorter/higher if it's longer?
Chris Russell wrote ...
For a moment, yes, current will flow, limited by the impedance of the antenna. Keep in mind that even if each leg of the dipole were 1km long, the current flow would only last three millionths of a second. If you could switch the polarity of the battery 300,000 times per second, then you could get a decent amount of current flowing in the antenna.
Chris, 3 nanoseconds is lots of time.
Chris Russell wrote ... Alternately, if you then removed the battery and shorted both legs of the dipole together, current would suge back and forth until it was dissipated as heat or radiation -- though the antenna would have to be charged to extremely high voltage to store any energy at all. Spark gap transmitters used to operate on a similar principle: charge each leg of the antenna to very high, opposite voltages, until a spark gap between both legs shorts them together. Current rushes one way and then changes direction over and over as the antenna "rings down." Doing this many times per second created a very rough, ugly, and broadband signal, but it did work.
Here it is again! So the halfwave dipole resonates like a tuned circuit but at frequency equal to c/2l. And without a single coil or cap in sight.
Chris Russell wrote ... You're thinking of Q. A dipole does have a Q factor that causes it to "ring" a bit, meaning that some of the cycles overlap, but the radiation resistance is so high that this is rarely a factor. In the case of a shortened dipole with loading coils, the Q can get fairly high. The result of this is that the bandwidth of the antenna suffers.
I meant surge impedance, but this too, what determines Q of it's Other kind of resonance if it's ohmic resistance is dominated by radiation resistances..
What determines how much energy can a dipole store this way at all? Apart from the voltage.
Chris Russell wrote ... It simply defines the relationship between electric and magnetic fields (E and H) in a vacuum. It doesn't have an area, and you'll have a heck of a time measuring it directly. If you can imagine two parallel conductors of infinite length, over any given length there is going to be a ratio of capacitance to inductance (governed by mu and epsilon), resulting in a characteristic impedance. In order for electromagnetic waves to propagate through space, thus acting as a transmission line of sorts, the impedance of free space has to have a real value. If it were 0 ohms, a changing magnetic field would not give rise to an electric field, since a potential cannot exist across a short. If it were infinite, no current could flow, and thus no magnetic fields could arise. This ties in to the whole "electric power always travels in the space between conductors" thing from earlier. We're not getting electricity out of the wires and into the air, all we're doing is trying to couple our electricity to the great big transmission line that surrounds us all.
NeilThomas wrote ... If they are open wires, then the caapcitiance will not be negligable. Everything's a transmission line, with the antenna wires being no exception. The input impedance, given by sqrt(L/C) (each per unit length) will typically be a few 10s of ohms, or a few 100s depending on the geomtery. As you connect the battery (it doesn't matter whather the battery is a theoretical zero output Z, finite like 50ohms, or a mess of its own internal transmission lines, but the simpler cases are easier to compute) then a current will flow into to chracterisitic impedance of the transmission line. So a 10v battery would send a 100mA pulse into a 100ohm line. The pulse of current with a voltage present will radiate some EM field.
I know I have the same Characteristic impedance with any length, of , for example, coax cable.
But I can shape the antenna into many ways to affect the ratio of it's true capacitance and inductance.
What does this mean to an antenna anyway, when it's 1/2pi*sqrtLC frequency is so high compared to operating fequency anyway?
When I have 50 ohms feedpoint impedance, what is it in the end? How can it be characteristic impedance if we said it was ohmic impedance before?
And isn't proper-length antenna just like a resistor?
Those two things are completely different.
Chris Russell wrote ...
A dipole is said to be "resonant" when the inductive and capacitive reactances are equal, thus cancelling eachother out. This leaves behind only the resistance of the dipole, which is composed of the ohmic resistance of the wire and the radiation resistance of the antenna. For a halfwave dipole, the ohmic resistance of the wire is usually negligible, and the radiation resistance is roughly 73 ohms. That is the impedance of a halfwave dipole in free space. Characteristic impedances are for transmission lines.
Now this seems to contradict the upper part. Characteristic and ohmic impedances are different things, why are they mixed all the time?
And these ''capacitive' and ''inductive'' parts are fictions, you use them to describe reactance and they really don't have anything with capacitance and inductance of dipole wires?
Please correct my wrongness on these.
NeilThomas wrote ... It's only after the energy has travelled along the line, "found out" whether there's a load, or open circuit, or short circuit at the end, and so been reflected or not in the appropriate phase, that the battery current eventually settles down to its long term value. In this case, a 100ohm load means no reflection. An o/c means reflection of an anti-phase current, which drops the current to zero.
A short circuit means reflection of in-phase current which doubles the battery output. If the battery is itself zero Z out, then there's a further reflection and the output current increases further. The current builds a t rate dependant on the length and impedance of the line, or in otherwords, its inductance.
I've described this as if it's a single bit of 100ohm line, but of course that's not how an antenna looks. If you "play wavefront" and walk down the wire, looking around you to see what the wavefront sees, then a dipole will start lowish impedance at the feedpoint where the conductors are close together and have high capacitance, then have a gradually increasing impedance towards the ends. This only chages the detail of the solution, not the form. You can regard each change in impedance of the line as generating its own reflection. As you segment the line into more and more smaller sections, or better still represent the impedance analytically as a function of distance, then it approaches the behaviour of a real dipole.
I don't think I'm understanding this. Am I missing something cruical by disbanding characteristic impedance?
Characteristic impedance can also be called surge impedance as it defines surge behavior of LC circuits... So if I drive series RLC with characteristic impedance of 100 ohms at resonant frequency it's impedance will be 100 ohms only during first half-wave, and after a number of cycles current will be limited only by R. (ay it's short), and at other frequencies it will show other impedances.
100 ohm resistor will show 100 ohms regardless of frequency and time.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Hi Marko,
wrote ... ''some Other (capital O) kind of resonance''
You are right to think of "some other kind of resonance". Distributed circuits can resonate too, even though they have no lumped capacitance or inductance that you could point to and say "here is the inductor" or "here is the capacitor".
The classic example is a length of coax, or other transmission line, with one end left open. It shows a short circuit at the other end, when the frequency is such that the line is 1/4 wave long (allowing for velocity factor.) Either side of this 1/4 wave resonance, it shows capacitive and inductive reactance exactly as if it were a lumped series-tuned circuit with Zo of 50 ohms, or 75, or whatever kind of line you used.
The Zo of a line is a similar kind of thing to the Zo of a lumped resonant circuit, and indeed quarter-wave lengths of line can be used for impedance matching, in the same way and using the same equations as for lumped L-match circuits. And, driving your stub of line at the 1/4 wave resonant frequency will build up a high voltage on the open end, in exactly the same way as a Tesla resonator. I think someone once built a 144MHz VTTC this way.
Of course, you can tell that it's not a lumped circuit by testing at higher frequencies, because it goes on to have a "parallel" resonance at 1/2 wavelength, then another "series" one at 3/4 wave and so on ad infinitum.
The wires that antennas are made of have distributed resonances in the same way, but their Q is lowered by the radiation resistance, because the wires are arranged to fling the RF out into space as generously as possible. The ends of a dipole do build up a higher voltage than the feedpoint, but you won't be shooting corona off them unless you're running broadcast powers!
The wires of a transmission line are arranged to trap the RF inside because that's what you want from a transmission line, and that's why it resonates more obviously.
BTW, you also asked about the requirements for turning electric energy into electromagnetic. You don't need to, because it's all electromagnetic all the time. You just have to decide whether you want to trap it inside your circuit or let it escape as radiation.
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
There is just too much very basic things I don't understand.
What is characteristic impedance, after all? I don't know why you guys keep mixing it with resistance?
My point is, if I have 50 ohm resistor, it will be 50 ohms at any frequency, and series LC still with 50 ohms characteristic impedance sqrtL/C will be a short at it's resonant frequency. You get what I mean?
I know that characteristic impedance will determine rate of resonant rise, which I assumed unimportant here. But not much more than that :/
Multiplied with series R^-1 it gives Q of the circuit.
Now what is really ''impedance matching'', how it works there? I know quite well of analogies with DC circuits, resistors and batteries, but how to apply this to transmission line impedances?
The classic example is a length of coax, or other transmission line, with one end left open. It shows a short circuit at the other end, when the frequency is such that the line is 1/4 wave long (allowing for velocity factor.) Either side of this 1/4 wave resonance, it shows capacitive and inductive reactance exactly as if it were a lumped series-tuned circuit with Zo of 50 ohms, or 75, or whatever kind of line you used.
Yes, I can't know capacitance or inductance of the cable if I don't know it's length, but have always constant sqrtL/C.
Again I need to go back to fundamentals.
I have one question unanswered, and that is, where is the energy stored?
Chris talked about ''charging up'' a dipole antenna with negligible capacitance and then letting it resonate.
So, voltage is applied from the source and current flows because of relativistic delay, but into what?
And I assume it is impossible to charge up a coax cable other way than charging it's capacitance.
Maybe I need a completely different approach.
We're not getting electricity out of the wires and into the air, all we're doing is trying to couple our electricity to the great big transmission line that surrounds us all.
Alright;
So is it OK for me, to just imagine vacuum as infinite network of inductances and capacitances, with characteristic impedance of 376ohms.
After all this, it is not something too hard to imagine.
Displacement current flows in a sort of ''conductor'' which has capacitance with everything around, plus it produces magnetic field as any current would, right?
Now is ''LC resonance'' of these what is linked to just the conductor length and is responsible of all this?
This also forces me to conclude, that in some way, speed of light itself is actually limited and defined by the propagation delay in this transmission line. c^2=1/e0*u0?
And electromagnetic wave itself is nothing but energy transferred endlessly by magnetic and electric coupling between these LC's.
Still all this is passing around my head like a dream and I can't continue further.
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Antennas and radiation - some questions...
Otherwise you will need to correct with a factor 1/sqrt(eps_r). eps_r is the relative permittivity of the dielectric.
I am going to bed.
