When you transmit signals at radio frequency why are the antennas open ended (open circuit)? I can understand that changing the electric field causes changes in the magnetic field and hence the wave will propagate. But what I don't understand is how you can have an electric field if the circuit is open ended in the first place because the impedance should be infinite?
It's a bit unclear what exactly you are asking. I suppose you have seen a antenna like a dipole and it looks at first glance like just two wires that aren't connected to anything, therefore the question is how can current flow and power be drawn? If so, it would help if you clarified that. Some antennas, like a loop or folded dipole, are exactly the opposite in that they appear to be dead shorts at first glance.
In any case (if my interpretation of your question is correct), what you are missing is that antenna is no longer a open or short at its intended frequency. Antennas are not lumped systems, which means that different parts will be at different phases of the signal at the same time. In fact, antennas exploit this to help produce the large voltages and current it takes to radiate significant power.
Often resonance is involved. Fill a bathtub partway with water. Now put your hand in the middle and move it back and forth only a short distance in the end to end direction. Once you find the right frequency, you will see that you can get a lot of water to slosh back and forth despite only a relatively small motion of your hand. Note that at the peaks, the water at one end of the tub is high and the other low, and nothing is flowing. In between the water is roughly level but is flowing strongly in the middle. Also note that it takes very little force from your hand to cause this and keep it going, but you have to be moving your hand at just the right frequency. A little faster or slower and it doesn't work anymore.
That was resonance, and is exactly what is happening in a dipole. In the case of a dipole, that water level becomes voltage and the flow rate of the water becomes current. The feedpoint in the middle of the dipole is where a little current of just the right frequency is fed in, which causes resonant sloshing of charges back and forth in the antenna. The voltages created at the ends of the antenna can therefore be a lot higher than anything you put in.
This sloshing of current back and forth makes high voltages at the ends. Together with the high current in the middle, the antenna disturbs the local E and B fields in such a way that power is lost from the antenna into those fields. That power eventually organizes itself into a self-propagating wave we call radio.
Since real power is lost, the antenna must appear to have a resistive component from the driving circuit point of view. A ideal antenna used at exactly the right frequency will appear to be purely resistive, meaning all the power dumped into it gets transmitted. This happens very close to the best resonant sloshing frequency for most antennas. That also explains why antennas often only work well for a narrow frequency range.
So getting back to your question, the problem is you are analyzing the antenna at DC, which is completely irrelevant. You have to analyze antennas at the frequencies they are intended to radiate at. At those frequencies, a lot of other stuff happens so that they don't look like opens or shorts as they do at DC.
There will be capacitance to ground, or the other element of a dipole, for instance. Taken as a whole, the antenna and ground constitute a tuned circuit.
There are plenty types of antennas with different geometries.
The resonating antennas (where the dipole you said stays), used for far field transmissions, take their properties from the occurrence of a standing wave inside of it.
In simple words, if you take two wires with the same length and you manage to have the two minimums of the cosine coincident with the end of the wire, this will create a standing wave with an high potential that can radiate (a standing wave is not still! Its magnitude changes with time but it does not not propagate on the wire...)
protected by W5VO♦ Apr 8 '13 at 15:11
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