But what about when the dipole is not a multiple of 0.5 wavelengths because it is NOT resonant? What would be the current distribution when the dipole is 3/4 wavelength? How can both ends be a high voltage? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?

But if you have any current flowing at the end of the wire, where is it flowing from/to?

If you don't have current flowing, then the impedance is high.

(Or, if you look at it another way, all the current flowing outwards at that point is reflected

from the open circuit at the end of the wire so it flows back the opposite direction. The

*net* current is zero.)

The answer to your conundrum is what happens at the feedpoint.

But let's look at the antenna first. Starting at the right end, we have high impedance at

the end, low impedance (maximum current) 1/4 wave in from the end, and 1/8 wave

later we hit the feedpoint at a point in the cycle where both the voltage and current

are at intermediate values. If we repeat the process from the left end, we have the

same current distribution, and end at the same intermediate point in the cycle.

If we were to plot the voltage distribution we'd see it high at the ends, dropping to zero

1/4 wave in from the end, then rising part way back up to the maximum at the feedpoint.

The current would be minimum at the ends, maximum 1/4 wave back, then dropping back

somewhat by the time you hit the feedpoint.

What happens at the feedpoint? We know that the current has to be in phase across

the feedpoint, so if we are plotting the current along the wire we can make the

sections in phase on each side of it. Since there is no point of minimum current

between the feedpoint and the end of the wire (which is where the current changes

phase along the wire) then all the current on the wire will be in the same phase.

Basically, the fact that the current must be 0 at the open end of a wire antenna, and

the sine / cosine distribution along the wire, force a particular condition at the feedpoint

regardless of the length of the wire. That's where you have any discontinuity in the

voltage or current distributions, and that's why the impedance can vary wildly in the

center of a doublet as you change frequency. If the feedpoint is at one end of the

wire rather than in the middle the current distribution will be different, but the same

rules apply. (Except that you wouldn't say that the current is zero there because

it isn't a

*free end* of the wire - it is connected to something, namely the feedline.)