Standing waves in antenna: why is the current wave symmetric with respect to the centre?

Question

In trasmitting antennas the current is described with a standing wave (in resonant configuration). Since the current must be zero at both ends of the antenna (it could not be otherwise) the mechanical analogy should be the standing wave on a rope fixed at both ends.

Nevertheless if the length of antenna is $d=2 \lambda$ the configuration of current is

(For this and other configurations: see last pages of http://www.amanogawa.com/archive/docs/antennas1.pdf)

While the configuration of a rope fixed at both ends when $d=2 \lambda$ is

(of course it's not "current" here, but "dispacement")

I'm quite confused about this difference. I see that the reason of this difference is that the antenna in the first picture is connected in the central points to an AC current generator, and that connection "creates a condition of specular symmetry between the two branches".

But I do not see why the AC generator should provide a symmetrical situation at both sides necessarily.

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On wikipedia: https://en.wikipedia.org/wiki/Antenna_(radio)#Resonant_antennas it is represented also the voltage wave for $d=\frac{\lambda}{2}$ (which is not the previous case).

It is clear that the voltage wave is not symmetric as the current wave is!

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So my question is: why does the AC generator create a specular symmetric current wave for the two sides? And why is the voltage wave a "normal" standing wave (not specular) instead?

Answer 1

Let's go back to basics and consider just a linear current source (say just a line). If you suppose a time-harmonic field (i.e. temporal sinusoidal variation), then you will see that there are natural "fundamental" modes of the structure (eigenmodes if you want to be fancy). These correspond to the standing wave modes, similar to a vibrating string. They exist in all of the same configurations that you see, including both the symmetric and asymmetric distributions you mentioned.

However, things change a bit when you consider what modes you can excite using a balanced transmission line. If you are feeding in the center with a balanced transmission line (i.e. an AC source where the top and bottom rails have the same magnitude current), then intuitively, it makes sense that the current on the structure be symmetric. As the top half of the line flows in and up, the bottom flows out at the same magnitude. Hence, you have a symmetric distribution.

This isn't the case if your feed-line were not balanced or if you feed it off center. If you wanted to excite a current distribution that's asymmetric, you could move the feed off center, preferably to a point where the impedance of the antenna is closely matched to the impedance of the feed line.

Furthermore, the feed-line is of some characteristic impedance $Z_0$, so you can't have current distributions that go to $0$ at the feed point. If you tried to do that, the line would see an infinite impedance ($Z_{ant} = V_{ant}/I_{ant} = \infty$), and so all of the input power would be reflected back down the line.

For a half-wave dipole, the input impedance is $73 +j42.5$ $\Omega$; thus, there is a reactive component, which causes a phase difference between the voltage and current. This is why you see that the standing waves are offset.