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

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.

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!

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?

• Since the current in a dipole antenna is just the displacement current from Maxwell's equations, you can see that it results from the time variation of an electric field. In a linear circuit, the potential/voltage is basically just a constant offset from the electric field magnitude. From your GIF, it is clear the red could be a half-period cosine wave and the blue a half-period sine wave (i.e., the blue is the derivative of the red). Aug 26, 2017 at 19:29

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.