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Above is a connection of half-wave dipole antenna. The book states that the current profile is same at any given time instant on both the radiators, each of quarter wavelength length.

Now my confusion is clearly the coaxial cable is connected to both radiators. Say I am sending the signal through centre wire and the braid of the coaxial cable is always at ground. Now at any given time instant, only left end of the cable is going to have non zero potential, while the right end is always at 0V. Then how come both radiators would have same current profile at any given time when the excitation is different for left and right radiators!

To be very specific, say at any given time instant the voltage at the left end is 2.3V, so standing wave is generated in the left radiator, but on the right radiator the voltage is zero (braid is at ground), so how any excitation would happen don the right radiator.

Support will be greatly appreciated. I am a total newbie in antenna systems.

  • \$\begingroup\$ Look again - the inner core is connected to the left and the braid to the right with an insulator between. \$\endgroup\$ Commented Jan 20, 2017 at 12:32
  • 1
    \$\begingroup\$ Your confusion comes from that your not considering transmission line effects. In non transmission line world a wire can not have an I/V distribution. The current and voltage through the entire wire is the same. In transmission line world the wire behaves differently. The wire can have an I/V distribution. Having a wire connected to ground means nothing at all. I have seen cell phone antennas where dozens of dipoles are stamped out of a single sheet of aluminum. Common sense says they should all short together but because of transmission line effects they work just fine. \$\endgroup\$
    – vini_i
    Commented Jan 20, 2017 at 13:02
  • \$\begingroup\$ Yes, I forgot to take into account the transmission line character at RF frequency. Thanks all for the great guidance! \$\endgroup\$
    – niki_t1
    Commented Jan 20, 2017 at 14:16

2 Answers 2


It seems the confusion is due to the shield of the coax being "grounded".

Yes, this may be grounded at the transmitter end. However, the antenna only "sees" the difference in potential between the two points it is fed at. It doesn't matter to the antenna that you consider one of these to be "ground".

In the case of a true earth ground, it can matter to the radiation pattern when the antenna is less than a wavelength or above the earth. This is why you often see a balun between a self-contained antenna like a dipole, and a non-symmetric transmission line like a coax cable. This balun essentially becomes a single-ended to differential voltage converter at the RF frequency.

A quick hack that can still make a effective balun is to wrap the coax around something magnetic for a few turns. This increase the impedance of the coax to common mode signals, while having no effect on differential mode signals. At the transmitter, the shield is at a fixed voltage (ground) and the center conductor varies. After the balun, both can vary up and down arbitrarily together, and only the difference between the two has low impedance and can carry any meaningful power at the RF frequency.


You are working under a misnomer, that is NOT a half-wave dipole. That is a 1/4 wave horizontally polarized vertical with a artificial GROUND-Plane. To make that a dipole you have to feed BOTH 1/4 wave sections. RULE: You cannot/should not feed a BALANCED MEDIUM with an UNBALANCED MEDIUM. A COAX is UN-BALANCED. Want to prove it? Physically touch the ground of the COAX. You will NOT get an RF Burn. DO NOT touch the center conductor or the part of the antenna that is connected to the center conductor - You WILL get a nice RF Burn or worse. Now, to answer the current question. There is NO current in the right leg. Get a hold of the ARRL book on antennas. It will explain what the radiation and current pattern for a dipole is and also how to build both a Toroild (Q1 material) 1:1 Balun and a COAX Balun. Also... You should be feeding the dipole with 50/52 Ohm (impedance) COAX. When you start with a 75 ohm COAX you are starting off with an SWR greater than 1.5:1.

  • \$\begingroup\$ Actually not such thing as equipotential ground exists in the world of radiowaves. It's only an approximation that is usable when distance is shorter than 10% of the wavelength. From necessity I operated 35 years ago a transportable about 100 W HF radio station with morsecode and temporary antennas that were dipoles with coaxial unbal feed. Both antenna branches made the fingers smoking, altough less when operated at lower frequency. \$\endgroup\$
    – user136077
    Commented Jan 21, 2017 at 5:04

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