I know that the directivity of the monopole is twice the directivity of the dipole because the transmitted power in case of the monopole is distributed only across a half of the space.

But why the input impedance of the monopole is half of the impedance of the dipole? Is there some proof to that?


1 Answer 1


It's a simple symmetry argument:

enter image description here

from https://en.wikipedia.org/wiki/Monopole_antenna

The fields from the two arrangements in the figure are the same above the ground. The monopole has the same current but half the voltage, hence has half the impedance. For the same current it also radiates half the power as it is radiating into half the space.

  • \$\begingroup\$ Hello Tesla,If we apply V voltage on dipole then it will be Z=V/I if we apply V on monople then by mirroring it will be Z=2V/I so its twice the impdeance of the dipole. Where did i go wrong in the logic? THanks. \$\endgroup\$
    – rocko445
    Jan 7, 2022 at 22:40
  • \$\begingroup\$ Z=V/I, same I in both, half V for the monopole \$\endgroup\$
    – Tesla23
    Jan 7, 2022 at 22:55
  • \$\begingroup\$ Hello,Why in monopole we have half the voltage? we have the same voltage source as the dipole and mirroring cause this voltage to double instad of being half \$\endgroup\$
    – gop664
    Jan 8, 2022 at 12:53
  • \$\begingroup\$ In the diagram I showed, you are driving the monopole against the groundplane with a voltage V, generating the current I (ignote the image shown - it is not real). Hence it's impedance is V/I. For the dipole you need to apply 2V across the terminals to generate I, hence the impedance is 2V/I, or twice that of the monopole. \$\endgroup\$
    – Tesla23
    Jan 10, 2022 at 1:58

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