The Monopole above a Ground Plane
The aim of this article is not to offer help with any practical antenna, but is more of a “thought exercise”.
The dipole in free space is usually taken as the “standard” or “reference” aerial when making comparisons of different types of aerial. How does the signal received from a quarter wavelength monopole above a perfect ground plane compare with that received from a full half wavelength dipole in free space, you may wonder? To establish this, we first need a quick recap on how a (vertical) half wave dipole radiates. Electric charge, (electrons), surge up and down from one end of the dipole to the other giving rise to the familiar “half sine wave distribution” of current, with its maximum in the centre, and zero at its ends. The centre of the current maximum is where the “radiation resistance” is defined, (about 72Ω for a half wave dipole), and this is where the feeder is usually connected.
Current in a straight wire generates a cylindrical magnetic field around the wire, with the wire forming the axis of the cylinder. The magnitude of the field is proportional to the current. The direction of this magnetic field at any point around the wire is tangential to the circumference of the cylinder, so for a vertical dipole the magnetic field is horizontal. Each tiny “elemental” length of the dipole caries a different current, so the strength of the magnetic field around the wire also follows a half sine wave form as indicated in figure 1, where the distance of the curved line from the dipole represents the strength of the magnetic field.
According to Maxwell’s laws, a changing magnetic field generates an electric field proportional to the rate of change of magnetic field and at right angles to it. In this case the magnetic field lines are horizontal so the induced electric field is vertical. Thus an “electro-magnetic” wave comprising a horizontal magnetic and a vertical electric field is launched from the vertical dipole.
At a distant point, the strength of the received signal is proportional to the sum of all the separate radiating elemental lengths of the transmitting aerial times the current flowing in each of them. (In the following it is assumed that the received signal is measured on the “peak of the beam”). The question you may ponder is, “Does a quarter wave monopole over a perfect ground plane produce less received signal than a half wave dipole in free space”? After all, it has only half the number of radiating elemental lengths. There are 2 ways, (‘a’ and ‘b’ below), to try to answer this question.
a) Although the quarter wave monopole above a perfect infinite ground plane has only half the number of elemental radiating elements, it produces only half the elevation pattern of a half wave dipole in free space, i.e. there is no radiation pattern below the ground plane. So, for the same transmitter power fed into the dipole and monopole-ground plane aerials, they should both deliver to same power to a receiving aerial.
b) The radiation resistance of the quarter wave monopole is only half that of the dipole, i.e. about 36Ω instead of 72Ω. Now power is proportional to “current squared times resistance”, (I2.R). Therefore, if the same transmitter power is fed into each type of aerial, the current in the quarter wave monopole must be the square root of two times as great as that fed to the dipole. But the voltage induced in a distant receiving aerial by the passing magnetic wave only “knows about” the sum of the elemental lengths of the transmitting aerial times the current in each of them. This is half the number of elemental lengths for the monopole, (as compared to the dipole), times the (square root of 2) times the current which would be in the elemental lengths of the transmitting dipole. The power of the received signal from the monopole and ground plane is therefore proportional to ½ times (the square root of 2) squared, which is the same as that received from the half wave dipole in free space. In summary: As far as the receiver is concerned, they are both the same.
Before you try to apply the above reasoning to a practical case, remember this is only a “thought exercise”, i.e. an interesting bit of theory. In “earth bound” practice, there is no such thing as an “aerial in free space”. Whatever height the aerial is, there is always an earth plane to reflect some of its radiation, either to add or subtract from its direct ray radiation. Also, don’t carry the “image in the ground plane” analogy too far. There is a significant difference between the vertical radiation patterns of a quarter wave aerial above a ground plane and a true half wave aerial in free space.