Big Wheel Antenna Engineering Letter

The two variables you referenced are the Greek letter ‘Theta’, E and H refer to the ‘electric’ field and the ‘magnetic’ field respectively. Theta in mathematics designates the angle of an argument. When you use horizontal polarization the ‘E’ field lies within the horizontal plane parallel to the surface of the Earth. The ‘H’ is perpendicular to it. When we use antennas, we use (generally) vertical or horizontally polarized elements. There are other polarization configurations but to keep it simple right now lets just entertain vertical and horizontal polarities. Theta ‘E’ and ‘H’ are used to measure the beamwidth of the antenna in degrees or the area of a beam front propagated by an antenna of so many elements and length. We calculate the gain of an antenna by using: Gain (in dBi) = 10 Log (41253/theta ‘E’ X theta ‘H’). If we look at a point source inside a sphere of radius 1 radian and that point source is illuminating it the gain is 1. e.g. Area of the surface of a sphere = 4 X PI X R^2 (radians), a radian we know to be 57.3 degrees, therefore the surface of the sphere contains 41,253 (square degrees, actually steradial degrees). So, if we have an antenna that measures say 60 degrees in the ‘E’ plane and 30 degrees in the ‘H’ plane; 30 X 60 = 1800 and 41,253/1800 = ~23 and 10 log 23 = 13.6 dBi (i being over an isotropic source).

The wavefront transmitted from a horizontally polarized antenna will therefore have an ‘E’ plane that is horizontal. In order to receive the maximum amount of energy from the wave front of that propagated energy, the receiving antenna must also be horizontally polarized. If the receive antenna is moved out of the horizontal toward vertical or perpendicular to the plane of the transmit antenna the amount of received energy falls off as the cosine of the angle. For example: Let the receive antenna be positioned 45 degrees out of the plane of the transmitting antenna. the cosine of 45 degrees is 0.707. To find the amount of energy lost in decibels we take 20 Log of 0.707 and see that the loss is -3.01 dB, or half the power (actually half the voltage component of the power transmitted). If the angle between the transmit and receive antennas become 90 degrees the loss becomes -27 dB, this only in theory as I have, at best, working at the National Laboratories in Boulder, CO been only able to measure a loss of -22 dB between cross polarized elements. The reason being that there is no perfect horizontal or vertical wavefront. Each have components of the other to some degree.

Now, maybe this is more than you asked for but it does lead to some rationality. In your experiment with the balloon borne payload carrying an ATV transmitter of say 1 watt of power, that 1 watt is spread out over 12 MHz under video modulation (plus the audio subcarrier, if used) as those transmitters are double sideband emitting devices. Granted you need only 6 MHz to transmit a standard NTSC TV signal but in order to do so requires a vestigial sideband filter which you don’t need in you payload weight budget. On the ground side, the receive side the band pass of the receiver has to be at least 6 MHz wide (and in reality is probably wider). The wider the bandpass of a receiving system the more prone it is to noise. Manmade noise, as from powerlines, from static discharges of a multitude of devices. The more noise you can keep out of the band pass of a receiver the less corrupted the TV picture will be. Manmade noise is predominately vertically polarized. This is the case for using horizontal polarization in ‘weak signal’ work.

The Wheel antenna is almost a perfect omni-directional radiator of horizontal polarization. They have been used for the past fifteen years that I know of for propagating ATV and TV transmissions from balloon borne platforms, amateur, military and aerospace.

In order to defend your position/belief on the use of the Wheel antenna and the project being a success you will have to present you professors with a link analysis or budget. Your payload will probably not be transmitting more than 20 miles of slant range before the balloon bursts at some altitude around 100,000 feet. Use 20 miles as your hypotenuse and 100K’ as your “Y” dimension to find your distance, “X”, across the Earth. The slant range will be the distance with which you will wish to compute your ‘space loss’. SL (in dB) = 36.6 + 20 Log frequency in MHz + 20 Log distance in miles, e.g., 20 log 426.25 = 52.6, 20 log 20 = 26. The result is a Space Loss in dB of -115 dB. This loss has to be made up in order to make link or have a usable picture. A -60 dBm signal and a 37 dB S/N (signal to noise) ratio at your receiver is required to get a snow free picture. Start whittling this space loss down with what you have a +’s. Antenna gain will be 2 X 2.14 dBi = 4.28, the transmitter is 1 watt = 30 dBm and your receiver will have a sensitivity of around -100 dBm. Add these up and you have 134.3 dBm system gain, add the -115 dB loss and you have 20 dB to the good or a 20 dB fade margin. This would be your worst case scenario. Hopefully you will run more than a 1 watt transmitter! Also, this was assuming Wheels at both ends of the link. At your ground station you will probably want as much gain in your antennas system as possible using yagi antennas to track the payload and a good low noise preamplifier will also help.

Dave W6OAL –

Olde Antenna Laboratory 41541 Dublin Drive Parker, CO 80138