Off Topic warning: This has absolutely nothing to do with fixing up my house. I wrote about my road trip to Gainesville, Florida a while back. This is the follow up trip, only this time it involves flying to Gainesville instead of driving.
The replacement tower has been put up, the new unipole antenna has been installed and now it is time to match the transmitters to the new tower. This involves using some math. At some point in radio history, someone decided that all transmitters should have an output impedance(Z) of 50Ω (ohms). Impedance in an alternating current (AC) circuit (all radio frequency is AC) is like resistance in a direct current (DC) circuit. The only difference is impedance requires the use of the Z axis to calculate. You remember 9th grade algebra and the Cartesian Coordinate graphing system, the X and the Y axis. Looking down on the X and Y axis, the Z axis would be stick straight up, which makes it a three dimensional problem. More challenging, and thus much more fun. So, here goes…
There are two radio stations combined onto the same tower, this makes the problem extra special as filter networks need to be installed to make sure that each frequency does not get back into the other station’s transmitter.
Station number one, broadcasts on 980 kHz at 5,000 watts. The tower is 240 feet tall which is close to 1/4 wave length, nearly ideal for an AM station. Using a Delta Operating Impedance Bridge (OIB-1) and a Potomac SD-31 frequency Generator, I measured the impedance at the base of the tower. On 980 khz it is 74Ω with +j160 reactance. Using the ohms law pie chart, we can figure out almost anything we need to know about electricity:
Therefore, the base current will be I=√(P/R) =√(5000/74) = 8.3 amps. The voltage will be E=√PxR =√(5000×74) = 608 Volts.
A bit about reactance; it is noted by using the letter j, which indicates it is an imaginary number. Basically in an AC circuit, there is inductance and capacitance. They are the reciprocal of each other, sort of (this could get into a long, long post if I have to explain the roles of inductance and capacitance in and AC circuit). Reactance is an undesired inductive or capacitive component that has to do with the lead or lag time between the voltage wave form peak and the current wave form peak. In standard utility company parlance it is know as the “Power Factor”. In RF circuits it causes inefficient power transfer and it needs to be canceled. A +j value indicates that the reactance is inductive, and therefore needs to be canceled out with a capacitor. A -j value indicates the reactance is capacitive and needs to be canceled out with an inductor.
Then there is the difference in impedance, the transmitter and transmission line is 50Ω and the tower is 74Ω. Enter the antenna tuning unit (ATU). The ATU matches the base impedance of the tower through the use of a T network:
To determine the value of each leg of the T network, we need to employ math again. Here is where the details will catch up with you. Remember, there are two stations on this tower, a 980 kHz and a 1430 kHz. We need to make two T networks, one for each station. There are a few characteristics of a T network that can be used to our advantage here. A T network can also function as a low pass or high pass filter depending on the relationship between capacitance and inductance. In an inductive circuit the phase is advanced and in a capacitive circuit the phase is retarded. If we can make the phases of the two stations 180° opposing, this makes an excellent start to a filter network. Therefore, one station should be +90° and the other should be -90°.
So, on 980 kHz we want to match 50Ω to 74Ω with a +90° phase shift. SImple. Each leg of the T network needs to have the following value
Z(leg)=√Z(antenna) x Z(transmitter) or Z=√(50 x 74) = √3700 = 60.8Ω
This is a highly simplified diagram that does not show the pass/reject filters employed between the ATU and the tower to properly combine both stations onto one antenna. That would be an extensive topic that I am not even sure I could adequately describe here:
So each leg needs to have an impedance of 60.8Ω. The input leg is inductive, the ground leg is capacitive and the output leg is inductive. Remember, the output leg is already inductive by +j163. The inductive reactance needs to be canceled out, but some of it can be used in the T network. To make the output network match the rest of the T network, it will need 102.4Ω capacitive reactance (163-60.8=102.4Ω). To calculate these values, we use the L and C formulas which are 980 KHz = .98 mHz):
C = 1/(2π f (mHz)Xc) or 1/(6.28 x .98 x 60.8) or 0.00267 uf (gound leg, 60.8Ω)
C= 1/(2π f (mHz)Xc) or 1/(6.28 x .98 x 102.4) or 0.00159 uf (output leg, 102.4Ω)
and
L= Xl / (2π f(mhz) or 60.8 / (6.28 x .98) or 9.88 uH (input leg, 60.8Ω)
This combination should get us close to the Z 50Ω impedance that the transmitter is looking for.
The next frequency is 1430 kHz with a power of 10,000 watts. This frequency should be retarded by -90 degrees, so the input will be capacitive with in inductive leg to ground and a capacitive output. The tower measures 165Ω with -j105. Perfect!
Again, the current and voltage at the base of the tower on this frequency will be I=√(P/R) = 7.78 amps and E=√PR = 1,285 volts.
Z= √(50×165) = √8250 = 90.82Ω
L = Xl / (2π f(mHz) = 90.82 / (6.28 x 1.43) = 10.11 uH (ground leg, 90.82Ω)
C= 1 / (2π f(mHz) Xc) = 1 / (6.28 x 1.43 x 90.82) = 0.00122 uf (input leg, 90.82Ω)
and
C= 1 / (2π f(mHz) Xc) = 1 / (6.28 x 1.43 x (-j105-90)) = 0.0074 uf (output leg, 75.8Ω)
Since the current and voltage for both stations are additive (with slight variations due to phasing on the two frequencies) the total current at the tower base will be 8.3 amps + 7.8 amps = 16.1 amps and the total voltage will be 608 volts + 1285 volts = 1,893 volts. Now you know why there is a fence around the bottom of the tower!
And there you have it, that is how an AM transmitter is coupled to the base of a transmitting tower.
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