Before reading this article, read and consider the question shown in the box.
CPBE Sample QuestionIn a negative 90 degree ‘T’ network ATU for a 90 degree AM series tower, the adjustment of what component would have the greatest effect on the input reactance?
a. The variable component (normally a tapped coil) in the shunt arm.
b. The variable component (normally a tapped coil) in the output arm.
c. The variable component (normally a tapped coil) in the input arm.
d. The value of the fixed output capacitor.
e. None of the above as intrinsically the input is always a resistance.
SBE certification is the emblem of professionalism in broadcast engineering. To help you get in the examination frame of mind, Radio World Engineering Extra, in cooperation of the SBE, will pose a typical question in each edition. Although similar in style and content to actual exam questions, our examples are not from past exams nor will they be on any future exams in this exact form; they are intended to help you prepare.
ANSWER AND DISCUSSION
The correct answer to the question posed in the box is c, the variable component (normally a tapped coil) in the input arm.
(click thumbnail)Fig. 1: The ‘T’ network is viewed and analyzed as a simple four-terminal network.The negative 90 degree “T” network ATU (antenna tuning unit) is ubiquitous in non-directional stations and a handful of “T” networks of varying phase shifts are in almost all DAs.
Why are they so prevalent? The math works out to be simple for making calculations at –90 degrees, the component count is low, the inductance in the arms create a low-pass filter for harmonic reduction and the circulating currents and voltage peaks are not extreme. The 90 degree T network also is easy to adjust.
The greatest power transfer (lowest loss) is achieved when the impedance of the generator (the transmitter/line) matches the impedance of the load (the antenna in this case). That’s what any ATU accomplishes; it optimizes the impedance conversion match between the line and the antenna.
The “T” network is viewed and analyzed as a simple four-terminal network (see Fig. 1).
A typical antenna resistance encountered in a real-world 90 degree tower is about 60 ohms + 70j (the positive sign indicates an inductive reactance component).
So in our specific case, R1 is 50 ohms + 0j, which is the characteristic impedance of the coax line and essentially the output of our transmitter. R2 is 60 ohms + 70j, which is the characteristic impedance of our antenna.
The formulas for calculating the values of the three legs are:
We have two configuration choices: capacitors in the series arms and an inductor in the shunt arm; or inductors in the series arms and a capacitor in the shunt arm (the industry standard).
As mentioned, this latter configuration has the bonus of being a low-pass filter attenuating any harmonics that might be in the original signal input.
(click thumbnail)Fig. 2: Twin caps in the shunt arm were used by the ATU vendor to split the current flow, and allow the use of cheaper small current capable capacitors as a cost function.Most likely then our ATU, as a practical matter, would look something like Fig. 2.
Twin caps in the shunt arm were used in this case by the ATU vendor to split the current flow, and so allow the use of cheaper small current capable capacitors as a cost function. Most often the large negative reactance value of the shunt capacitor will be adjusted exactly to the desired network value by a small series coil providing positive reactance.
We mentioned simplicity, and addressing the antenna reactance first really simplifies matters. That reactance is nulled out by a contrary negative reactance supplied above by a capacitor labeled C1. The bonus here is that this capacitor eliminates the DC path that lightning likes to take into your transmission plant.
What we have left to address then is the resistive components of the input and the antenna. The –90 degree angle cancels most of the trigonometry associated with this network. If we substitute into the above formulas just the trigonometric values of angle µ(beta) at –90 degrees (sin = –1 and cos = zero) and the resistive elements, we are left with the reactance arms of the network being determined by taking the square root of the product of the line resistance times the load resistance: Z1 = Z2 = –Z3 = v R1 * R2 (resistance only)
In this case the resultant would be 54.772j ohms (pure reactance) and to make it easy on ourselves, we’ll round that to 54.8js.
Now that we’ve addressed the reactance in the antenna, let’s restructure the basic leg calculations to solve for the input impedance. We’ll use some incremental reactance values substituting each in turn for the value of the three network legs. This will identify which one has the greatest effect on the input reactance when adjusted.
Below is a tabulation of the input reactance using our antenna load resistance of 60 + 0j ohms (reactance cancelled out):
Change in leg reactanceZ1 leg valueZ2 leg valueZ3 leg value(– other legs at 54.8j)= input reactance= input reactance= input reactance +5 59.8 = +5 59.8 = –4.14 –59.8 = –0.07 +3 57.8 = +1 57.8 = –2.5 –57.8 = –0.22 +1 55.8 = +1 55.8 = –0.83 –55.8 = –0.14 0 54.8 = 0 54.8 = 0 –54.8 = 0 –1 53.8 = –1 53.8 = 0.83 –53.8 = 0.2 –3 51.8 = –3 51.8 = 2.5 –51.8 = 0.77 –5 49.8 = –5 49.8 = 4.14 –49.8 = 1.58
Evident from the above numbers, changing the value of Z1 (the input leg) causes the greatest change in input reactance.
Changing the value of any of these legs also alters the phase shift somewhat through the network as well but we’ll leave that discussion for another SBE Certification Corner.
Broadcast engineering is a practice art and this question could be answered simply from experience. The values of the networks are usually set in advance before applying power by either using an RF bridge or rough calculation.
Here’s your question to mull for the next issue: “How are transmitter final stage efficiency factors determined?
“Electronic and Radio Engineering” — Terman 4th edition, pages 112–115
“Radio Engineering Handbook” — Terman 1943, pages 210–215