18

BEYOND THE BASICS

R1 ≔ 1 R2 ≔ 1 C ≔ 1000 ⋅ pF

Rresult ≔ 0 XL ≔ 1

Zresult ≔ 0

Z1 ≔ 1

Z2 ≔ 1

R1 ≔the 1 above. R2 ≔Required 1 Rresult ≔ MathCad 0 Z ≔ right 0 Z1 ≔ 1 Ignore to get toresult work C ≔ 1000 ⋅ pF XL ≔ 1

Don’t Let Complex Numbers Scare You

Z2 ≔ 1

f ≔ 1450 ⋅

June 13, 2018 f ≔ 1450 ⋅

Ignore the above. Required to get MathCad to work righttors in series and can do the same with

Anyone who can do Ohm’s Law calculations can handle the math of the complex load BY FRANK McCOY

While perusing one of the Facebook groups that post pictures of various broadcast assets (it’s a guilty pleasure we all share, I think), I came across a thread about an impedance sweep of the common point of some AM station’s antenna somewhere. Reading the comment thread made me realize that there are many engineers who find complex impedance calculations to be a deep mystery. Something about that “J-factor” is black magic, understood only by those with engineering degrees. Well, I have one, and I believe anyone who can do plain old Ohm’s Law calculations can handle the math of the complex load. Part of the problem is that there are multiple, confusing and different systems of notation to represent complex impedances. There is the magnitude and phase style, which looks like 20

∠+45°. There’s the lower-case j style that looks like 13.9 + j 13.9. And there are weird combinations of letters and numbers like 20 ejπ/4 that we’ll ignore for now. These are all different ways of representing the same fundamental properties and, in the examples above, the same actual impedance value. Ohm’s Law applies universally. Even with reactive loads. In exactly the same way as with resistive loads. Yep, E still equals I times R. If we parallel two identical resistors, we know the resulting resistance is half. And if you parallel two unequal resistors, the resistance is the reciprocal of the sum of one over the resistance of each resistor. Actually, what those “one over the …” calculations are doing is converting to admittance. The last reciprocal calculation just gets back to resistance and good old ohms. But you don’t need to know that to perform the calculation.

1 Rresult ≔ ―――― ⎛1 1⎞ 1 + ― ― ⎜ ⎟ Rresult ≔ ―――― ⎛⎝ R11 R12 ⎠⎞ + ―⎟ ⎜― R ⎝ when 1 1 RRbecomes 2⎠ But what happens Z? The formula now looks like this: Zresult ≔ ――― ⎛1 1⎞ 1 + ― ― ⎜ ⎟ Zresult ≔ ――― ⎛⎝ Z11 Z12 ⎞⎠ +― ⎜― Z1 Z2 ⎟⎠ ⎝1 XC ≔me,――― Believe in exactly the 2this ⋅ works same way. I know, those Z terms seem 1⋅ f ⋅ C XC ≔but ――― mysterious, they really aren’t. 2 ⋅ f ⋅two C systems of Earlier, I showed XL It ⋅you complex notation. turns L ≔ ―― out that both systems are useful in different contexts. 2 ⋅X and ⋅f When you add L subtract complex L ≔use―― values, our R + jX notation style. Then add 2 (or⋅ subtract) ⋅ f the resistive to (or from) the resistive and do the same with j to get the result. We add resis-

complex values. 1 + j2 in series with 3 + j4 equals the expected 4 + j6. If you use MS Excel’s complex number functions, be aware that MS wants the value to look like 4 + 6j. The sign of j is something we all probably know. Inductors are positive and capacitors are negative. Notice that we haven’t involved frequency at all. We won’t until we have a result using these methods. This is what is known as “phasor math,” which is probably where that big box at the AM transmitter with all the cranks on it got its name. The design process almost certainly used phasor math, since line lengths and tower spacings are invariably in angular values. Converting between R + jX and Z ∠° is key to successful application of phasor math techniques. Remember that I said the two earlier values were equivalent? They are. Picture a rectangular grid (Cartesian coordinates to be precise) with the R part on the horizontal axis and the j part on the vertical axis. Measure from the center (the “origin” in traditional mathematical terms) horizontally 13.9 units to the (continued on page 20)

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