Curve of Equal SWR

I'm trying to figure out how radio frequency "matching stubs" work. In order to fully understand the problem, I need to know how the "curve of equal SWR" looks like.

I did a few plots, and it looks like it's a circle, or something close.

Whenever I approach the matter analytically, the equations explode and become very ugly.

I hope there's an easy way to prove that the following equations describes a circle, or ellipse on the complex plane:

$$VSWR = \frac{1+|\Gamma|}{1-|\Gamma|}$$

where VSWR is a constant, and $\Gamma$ (capital gamma) is the reflection coefficient:

$$\Gamma = \frac{Z_{load}-Z_{line}}{Z_{load}+Z_{line}}$$

$Z_{line}$ is a constant, normally a real number, actually equals to 50 Ohms in my specific case. (However, it would be interesting to see what happens if this can be a complex value, but this is really unimportant right now).

$Z_{load}$ is the variable, I suspect that these can be found on a circle or ellipse on the complex plane.

So can you girls/guys see an easy way to prove that $Z_{load}$ values are on a circle or ellipse? I do not seek absolute precise mathematical proof (it would be nice though), I just want to be "reasonably sure" that this is ture.

Thank you very much,

Tamás.

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Your variable $Z_{load}$ does indeed lie on a circle, provided that $VSWR>1$.

I'll use more typical mathematical names for the unknowns, since most here are not into radio frequency terminology. So let $$z=Z_{load},\qquad z_0=Z_{line}, \qquad C=VSWR$$ Then your equation is given by $${1+\Bigl|{z-z_0\over z+z_0}\Bigr| \over 1-\Bigl|{z-z_0\over z+z_0}\Bigr|}=C$$ A little algebra turns this into $$\Bigl|{z-z_0\over z+z_0}\Bigr|={C-1\over C+1}$$ The map $w=\phi(z)={z-z_0\over z+z_0}$ is a linear fractional transformation, and both $\phi$ and $\phi^{-1}$ will map circles to circles (where some circles may degenerate into lines). Since the curve you are interested in consists of exactly those $z$ which are mapped to the circle $|w|={C-1\over C+1}$, your curve will be a circle.

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Thank you! VSWR is indeed > 1 in my case. If VSWR = 1, then it is a "perfect match" in rf terminology, and there's no need to use matching stubs (or any kind of matching).In this case the "curve" is a single point, since $\Gamma$ - the reflection coefficien - is zero (no reflection at all). If the match is perfect, $z_o = z$, and all rf energy is transferred to the terminating load, no reflection indeed. I'll review the solutions and decide wich one to accept. Thanks again! – netom Nov 20 '12 at 10:00
A more complete answer, nice 8) – Alexei Averchenko Nov 20 '12 at 10:14

Your equation uniquely determines $|\Gamma|$, which is real positive, and complex numbers with length $|\Gamma|$ form a circle of radius $|\Gamma|$.

The details: let $x = |\Gamma|$. We have the equation of the form $$C = \frac{1 + x}{1 - x}.$$ When $x \neq 1$ it is equivalent to $C - Cx = 1 + x$, or $(1 + C)x = C - 1$, or $$x = \frac{C - 1}{C + 1}.$$ It is positive whenever $C > 1$. When $x = 1$, the equation makes no sense, but we can still ask what happens when $x$ is arbitrarily close to $1$. $\lim_{x \to 1} \frac{1 + x}{1 - x}$ does not exist because it approaches $+\infty$ or $-\infty$ depending on where you are approaching $1$ from. You can still say it's the infinity in the projective line sense (in that it's infinitely far away from any real number), but I doubt it's useful in your case.

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Good observation with $x = 1$. $|\Gamma|$ is the so-called "reflection coefficient", and it determines how much energy is reflected from a poorly matched load at the end of a transmission line (like an antenna tied to a coaxial cable). When $|\Gamma| = 1$, it means that all power is being reflected, and we say that the SWR is "infinite" (yes, in the projective sense as you've said), which is a terrible situation. :) In this case, no matching stub will help, so we do not have to worry about that. Thanks again, I'll review the answers and choose one to accept. – netom Nov 20 '12 at 10:12