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Calculus not needed (was: Reflection Coefficient Smoke Clears a Bit)
I've enjoyed reading this and related threads. Some comments have been
made about using calculus. Though I spent a significant portion of my pre-retirement life attempting to teach that subject to undergraduates, I don't believe in using calculus whenever simple geometry and/or algebra makes it unnecessary. A proof that avoids calculus can be meaningful for those who don't know calculus, or who haven't used it for a while. With that said, I have a few comments to make about some of the assertions I have read here recently, some of which have appeared without explicit proof. (1) The surge impedance of a (lossy) transmission line cannot have an angle more than 45 degrees away from the real axis. This is true. Z_0 = sqrt((R + jwL)/(G + jwC)) (here I am using "w" instead of omega). Both the numerator and denominator lie in the first quadrant, so their quotient lies in the right half plane (angles subtract when one divides), and the square root of that has an angle lying between - 45 degrees and + 45 degrees. (The branch of the square root with positive real part has to be taken; if you can find coax whose surge impedance has a negative real component, I'll pay you good money for it.) [Since this angle lies between - 45 degrees and + 45 degrees, peculiar consequences deduced from calculations involving surge impedances such as 50 - j200 can be ignored.] (2) There is a nice geometrical interpretation for the reflection coefficient, or rather for its magnitude. Since the coefficient is (Z_L - Z_0)/(Z_L + Z_0), its magnitude expresses how much further Z_L is from the surge impedance Z_0 than it is from the negative, - Z_0, of the surge impedance. If Z_L is equidistant from Z_0 and - Z_0, then the magnitude of the reflection coefficient is 1. If Z_L is closer in the complex plane to Z_0 than it is to - Z_0, then the magnitude of the reflection coefficient is less than 1. If Z_L is closer to - Z_0 than to Z_0, then the reflection coefficient's magnitude exceeds 1. Now plot the points Z_0 and - Z_0 and draw the perpendicular bisector of the segment joining them. If Z_L is on that perpendicular bisector, the magnitude of the reflection coefficient is 1; if it is on Z_0's side of the bisector, the magnitude is less than 1; if it is on - Z_0's side, the magnitude exceeds 1. Of course Z_L has to stay in the right half plane; if it didn't have to do this, you could take Z_L very close to - Z_0 and get enormous reflection coefficient magnitudes. (3) Consider an ellipse having Z_0 and - Z_0 as its foci. There are infinitely many such ellipses, including a degenerate one (just the segment between the "foci"). All these different ellipses fill up the complex plane, and no point in the plane is on more than one of them. On any one such ellipse, the sum of the distances from a point on the ellipse to the two foci Z_0 and - Z_0 is constant (definition of an ellipse), the value of that constant depending upon which ellipse it is but the constant has to be at least as large as the interfocal distance. We should ignore points on the ellipse that are in the left half plane. A portion of the ellipse will be in the same quadrant as Z_0, and a portion will be in the quadrant that contains the conjugate of Z_0. (Remember we are ignoring the points in the left half plane.) All of the points on the ellipse that are in Z_0's quadrant are closer to Z_0 than to - Z_0, so they'll give reflection coefficients with magnitude less than 1. So will those of the points on the ellipse in the other quadrant under consideration that are between the real axis and the perpendicular bisector. But those that are between the perpendicular bisector and the imaginary axis will be closer to - Z_0 than to Z_0 and thus will yield reflection coefficients with magnitudes greater than 1. It should be obvious that, along any given ellipse, the one for which the magnitude of the reflection coefficient is greatest is the one on the imaginary axis, since as we move along the ellipse towards that point, the distance to - Z_0 decreases and the distance from Z_0 increases (remember, their sum is constant along the ellipse). So on any *one* ellipse, the largest reflection coefficient magnitude occurs where the ellipse meets the imaginary axis, and thus Z_L has real part 0 and imaginary part of opposite sign to that of Z_0. (4) Start with Z_0 real, and slowly rotate Z_0 into either the first or fourth quadrant, but not more than 45 degrees in either direction, keeping the same magnitude while you rotate. The segments joining Z_0 and - Z_0, their perpendicular bisectors, and the various ellipses will all simultaneously rotate. It's now obvious that for ellipses of any fixed size, the one producing the largest magnitude for the reflection coefficient will occur when Z_0 is at + 45 degrees or - 45 degrees. So if we want to maximize the reflection coefficient magnitude, we can restrict attention to those two cases. The - 45 degree case (capacitive surge impedance) is the more familiar one, but the math is the same either way. The only question is, which one of the ellipses should we use, if we wish to maximize the magnitude of the reflection coefficient? (5) So now we're going to assume Z_0 = k(1 - j), and thus - Z_0 = k(-1 + j), while Z_L = ktj. It's clear that the factor k is going to cancel out when calculating the reflection coefficient, so I will henceforth ignore it (i.e., normalize it to k = 1 by appropriate choice of units). If you are of a geometrical turn of mind, you can produce a geometrical argument showing that the best one can do is to make sure the ellipse meets the imaginary axis at the same distance from the origin as the two foci, i.e. at j*sqrt(2). If you are of an algebraic turn of mind, you can make an algebraic argument involving completing the square to demonstrate the same thing. If you insist on using calculus, it's now just one variable calculus, not multivariable calculus, since the only independent variable is t, which will turn out to be sqrt(2) at the maximum. (Hint: don't look at the ratio of distances; look at the square of that ratio, so as to get rid of all those square roots.) (6) Once all that is done, it's just a bit of algebra to show that when t = sqrt(2) then the magnitude of the reflection coefficient is 1 + sqrt(2). That's the best (worst?) you can do. And if you can find some coax whose surge impedance angle is - 45 degrees, you can indeed do it. All the above was done from first principles. I am not fortunate enough to own a copy of Chipman, though I wish I were, but if this is what he says, then I am in full agreement with him. David, ex-W8EZE, willing to part with some of my pension money for a copy of Chipman if you know where one can be found (Powell's doesn't have any copies) -- David or Jo Anne Ryeburn To send e-mail, remove the letter "z" from this address. |
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