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Published byRodney Walters Modified over 9 years ago
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Look at a distance z = - L toward the generator -z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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-z ZLZL Z c1 z = 0 Z c2
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-z ZLZL ZcZc z = 0
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X = Z c tan kL L
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Should I add something in series or in parallel?
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Eureka! Use a stub. -z ZLZL ZcZc z = 0 -z ZLZL ZcZc z = 0 -z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0 2 1.5 1
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-z ZLZL ZcZc z = 0 Not matched! -z ZLZL ZcZc z = 0 Needs some -jB somewhere, but where? -z ZLZL ZcZc z = 0 Eureka! Use a stub, somewhere. How long should it be?
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An article appeared in the January, 1939 issue of Electronics that changed forever the way radio engineers think about transmission lines. Phil Smith [1907-1985] devised an extraordinarily clever circular chart that revealed graphically the complex impedance anywhere along a line. No math and minimum fuss. There's a marvelous symmetry in its design - everything fits together neatly. So ingenious was his invention that it has been the standard of the industry - for over sixty years.
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-z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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-z ZLZL ZcZc z = 0
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a x b y r Equation of circle (x - a) 2 + (y - b) 2 = r 2
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clf; clear; plot([-1 1], [0 0], ’y’) axis equal axis off hold on
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for r = [0.2.5 1 2 5] rr = 1 / (r + 1); cr = 1 - rr; tr = 2 * pi * (0 :.01: 1); plot(cr + rr * cos (tr), rr * sin (tr), ‘y’); end
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RrRr RiRi R r = 1
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for x = [.2.5 1 2 5] rx = 1 / x; cx = rx; tx = 2 * pi * (0 :.01: 1); plot(1 - rx * sin (tx), cx - rx * cos (tx), ‘y’); plot(1 - rx * sin (tx),- cx + rx * cos (tx), ‘y’); end
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RrRr RiRi R r = 1
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for x = [.2.5 1 2 5] rx = 1 / x; cx = rx; tx = 2 * atan(x) * (0 :.01: 1); plot(1 - rx * sin (tx), cx - rx * cos (tx), ‘y’); plot(1 - rx * sin (tx),- cx + rx * cos (tx), ‘y’); end
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RrRr RiRi R r = 1
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RrRr RiRi
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To load To generator
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|R| 1 0
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Movie to illustrate the frequency dependence of the impedance of a series resonant circuit
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Movie to illustrate the frequency dependence of the impedance of a parallel resonant circuit
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Movie to illustrate the transformation of a load impedance at various locations on the transmission line
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Movie to illustrate the transformation of an impedance to an admittance
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y = z = 0
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Z c = 50 Z in = ?
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z = j1 z = 0
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Z c = 50 Z in = ?
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0.434 Z c = 100 Z in = ?
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z L = 2.6 + j1.8 z = 0.7 + j1.2
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0.434 Z c = 100 Z in = ?
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Single stub matching
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y L = 2 + j1 y = 1 - j1 y = 1 + j1
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y L = y = - j1 y = + j1
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