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Unformatted text preview: 1 C:\User\Teaching\505506\S11\Mid'sol.doc PHY 506 Classical Electrodynamics II Spring 2011 Midterm exam with solutions Problem M.1 (to be graded of 200 points). Find the lumped ac circuit equivalent to a TEM transmission line of length l ~ , with small crosssection area A << 2 , as “seen” (measured) from one end, if the line’s conductors are connected (“shortened”) at the other end. Discuss the result’s dependence on the signal frequency. Solution : Neglecting losses in the transmission line, we can describe any “global” variable, e.g., voltage V ( z , t ) or current I ( z , t ), by a sum of two waves: one traveling to the shortened end, and another one reflected from it. Taking the shortened end’s position for z = 0, for the complex amplitudes of these variables we can write (cf. Eq. (7.61) for plane waves): ikz R ikz ikz R ikz e e Z V z I e e V z V W ) ( , ) ( . Now requiring the ratio V / I to vanish at the shorted end (at z = 0), we get R = 1, so that at the other end of the line (at z = – l ) we get kl iZ e e e e Z I V Z W W l z ikl ikl ikl ikl tan ) ( . This result shows that in contrast to the line impedance Z W = ( L / C ) 1/2 , which does not depend on frequency, the effective ac impedance Z ( ) of the shorted line segment is a strong function of . In particular, at low frequencies << v / l (i.e. kl << 1) this expression reduces to L i l L i l C L C L i kl iZ Z W 2 / 1 2 / 1 ) ( , where L = L l is the segment’s inductance. This is a wellknown expression for the complex impedance is the segment’s inductance....
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 Spring '08
 Stephens,P
 Alternating Current, Complex number, Impedance matching, dielectric

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