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week3 - Impedance on lossless lines For lossless lines = j...

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EE 311 - Lecture 7 Impedance on lossless lines Reflection coefficient Impedance equation Shorted line example Assigned reading: Sec 2.5.1 of Ulaby 1 Impedance on lossless lines For lossless lines, γ = L 0 C 0 = ; we will focus on lossless lines almost always from here on out Thus ˜ V ( z ) = V + 0 e - jβz + V - 0 e jβz (1) ˜ I ( z ) = 1 Z 0 ( V + 0 e - jβz - V - 0 e jβz ) (2) where V + 0 and V - 0 are two phasor constants and Z 0 = p L/C for a lossless line The impedance on a lossless line is then Z ( z ) = ˜ V ( z ) ˜ I ( z ) = Z 0 V + 0 e - jβz + V - 0 e jβz V + 0 e - jβz - V - 0 e jβz which is a function of z ! Thus, the impedance on a transmission line changes depending where you measure it on the line 2 Alternate form Note that we can factor out V + 0 e - jβz on the top and bottom to get: Z ( z ) = Z 0 V + 0 e - jβz V + 0 e - jβz 1 + V - 0 V + 0 e 2 jβz 1 - V - 0 V + 0 e 2 jβz = Z 0 1 + Γ e 2 jβz 1 - Γ e 2 jβz where we have defined Γ = V - 0 V + 0 as a “voltage reflection coefficient” Note that if Γ = 0 (we have no V - 0 wave), Z ( z ) = Z 0 meaning that the impedance is no longer a function of z This is usually the case when we talk about infinitely long lines, where + propagating waves can propagate forever without generating a - traveling wave Finite length lines terminated in an impedance, however, will typically have Γ 6 = 0 as we will see next... 3 Finding gamma Consider a Xmission line of length l terminated in an impedance Z L . A picture is below The two parallel lines in this picture represent a transmission line. It could be any type of line (parallel plate, coax, etc.) but we are told the Z 0 and the μ, ² inside the line which is all we need to know to talk about voltages, currents, and impedances Since we’ve drawn a circuit impedance Z L at z = 0, the impedance on the transmission line at z = 0 must be Z L V g I i Z g Z in Z 0 Z L ~ V i ~ ~ + + - - V L ~ I L ~ + - Transmission line Generator Load z = - l z = 0 4
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Using our previous formula for Z ( z ) and plugging in z = 0 we get Z (0) = Z 0 1 + Γ 1 - Γ = Z L which can be solved to obtain Γ = Z L - Z 0 Z L + Z 0 Note Γ in general is complex; Ulaby writes Γ = | Γ | e r Knowing Γ we can find the impedance anywhere else on the line, for example at z = - l , by using Z ( z ) = Z 0 1 + Γ e 2 jβz 1 - Γ e 2 jβz since all quantities in this equation are now known (note β = ω LC = ω μ² ) Another way of writing this equation is Z ( z ) = Z 0 1 + Γ( z ) 1 - Γ( z ) where Γ( z ) = Γ e 2 jβz 5 Impedance equation without Γ We can also plug in Γ = Z L - Z 0 Z L + Z 0 and simplify some to get Z ( z ) = Z 0 Z L - jZ 0 tan( βz ) Z 0 - jZ L tan( βz ) so that Z ( - l ) = Z 0 Z L + jZ 0 tan( βl ) Z 0 + jZ L tan( βl ) This last formula is probably the most useful of all the equations since it directly gives the impedance at z = - l if we know Z 0 , Z L , β , and l Notice the dependence on l occurs only in the tan( βl ) terms, which are periodic in βl with period π . Since β = 2 π/λ , the impedance is periodic in l with period λ/ 2.
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