Chapter2_Notes

# Z tc 16 z t and let z approach zero so we

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Unformatted text preview: econd, ﬁrst-order equation) − • These are transmission line equations a.k.a. telegrapher’s equations in the time domain: ∂v (z, t) ∂ i(z, t) = R i(z, t) + L ∂z ∂t ∂i(z, t) ∂ v (z, t) = G v (z, t) + C − ∂z ∂t − Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (18) (19) Electromagnetics I: Transmission lines 21 • Further simpliﬁcation is possible for sinusoidal sources, v (z, t) = ˜ V (z )ejωt (20) i(z, t) = ˜ I (z )ejωt . (21) • In that case we have a single frequency (time harmonic) and can go to phasor representation: ˜ ∂v (z, t) dV (z ) ∂ i(z, t) ˜ = R i(z, t)+ L →− = (R + jωL )I (z ) ∂z ∂t dz (22) ˜ ∂ v (z, t) dI (z ) ∂i(z, t) ˜ = G v (z, t)+C →− = (G +jωC )V (z ) − ∂z ∂t dz (23) − These are equations 2.18a and 2.18b in the book. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 22 1.4. Wave propagation on a transmission line We have the transmission line equations– how do we solve them? • First combine two ﬁrst-order coupled equations into two uncoupled second order equations. How? Take d/dz to get: ˜ d dV (z ) d ˜ (R + jωL )I (z ) = dz dz dz (24) ˜ ˜ dI (z ) d2 V (z ) = (R + jωL ) 2 dz dz (25) ˜ dI (z ) ˜ = (G + jωC )V (z ) dz (26) − On RHS plug in: − ˜ d2 V (z ) ˜ = (R + jωL )(G + jωC )V (z ) dz 2 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (27) Electromagnetics I: Transmission lines ˜ d2 V (z ) ˜ − (R + jωL )(G + jωC )V (z ) = 0 dz 2 use γ = (R + jωL )(G + jωC ) ˜ d2 V (z ) ˜ ⇒ − γ 2 V (z ) = 0 dz 2 • Same can be done for the current ˜ d2 I (z ) ˜ − γ 2 I (z ) = 0 dz 2 23 (28) (29) (30) (31) • These equations (30 and 31 ) are called wave equations. • Parameter γ is called the complex propagation constant. It consists of the real part α = attenuation constant, and imaginary part β = phase constant α = (γ ) = (R + jωL )(G + jωC ) (Np/m) (32) β = (γ ) = (R + jωL )(G + jωC ) (rad/m) (33) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 24 • Note that we’ve assumed α and β to be positive. This will turn out to be a necessity for a propagating wave with possibly some decay. We still don’t have a solution, but with equations in the form ˜ d2 V (z ) ˜ − γ 2 V (z ) = 0 dz 2 (34) ˜ d2 V (z ) ˜ − γ 2 V (z ) = 0 (35) dz 2 it is easy to see (how?) that the solutions will be exponentials like, ˜ V (z ) = V0 eγz (36) But, we can also have a solution with a negative exponent, ˜ V (z ) = V0 e−γz Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (37) Electromagnetics I: Transmission lines 25 So, we can have a total solution that contains both. We use the superscripts + and − on the amplitude to denote waves going in the positive and negat...
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