Chapter2_Notes

# On fundamentals of applied electromagnetics ulaby et

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Unformatted text preview: ive z directions. The solution then looks like: ˜ V (z ) = V0+ e−γz + V0− eγz (38) + − ˜ I (z ) = I0 e−γz + I0 eγz (39) • Each term represents a wave, so that we have two waves, but going in opposite directions! • Convention has it that e−γz is a wave in +z direction • Still no solution; how many unknowns? Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 26 • We can take the derivative of, ˜ V (z ) = V0+ e−γz + V0− eγz ˜ dV (z ) = −γV0+ e−γz + γV0− eγz dz ˜ dV (z ) = γ V0+ e−γz − V0− eγz − dz (40) (41) (42) and plug into, − ˜ dV (z ) ˜ = (R + jωL )I (z ) dz ˜ γ V0+ e−γz − V0− eγz = (R + jωL )I (z ) γ ˜ V0+ e−γz − V0− eγz I (z ) = R + jωL Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (43) (44) (45) Electromagnetics I: Transmission lines 27 • Does this look familiar? Compare each of the terms of, γ V0+ e−γz − V0− eγz R + jωL (46) + − ˜ I (z ) = I0 e−γz + I0 eγz ˜ I (z ) = (47) with, And we can see that, + I0 = γV0+ R + jωL (48) or, V0+ R + jωL = Z0 . += γ I0 (49) −V0− R + jωL = Z0 . −= γ I0 (50) Similarly, Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 28 That is, V0+ −V0− = Z0 = − + I0 I0 (51) • The parameter Z0 is the characteristic impedance of the transmission line, and is given by (recalling the deﬁnition of γ ), γ= Z0 = (R + jωL )(G + jωC ) R + jωL = γ R + jωL (Ω) G + jωC (52) (53) • An important point is that Z0 is the ratio of the voltage and current amplitudes for each of the propagating waves separately. ˜ The total voltage V (z ) is the sum of the two waves traveling in opposite directions. Therefore, Z0 is not the ratio of the total voltage and current. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 29 • Note that V0+ and V0− are complex numbers, so each has a + − magnitude and phase, i.e. V0+ = |V0+ |ejφ , V0− = |V0− |ejφ • With the boundary conditions we will be able to solve for the voltage along the transmission line in the phasor domain. Remember the process for returning back to the time domain. – multiply phasor solution by ejωt – take the real part • The time domain solution will have the form v (z, t) = V0+ e−γz + V0− eγz ejωt (54) − + |V0+ |ejφ e−(α+jβ )z ejωt + |V0− |ejφ e(α+jβ )z ejωt (55) Rearrange a little, |V0+ |e−αz ej (ωt−βz+φ + ) − + |V0− |eαz ej (ωt+βz+φ ) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (56) Electromagnetics I: Transmission lines 30 v (z, t) = |V0+ |e−αz cos(ωt − βz + φ+ )+ |V0− |eαz cos(ωt + βz + φ− ) (57) • From before, the ﬁrst term is +z traveling wave (why?), while the 2nd one i...
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