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CHAPTER 10   Transmission Lines           315

                     Solution. Because the line is lossless, both R and G are zero. The characteristic
                     impedance is


                                              L      0.25 × 10 −6

                                        Z 0 =    =             = 50
                                              C      100 × 10 −12
                                         √                        √
                     Because γ = α + jβ =  (R + jωL)(G + jωC) = jω LC,we see that
                             √
                                               6
                                                           −6
                        β = ω LC = 2π(600 × 10 ) (0.25 × 10 )(100 × 10 −12 ) = 18.85 rad/m
                     Also,
                                                         6
                                          ω    2π(600 × 10 )       8
                                      ν p =  =              = 2 × 10 m/s
                                          β        18.85

                     10.7 LOW-LOSS PROPAGATION
                     Having obtained the phasor forms of voltage and current in a general transmission
                     line [Eqs. (42a) and (42b)], we can now look more closely at the significance of these
                     results. First we incorporate (41) into (42a)to obtain


                                                 + −αz − jβz
                                                              − αz jβz
                                        V s (z) = V e  e  + V e e                    (48)
                                                0
                                                             0
                     Next, multiplying (48) by e jωt  and taking the real part gives the real instantaneous
                     voltage:
                                          + −αz
                                                              − αz
                                V(z, t) = V e  cos(ωt − βz) + V e cos(ωt + βz)       (49)
                                                              0
                                          0
                     In this exercise, we have assigned V 0 +  and V 0 −  to be real. Eq. (49) is recognized
                     as describing forward- and backward-propagating waves that diminish in amplitude
                     with distance according to e −αz  for the forward wave, and e αz  for the backward wave.
                     Both waves are said to attenuate with propagation distance at a rate determined by
                     the attenuation coefficient, α,expressed in units of nepers/m [Np/m]. 2
                         The phase constant, β, found by taking the imaginary part of (41), is likely to be a
                     somewhat complicated function, and will in general depend on R and G.Nevertheless,
                     β is still defined as the ratio ω/ν p , and the wavelength is still defined as the distance
                     that provides a phase shift of 2π rad, so that λ = 2π/β.By inspecting (41), we
                     observe that losses in propagation are avoided (or α = 0) only when R = G = 0. In
                                                                  √
                                                 √
                     that case, (41) gives γ = jβ = jω LC, and so ν p = 1/ LC,aswe found before.
                         Expressions for α and β when losses are small can be readily obtained from (41).
                     In the low-loss approximation, we require R   ωL and G   ωC,a condition that



                     2  The term neper was selected (by some poor speller) to honor John Napier, a Scottish mathematician
                     who first proposed the use of logarithms.
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