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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.