Page 332 - Engineering Electromagnetics, 8th Edition

314 ENGINEERING ELECTROMAGNETICS

The wave equation for current will be identical in form to (40). We therefore expect
the phasor current to be in the form:

+ −γ z
− γ z
I s (z) = I e + I e (42b)
0
0
The relation between the current and voltage waves is now found, as before,
through the telegraphist’s equations, (5) and (8). In a manner consistent with Eq. (37b),
we write the sinusoidal current as
1 1
e
jξ
I(z, t) =|I 0 | cos(ωt ± βz + ξ) = (|I 0 |e ) e ± jβz jωt + c.c. = I s (z)e jωt + c.c.
2
2
I 0
(43)
Substituting the far right-hand sides of (37b) and (43) into (5) and (8) transforms the
latter equations as follows:

∂V =− RI + L ∂I dV s (44a)
∂z ∂t ⇒ dz =−(R + jωL)I s =−ZI s
and

∂I =− GV + C ∂V dI s (44b)
∂z ∂t ⇒ dz =−(G + jωC)V s =−YV s

We can now substitute (42a) and (42b) into either (44a)or (44b) [we will use (44a)]
to find:
− γ z
+ −γ z
− γ z
+ −γ z
−γ V e + γ V e =−Z(I e + I e ) (45)
0
0
0
0
γ z
Next, equating coefficients of e −γ z and e ,we find the general expression for the
line characteristic impedance:
V + V − Z Z Z
Z 0 = 0 =− 0 = = √ = (46)
I 0 + I 0 − γ ZY Y
Incorporating the expressions for Z and Y,we find the characteristic impedance in
terms of our known line parameters:

R + jωL jθ
Z 0 = =|Z 0 |e (47)
G + jωC
Note that with the voltage and current as given in (37b) and (43), we would identify
the phase of the characteristic impedance, θ = φ − ξ.

EXAMPLE 10.2
A lossless transmission line is 80 cm long and operates at a frequency of 600 MHz.
The line parameters are L = 0.25 µH/m and C = 100 pF/m. Find the characteristic
impedance, the phase constant, and the phase velocity.