Page 325 - Engineering Electromagnetics, 8th Edition

CHAPTER 10 Transmission Lines 307

Under this condition, only the first term on the right-hand side of either Eq. (11) or
Eq. (12) survives. Eq. (11), for example, becomes

2
2
∂ V = LC ∂ V (13)
∂z 2 ∂t 2
In considering the voltage function that will satisfy (13), it is most expedient to
simply state the solution, and then show that it is correct. The solution of (13) is of
the form:
z z
+
V (z, t) = f 1 t − + f 2 t + = V + V − (14)
ν ν
where ν, the wave velocity, is a constant. The expressions (t ± z/ν) are the arguments
of functions f 1 and f 2 . The identities of the functions themselves are not critical to
the solution of (13). Therefore, f 1 and f 2 can be any function.
The arguments of f 1 and f 2 indicate, respectively, travel of the functions in the
forward and backward z directions. We assign the symbols V + and V − to identify
the forward and backward voltage wave components. To understand the behavior,
consider for example the value of f 1 (whatever this might be) at the zero value of its
argument, occurring when z = t = 0. Now, as time increases to positive values (as
it must), and if we are to keep track of f 1 (0), then the value of z must also increase
to keep the argument (t − z/ν) equal to zero. The function f 1 therefore moves (or
propagates) in the positive z direction. Using similar reasoning, the function f 2 will
propagate in the negative z direction, as z in the argument (t + z/ν) must decrease to
offset the increase in t. Therefore we associate the argument (t − z/ν) with forward
z propagation, and the argument (t + z/ν) with backward z travel. This behavior
occurs irrespective of what f 1 and f 2 are. As is evident in the argument forms, the
propagation velocity is ν in both cases.
We next verify that functions having the argument forms expressed in (14) are
solutions to (13). First, we take partial derivatives of f 1 , for example with respect to
z and t. Using the chain rule, the z partial derivative is

∂ f 1 ∂ f 1 ∂(t − z/ν) 1
= =− f 1 (15)
∂z ∂(t − z/ν) ∂z ν
where it is apparent that the primed function, f , denotes the derivative of f 1 with

1
respect to its argument. The partial derivative with respect to time is
∂ f 1 ∂ f 1 ∂(t − z/ν) = f (16)
∂t = ∂(t − z/ν) ∂t 1

Next, the second partial derivatives with respect to z and t can be taken using similar
reasoning:

2 2
∂ f 1 1 ∂ f 1
= f 1 and = f 1 (17)
∂z 2 ν 2 ∂t 2