IDEAL TRANSMISSION-LINE FUNDAMENTALS - Advanced Signal Integrity for High-Speed Digital Design - page 94

The solution to this partial differential equation can be found in terms of two
ordinary differential equations, assuming that the potential can be represented by
the product of a function for each coordinate [Jackson, 1999]:

(x,y) = X(x)Y(y) (3-49)

If equation (3-49) is substituted back into (3-48) and then divided by , the
result is

d 2 XY
dx 2

d 2 XY
dy 2

d 2 X
dx 2

d 2 Y
dy 2

1
XY

1
XY

1
X

1
Y

+ = + = 0 (3-50)

where the partial derivatives are replaced by total derivatives because each term
involves only one variable. This allows us to write two separate ordinary differ-
ential equations:

d 2 X
dx 2

1
X

β 2 =− (3-51)

d 2 Y
dy 2

1
Y

= β 2 (3-52)

The constant β must satisfy the conditions β 2 + ( − β 2 ) =

0 to satisfy Laplace's
equation. The differential equations in (3-51) and (3-52) are solved by finding
the roots of the characteristic equations:

X(x) = C 1 n e jβx + C 2 n e − jβx = C 1 n ( cos βx + j sin βx)

+ C 2 n ( cos βx − j sin βx) (3-53)

Y(y) = C 1 n e βy + C 2 n e − βy = C 1 n ( cosh βy + sinh βy)

+ C 2 n ( cosh βy −

sinh βy) (3-54)

The potential is then calculated by substituting (3-53) and (3-54) into (3-49):

(x,y) = X(x)Y(y) = [ C 1 n ( cos βx + j sin βx) + C 2 n ( cos βx − j sin βx) ]

[ C 1 n ( cosh βy + sinh βy) + C 2 n ( cosh βy −

· sinh βy) ] (3-55)

Note that since the solution is periodic, β must be an integer. The boundary
conditions that must be applied to solve (3-55) are

d
2

(x, y) = 0 when x =± (3-56)

which is the potential at the sidewalls and

(x, y) = 0 when y = 0 , ∞ (3-57)