CROSSTALK - Advanced Signal Integrity for High-Speed Digital Design - page 155

L 0

i 1 ( z ) i 1 ( z + dz )

+ +

L M C M C g

i 2 ( z ) L 0 i 2 ( z + dz )

v 1 (z) v 1 ( z + dz )

+ +

v 2 ( z )

v 2 ( z + dz )

C g

− −

−

−

dz

Figure 4-8 Differential circuit subsection for two lossless coupled transmission lines.

i 2 (z) − i 2 (z + dz) =− jω [ (C g + C M )v 2 (z) − C M v 1 (z) ] dz (4-30)

d i
dz =− jω Cv

(4-31)

where

i 1 (z) − i 1 (z + dz)
i 2 (z) − i 2 (z + dz)

d i
dz =

1
dz

taken in the limit as dz → 0

C g + C M

− C M

C =

− C M

C g + C M

v 1 (z)
v 2 (z)

v =

Notice in equations (4-29) and (4-30) that the current change on line 1 caused
by v 1 is proportional to the sum of the capacitance to ground C g and the mutual
capacitance between lines C M . This is consistent with our earlier discussion of
mutual capacitance. We can also reassure ourselves that this is correct by con-
sidering the response to a potential stimulus. Let us first assume that a potential,
v , is applied to line 1 (relative to ground), while no potential is applied to line 2.
In this case, we must charge up both the capacitance to ground and the mutual
capacitance between lines, a result that (4-29) predicts. In the second case we
assume that the same potential is applied to both lines. In this situation, lines 1
and 2 remain at the same potential, so that no charge is stored in the electric
field between the lines. Therefore, line 1 need charge up only the capacitance to
ground, a result that is also predicted by (4-29).
Returning to our derivation, we differentiate (4-28) with respect to z to get

d 2 v
dz 2

d i
dz

=− jωL (4-32)

Substituting for d i /dz from (4-31) results in the coupled voltage wave equation

d 2 v
dz 2

= ω 2 LCv

(4-33)