NETWORK ANALYSIS FOR DIGITAL ENGINEERS - Advanced Signal Integrity for High-Speed Digital Design - page 370

is zero are given by

1
2 (v even

1
2 i drive (Z even

v line1 = + v odd ) = + Z odd ) (9-31a)

1 1

v line2 =

2 (v even

− v odd ) =

2 i drive (Z even

− Z odd ) (9-31b)

Since the voltage across the 50- resistor at port 2 is v port2 = 50 i drive , the insertion

loss can be calculated:

v port2
v line1

100
Z even

S 21 = = (9-32a)

+ Z odd

and the peak value of the reverse crosstalk is the ratio of the voltage coupled
onto line 2 and the voltage propagating on line 1:

v line2
v line1

Z even − Z odd

S 31 = = (9-32b)

Z even + Z odd

Example 9-6 Calculate the frequency where the insertion loss will be minimum
and the reverse crosstalk will be maximum for a 10-in. loss-free homogeneous
transmission line where ε r = 4 . 0, with a circuit as shown in Figure 9-15. Assume

that Z even = 100 , Z odd = 25 , and all ports are terminated in 50 .

SOLUTION
Step 1: Calculate the propagation delay for the odd and even modes where
10 in.

10 8 m/s:

= 0 . 254 m and c = 3 ×

0 . 254 √ 4 . 0
3

l √ ε r
c

10 − 9 τ p = = = 1 . 69 × s

× 10 8

Step 2: Use equation (9-3a) to calculate the frequency of the first hump in

S 31 , where τ p = l √ LC :

n = 1 =

1
4 τ p

10 6 Hz

f (real) = 147 ×

Step 3: Calculate the maximum value of the crosstalk:

Z even − Z odd 100 − 25

S 31 = =

25 =

0 . 6

Z even + Z odd 100 +

Step 4: Calculate the minimum value of the insertion loss. Since this is a
loss-free transmission line, the insertion loss ( S 21 ) will be minimum when the
input reflections and crosstalk are maximum. Both the input reflections ( S 11 ) and
the reverse crosstalk ( S 31 ) will peak when the frequency is equal to f ( real ) ,as