Crosstalk
Tzong-Lin Wu, Ph.D.
Department of Electrical Engineering
National Taiwan University
1
Topics:
Introduction
Transmission lines equations for coupled lines




symmetrical lines and symmetrically driven
homogeneous medium
symmetrical lines in homogeneous medium
symmetrical lines and asymmetrical driven
Qualitative description of crosstalk

Terminations
Measurement of crosstalk parameters
How crosstalk noise can be reduced
2
Introduction
Three characteristics:
1. Polarity of noise at near end
is opposite to that at the far end.
2. Noise duration at near end is
larger than that at the far end.
3. Noise magnitude at near end
is smaller than that at the far
end.
3
Introduction
: Physical Geometry of Coupled Lines
Two kind of coupling:
1. Inductive coupling
2. Capacitive coupling
4
Introduction: cross-talk mechanism of the inductive coupling
Qualitative understanding of the coupling mechanism
• How about the cross-talk mechanism of the capacitive coupling ??
5
Driving signal
A
Cm Lm
Lm
B
Cm Lm
Cm Lm
C
D
Reverse
coupling
Tp
Inductive
A
Derivative of input
signal (negative)
B
Tr
D
C
2Tp
Forward
coupling
Tp
Capacitive
A
B
Tr
Total coupled areas D
are the same
Rise and fall times
same as input
C
2Tp
6
Coupled lines equations


I 1 ( z )  ( Lm z) I 2 ( z )
t
t

' 
I 1 ( z)  I 1 ( z  z)  C1 V1 ( z  z )  Cm [V1 ( z  z)  V2 ( z  z)]
t
t
V1 ( z)  V1 ( z  z)  ( L1 z)
7
Coupled lines equations
• Matrix forms
where
dV ( z)

  L I ( z)
dz
t
dI ( z)

 C V ( z)
dz
t
l
L
LM
Nl
c
L
C M
Nc
11
21
11
12
OP LM L OP
Q N LQ
c O LC  C
M
P
c Q N C
l12
L1

l22
Lm
21
22
m
2
'
1
m
m
OP L
M
Q N
C1
Cm

'
Cm
C2  Cm
Cm
C2
OP
Q
L 1, 2 : self-inductance per unit length of line 1 and 2 when isolated
Lm : mutual inductance between line 1 and 2
C’ 1,2 : capacitance (to GND) of line 1 and 2 when isolated **
Cm : mutual capacitance between line 1 and 2
8
Coupled lines equations
 C C12 
Capacitance matrix=  11

C21 C22 
C11  C1g  C12
 L11
inductance matrix= 
 L21
L12 
L22 
For multi-conductor coupled lines
 C11 C12
C
C22
Capacitance matrix=  21


CN 1
C1N 




CNN 
 L11
L
Inductance matrix=  21


 LN 1
L12
L22
L1N 




LNN 
9
Coupled lines equations
V1 ( z )  V10 e j t  z
V2 ( z )  V20 e
j t  z
I1 ( z )  I10 e j t  z
 V0 ( z )   j LI 0 ( z )
 I 0 ( z )   j CV0 ( z )
2
( A  2 )V0  0

I 2 ( z )  I 20 e j t  z
L1C1  LmCm
A  LC  [
LmC1  L2Cm
2
det( A  2 )  0

LmC2  L1Cm
]
L2C2  LmCm
for nontrivial solutions
The γcan be solved with two possible solutions
10
Symmetrical lines and symmetrical driven
Conditions:
• L1 = L2 (balanced lines)
C1 = C2
• The balanced lines are driven by
common mode or differential mode only
Characteristics:
• Propagation delay per-unit length:
   ( L1  Lm )(C1  Cm )
• Characteristic impedance
Z 0e 
L1  Lm
1  km
 Z1
C1  Cm
1  kc
Z 0o 
L1  Lm
1  km
 Z1
C1  Cm
1  kc
where
L
k m  m : magnetic coupling coefficient
L1
C
k c  m : capacitive coupling coefficient
C1
11
Symmetrical lines and symmetrical driven
Z 0e ,  
Z 0o ,  
12
Symmetrical lines and symmetrical driven (Microstrip example)
13
Symmetrical lines and symmetrical driven (another definition for even
and odd mode)
Coupling Factor = 20log(Z12)
Z12 = (Ze – Zo)/(Ze + Zo)
Note:
14
Physical concept for the even-mode and odd mode impedance
Odd mode I1 = -I2 and V1 = -V2
Lodd  L11  Lm  L11  L12
Z odd 
Lodd

Codd
TDodd  LoddCodd 
Codd  C1g  2Cm  C11  Cm
L11  L12
C11  C12
 L11  L12 C11  C12 
15
Physical concept for the even-mode and odd mode impedance
Even mode I1 = I2 and V1 = V2
Ceven  C0  C11  Cm
Leven  L11  Lm
Z even =
Leven
L L
 11 12
Ceven
C11  C12
TDodd  Leven Ceven 
 L11  L12  C11  C12 
16
Impedance trends for the even- and odd-modes
What impedance behaviors do you see ?
17
Impedance trends for the even- and odd-modes
1.
2.
3.
High-impedance traces exhibit significantly more impedance
variation than do the lower-impedance trace because the reference
planes are much farther in relation to the signal spacing.
At smaller spacing, the single-line impedance is lower than the
target. This is because the adjacent traces increase the selfcapacitance of the trace and effectively lower its impedance even
when they are not active.
Higher coupling the traces behave, the larger variation even- and
odd- impedances will have.
18
Mutual coupling trends for the microstrip and stripline
Mutual parasitic fall off exponentially with trace-trace spacing
19
Example by simulation:
Please compare the coupling characteristics
for the three cases:
•
•
Ze
Zo
20
21
Effect of switching patterns on transmission line
Crosstalk induced flight time and signal integrity variations
What phenomena do you see ??
Switching pattern
22
Effect of switching patterns on transmission line
As the switching patterns are in even-mode:


Over-driven behavior is seen (Overshooting).
Longer propagation delay is seen.
As the switching patters are in odd-mode:


Under-driven behavior is seen (Undershooting).
Shorter propagation delay is seen.
Why ??
23
Effect of switching patterns on transmission line
• Even-mode pattern
impedance changed from Z0 to Ze > Z0 .
overdriven when Ze > Zs.
Slower propagation velocity.
• Odd-mode pattern
impedance changed from Z0 to Zodd < Z0 .
under driven when Zodd < Zs.
Faster propagation velocity.
24
Effect of switching patterns on transmission line
Simulating traces in a multiconductor system using a single-line
equivalent model (SLEM)
Target trace is line 2.
Please find the equivalent impedance and time delay
Z 2,eff 
TD2,eff 
L22  L12  L23
C22  C12  C23
 L22  L12  L23 C22  C12  C23 
??
25
Effect of switching patterns on transmission line
TD2,eff 
 L22  L12  L23 C22  C12  C23 
Z 2,eff 
L22  L12  L23
C22  C12  C23
26
Example by TDR measurement:
27
Example by TDR measurement:
question one ??
28
Example by TDR measurement:
question one??
Differential impedance (D.I.) is 150Ω without GND plane
And about 100Ωwith GND plane. Why ??
29
First half: two traces refer each other like a twisted line, L↑ and C↓
Second half: two traces refer the GND plane, mutual coupling is
quite small (km ≒ kc ≒ 0), so D.I.=2*Zodd ≒ 2Z0=100
30
Example by TDR measurement:
question two??
What’s the variation of the differential impedance
For the differential pair crossing the slot ?
31
Example by TDR measurement:
question two??
The gap behaves like a high impedance discontinuity which
Will cause reflections.
32
Homogeneous Medium
Conditions:
• Real (not Quasi-) TEM mode propagating
on the transmission lines
Characteristics:
• Propagation delay per-unit length:
    0 1 k 2
where k  km  kc (Coupling coefficient)
km 
kc 
Lm
L1 L2
Cm
C1C2
: For mutual inductance coupling
: For mutual capacitance coupling
33
Symmetrical Lines on Homogeneous Medium
Conditions:
• L1 = L2 (balanced lines)
C1 = C2
• The balanced lines are driven by
common mode or differential mode only
•
 
Characteristics:
• Characteristic impedance
Z 0e Z 0o  Z12
where Z1 
L1
C1
34
Example: homogeneous coupled line
Please compare the coupling characteristics
for the three cases:
1. K
2. Ze
3. Zo
35
36
A Quiz:
Please compare their Ze, Zo, Ve, Vo, and C.
(b)
(a)






37
Symmetrical Lines and asymmetrically driven
Asymmetrical driven
P-Terminations:
V  0.5(V1  V2 )
V  0.5(V1  V2 )
R1  R2  Z 0e (terminate even mode)
2 Z 0e Z 0o
R3 
( terminate odd mode)
Z 0e  Z 0o
1
( Z 0o  ( R1 / / R3 ))
2
38
Symmetrical Lines and asymmetrically driven
T - terminations
R1  R2  Z 0o (terminate odd mode)
1
R3  ( Z 0e  Z 0o ) (terminate even mode)
2
( R 2  2 R3  Z 0e )
39
Imperfect Termination for Differential Transmission Line
1.
2.
What are their values of the resistance?
What are their advantages and disadvantages ?
Bridge Termination
AC Termination
Single ended termination
40
Imperfect Termination for Differential Transmission Line
2Zo
Bridge Termination
Zo
Zo
Zo
Zo
AC Termination
Single ended termination
41
Imperfect Termination for Differential Transmission Line
Bridge:
1. Odd mode is completely absorbed
2. Even mode is completely reflected
3. Only one resistance is used.
Single-ended
1. Odd mode is completely absorbed
2. Even mode has the reflection coefficient T= (Zo–Ze)/(Zo+Ze)
3. The extra damping of the even mode costs an additional resistance.
AC
•
Odd mode is completely absorbed
•
Even mode is completely reflected at low frequency, but has the
reflection T = (Zo–Ze)/(Zo+Ze) at high frequency.
3. Compared to bridge type, even mode is attenuated without increasing
static power dissipation.
4. Needing three components.
42
Example: Termination for Differential Transmission Line
43
Example: Termination for Differential Transmission Line
A simple bridge termination
44
Example: Termination for Differential Transmission Line
A simple bridge termination, but
Unbalance driving
Differential signal is still good
Common signal on each line is oscillating
45
Example: Termination for Differential Transmission Line
A Pi-termination for
unbalance driving
Common-mode is damped
46
Example: Termination for Differential Transmission Line
Common-mode resonance
Differential pair excited by common-mode signal
47
Example: Termination for Differential Transmission Line
Single-ended termination for
Unbalanced driving
Better than bridge termination
48
Example: Termination for Differential Transmission Line
AC T-termination for
Unbalanced driving
49
Example: Termination for Differential Transmission Line
Bridge termination at the far end and
Series termination at the near end
1.
2.
Power saving
Idea of LVDS
50
Example: Termination for Differential Transmission Line
Bridge termination at the far end and
Series termination at the near end
For LVDS case
51
Example: Effects of Termination Networks on Signal-Induced EMI from
the Shields of Fibre Channel Cables Operating in the Gb/s Regime
2000 IEEE EMC Symposium
52
Example: Effects of Termination Networks on Signal-Induced EMI from
the Shields of Fibre Channel Cables Operating in the Gb/s Regime
Measurement of S11 for both differential mode and common mode
Termination effect is reduced for
high frequency range
53
Example: Effects of Termination Networks on Signal-Induced EMI from
the Shields of Fibre Channel Cables Operating in the Gb/s Regime
EMI
differential
Why the termination effect is degraded
at high frequency range ?
54
Quantitative description of crosstalk
Capacitive coupling
V f ( x, t )  K f x
d
Vin (t  x)
dt
Inductive Coupling
(see appendix 7.1)
Vb ( x, t )  Kb [Vin (t  x)  Vin (t  x  2Td )]
If the coupled lines are symmetrical
1 L
K f   ( m  Cm Z1 )
2 Z1
1 Lm
Kb 
(
 Cm Z1 )
4 Z1
(ns / cm) : Forward coupling coefficient
(dimensionless) : Backward coupling coefficient
55
Quantitative description of crosstalk
In the loosely coupled system (km, kc << 1)
Z1 
L1
and  
C1
L1C1 is accurate enough
C Z
 L

 K f   ( m  m 1 )   (km  kc )
2 Z1

2
1 Lm
1
Kb 
(
 Cm Z1 )  ( k m  k c )
4 Z1
4
1.
2.
It is clear that the backward coupling can be
reduced by decreasing the mutual coupling
(Lm and Cm) or by increasing the coupling to
ground(L1, L2, C1, and C2).
The forward crosstalk is zero for two identical
lines in homogeneous medium. (not easy in
practical circuits)
56
Crosstalk for a ramp step driver
Vin (t )  Vm [ f (t )  f (t  Tr )]
57
Crosstalk for a ramp step driver
Forward crosstalk at the far end:
d
Vin (t  Td )
dt
V
= K f l m , Td  t  Td  Tr
Tr
= 0 elsewhere
V f (l , t )  K f l
Backward crosstalk at the near end
• If Tr < 2Td
Vb (0, t )  Kb [Vin (t )  Vin (t  2Td )]
Vb ,max  Kb
• If Tr > 2Td
Vb ,max  2(
l
Tr
) KbVm 
Kb
Tr
58
SPICE simulation of crosstalk
59
Measurement of coupling coefficient (Frequency domain)
: (example) RAMBUS RIMM Connector
•Test module design
60
Measurement of coupling coefficient (Frequency domain)
: RAMBUS RIMM Connector
• Calibration (Open, short, load )
61
Measurement of coupling coefficient (Frequency domain)
: (example) RAMBUS RIMM Connector
• Theory (mutual capacitance)
Open circuit !!
Low freq. assumption
The slope of S21(f) at
low frequency is
taking to compute the
Cm
62
Measurement of coupling coefficient (Frequency domain)
: (example) RAMBUS RIMM Connector
• Theory (mutual inductance)
Short circuit !!
The slope of S21(f) at
low frequency is
taking to compute the
Lm
63
Measurement of coupling coefficient (Frequency domain)
: (example) RAMBUS RIMM Connector
• Measurement setup
64
Differential Impedance Measurement by TDR/TDT
They have good EMS (Immunity) and EMI (Interference)
Performance.
65
Differential Impedance Measurement by TDR/TDT
Equivalent circuits of the differential interconnects: (Type I)
66
Differential Impedance Measurement by TDR/TDT
Equivalent circuits of the differential interconnects: (Type II)
Negative impedance
(not practical)
Advantage is it is a rigorous equivalent circuit.
Disadvantage is only valid in three conductor interconnects.
67
Differential Impedance Measurement by TDR/TDT
Equivalent circuits of the differential interconnects: (Type III)
68
Differential Impedance Measurement by TDR/TDT
Measurement Principle
69
Differential Impedance Measurement by TDR/TDT
Comparison of the equivalent model by the SPICE
70
A Final Question:
Can you explain why
71
How to reduce crosstalk noise
• Space out signal routing
• Route-void channels adjacent to critical nets
• Increase coupling to ground
• Run orthogonal rather than parallel
• Provide the homogeneous medium
• Using the slowest risetime and falling time
72
§10.1 Three conductor lines and crosstalk
Time Domain Crosstalk VS Frequency-domain Crosstalk
a. Two conductors transmission lines has no crosstalk problem. In order
to have crosstalk, we need to have three or more conductors.
b.
Generator
RS
Vs (t ) 

RNE
VNE
_
Receptor
I G ( z, t )


I R ( z, t )

RL
RFE VFE
_

z0
zL
z
73
c. The generator circuit consists of the generator conductor and reference
conductor and has current IG(z,t) along the conductors and voltage VG(z,t)
between them. The IG and VG will generate EM fields and interact with the
receptor circuit.
d. The objective in a crosstalk analysis is to determine the near-end voltage
VNE(t) and far-end voltage VFE(t) given the cross-sectional dimension and
the termination characteristics VS(t), RS, RL, RNE, RFE.
74
e. Two type of crosstalk analysis
(1) Time-domain crosstalk :
predict VFE (t ) and VNE (t ) given VS (t ).
(2) Frequency-domain crosstalk :
predict VˆFE ( jw) and VˆNE ( jw) given VS (t )  VS cos( wt   ).
f. (1) For simplicity, we will assume that the medium surrounding the
conductors for these configurations is homogeneous.
(2) Closed-form expressions for the per-unit-length parameters for
the lines in inhomogeneous medium, such as microstrip lines, are
difficult to determine.
75
Transmission Line Equations
a. Assume only TEM mode present on the materials, so line voltages
VG(z,t) and VR(z,t), as well as line currents IG(z,t) and IR(z,t) can be
uniquely defined.
b. Equivalent circuit of TEM-mode on 3-conductor material :
RG z
LG z
I G ( z, t )
I R ( z, t )
RR z
RO z
LR z
I G ( z  z )
Gm z
Cm z

GG z CG z VG ( z  z )
_
GR z
CR z

I R ( z  z )
VR ( z  z )
_
I G ( z, t )  I R ( z, t )
z
76

VG ( z , t )
I ( z , t )
I ( z , t )
 RG  RO I G ( z , t )  RO I R ( z , t )  LG G
 Lm R
z
t
t
I ( z , t )
VR ( z , t )
I ( z , t )
  RO I G ( z , t )  RR  RO I R ( z , t )  Lm G
 LR R
z
t
t
I G ( z , t )
V ( z , t )
V ( z , t )
 GG  Gm VG ( z , t )  GmVR ( z , t )  CG  Cm  G
 Cm R
z
t
t
V ( z , t )
I R ( z , t )
V ( z , t )
 GmVG ( z , t )  GR  Gm VR ( z , t )  Cm G
 C R  Cm  R
z
t
t
In matrix terms :
V ( z , t )
I ( z , t )

 RI( z , t )  L
z
t
I ( z , t )
V ( z , t )

 GV ( z , t )  C
z
t
77
where

RO 
 RG  RO
VG ( z, t )
 I G ( z , t )
V( z, t )  
, I( z, t )  
,R  



R
R

R
V
(
z
,
t
)
I
(
z
,
t
)
 R

 R

 O
R
O
Gm 
 Cm 
 LG Lm 
GG  Gm
CG  Cm
L
,
G

,
C


 G
 C
GR  Gm 
CR  Cm 
 Lm LR 
 m
 m
c. Solutions
(1)In time domain, it is a difficult problem.
When R=G=0 (lossless lines), there are exact solutions in SPICE.
(2)In frequency domain, the exact solutions for above material equations
are possible, we will show the exact solution for lossless case.
78
Note: In frequency domain, the equations =>
ˆ ( z)
 dV
ˆ Iˆ ( z )
 Z

dz

 dIˆ ( z )   Y
ˆV
ˆ ( z)
 dz
ˆ  R  jwL
Z
where 
ˆ  G  jw C
Y
ˆ ( z )e jwt
V ( z , t )  Re V

I ( z , t )  Re Iˆ ( z )e jwt




79
§ 10.3.1 The per-unit-length parameters
Internal parameters : RG, RR, RO and internal inductance are not dependent on
the line configurations.
a. For material in homogeneous medium (μ,ε)
1
(1)
LC  CL   1  C   L1  2 L1
v
Example : Three-conductor line
 Cm 
 Lm 
CG  Cm
 LR
1

 C
CR  Cm  v 2 ( LG LR  Lm 2 )   Lm LG 
 m

Lm
Cm  2
v ( LG LR  Lm 2 )


LR
 CG  Cm  2
2
v
(
L
L

L
)
G
R
m


LG
CR  Cm  2
v ( LG LR  Lm 2 )

(2)
LG  GL   1  G   L1
80
§ Wide-Separation Approximation for Wires
a. (1) Magnetic flux of a current-carrying wire through a surface
(2) voltage between two points for a charge-carrying wire
b
R1
R2
μ0 I  R2 

ln  
2  R1 
Vba
R1
q
R2

a
q
R2
Vba 
ln
20 R1
81
b. Parameters of three-conductor transmission lines
(1) Inductance :
G   LG Lm   I G 
  LI     
 I 
L
L

 R   m R  R 
 LG 
Lm 
G
IG
, LR 
I R o
G
IR

IG 0
R
IR
IG 0
R
IG
I R 0
82
1.
generator
dGR
receptor
rWG
rWR
dR
GND
rWO
2. Self inductance
IG
 0  d G   0  d G   0  d G 2 
 
 
LG 
ln 
ln 
ln 
2  rWG  2  rWO  2  rWG rWO 
rWG
G
rWO
 0  d R 2 
LR 
ln 
2  rWRrWO 

 IG
83
3. Mutual inductance
IG
dGR
rWG
rWR
dR
rWO 
R
Lm 

0  dG  0  d R 
 

ln 
ln 
2  d GR  2  rWO 
0  dG d R 

ln 
2  d GR rWO 
 IG
(2) Capacitance
 Cm  VG 
CG
q  CV  
 V 

C
C

C
 m R
m  R 
 PG pm  qG 
V  C q  pq  
 q 
p
p
 m R  R 
1
84
VG
 PG 
qG
qR 0
VR
, PR 
qR
qR 0
VR
, Pm 
qG
qR 0
VG

qR
qG  0
1. Self capacitance
qG
rWG
 dG 
 dG 
1
 

PG 
ln 
ln 
20  rWG  20  rWO 
 dG 2 
1


ln 
20  rWG rWO 
1

V
_
rWO
 qG
85
2. Mutual capacitance
qG
rWG
dGR

rWR
VR
_
dR
rWO
 dG 
 dR 
1
 

Pm 
ln 
ln 
20  d GR  20  rWO 
 dG d R 
1


ln 
20  d GR rWO 
1
reference
 qG
86
rWR
c. Example

LG  G
IG I 0
IG
S
rWG
S2
R
 0  hG   0  2hG 
 


ln 
ln 
2  rWG  2  hG 
 0  2hG 


ln 
2  rWG 
Lm 
S1
R
hG
hG
S3
hR
R
IG
I R 0
 R hR

 IG
 0  S1   0  S 2 

ln   
ln   ; ( S1  S3 )
2  S  2  S3 
 S   
h h 
 0 ln  2   0 ln 1  4 G 2 R 
2  S  4 
S 
87
§10.1.4 Frequency-Domain (Steady State) Crosstalk
§ General Solution :
a.
ˆ
dV ( z )
ˆ Iˆ ( z )
 Z
dz
b.
VˆG ( z )
 IˆG ( z )
ˆ ( z)  
V
 , Iˆ ( z )  

ˆ
ˆ
VR ( z ) 
 I R ( z ) 
dIˆ ( z )
ˆV
ˆ ( z)
 Y
dz
ˆ  R  jwL , Y
ˆ  G  jwC
where Z
ˆ ( z)
 d 2V
 ZYV ( z )

2
 dz

2ˆ
d
 I ( z )  YZI ( z )
 dz 2
Note : in general, ZY  YZ
88
(1) if we can find a transformation matrix Tˆ such that :
2

ˆ
r
0 
G

1
2
ˆ Y
ˆZ
ˆT
ˆ  rˆ  
T
  diagonalized
2
0 rˆR 
ˆ is the eigenvectors of Y
ˆZ
ˆ
where T
ˆZ
ˆ
rˆ is the eigenvalues of Y
2
(2) d Tˆ 1Iˆ ( z )  Tˆ 1Yˆ Zˆ Iˆ ( z )
dz 2
 IˆmG ( z )
ˆ I ( z )  Iˆ m ( z )  modal current  
let T

ˆ
 I mR ( z ) 
1 ˆ
d2 ˆ
ˆ 1Y
ˆZ
ˆT
ˆ Iˆ m ( z )
 2 I m ( z)  T
dz
rˆG 2 0   IˆmG ( z )



2
ˆ
0 rˆR   I mR ( z ) 
 rˆG z ˆ
rˆG z ˆ


 Iˆ  ( z )  rˆ 2 Iˆ ( z )
ˆ

I
(
z
)

e
I
(
z
)

e
I
(
z
)
 mG
mG
G mG
mG
mG

 

 rˆR z ˆ
rˆR z ˆ

2
ˆ
ˆ
ˆ

I
(
z
)

e
I
(
z
)

e
I mR ( z ) 89
 I mR ( z )  rˆR I mR ( z )
 mR
mR
(3)
Iˆ m ( z )  e  rˆz Iˆ m   e rˆz Iˆ m  in matrix form.
where e  rˆz
c.
ˆmG  

e  rˆG z 0 
I
ˆ m  

,
I


 rˆG z

e
 IˆmR 
0

ˆ Iˆ m ( z )  T
ˆ (e  rz Iˆ m   e rz Iˆ m  )
Iˆ ( z )  T
1 d ˆ
ˆ
ˆ
ˆ 1T
ˆ rˆ(e  rˆz Iˆ m   e rˆz Iˆ m  )
V ( z )  Y
I( z )  Y
dz
ˆ 1Y
ˆZ
ˆT
ˆ  rˆ 2  Z
ˆT
ˆ rˆ 1  Y
ˆ 1T
ˆ rˆ
by T


ˆ ( z)  Z
ˆT
ˆ rˆ 1T
ˆ 1 T
ˆ (e  rˆz Iˆ m   e rˆz Iˆ m  )
V
d.


ˆc Z
ˆ Tr
ˆ ˆ 1T
ˆ 1  Z(
ˆ YZ
ˆ ˆ )1
Z
ˆ 1 YZ
ˆ ˆ see below for detail
=Y
90

1 ˆ 1
ˆˆ T
上式的 Tr

ˆˆ
YZ

1
ˆ 1YZT
ˆ ˆ ˆ  rˆ 2  rˆ 1Tˆ 1YZTr
ˆ ˆ ˆ ˆ 1 =1
T

 

ˆ ˆ 1T
ˆ 1 YZ
ˆ ˆ Tr
ˆ ˆ 1Tˆ 1 =1
 Tr

ˆ ˆ  Tr
ˆ ˆ 1T
ˆ 1
 YZ
1 ˆ 1
ˆˆ T =
 Tr


ˆˆ
YZ
-2

1
similar to the scalar results
Zˆ 
zˆ ˆ ˆ ˆ 1
 Z ( YZ )
yˆ
e. Solving the four unknowns Im+ and Im- by the
termination condition
(1) V (0)  VS  Z S I(0)

 V ( L)  Z L I ( L)
 RS 0 
 RL 0 
VS 
where VS    , Z S  
, ZL  


0
R
0
R
0
 


NE 
FE 
91
(2) at z  0
ˆ (0)  Zˆ C Tˆ (Iˆ m   Iˆ m - )
V
Iˆ (0)  Tˆ (Iˆ m   Iˆ m - )
ˆ CT
ˆ (Iˆ m   Iˆ m - )  V
ˆ S  Zˆ S Tˆ (Iˆ m   Iˆ m - )
Z
at z  L
ˆ ˆ 
ˆ ˆ ˆ ( L)  Zˆ C Tˆ (e  rL
V
I m  e rL
Im )
ˆ ˆ 
ˆ ˆ ˆ (e  rL
Iˆ ( L)  T
I m  e rL
Im )
ˆ ˆ 
ˆ ˆ ˆ ˆ 
ˆ ˆ ˆ CT
ˆ (e  rL
Z
I m  e rL
I m )  Zˆ L Tˆ (e  rL
I m  e rL
Im )




 Zˆ
 Zˆ

 Zˆ
 Tˆ e  Zˆ
C
ˆS T
ˆ
Z
C
 Zˆ L

 Tˆ e
C
 Zˆ S Tˆ
C
 Zˆ L
ˆ
 rL
 Iˆ    ˆ 
  m    VS 
ˆ  ˆ rL
0
I
  m   
 Iˆ m  , Iˆ m - , Iˆ m  and Iˆ m - can be solved .
92
§ Exact Solution for Lossless Lines in Homogeneous Media
a. for lossless line :
ˆ  jwL, Y
ˆ  jwC
R  G  0 and Z
ˆ ˆ   w2 LC   w2  1    2 1  diagonal matrix
 YZ
where LC   1 for hom ogeneous transmission lines.
ˆ  1, rˆ  j  1  j w 1
T
v
b. It can be shown that :
j 2 L



R
S
  S  Iˆ
NE
VˆR (0)  VˆNE  
jwLm L  C 
LG
GDC
2


D
R

R


 en NE
FE
1 k


j 2 L


RNE RFE
1


jwCm L  C 
S VˆGDC 

RNE  RFE
1  k 2  LG 



RNE RFE
RFE
S 
ˆ
ˆ
ˆ
ˆ
VR ( L)  VFE 

jwLm LI GDC 
jwCm LVGDC 

Den  RNE  RFE
RNE  RFE

93
where :


2 1   SG LR 1   LG SR 
Den  C  S w  G R 1  k
  jwCS  G   R 
1   SR LR 1   SG LG  

sin(  L)
C  cos(  L), S 
L
Lm
Cm
k (coupling coefficient ) 

, k 1
LG LR
 CG  Cm  CR  Cm 
2
2
2
LG L
RS RL
 G (time constant ) 
  CG  Cm  L
RS  RL
RS  RL
 R (time constant ) 
R R
LR L
  CR  Cm  L NE FE
RNE  RFE
RNE  RFE
94
VˆGDC 
VˆS
RL
ˆ
ˆ
VS , I GDC 
RS  RL
RS  RL
ZˆCG 
LG
 vLG 1  k 2
CG  Cm
ZˆCR 
LR
 vLR 1  k 2
C R  Cm
 SG
 DC excitation at generator circuit
Characteristic impedances
of each circuit in the presence
of the other circuit.
RS
RNE
RL
RFE

,  LG 
,  SR 
,  LR 
Z CG
Z CG
Z CR
Z CR
if   1  low impedance load .
  1  high impedance load .
95
§ Inductive and Capacitive coupling
a. Two reasonable assumptions
(1)The line is electrically short . i.e. L

 C  cos   L   1,   1
(2)The generator and receptor circuits are weakly coupled .
k 1
 Den  (1  jw G )(1  jw R )
if w is small , then Den  1
96
b.
VˆNE 
RNE
R R
jwLm LIˆGDC  NE FE jwCm LVˆGDC
RNE  RFE
RNE  RFE
VˆFE  
(1)
R R
RFE
jwLm LIˆGDC  NE FE jwCm LVˆGDC
RNE  RFE
RNE  RFE
jwLm LIˆGDC

RNE
VNE
_
(2)


jwCm LVˆGDC
RFE VFE
_
d

jwLm LIˆGDC    ( Lm L)  IˆGDC   emf
 dt

d

d

jwCm LVˆGDC    (Cm L)  VˆGDC     (CV ) 
 dt

 dt

which is the independent current source due
to the voltage of generator.
97
(3) Crosstalk is the superposition of two components; One is inductive
coupling, the other one is capacitive coupling.
c. Intuitively :
(1) When low-impedance load (high current, low voltage), the
inductive coupling is dominant.
(2) When high-impedance load (low current, high voltage), the
capacitive coupling is dominant.
98