Download Class2_7_8_9_frequency_domain

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Instructor: Richard Mellitz
Introduction to Frequency
Domain Analysis (3 Classes)
Many thanks to Steve Hall, Intel for the use of his slides
Reference Reading: Posar Ch 4.5
http://cp.literature.agilent.com/litweb/pdf/5952-1087.pdf
Slide content from Stephen Hall
Differential Signaling
Outline
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Motivation: Why Use Frequency Domain Analysis
2-Port Network Analysis Theory
Impedance and Admittance Matrix
Scattering Matrix
Transmission (ABCD) Matrix
Mason’s Rule
Cascading S-Matrices and Voltage Transfer Function
Differential (4-port) Scattering Matrix
Differential Signaling
Motivation: Why Frequency Domain Analysis?
Time Domain signals on T-lines lines are hard to analyze
Many properties, which can dominate performance, are frequency
dependent, and difficult to directly observe in the time domain
• Skin effect, Dielectric losses, dispersion, resonance
Frequency Domain Analysis allows discrete
characterization of a linear network at each frequency
Characterization at a single frequency is much easier
Frequency Analysis is beneficial for Three reasons
Ease and accuracy of measurement at high frequencies
Simplified mathematics
Allows separation of electrical phenomena (loss, resonance … etc)
Differential Signaling
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Key Concepts
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Here are the key concepts that you should retain from this
class
The input impedance & the input reflection coefficient of a
transmission line is dependent on:
Termination and characteristic impedance
Delay
Frequency
S-Parameters are used to extract electrical parameters
Transmission line parameters (R,L,C,G, TD and Zo) can be
extracted from S parameters
Vias, connectors, socket s-parameters can be used to create
equivalent circuits=
The behavior of S-parameters can be used to gain
intuition of signal integrity problems
Differential Signaling
Review – Important Concepts
 The impedance looking into a terminated transmission
line changes with frequency and line length
 The input reflection coefficient looking into a
terminated transmission line also changes with
frequency and line length
 If the input reflection of a transmission line is known,
then the line length can be determined by observing
the periodicity of the reflection
 The peak of the input reflection can be used to
determine line and load impedance values
Differential Signaling
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Two Port Network Theory
Network theory is based on the property that a linear
system can be completely characterized by
parameters measured ONLY at the input & output
ports without regard to the content of the system
Networks can have any number of ports, however,
consideration of a 2-port network is sufficient to
explain the theory
A 2-port network has 1 input and 1 output port.
The ports can be characterized with many parameters, each
parameter has a specific advantage
Each Parameter set is related to 4 variables
2 independent variables for excitation
2 dependent variables for response
Differential Signaling
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Network characterized with Port Impedance
Measuring the port impedance is network is the most
simplistic and intuitive method of characterizing a network
I1
Port 1
I2
V1 +
-
2- port
2-port
Networ
Network
k
+ V
2
-
Port 2
Case 1: Inject current I1 into port 1 and measure the open circuit voltage at
port 2 and calculate the resultant impedance from port 1 to port 2
Z 21 
Vopen, port2
I port1
Case 2: Inject current I1 into port 1 and measure the voltage at port 1
and calculate the resultant input impedance
Z11 
Vopen, port1
I
1
Differentialport
Signaling
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Impedance Matrix
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 A set of linear equations can be written to describe the network in
terms of its port impedances
V1  Z11I1  Z12 I 2
V2  Z 21I1  Z 22 I 2
Or
V1 Z11 Z12 I1


V2 Z 21 Z 22 I 2
Where:
Vi
Z ij  
Ij
Open Circuit Voltage measured at Port i
Current Injected at Port j
Zii  the impedance looking into port i
Zij  the impedance between port i and j
If the impedance matrix is known, the response of the
system can be predicted for any input
Differential Signaling
Impedance Matrix: Example #2
Calculate the impedance matrix for the following circuit:
R2
R1
Port 1
R3
Differential Signaling
Port 2
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Impedance Matrix: Example #2
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Step 1: Calculate the input impedance
R2
R1
+
I1
V1
V1  I1 ( R1  R3 )
V1
Z11   R1  R3
I1
R3
Step 2: Calculate the impedance across the network
R1
R2
+
I1
R3
V2
-
R3
V2  V1
R1  R3
R3
 I1 ( R3  R1 )
 I1 R3
R1  R3
V2
Z 
 R3
Differential Signaling 21
I1
Impedance Matrix: Example #2
Step 3: Calculate the Impedance matrix
Assume: R1 = R2 = 30 ohms
R3=150 ohms
Z11  R1  R3  180
Z 21  30
180 30
Z Matrix 
30 180
Differential Signaling
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Measuring the impedance matrix
Question:
 What obstacles are expected when measuring the impedance
matrix of the following transmission line structure assuming that
the micro-probes have the following parasitics?
 Lprobe=0.1nH
 Cprobe=0.3pF
Assume F=5 GHz
Port 1
T-line
0.1nH
0.3pF
0.1nH
0.3pF
Zo=50 ohms, length=5 in
Differential Signaling
Port 2
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Measuring the impedance matrix
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Answer:
 Open circuit voltages are very hard to measure at high frequencies
because they generally do not exist for small dimensions
 Open circuit  capacitance = impedance at high frequencies
 Probe and via impedance not insignificant
Without Probe Capacitance
Port 1
0.1nH
0.3pF
0.1nH
0.1nH
T-line
T-line
Zo = 50
Port 2
Port
0.3pF
Port 1
Port 2
Z21 = 50 ohms
Zo=50 ohms, length=5 in
With Probe Capacitance @ 5 GHz
Zo = 50
Z probe_ L  2fL  3
1
Z probe_ C 
 106
2fC
Port 2
Port 1
106 ohms
Differential Signaling
106 ohms
Z21 = 63 ohms
Advantages/Disadvantages of Impedance Matrix
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Advantages:
The impedance matrix is very intuitive
Relates all ports to an impedance
Easy to calculate
Disadvantages:
Requires open circuit voltage measurements
Difficult to measure
Open circuit reflections cause measurement noise
Open circuit capacitance not trivial at high frequencies
Note: The Admittance Matrix is very similar, however, it is characterized
with short circuit currents instead of open circuit voltages
Differential Signaling
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Scattering Matrix (S-parameters)
Measuring the “power” at each port across a well
characterized impedance circumvents the problems
measuring high frequency “opens” & “shorts”
The scattering matrix, or (S-parameters), characterizes the
network by observing transmitted & reflected power waves
a2
a1
Port 1
2-port
Network
R
R
Port 2
b2
b1
ai represents the square root of the power wave injected into port i

Vi
V2
P
 P  ai 
R
R
bj represents the power wave coming out of port j
Differential Signaling
bj 
Vj

R
Scattering Matrix
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 A set of linear equations can be written to describe the network in
terms of injected and transmitted power waves
b1  S11a1  S12a2
b2  S 21a1  S 22a2
b1 S11 S12 a1


b2 S 21 S 22 a2
Where:
bi
Sij 

aj
Power measured at port i
Power injected at port j
Sii = the ratio of the reflected power to the injected power at port i
Sij = the ratio of the power measured at port j to the power injected at
port i
Differential Signaling
Making sense of S-Parameters – Return Loss
When there is no reflection from the load, or the line length
is zero, S11 = Reflection coefficient
R=50
Zo
Z=-l
b1
S11 
a1
a 20
R=Zo
Z=0

V1

Vreflected
Z o  50
V1
R
    
 o 
Vincident
Z o  50
V1
V1
R
S11 is measure of the power returned to the source,
and is called the “Return Loss”
Differential Signaling
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Making sense of S-Parameters – Return Loss
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When there is a reflection from the load, S11 will be
composed of multiple reflections due to the standing waves
1  (l )
Z in  Z (l )  Z o
1  (l )
Zo
Z=-l
RL
Z=0
If the network is driven with a 50 ohm source, then S11 is
calculated using the input impedance instead of Zo
50 ohms
S 
11
Z in  50 Z
in

Z in  50
S11 of a transmission line
will exhibit periodic effects
due to the standing waves
Differential Signaling
Example #3 – Interpreting the return loss
Based on the S11 plot shown below, calculate both the
impedance and dielectric constant
R=50
Zo
R=50
L=5 inches
0.45
S11, Magnitude
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1.0
1.5
2.0
2.5
3..0 3.5
Differential Signaling
Frequency, GHz
4.0
4.5
5.0
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Example – Interpreting the return loss
0.45
1.76GHz
2.94GHz
0.4
S11, Magnitude
20
Peak=0.384
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
Frequency, GHz
 Step 1: Calculate the time delay  Step 2: Calculate Er using the velocity
of the t-line using the peaks
1
c
3 108 m / s
v

1
f peaks  2.94GHz  1.76GHz 
TD
Er
Er
2TD
1
TD  423.7 ps  TD / inch

84.7 ps / inch  39.37inch / m)
TD / inch  423.7 ps / 5"  84.7 ps / inch
Differential Signaling Er  1.0
Example – Interpreting the return loss
 Step 3: Calculate the input
impedance to the
transmission line based on
the peak S11 at 1.76GHz
S11 
21
Z in  50
 0.384
Z in  50
Z in  112.33
Note: The phase of the
50  Z o j 4fl LC
reflection should be either
( x)  o e 2l 
e
50  Z o
+1 or -1 at 1.76 GHz because
it is aligned with the
e j 4  fl LC  e j 4 1.76GHz( 5)84.7 ps  e  j 9.366  1
incident
 Step 4: Calculate the
characteristic impedance
based on the input
impedance for x=-5 inches
50  Z o
(1)
50  Z o
1  ( x  5)
Z in  Z o
 112.33  Z o
50  Z o
1  ( x  5)
1
(1)
50  Z o
1
Z o  74.9
Er=1.0 and Zo=75 ohms
Differential Signaling
Making sense of S-Parameters – Insertion Loss
When power is injected into Port 1 with source impedance Z0
and measured at Port 2 with measurement load impedance
Z0, the power ratio reduces to
a voltage ratio

b2
S 21 
a1
a 20
a1
V1
Zo
V2

Z o V2
Vtransmitted
    
Vincident
V1
V1
a =0
Zo
2
2-port
Network
Zo
V2
b2
b1
S21 is measure of the power transmitted from
port 1 to port 2, and is called the “Insertion Loss”
Differential Signaling
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Loss free networks
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 For a loss free network, the total power exiting the N ports must
equal the total incident power
Pincident  Pexit
 If there is no loss in the network, the total power leaving the
network must be accounted for in the power reflected from the
incident port and the power transmitted through network
Preflected _ port1
Pincident

Ptransmitted _ port1 port2
Pincident
1
 Since s-parameters are the square root of power ratios, the
following is true for loss-free networks
S11   S21 
2
2
1
 If the above relationship does not equal 1, then there is loss in the
network, and the difference is proportional to the power dissipated
by the network
Differential Signaling
Insertion loss example
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Question:
 What percentage of the total power is dissipated by the
transmission line?
 Estimate the magnitude of Zo (bound it)
S-parameters; 5 inch microstrip
1.2
1
Magnitude
0.8
S(1,1)
S(1,2)
0.6
0.4
0.2
0
0.E+00
2.E+09
4.E+09
6.E+09
8.E+09
Frequency, Hz
Differential Signaling
1.E+10
1.E+10
Insertion loss example
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 What percentage of the total power is dissipated by the transmission
line ?
 What is the approximate Zo?
 How much amplitude degradation will this t-line contribute to a 8 GT/s
signal?
 If the transmission line is placed in a 28 ohm system (such as Rambus),
will the amplitude degradation estimated above remain constant?
 Estimate alpha for 8 GT/s signal
S-parameters; 5 inch microstrip;
1.2
1
S(1,1)
Magnitude
0.8
S(1,2)
0.6
0.4
0.2
0
0.E+00
2.E+09
4.E+09
6.E+09
Differential
Signaling
Frequency, Hz
8.E+09
1.E+10
Insertion loss example
Answer:
 Since there are minimal reflections on this line, alpha can be
estimated directly from the insertion loss
 S21~0.75 at 4 GHz (8 GT/s)
When the reflections are minimal, alpha can be estimated
S21  el  0.75  e (5)    0.057
 If S11 < ~ 0.2 (-14 dB), then the above approximation is valid
 If the reflections are NOT small, alpha must be extracted
with ABCD parameters (which are reviewed later)
 The loss parameter is “1/A” for ABCD parameters
 ABCE will be discussed later.
Differential Signaling
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Important concepts demonstrated
 The impedance can be determined by the
magnitude of S11
 The electrical delay can be determined by the
phase, or periodicity of S11
 The magnitude of the signal degradation can be
determined by observing S21
 The total power dissipated by the network can be
determined by adding the square of the insertion
and return losses
Differential Signaling
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A note about the term “Loss”
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True losses come from physical energy losses
Ohmic (I.e., skin effect)
Field dampening effects (Loss Tangent)
Radiation (EMI)
Insertion and Return losses include effects such as
impedance discontinuities and resonance effects, which are
not true losses
Loss free networks can still exhibit significant insertion and
return losses due to impedance discontinuities
Differential Signaling
Advantages/Disadvantages of S-parameters
Advantages:
Ease of measurement
Much easier to measure power at high frequencies than open/short
current and voltage
S-parameters can be used to extract the transmission line
parameters
n parameters and n Unknowns
Disadvantages:
Most digital circuit operate using voltage thresholds. This
suggest that analysis should ultimately be related to the time
domain.
Many silicon loads are non-linear which make the job of
converting s-parameters back into time domain non-trivial.
Conversion between time and frequency domain introduces
errors
Differential Signaling
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Cascading S parameter
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3 cascaded s parameter blocks
a11
a21 b12
b22 a13
s111 s121
s113 s123
s211 s221
b11
a13
s213 s223
 s11 s12 
s112 s122
b21 a12
s212 s222
a22 b13


 s21 s22 
b13
 While it is possible to cascade s-parameters, it gets
messy.
 Graphically we just flip every other matrix.
 Mathematically there is a better way… ABCD parameters
 We will analyzed this later with signal flow graphs
Differential Signaling
ABCD Parameters
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The transmission matrix describes the network in terms
of both voltage and current waves
I1
V1
V1  AV2  BI 2
I2
2-port
Network
V2
The coefficients can be defined
using superposition
V1
A
V2
I 2 0
V1
I1
B
C
I 2 V 0
V2
2
Differential Signaling
I1  CV2  DI 2
V1 A B V2

I1 C D I 2
I 2 0
I1
D
I2
V2  0
Transmission (ABCD) Matrix
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Since the ABCD matrix represents the ports in terms of
currents and voltages, it is well suited for cascading
elements
I3
I2
I1
A B
C D1
V1
V2
A B
C D2
V3
The matrices can be cascaded by multiplication
V1
I1
V2
I2


V2
A
B
C
D 1 I2
A
V3

D 2 I3
C
B

V1 A B A B V3



I1 C D 1 C D 2 I 3
This is the best way to cascade elements in the frequency domain.
Differential Signaling
It is accurate,
intuitive and simplistic.
Relating the ABCD Matrix to Common
Circuits
Z
Port 1
Y
Port 1
Z1
Port 1
A 1 B  Z
C  0 D 1
Port 2
A 1
C Y
Port 2
Z2
Z3
Port 2
Y3
Port 1 Y1
Port 1
Y2
Zo ,
l
Port 2
Port 2
Assignment 6:
Convert these
to s-parameters
B0
D 1
A  1  Z1 / Z 3
B  Z1  Z 2  Z1Z 2 / Z 3
C  1 / Z3
D  1  Z 2 / Z3
A  1  Y2 / Y3
B  1 / Y3
C  Y1  Y2  Y1Y2 / Y3
D  1  Y1 / Y3
A  cosh( l )
B  Z o sinh( l )
C  (1 / Z o ) sinh( l )
D  cosh( l )
Differential Signaling
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Converting to and from the S-Matrix
 The S-parameters can be measured with a VNA, and converted
back and forth into ABCD the Matrix
Allows conversion into a more intuitive matrix
Allows conversion to ABCD for cascading
ABCD matrix can be directly related to several useful circuit topologies
(1  S11 )(1  S 22 )  S12 S 21
A
2 S 21
S11 
A  B / Z o  CZ o  D
A  B / Z o  CZ o  D
(1  S11 )(1  S 22 )  S12 S 21
B  Zo
2 S 21
2( AD  BC )
S12 
A  B / Z o  CZ o  D
1 (1  S11 )(1  S 22 )  S12 S 21
C
2S 21
Zo
2
S 21 
A  B / Z o  CZ o  D
(1  S11 )(1  S 22 )  S12 S 21
D
2S 21
S11 
Differential Signaling
 A  B / Z o  CZ o  D
A  B / Z o  CZ o  D
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ABCD Matrix – Example #1
35
 Create a model of a via from the
measured s-parameters
Port 1
Port 2
S11 S12  0.110  j 0.153 0.798  j 0.572

S 21 S 22
0.798  j 0.572  0.110  j 0.153
Differential Signaling
ABCD Matrix – Example #1
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The model can be extracted as either a Pi or a T network
L1
L2
Port 1
CVIA
Port 2
The inductance values will include the L of the trace and the via
barrel (it is assumed that the test setup minimizes the trace
length, and subsequently the trace capacitance is minimal
The capacitance represents the via pads
Differential Signaling
ABCD Matrix – Example #1
Assume the following s-matrix measured at 5 GHz
S11 S12  0.110  j 0.153 0.798  j 0.572

S 21 S 22
0.798  j 0.572  0.110  j 0.153
Differential Signaling
37
ABCD Matrix – Example #1
Assume the following s-matrix measured at 5 GHz
S11 S12  0.110  j 0.153 0.798  j 0.572

S 21 S 22
0.798  j 0.572  0.110  j 0.153
Convert to ABCD parameters
A B
0.827

C D j 0.0157
j 20.08
0.827
Differential Signaling
38
ABCD Matrix – Example #1
39
Assume the following s-matrix measured at 5 GHz
S11 S12  0.110  j 0.153 0.798  j 0.572

S 21 S 22
0.798  j 0.572  0.110  j 0.153
Convert to ABCD parameters
A B
0.827

C D j 0.0157
j 20.08
0.827
Relating the ABCD parameters to the T circuit topology,
the capacitance and inductance is extracted from C & A
Z1
Port 1
Z2
Z3
C  j 0.0157 
Port 2
1

Z3
1
 CVIA  0.5 pF
1
j  2fCVIA
Z1
j  2fL
 0.827  1 
 L1  L2  0.35nH
Z3
1 /( j  2fCVIA )
Differential
Signaling
A  1
ABCD Matrix – Example #2
40
Calculate the resulting s-parameter matrix if the two
circuits shown below are cascaded
Port 1
Port 2
2-port
Network X
Network
50
Port 1
50
Port 2
2-port
Network Y
Network
50
50
Port 1
50
S X 11 S X 12
SX 
S X 21 S X 22
SY 11 SY 12
SY 
SY 21 SY 22
2-port
Network Y
Network
2-port
Network X
Network
S XY  ?
Differential Signaling
50
Port 2
ABCD Matrix – Example #2
41
Step 1: Convert each measured S-Matrix to ABCD
Parameters using the conversions presented earlier
AX
S X  TX 
CX
BX
DX
AY
SY  TY 
CY
BY
DY
Step 2: Multiply the converted T-matrices
TXY
AX
 TX  TY 
CX
BX AY

DX CY
BY
AXY

DY C XY
BXY
DXY
Step 3: Convert the resulting Matrix back into Sparameters using thee conversions presented earlier
TXY  S XY
S X 11 S X 12

S X 21 S X 22
Differential Signaling
Advantages/Disadvantages of ABCD Matrix
Advantages:
The ABCD matrix is very intuitive
Describes all ports with voltages and currents
Allows easy cascading of networks
Easy conversion to and from S-parameters
Easy to relate to common circuit topologies
Disadvantages:
Difficult to directly measure
Must convert from measured scattering matrix
Differential Signaling
42
Signal flow graphs – Start with 2 port first
The wave functions (a,b) used to define s-parameters for a
two-port network are shown below. The incident waves is a1,
a2 on port 1 and port 2 respectively. The reflected waves b1
and b2 are on port 1 and port 2. We will use a’s and b’s in the
s-parameter follow slides
Differential Signaling
43
Signal Flow Graphs of S Parameters
“In a signal flow graph, each port is represented by two
nodes. Node an represents the wave coming into the device
from another device at port n, and node bn represents the
wave leaving the device at port n. The complex scattering
coefficients are then represented as multipliers (gains) on
branches connecting the nodes within the network and in
adjacent networks.”*
Example
a1
s21
b2
S
s22
s11
L
s12
b1
a2
s11 
a1 S L 0
b1 a2 0
Measurement equipment
strives to be match i.e.
reflection coefficient is 0
See: http://cp.literature.agilent.com/litweb/pdf/5952-1087.pdf
Differential Signaling
44
Mason’s Rule ~ Non-Touching Loop Rule
 T (1   (1)
T
(1   (1)
mk
k
k
L(mk )
(k )
45
)
mk
mk
L(mk ))
mk
 T is the transfer function (often called gain)
 Tk is the transfer function of the kth forward path
 L(mk) is the product of non touching loop gains on path k
taken mk at time.
 L(mk)|(k) is the product of non touching loop gains on path k
taken mk at a time but not touching path k.
 mk=1 means all individual loops
Differential Signaling
Voltage Transfer function
46
 What is really of most relevance to time domain analysis is
the voltage transfer function.
 It includes the effect of non-perfect loads.
 We will show how the voltage transfer functions for a 2 port
network is given by the following equation.
s21
2


 L  1  1  s

1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
 Notice it is not S21
Differential Signaling
Forward Wave Path
47
Z0
Vs
( ZS  Z0)
S
a1
s21
b2
s22
s11
L
s12
b1
b2
Vs
s21
a2
Z0
( ZS  Z0)
1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
Differential Signaling
Reflected Wave Path
48
Z0
Vs
( ZS  Z0)
S
a1
s21
b2
s22
s11
L
s12
b1
a2
Vs
s21  L
a2
Z0
( ZS  Z0)
1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
Differential Signaling
Combine b2 and a2
s21
b2
Vs
49
( ZS  Z0)
1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
Vs
b2
Z0

a2
Vs


s21 1   L 
Z0
( ZS  Z0)
1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
Differential Signaling
Convert Wave to Voltage
- Multiply by sqrt(Z0)

s
Vo
Vs

s21 1   L 
Vo
Vs
50
Z0
( ZS  Z0)
1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
ZS  Z0
ZS  Z0
1  s

1
ZS  Z0
ZS  Z0

s21 1   L 
1  s
2
Z0
( ZS  Z0)
 1   s
2
1  s11  s  s22  L  s21 s12  L  s  s11 s22  L  s
Differential Signaling
Voltage transfer function using ABCD
Let’s see if we can get this results another way
 ( 1  s11) ( 1  s22)  s12 s21 [ ( 1  s11) ( 1  s22)  s12 s21]  Z0 


2 s21
2 s21
ABCD_CHANNEL 

( 1  s11) ( 1  s22)  s12 s21
( 1  s11) ( 1  s22)  s12 s21


2 s21 Z0
2 s21


ABCD_SOURCE
ZL
Z0
1  L
1  L
 1 Zs 


0
1


 1 0

ABCD_LOAD  1

1
ZL


Zs
Z0
1  s
1   s
Differential Signaling
51
Cascade [ABCD] to determine system [ABCD]
VOLTAGE_TRANSFER_FUNCTION
1  s

 1 Z0
1   s


1
0
52
ABCD_SOURCE ABCD_CHANNEL ABCD_LOAD
1
0
  ( 1  s11) ( 1  s22)  s12 s21 [ ( 1  s11) ( 1  s22)  s12 s21]  Z0  



2 s21
2 s21
1
1



( 1  s11) ( 1  s22)  s12 s21
  ( 1  s11) ( 1  s22)  s12 s21
  Z0 1  L


2 s21 Z0
2 s21
1  L

Simplify
1  s22  s11  s  s11 s22  s  s12 s21  s
 1  s11  s  s22 L  s11 s22  s  L  s12 s21  s  L

2

Z0


1   s  s21 1  L
1   s  s21


1  s22 L  s11  s11 s22 L  s12 s21 L
1 1  s22  s11  s11 s22  s12 s21




2
s21
s21

Z0

1

L

Differential Signaling






Extract the voltage transfer function
"A" parameter which input over output transfer. We are
looking for "1/A" which is output over input
 1  s11  s  s22 L  s11 s22  s  L  s12 s21  s  L 
2


1   s  s21 1  L
1
Simplify and re-arange
s21  1  L 
1   s 
2
1  s11  s  s22 L  s12 s21  s  L  s11 s22  s  L
Same as with flow graph analysis
Differential Signaling
53
Cascading S-Parameter
54
 As promised we will now look at how to cascade sparameters and solve with Mason’s rule
 The problem we will use is what was presented earlier
 The assertion is that the loss of cascade channel can be
determine just by adding up the losses in dB.
 We will show how we can gain insight about this
assertion from the equation and graphic form of a
solution.
a11
a21 b12
b22 a13
s111 s121
s113 s123
s211 s221
s213 s223
s112 s122
b11
a13
b21 a12
s212 s222
a22 b13
Differential Signaling
 s11 s12 


s21
s22


b13
Creating the signal flow graph
a11
a21 b12
b22 a13
s113 s123
s211 s221
s213 s223
b11
s212 s222
b21 a12
s211
B21
1
s221
B11
a13
s111 s121
s112 s122
A11
55
s121
A21
A12
s212
a22 b13
B22
1
 s11 s12 


s21
s22


s213
A13
s112 s222
1
B12
s122
A22
B23
s113
1
s223
s123
B13
 We map output a to input b and visa versa.
 Next we define all the loops
 Loop “A” and “B” do not touch each other
Differential Signaling
b13
A23
Use Mason’s rule
A11
s211
B21
1
56
A12
s221
B11
b6
a1
s121
s212
s213
1
B22
A13
s112 s222
A21
1
B12
s122
s113
B13
s21  s21  s21

2
1
3
2
3
1
3
2
  s221s112s222s113
2
 There is only one forward path a11 to b23.
 There are 2 non touching looks
Differential Signaling
A23
Mason’s Rule
1  s22  s11  s22  s11  s11  s22  s12  s21
2
s223
s123
1
A22
1
B23
Evaluate the nature of the transfer function
57
Assumption is that these are ~ 0
b6
a1
s21  s21  s21

1
2
3

1  s22  s11  s22  s11  s11  s22  s12  s21  s22  s11  s22  s11
2
1
3
2
3
1
2
2
2
1
• If response is relatively flat and reflection
is relatively low
– Response through a channel is s211*s212*213…
Differential Signaling
3
4
Jitter and dB Budgeting
S211  e
 j  211
 S21  e
 j  211
2
S21  S21
1
Smag e
S
 Change s21 into a phasor
58
2
 j 
 j  213
 S213  e
 j   211 21
=2 213
s21  e
3
 Insertion loss in db

20 log S21  S21
1

2
20 log s21
s21
3

=
  20logs212  20logs213
1
i.e. For a budget just add up the db’s and jitter
db
sys
 dbi i1
delay
n
 i i1
n
Differential Signaling
Differential S-Parameters
59
Differential S-Parameters are derived from a 4-port
measurement
a2
a1
b1
4-port
b2
b1
S11
S12
S13
S14
b2
S21
S22
S23
S24
S31
S32
S33
S34
S41
S42
S43
S44
a3
a4
b3
b3
b4
b4
=
a1
a2
a3
a4
Traditional 4-port measurements are taken by driving
each port, and recording the response at all other ports
while terminated in 50 ohms
Although, it is perfectly adequate to describe a differential
pair with 4-port single ended s-parameters, it is more
useful to convert to a multi-mode port
Differential Signaling
60
Differential S-Parameters
 It is useful to specify the differential S-parameters in terms of differential
and common mode responses
Differential stimulus, differential response
Common mode stimulus, Common mode response
Differential stimulus, common mode response (aka ACCM Noise)
Common mode stimulus, differential response
 This can be done either by driving the network with differential and
common mode stimulus, or by converting the traditional 4-port s-matrix
adm2
bdm1
acm1
bcm1
Multi-Mode
Port
bdm2
acm2
Multi-Mode Port 2
Multi-Mode Port 1
adm1
bdm1
DS11 DS12 DCS11 DCS12
adm1
bdm2
DS21 DS
22 DCS21 DCS22
adm2
CDS11 CDS12 CS11 CS12
acm1
CS21 CS
22
acm2
bcm1
bcm2
=
CDS21 CDS
22
bcm2
Matrix assumes differential
and common mode stimulus
Differential Signaling
Explanation of the Multi-Mode Port
Differential Matrix:
Differential Stimulus, differential response
i.e., DS21 = differential signal [(D+)-(D-)]
inserted at port 1 and diff signal measured at port 2
Common mode conversion Matrix:
Differential Stimulus, Common mode
response. i.e., DCS21 = differential
signal [(D+)-(D-)] inserted at port 1
and common mode signal [(D+)+(D-)]
measured at port 2
bdm1
DS11 DS12 DCS11 DCS12
adm1
bdm2
DS21 DS
22 DCS21 DCS22
adm2
CDS11 CDS12 CS11 CS12
acm1
CS21 CS
22
acm2
bcm1
bcm2
=
CDS21 CDS
22
61
differential mode conversion Matrix:
Common mode Matrix:
Common mode Stimulus, differential
Common mode stimulus, common mode
mode response. i.e., DCS21 = common
Response. i.e., CS21 = Com. mode signal
mode signal [(D+)+(D-)] inserted at port
[(D+)+(D-)] inserted at port 1 and Com. mode
1 and differential mode signal [(D+)-(D-)]
Differential
Signaling
signal
measured at port 2
measured at port 2
Differential S-Parameters
62
 Converting the S-parameters into the multi-mode requires just a little algebra
Example Calculation, Differential Return Loss
The stimulus is equal, but opposite, therefore:
a3  a1 ; a4  a2
b
DS11  dm1
adm1
adm2  0;acm  0
b b
 1 3
a1  a3
1
a2  a4  0
b1  S11a1  S12a2  S13a3  S14a4
4-port
2-port
Network
3
2
4
b3  S31a1  S32a2  S33a3  S34a4
b1  b3  a1 ( S11  S31 )  a2 ( S12  S32 )  a3 ( S13  S33 )  a4 ( S14  S34 )
Assume a symmetrical network and substitute
 DS11 
a3  a1 ; a4  a2
S12  S34 ; S32  S14
1
S11  S31  S13  S33 
2
Other conversions that are useful for a differential bus are shown
Differential Insertion Loss:
DS 21 
1
S 21  S 41  S 23  S 43 
2
Differential to Common Mode Conversion (ACCM):
CDS 21 
1
S21  S43  S23  S41 
2
Similar techniquesDifferential
can be used
for all multi-mode Parameters
Signaling
Next class we will develop more differential concepts
Differential Signaling
63
backup review
64
Differential Signaling
Advantages/Disadvantages of Multi-Mode
Matrix over Traditional 4-port
Advantages:
Describes 4-port network in terms of 4 two port matrices
Differential
Common mode
Differential to common mode
Common mode to differential
Easier to relate to system specifications
ACCM noise, differential impedance
Disadvantages:
Must convert from measured 4-port scattering matrix
Differential Signaling
65
High Frequency Electromagnetic Waves
 In order to understand the frequency domain analysis, it is
necessary to explore how high frequency sinusoid signals behave
on transmission lines
 The equations that govern signals propagating on a transmission
line can be derived from Amperes and Faradays laws assumimng
a uniform plane wave
The fields are constrained so that there is no variation in the X and Y
axis and the propagation is in the Z direction
 This assumption holds true for
transmission lines as long as the
wavelength of the signal is much
greater than the trace width
X
Direction of
propagation
Z

m
in 
  3 108 39.4 
s
m1



 W


f
r




For typical PCBs at 10 GHz with 5 mil traces (W=0.005”)
Y
  0.59"  0.005"
Differential Signaling
66
High Frequency Electromagnetic Waves
 For sinusoidal time varying uniform plane waves,
Amperes and Faradays laws reduce to:
Amperes Law:
A magnetic Field will be
induced by an electric current
or a time varying electric field
By
z
Faradays Law:
An electric field will be
generated by a time varying
magnetic flux
  jE x
E x
  j B y
z
 Note that the electric (Ex) field and the magnetic
(By) are orthogonal
Differential Signaling
67
High Frequency Electromagnetic Waves
 If Amperes and Faradays laws are differentiated with
respect to z and the equations are written in terms of the
E field, the transmission line wave equation is derived
B y
B y
 2 Ex
 2 Ex 1
  j

 2
  jE x
2
z
z
z
z j
 2 Ex
2 2

j
 E x  0
2
z
This differential equation is easily solvable for Ex:
Ex ( z )  C1e
 j (  ) z
 C2e
Differential Signaling
j (  ) z
68
High Frequency Electromagnetic Waves
69
The equation describes the sinusoidal E field for a plane
wave in free space
Note the positive exponent
  j (  ) z
M
Ex ( z )  E e
Portion of wave traveling
In the +z direction

 j (  ) z
M
E e
is because the wave is
traveling in the opposite
direction
Portion of wave traveling
In the -z direction
= permittivity in Farads/meter (8.85 pF/m for free space)
(determines the speed of light in a material)
permeability in Henries/meter (1.256 uH/m for free space
 = and
non-magnetic materials)
Since inductance is proportional to  & capacitance is proportional
to  , then  is analogous to LC in a transmission line, which
is the propagation delay
Differential Signaling
High Frequency Voltage and Current Waves
70
 The same equation applies to voltage and current waves on a
transmission line
Incident sinusoid
Reflected sinusoid
z=-l
RL
z=0
If a sinusoid is injected onto a transmission line, the resulting voltage
is a function of time and distance from the load (z). It is the sum of the
incident and reflected values
z
V ( z, t )  Vine e
Voltage wave traveling
towards the load
jt
z
 Vref e e
jt
j t
Note:
is added to
specifically represent
the time varying
Sinusoid, which was implied
in the previous derivation
Voltage wave reflecting
off the Load and traveling
towards the source
Differential Signaling
e
High Frequency Voltage and Current Waves
71
The parameters in this equation completely describe the
voltage on a typical transmission line
z
V ( z, t )  Vine e
    j
  j LC

jt
z
 Vref e e
jt
= Complex propagation constant – includes all the
transmission line parameters (R, L C and G)
(For the loss free case)
  ( R  jL)(G  jC )
(lossy case)
= Attenuation Constant (attenuation of the signal due to transmission line losses)
1
C
L

 (For good conductors)
  R
G
2
L
C 

= Phase Constant (related to the propagation delay across the transmission line)
   LC (For good conductors and good dielectrics)
Differential Signaling
High Frequency Voltage and Current Waves
The voltage wave equation can be put into more intuitive
terms by applying the following identity:
j
e  cos( )  j sin(  )
V ( z , t )  Vine  (  j ) z e jt  Vref e (  j ) z e jt
Subsequently:
z
z 

 e Vin cos  (t  )  j sin  (t  )

 

z
z 

 ezVref cos  (t  )  j sin  (t  )

 

z
The amplitude is degraded by z
e
The waveform is dependent on the driving function
(cos t  j sin t ) & the delay of the line    LC
Differential Signaling
72
Interaction: transmission line and a load
73
The reflection coefficient is now a function of the Zo
discontinuities AND line length
Influenced by constructive & destructive combinations of the
forward & reverse waveforms
Zo
(l )
Zl
(Assume a line length of l (z=-l))
Z=-l

Z=0

V (l )  Vinel  Vref el  Vin el  oel  Vinel 1  (l )
(l ) 
Vref el
Vine
l
 o e
2l
Z l  Z o 2l

e
Zl  Zo
This is the reflection coefficient looking into a t-line of length l
Differential Signaling
Interaction: transmission line and a load
74
If the reflection coefficient is a function of line length, then
the input impedance must also be a function of length
Zin
RL
Z=-l
Z=0


V (l )  Vine l  Vref el  Vin e l  o el  Vine l 1  (l )




1
1
l
l
I (l ) 
Vine  Vref e 
Vine l 1  (l )
Zo
Zo
Vine l 1  (l )
V (l )
1  (l )
Z in  Z (l ) 

 Zo
1
I (l )
1  (l )
Vine l 1  (l )
Zo


Note: (l ) is
dependent on
and
l
This is the input impedance looking into a t-line of length l
Differential Signaling
Line & load interactions
 In chapter 2, you learned how to calculate waveforms
in a multi-reflective system using lattice diagrams
 Period of transmission line “ringing” proportional to the line delay
 Remember, the line delay is proportional to the phase constant
 In frequency domain analysis, the same principles
apply, however, it is more useful to calculate the
frequency when the reflection coefficient is either
maximum or minimum
 This will become more evident as the class progresses
To demonstrate, lets assume a loss free transmission line
    j  R  jL G  jC 
 0 R G 0
j 
j 2 2 LC  j LC
Differential Signaling
75
Line & load interactions
76
Remember, the input reflection takes the form (l )  o e 2l
The frequency where the values of the real & imaginary
reflections are zero can be calculated based on the line length
o e 2 ( l )  o e j 2  ( l )  o e  j 2l
4fl LC 
n
2
LC
 
Term 1


Term 2
 o cos 4fl LC  j sin 4fl LC
4fl LC  n
n
Term 1=0
n
Term 2=0
f 
Term 2 = 
f 
Term 1 = 
4l LC
o
o
8l LC
n  1,2,3...
n  1,3,5...
Note that when the imaginary portion is zero, it means the phase
of the incident & reflected waveforms at the input are aligned. Also
notice that value of “8” and “4” in the terms.
Differential Signaling

Example #1: Periodic Reflections
Calculate:
1. Line length
2. RL
(l )
(assume a very low loss line)
77
Er_eff=1.0
Zo=75
Z=-l
Z=0
2.5E-01
Imaginary
2.0E-01
Real
Reflection Coeff.
1.5E-01
1.0E-01
5.0E-02
0.0E+00
-5.0E-02
-1.0E-01
-1.5E-01
-2.0E-01
-2.5E-01
0.0E+00
5.0E+08
1.0E+09 1.5E+09
Frequency
RL
2.0E+09
Differential Signaling
2.5E+09
3.0E+09
Example #1: Solution
Step 1: Determine the periodicity zero crossings or peaks & use the
relationships on page 15 to calculate the electrical length
3
1
1
f n 3  f n 1 


 1.76GHz  588MHz  1.176GHz
4l LC 4l LC 2l LC
1
1
TD  l LC 

 425 ps
2( f n 3  f n 1 ) 2.35GHz
Imaginary
Differential Signaling
78
Example #1: Solution (cont.)
79
 Note the relationship between the peaks and the electrical
length
1
TD  l LC 
2( f n 3  f n 1 )
 This leads to a very useful equation for transmission lines
Fpeaks
1

2TD
 Since TD and the effective Er is known, the line
length can be calculated as in chapter 2
length 
TD
425 ps

 0.127m  5in
Er _ eff
1 s
8
3 10Signaling
m
c Differential
Example #1: Solution (cont.)
 The load impedance can be calculated by observing the
peak values of the reflection
 When the imaginary term is zero, the real term will peak, and the
maximum reflection will occur
 If the imaginary term is zero, the reflected wave is aligned with
the incident wave and the phase term = 1
RL  Z o 2l RL  75
(l ) 
e 
(1)  0.2
RL  Z o
RL  75
RL  50
Important Concepts demonstrated
 The impedance can be determined by the magnitude of
the reflection
 The line length can be determined by the phase, or
periodicity of the reflection
Differential Signaling
80