Transcript Model Order Reduction I

CSE245: Computer-Aided
Circuit Simulation and
Verification
Lecture Note 3
Model Order Reduction (1)
Spring 2008
Prof. Chung-Kuan Cheng
Outline
• Introduction
• Formulation
• Linear System
– Time Domain Analysis
– Frequency Domain Analysis
– Moments
– Stability and Passivity
– Model Order Reduction
Transfer Function
State Equation in Frequency Domain
( suppose zero initial condition)
Solve X
Express Y(s) as
a function of U(s)
Transfer Function:
Stability
• A network is stable if, for all bounded
inputs, the output is bounded.
• For a stable network, transfer function
H(s)=N(s)/D(s)
– Should have only negative poles pj, i.e.
Re(pj)  0
– If pole falls on the imaginary axis, i.e.
Re(pi) = 0, it must be a simple pole.
Passivity
• Passivity
– Passive system doesn’t generate energy
– A one-port network is said to be passive if the
total power dissipation is nonnegative for all
initial time t0, for all time t>t0, and for all possible
input waveforms, that is,
where E(t0) is the energy stored at time t0
• Passivity of a multi-port network
– If all elements of the network are passive, the
network is passive
Passivity in Complex Space Representation
• For steady state response of a one-port
• The complex power delivered to this one-port is
• For a passive network, the average power
delivered to the network should be nonnegative
Linear Multi-Port Passivity
• For multi-port, suppose each port is either a voltage
source or a current source
– For a voltage source port, the input is the voltage and the
output is a current
– For a current source port, the input is the current and the
output is a voltage
– Then we will have D=BT in the state equation
– Let U(s) be the input vector of all ports, and H(s) be the
transfer function, thus the output vector Y(s) = H(s)U(s)
• Average power delivered to this multi-port network is
• For a passive network, we should have
Linear System Passivity
• State Equation (s domain)
• We have shown that transfer function is
where
• We will show that this network is passive, that is
Passivity Proof
• To show
• Is equivalent to show
where
• We have
• Thus
Passivity and Stability
• A passive network is stable.
• However, a stable network is not necessarily
passive.
• A interconnect network of stable components is
not necessarily stable.
• The interconnection of passive components is
passive.
Model Order Reduction (MOR)
• MOR techniques are used to build a reduced
order model to approximate the original circuit
i1(t)
R
L
R
L
R
L
R
L
v1(t)
C
G
C
Huge
Network







C
G
C
G
Formulation
   
   
   
 s  x  
   
   
   
A
C
i2(t)
i1(t)
v2(t)
v1(t)
L
R
i2(t)
C
G
Small
Network
   
   
     MOR
  x   b  u
   
   
    







C'
G
Realization
   
   
   
 s  x '  
   
   
   
A'
   
   
   
  x '   b ' u '
   
   
    
Model Order Reduction: Overview
• Explicit Moment Matching
– AWE, Pade Approximation
• Implicit Moment Matching
– Krylov Subspace Methods
• PRIMA, SPRIM
• Gaussian Elimination
– TICER
– Y-Delta Transformation
Moments Review
• Transfer function

H ( s)   e st h(t )dt
0


 


1 2
   h(t )dt     th(t )dt  s    t h(t )dt  s 2 
2!  0
0
 0



 2 q 1
(1) 2 q 1  2 q 1

t
h
(
t
)
dt
 O (s 2 q )

s
(2q  1)!  0

• Compare
H (s)  m0  m1s  m2 s 2 
• Moments
 m2q1s 2q1  O(s 2q )
Moments Matching: Pade Approximation
H (s)  m0  m1s  m2 s 2 
Hˆ ( s ) 
 m2q1s 2q1  O(s 2q )
b0  b1s    bq1s q1
1  a1s    aq s q
Choose the 2q rational function coefficients a1 , a2 , aq 1 , b1 , b2 , bq 1 ,
So that the reduced rational function Hˆ ( s ) matches the first 2q moments
of the original transfer function H(s).
Moments Matching: Pade Approximation
– Step 1: calculate the first 2q moments of
H(s)
– Step 2: calculate the 2q coeff. of the Pade’
approximation, matching the first 2q
moments of H(s) b0 , b1 ,, bq 1 , a1 ,, aq
Pade Approximation: Coefficients
b0  b1s    bq1s q1
1  a1s    aq s q
 m0  m1s  m2 s 2    m2 q1s 2 q1
For a1 a2,…, aq solve the following linear system:
 m0 m1 m2  mq 1   aq 
 mq 
m
 a 
m 
m

2
 1
  q1 
 q 1 
 m2
 aq 2     mq 2 





 
  
  
mq 1 
m2 q1 
m2 q 2   a1 
For b0 b1, …, bq-1 calculate:
b0  m0
b1  m1  a1m0

bq1  mq1  a1mq 2    aq1m0
Pade Approximation: Drawbacks
• Numerically unstable
– Higher order moments
– Matrix powers converge to the eigenvector corresponding to
the largest eigenvalue.
 m0 m1 m2  mq1   aq 
 mq 
m
m 
m2
   aq1 
1


 q1 

 m2
 aq2     mq2 













mq1 
m2 q1 
m2 q2   a1 
– Columns become linear dependent for large q. The problem is
numerically very ill-conditioned.
• Passivity is not always preserved.
– Pade may generate positive poles