CSE245: Computer-Aided Circuit Simulation and Verification Lecture Notes 3 Model Order Reduction (1)...
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Transcript of CSE245: Computer-Aided Circuit Simulation and Verification Lecture Notes 3 Model Order Reduction (1)...
CSE245: Computer-Aided Circuit Simulation and
VerificationLecture Notes 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 FunctionState Equation in Frequency Domain (
suppose zero initial condition)
Express Y(s) as
a function of U(s)
Solve X
Transfer Function:
Recall:
• U is 1xn vector of inputs (voltage, current sources)
• X is state variable (voltage or current of inductor, cap., etc.)
• Y is voltages of current sources & currents of voltage sources.
• C is defined as
• A is defined as
• Will show that D = BT
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) (general form)
(Recall: H(s) = Y(s)/U(s) in our case)
– 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
and
Plug conjugate H(s) in here
…Why is this important?
After Model Order Reduction (MOR) (in later slides), this condition must still be true.
Passivity Proof (part 1)• To show
• Is equivalent to show
where
Re[ U*(s)H(s)*U(s) ] ≥ 0
Set U*(s)H(s)*U(s) = F
½ Re(F* + F) ≥ 0
F* + F ≥ 0
Plugging F back in:
(U*(s)H*(s)U(s))* + U*(s)H*(s)U(s)
= U*(s)H(s)U(s) + U*(s)H*(s)U(s)
= U*(s) (H(s) + H*(s)) U(s) ≥ 0
Here’s how…
Passivity and Stability
• A passive network is stable.
• However, a stable network is not necessarily passive.
• All poles could be on LHS, but some could be negative!
• 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 circuitR L
C G
i2(t)i1(t)
v1(t)
R L
C G
R L
C G
R L
C G
i1(t)
v1(t)
i2(t)
v2(t)
R L
C G
s uC x A x b
'' ' ' ' 's uC x A x b
Huge
Network Small
Network
MOR
Formulation
Realization
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
• Compare
• Moments
0
2 12 2 2 1 2 1 2
0 0 0 0
( ) ( )
1 ( 1)( ) ( ) ( ) ( ) ( )
2! (2 1)!
st
qq q q
H s e h t dt
h t dt th t dt s t h t dt s t h t dt s O sq
2 2 1 20 1 2 2 1( ) ( )q q
qH s m m s m s m s O s
Moments Matching: Pade Approximation
sasa
sbsbbsH
1
1110
1)(ˆ
Choose the 2q rational function coefficientsSo that the reduced rational function matches the first 2q moments of the original transfer function H(s).
,,,,,, 121121 qq bbbaaa ˆ ( )H s
2 2 1 20 1 2 2 1( ) ( )q q
qH s m m s m s m s O 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) qq aabbb ,,,,,, 1110
Pade Approximation: Coefficients12
122
2101
1110
1
q
qqq
qq smsmsmmsasa
sbsbb
012111
0111
00
mamamb
mamb
mb
qqqq
For aFor a11 a a22,…, a,…, aqq solve the following linear system: solve the following linear system:
12
2
1
1
2
1
221
2
21
1210
q
q
q
q
q
q
q
q
m
m
m
m
a
a
a
a
mm
m
mm
mmmm
Then, use the aThen, use the a11 a a22 … a … aqq to calculate b to calculate b00 b b11 … b … bq-1q-1 : :
Pade Approximation: Drawbacks• Numerically unstable
– Higher order moments
– Matrix powers converge to the eigenvector corresponding to the largest eigenvalue.
12
2
1
1
2
1
221
2
21
1210
q
q
q
q
q
q
q
q
m
m
m
m
a
a
a
a
mm
m
mm
mmmm
– Columns become linear dependent for large q. The problem is numerically very ill-conditioned.
• Passivity is not always preserved.– Pade may generate positive poles