lecture 3: model order reduction i
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TRANSCRIPT
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CSE245: Computer-Aided Circuit Simulation and
VerificationLecture Note 3
Model Order Reduction (1)
Spring 2008
Prof. Chung-Kuan Cheng
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Outline
• Introduction
• Formulation
• Linear System– Time Domain Analysis– Frequency Domain Analysis– Moments– Stability and Passivity– Model Order Reduction
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Transfer FunctionState Equation in Frequency Domain (
suppose zero initial condition)
Express Y(s) as
a function of U(s)
Solve X
Transfer Function:
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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.
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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
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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
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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
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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
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Passivity Proof• To show
• Is equivalent to show
• We have
where
• Thus
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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.
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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
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Model Order Reduction: Overview
• Explicit Moment Matching– AWE, Pade Approximation
• Implicit Moment Matching– Krylov Subspace Methods
• PRIMA, SPRIM
• Gaussian Elimination– TICER– Y-Delta Transformation
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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
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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
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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
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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
For bFor b00 b b11, …, b, …, bq-1q-1 calculate: calculate:
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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