Model Order Reduction - Circuit Simulation - Lecture Slides, Slides of Computer Science

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CSE245: Computer-Aided CircuitSimulation and Verification

Model Order Reduction (1)^ Lecture Notes 3

• Introduction Outline
• • FormulationLinear System
  • – Time Domain AnalysisFrequency Domain Analysis
  • – MomentsStability and Passivity
  • Model Order Reduction

• A network is stable if, for all bounded inputs,Stability
• the output is bounded.For a stable network, transfer function

H(s)=N(s)/D(s) (Recall: H(s) = Y(s)/U(s) in our case) (general form)

• Should have only negative poles p  0 j, i.e. Re(pj)
• If pole falls on the imaginary axis, i.e. Re(p = 0, it must be a simple pole. i)

• Passivity Passivity
  • – Passive system doesnA one-port network is said to be passive if the’t generate energy

total power dissipation is nonnegative for all initialtime twaveforms, that is, 0 , for all time t>t 0 , and for all possible input

• Passivity of a multi-port network^ where E(t^0 ) is the energy stored at time t^0
  • If all elements of the network are passive, thenetwork is passive

• For multi-port, suppose each port is either a voltage sourceLinear Multi-Port Passivity

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 isa voltageThen we will have D=BT (^) in the state equation
• Let U(s) be the input vector of all ports,function , thus the output vector Y(s) = H(s)U(s) and H(s) be the transfer

7

• Average power delivered to this multi-port network is
• For a passive network, we should have

Linear System Passivity

8

• State Equation (s domain)
  • We have shown that transfer function is
    • We will show that this^ where^ and network is passive , that is Plug conjugateH(s) in here After Model Order Reduction (MOR) (in laterslides), this condition^ …Why is this important? must still be true.

Passivity and Stability

10

• • 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 isnot necessarily stable.
• The interconnection of passive components ispassive.

Model Order Reduction (MOR)

11

• MOR techniques are used to build a reducedorder model to approximate the original circuit iv 1 (t) 1 (t) R LC (^) G R LC GR LC G R LC (^) G iv 22 (t)(t) iv 11 (t)(t) R LC (^) G i 2 (t)

C s^ x A x b u

                                                  

 (^) C '  (^) s  (^) x '  (^)   (^) A '          x (^) '  (^)  b (^) ' u '

HugeNetwork SmallNetwork MOR

Formulation Realization

Moments Review

13

• Transfer function
• Compare
• Moments

H s ( )    (^0)  0 e (^) h t dt ^ st ( ) h t dt ( )  (^)    0 th t dt ( )  (^) s  (^) 2! 1   0 t h t dt (^2) ( )  (^) s (^2)   (^) (2( 1) q  (^2) 1)! q  (^1)   0 t 2 q  (^1) h t dt ( )  s 2 q  (^1)  O s ( 2 q ) H s ( )  m 0 (^)  m s 1  m s 2^2   m 2 (^) q  1 s^2 q^ ^1  O s ( 2 q )

Moments Matching: Pade Approximation

14

Choose the 2q rational function coefficients H ˆ^ ( s )  b^01  ba^1 s^1 s ^  baq  q^1^ ssqq^ ^1 So that the reduced rational function of the original transfer function H(s). (^) H s ˆ^ ( ) matches the first 2q moments a^1^^ ,^ a^2 , aq^ ^1 , b^1 , b^2 , bq ^1 ,

H s ( )  m 0 (^)  m s 1  m s 2^2   m 2 (^) q  1 s^2 q^ ^1  O s ( 2 q )

Pade Approximation: Coefficients

16

b 0 1  ba 11 ss  baq  q 1 ssqq ^1  m 0  m 1 s  m 2 s 2  m 2 q  1 s 2 q  1

11 1 1 1 10 2 1 0

0 0 bb mm aamm a m

b m q   q   q    q 

  

For a 1 a 2 ,…, aq solve the following linear system:





   

 2 1

21 1

21 1 2 2

21 2

0 1 2 1 q

qq

q qq

q q q

q m

mm

m a

aa

a m m

mm m

m m m m    

Then, use the a 1 a 2 … aq to calculate b 0 b 1 … bq-1 :

Pade Approximation: Drawbacks

17

• Numerically unstable – Higher order moments
  • Matrix powers converge to the eigenvector corresponding to the largest eigenvalue. 

 

 

   

 

 

 

 

 

 

 

      

 2 21

1 12

1 (^2122) 10 21 2 1 q q qqq qq q q

q mm

mm aa aa mm m

mm mm m m    

 

• Passivity is not always preserved.^ –^ Columns become linear dependent for large q. The problem isnumerically very ill-conditioned.
  • Pade may generate positive poles Docsity.com