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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 m2q1s 2q1 O(s 2q ) Moments Matching: Pade Approximation H (s) m0 m1s m2 s 2 Hˆ ( s ) m2q1s 2q1 O(s 2q ) b0 b1s bq1s q1 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 bq1s q1 1 a1s aq s q m0 m1s m2 s 2 m2 q1s 2 q1 For a1 a2,…, aq solve the following linear system: m0 m1 m2 mq 1 aq mq m a m m 2 1 q1 q 1 m2 aq2 mq 2 mq 1 m2 q1 m2 q 2 a1 For b0 b1, …, bq-1 calculate: b0 m0 b1 m1 a1m0 bq1 mq1 a1mq2 aq1m0 Pade Approximation: Drawbacks • Numerically unstable – Higher order moments – Matrix powers converge to the eigenvector corresponding to the largest eigenvalue. m0 m1 m2 mq1 aq mq m m m2 aq1 1 q1 m2 aq2 mq2 mq1 m2 q1 m2 q2 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