CSE 245: Computer Aided Circuit Simulation and Verification · Impulse Response The Impulse Response of a system is defined as the Zero State Response resulting from an impulse excitation

CSE 245: Computer Aided Circuit Simulation and Verification

Fall 2004, Sep 28

Lecture 2:

Dynamic Linear System

Lecture2.2

OutlineTime Domain Analysis

State Equations RLC Network Analysis by Taylor ExpansionImpulse Response in time domain

Frequency Domain AnalysisFrom time domain to Frequency domainCorrespondence between time domain and frequency domainSerial expansion of (sI-A)-1

Lecture2.3

Outline (Cont’)Model Order Reduction

Moments Passivity, Stability and Realizability

Symbolic Analysis Y-Delta TransformationBDD Analysis

Lecture2.4

State of a systemThe state of a system is a set of data, the value of which at any time t, together with the input to the system at time t, determine uniquely the value of any network variable at time t. We can express the state in vector form

x =

Where xi(t) is the state variables of the system

)(...

)()(

2

1

tx

txtx

k

Lecture2.5

State VariableHow to Choose State Variable?

The knowledge of the instantaneous values of all branch currents and voltages determines this instantaneous stateBut NOT ALL these values are required in order to determine the instantaneous state, some can be derived from others.choose capacitor voltages and inductor currents as the state variables! But not all of them are chosen

Lecture2.6

Degenerate NetworkA network that has a cut-set composed only of inductors and/or current sources or a loop that contains only of capacitors and/or voltage sources is called a degenerate networkExample: The following network is a degenerate network since C1, C2 and C5 form a degenerate capacitor loop

Lecture2.7

Degenerate NetworkIn a degenerated network, not all the capacitors and inductors can be chosen as state variables since there are some redundancy

On the other hand, we choose all the capacitor voltages and inductors currents as state variable in a nondegenerate network

We will give an example of how to choose state variable in the following section

Lecture2.8

Order of Circuitn = bLC – nC - nL

n the order of circuit, total number of independent state variablesbLC total number of capacitors and inductors in the networknC number of degenerate loops (C-E loops)nL number of degenerate cut-sets (L-J cut-sets)n = 4 – 1 = 3

In a nondegenerate network, n equals to the total number of energy storage elements

Lecture2.9

State Equations

StateInput

Output

)()()(

)()(

tDutQxty

tButAxdtdx

+=

+=

Linear system of ordinary differential equations

Lecture2.10

State Equation for RLC Circuits

The state equation is of the form

Or

vt: voltage in the trunk, capacitor voltageil: current in the loop, inductor current.Y and R are the admittance matrix and impedance matrix of cut-set and meshE covers the co-tree branches in the cut-set–ET covers the tree trunks in the mesh analysis

L

C0

0

l

t

iv&&

− RE

EYT

l

t

iv

= - + Pu

= Gx(t) + Pu(t))(tx&M

Lecture2.11

State Equations

If we shift the matrix M to the right hand side, we have

Let A = M-1G and B = M-1P, we have the state equation

Together with the output equation

are called the State Equations of the linear system

= Gx(t) + Pu(t))(tx&M

)(tx& = M-1Ax(t) + M-1Bu(t)

= Ax(t) + Bu(t))(tx&

)(ty = Qx(t) + Du(t)

Lecture2.12

RLC Network AnalysisA given RLC network

Degenerate Network, Choose only voltages of C1 and C5, current of L6 as our state variable

Vs

g3

g4C1

C2

C5

L6

1 2

0

Lecture2.13

Tree StructureTake into tree as many capacitors as possible and,as less inductors as possibleResistors can be chosen as either tree branches or co-tree branches

Vs

g3

g4C1

C2

C5

L6

1 2

0

g3C1 C5 g4

1 2C2/L6

0

Vs

Lecture2.14

Linear State EquationBy a mixed cut-set and mesh analysis, consider capacitor cut-sets and inductor loops only. we can write the linear state equation as follows

M = Gx(t) + Pu(t))(tx&

Cut-set KCL

Loop KVLCut-set KCL

+−

−+

6

522

221

0000

LCCC

CCC

6

2

1

ivv

&

&

&

−−01110

10

4

3

gg

6

2

1

ivv

003g

=- + Vs

Lecture2.15




Lecture2.16

Solving RCL Equation by Taylor Expansion(1)General Circuit Equation

Consider homogeneous form first

BUAXX +=•

AXX =•

0XeX At=

...!

...!2!1

22

+++++=ktAtAAtIe

kkAt

Q: How to Compute Ak ?

and

Lecture2.17

Assume A has non-degenerate eigenvalues and corresponding linearly

independent eigenvectors , then A can be decomposed as

where and

Solving RCL Equation by Taylor Expansion (2)

1−ΧΛΧ=A

kΧΧΧ ,...,, 21

kλλλ ,...,, 21

=Λ

kλ

λλ

LL

MOLM

ML

L

0

000

2

1

[ ]kΧΧΧ=Χ ,...,, 21

Lecture2.18

What’s the implication then?

To compute the eigenvalues:

1−ΧΛΧ=A

01

1 ...)det( ccAI nn

n +++=− −− λλλ

0)...(...))(( 212

0 =++−= ppp λλλ

realeigenvalue

ConjugativeComplexeigenvalue

122 −ΧΧΛ=A

1−Λ ΧΧ= tAt ee where

=Λ

ke

ee

e t

λ

λ

λ

LL

MOLM

ML

L

0

000

2

1


Lecture2.19

In the previous example

==

−−

)0()0(

1

2//1/10

01

2

iv

eXeiv t

lrlc

At

11

00

1110 −

−

+− Χ

=ΛΧ=

−−

=λ

λXXA

231

231

j

j

−−=

+−=

−

+

λ

λwhere

−

−−−

=Χ1

231

12

31

3 j

jj

−−+−=Χ−

231

231

111 jj

hence 1

00 −Χ

Χ=

−

+

λ

λ

ee

eAt

Let c=r=l=1, we have


Lecture2.20

What if matrix A has degenerated eigenvalues? Jordan decomposition !

1−ΧΧ= JAJ is in the Jordan Canonical form

And still 1−ΧΧ= JtAt ee


Lecture2.21

Jordan Decomposition

=

λλ0

1J

=+

+

= t

ttJt

etee

teλ

λλ

λλ

001

1001

L

=

λλ

λ

001001

J

=+

+

=

t

tt

ttt

Jt

etee

ettee

teλ

λλ

λλλ

λλ

λ

000

!2

001001

100010001

2

L

similarly

Lecture2.22


State Equations RLC Network Analysis by Taylor ExpansionResponse in time domain


Lecture2.23

Response in time domainWe can solve the state equation and get the closed form expression

The output equation can be expressed as

Note: * denotes convolution

Lecture2.24

Impulse ResponseThe Impulse Response of a system is defined as the Zero State Response resulting from an impulse excitation

Thus, in the output equation, replace u(t) by the impulse function δ(t), and let x(t0)=0 we have

h(t) = y(t) = QeAt B

Lecture2.25




Lecture2.26

Solutions in S domainBy solving the state equation in s domain, we have

Suppose the network has zero state and the output vector depends only on the state vector x, that is, x(t0) = 0 and D = 0, we can derive the transfer function of the networkH(s) = = Q(sI-A)-1B

x(s) = (sI-A)-1 x(t0)+ (sI-A)-1 Bu(s)y(s) = Qx(s) +Du(s) = Q(sI-A)-1(x(t0) + Bu(s)) +Du(s)

)()(

ss

uy

Lecture2.27

Key Key Transform Transform Property:Property:

Bilateral Bilateral LaplaceLaplaceTransform:Transform:

)()()()()(

sQxsysBusAxssx

=+=

)()(

)()()(

tQxty

tButAxdt

tdx

=

+=

dtdx )(ssx

dtetxsx st∫∞

∞−

−= )()()(tx

Frequency Domain Representation

Lecture2.28

Express y(s) as a Express y(s) as a function of u(s)function of u(s)

)()()()()(

sQxsysBusAxssx

=+=

Transfer Function:Transfer Function: )(sH

)()()( 1 sBuAsIQsy −−−=

System Transfer Function

Lecture2.29

Transfer FunctionTime Domain Impulse ResponseFrequency domain representationFrequency domain representation

H(s)u(s) y(s) = H(s) u(s)

Linear systemLinear system

h(t)u(t) ∫ −=t

duthty0

)()()( τττ

Linear systemLinear system

Time domain representationTime domain representation

The transfer function H(s) is the The transfer function H(s) is the LaplaceLaplace Transform Transform of the impulse response h(t)of the impulse response h(t)

Lecture2.30




Lecture2.31

Correspondence between time domain and frequency domain

We can derive the time domain solutions of the network from the s domain solutions by inverse Laplace Transformation of the s domain solutions.

State Equations in S domain

State Equations in time Domain

Inverse LaplaceTransform

sx(s) – x(t0)= Ax(s) +Bu(s)

y(s) = Qx(s) +Du(s)

x(t) = L-1[(sI-A)-1x(t0) + (sI-A)-1 Bu(s)]

= L-1[(sI-A)-1]x(t0) + L-1[(sI-A)-1]B*u(t)

y(t) = L-1[Q(sI-A)-1(x(t0) + Bu(s)) +Du(s)]

= Q L-1[(sI-A)-1] x(t0) + {QL-1 [(sI-A)-1]B +Dδ(t)}* u(s)

Lecture2.32

Correspondence between time domain and frequency domain

(sI-A)-1 eAt

multiplication of u(s) in s domain corresponds to the convolution in time domain

Solution from time domain analysis

Solution by inverse Laplacetransform

x(t) = L-1[(sI-A)-1x(t0) + (sI-A)-1 Bu(s)]

= L-1[(sI-A)-1]x(t0) + L-1[(sI-A)-1]B*u(t)

y(t) = L-1[Q(sI-A)-1(x(t0) + Bu(s)) +Du(s)]

= Q L-1[(sI-A)-1] x(t0) + {QL-1 [(sI-A)-1]B +Dδ(t)}* u(s)

Lecture2.33




Lecture2.34

Serial expansion of (sI-A)-1

When s 0 we can write (sI-A)-1 as

Thus, the transfer function can be written as

When s ∞ we can write (sI-A)-1 as

The transfer function can be written as

(sI-A)-1 = -A-1(I – sA-1) = -A-1(I + sA-1 + s2A-2 + … + skA-k + …)

H(s) = Q(sI-A)-1B = -QA-1(I + sA-1 + s2A-2 + … + skA-k + …)B

(sI-A)-1 = s-1(I – s-1A)-1 = s-1(I + s-1A + s-2A2 + … + s-kAk + …)

H(s) = Q(sI-A)-1B = s-1(I + s-1A + s-2A2 + … + s-kAk + …)B

Lecture2.35

Assume A has non-degenerate eigenvalues and corresponding linearly

independent eigenvectors , then A can be decomposed as

where and

Matrix Decomposition

1−ΧΛΧ=A

kΧΧΧ ,...,, 21

kλλλ ,...,, 21

=Λ

kλ

λλ

LL

MOLM

ML

L

0

000

2

1

[ ]kΧΧΧ=Χ ,...,, 21

Lecture2.36

Matrix DecompositionThen we can write (sI-A)-1 in the following form

(sI-A)-1 in s domain corresponds to the exponential function eAt in time domain, we can write eAt as

(sI-A)-1 = (SI – XΛX-1)-1 = X-1(sI – Λ)-1X = X-1

−

−

−

ns

s

s

λ

λ

λ

1.

.2

1

1

1

X

eAt = X-1

t

t

t

ne

ee

λ

λ

λ

..

2

1

X

CSE 245: Computer Aided Circuit Simulation and Verification · Impulse Response The Impulse Response of a system is defined as the Zero State Response resulting from an impulse excitation

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