Solutions of Linear Systems

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Analysis of autonomous linear systems Stability Lyapunov theory Variation-of-Constants Formula Transfer Matrices

Related Reading

[KK]: 2.5-2.6 and [AM]: 4.3-4.4, 5.1-5.3, 8.1-8.3

Autonomous Systems

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A dynamical state-space system without inputs/outputs is described as

x = f(x) with f : X Rn, X Rn.

Understanding the response of such systems to initial conditions Xis a rich field called dynamical systems theory. For us the main tool is

numerical simulation.

The system is linear if f is a linear map. All such maps are described as

f(x) = Ax with some matrix A Rnn. Hence a linear autonomoussystem is described as

x = Ax with A Rnn.

The responses of such systems can be completely understood by

using tools from linear algebra.

Diagonal A

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Suppose that A is a diagonal matrix:

A =

1 0... . . . ...0 n

with 1, . . . , n R.Then x = Ax splits up into the n scalar differential equations

x1 = 1x1, . . . , xn = nxn.

These can be solved independently. All solutions are

x1(t) = e1t1, . . . , xn(t) = e

ntn with 1, . . . , n R.This can be compactly written as

x(t) =

e1t 0... . . . ...0 ent

with = 1...n

.Observe that x(0) = .

State-Coordinate Change

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In practice A will not be diagonal. Can we still exploit what we have

seen? For this this purpose we introduce the following key concept.

Definition 1 Any non-singular (i.e. invertible) matrix T Rnndefines a state-coordinate transformation z = Tx.

Suppose that x(t) satisfies x(t) = Ax(t). Then z(t) = Tx(t) satisfiesz(t) = T x(t) = TAx(t) = (TAT1)z(t) = Az(t).

In the new coordinates the system is described with A = TAT1.

Conversely suppose that z(t) is any trajectory of z(t) = Az(t). Thenx(t) = T1z(t) satisfies x(t) = Ax(t).

The solution set of z = Az is linearly transformed by T1 into the

solution set of x = Ax.

Diagonizable A

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In many practical cases A can be transformed into a diagonal matrix A

by a suitably chosen state-coordinate change.

Theorem 2 Suppose that TAT1 = diag(1, . . . , n) Rnn.Then the unique solution of x = Ax, x(0) = is given by

x(t) = [T1 diag(e1t, . . . , ent)T ].

Proof. Just combine the two previous slides. x(t) satisfies

x(t) = Ax(t) with x(0) =

if and only if z(t) = Tx(t) and = T satisfy

z(t) = diag(1, . . . , n)z(t) with z(0) = .

The latter ivp has the unique solution

z(t) = diag(e1t, . . . , ent).

How to Diagonalize A?

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Matlab does it for us with [S,La]=eig(A).

Linearize Segway in upright position: Leads to

x = Ax with A =

0 0 1 00 0 0 10 3.92 2 0.320 22.1 3.23 1.82

.Get = S1AS with

S =

1 0.03 0.04 0.580 0.26 0.16 0.110 0.12 0.22 0.790 0.96 0.96 0.16

, =

0 0 0 00 3.72 0 00 0 6.16 00 0 0 1.38

.Qualitative insight. Each component of any solution x(.) of x =

Ax is a linear combination of the exponential functions

e0 t = 1, e3.72 t, e6.16 t, e1.38 t.

Example

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We can now answer all kinds of questions

about all or particular solutions of the linearization.

If equals either the first, the second, the third or

the fourth column of S, the respective solutions are1000

e0 t,

0.030.260.120.96

e3.72 t,0.040.16

0.220.96

e6.16 t,

0.580.110.79

0.16

e1.38 t.All other solutions are linear combinations thereof. Please note

that we have complete insight into how solutions behave for t!

Question: For which initial conditions does the solution not blow up?

Answer: For all in the subspace

S10

34

| 1, 3, 4 R.

Example

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For the linearization in the downright position

suitable matrices S, T = S1 and are given by

T =

1 0 0.7 0.120 0.03 + 2.58i 0.12 0.37i 0.54 + 0.08i0 0.03 2.58i 0.12 + 0.37i 0.54 0.08i0 0.36 1.33 0.21

=r1r2r3r4

S =

1 0.04i 0.04i 0.540 0.05 0.2i 0.05 + 0.2i 0.120 0.17 0.05i 0.17 + 0.05i 0.810 0.96 0.96 0.18

= ( c1 c2 c3 c4 )

=

0 0 0 00 1.17 + 4.46i 0 00 0 1.17 4.46i 00 0 0 1.49

= diag(1, 2, 3, 4).

Complex Transformations and Diagonal Matrices

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Despite the fact that T and are complex, the main insight on

slide 5 stays true without change!

In fact observe that 2 = 3, c2 = c3, r2 = r3. Therefore

T1 diag(e1t, e2t, e3t, e4t)T =

=(c1e

1t c2e2t c2e

2t c4e4t)

r1r2r2r4

== e1tc1r1 + e

2tc2r2 + e2tc2r2 + e

4tc4r4 =

= e1tc1r1 + 2Re[e2tc2r2] + e

4tc4r4

is always a real matrix. The response to the initial condition R4 ise1tc1(r1) + 2Re[e

2tc2(r2)] + e4tc4(r4).

Complex Transformations and Diagonal Matrices

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For a complex number = + i ( = Re(), = Im()) recall

et = e(+i)t = et[cos(t) + i sin(t)].

If c is a complex column and r a complex row then

cr = [Re(c) + iIm(c)][Re(r) + iIm(r)] = M + iN

with the real matrices

M = [Re(c)Re(r) Im(c)Im(r)], N = [Re(c)Im(r) + Im(c)Re(r)].

This leads to the explicit formula

Re[etcr] = et [cos(t)M sin(t)N ] .

The components of Re[etcr] hence are, qualitatively,

sustained ( = 0), growing ( > 0) or decaying ( < 0) oscillations.

Example: Linearization of Segway

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Solution with = col(1, 0, 0, 0) is x(t) = col(1, 0, 0, 0) for all t 0.

Solution with = col(0, 1, 0, 0) is

2 e1.17 t

cos(4.46 t)0.10.520.150.03

sin(4.46 t)0.010.130.43

2.48

+e1.49 t

0.2

0.040.29

0.06

and hence decaying in an oscillatory fashion:

0 1 2 3 4 50.5

0

0.5

1p and

Recap

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The complex number C is an eigenvalue of A Rnn if it is azero of the characteristic polynomial of A:

det(I A) = 0.If is an eigenvalue of A then any non-zero vector v Cn with

(I A)v = 0 or equivalently Av = vis a corresponding eigenvector of A.

Some very simple facts

If (, v) is a pair of eigenvalue/eigenvector then so is (, v). Any matrix A Rnn does have at most n different eigenvalues. For each eigenvalue one can compute at most n linearly independent

eigenvectors (by computing a basis of the null space of I A). Eigenvalues are invariant under a coordinate change.

How to Diagonalize A?

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For each eigenvalue of A Rnn we can hence compute a set of linearlyindependent eigenvectors; if done for , take the complex conjugate

vectors for ; collect these vectors in the list v1, . . . , vg and denote by

1, . . . , g the corresponding (not necessarily different) eigenvalues.

Theorem 3 The vectors v1, . . . , vg are linearly independent and

hence g n. In case that g = n then A can be diagonalized:S1AS = diag(1, . . . , n) with S =

(v1 vn

).

If g < n then A cannot be diagonalized by a coordinate change.

The formula for S1AS is very easy to see:

AS = A(v1 vn

)=

=(Av1 Avn

)=(1v1 nvn

)=

=(v1 vn

)diag(1, . . . , n) = S diag(1, . . . , n).

Example 1

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Consider the matrix

A =

3 4 6 4 82 3 4 2 60 0 1 0 00 2 2 1 22 2 4 2 5

The characteristic polynomial of A is computed with poly(A) to

det(I A) = (+ 1)2( 1)3.The LU-factorization (with command lu) of 1I A and 1I A give

U1 =

2 4 6 4 8

0 2 2 2 20 0 0 0 00 0 0 0 00 0 0 0 0

, U2 =4 4 6 4 8

0 2 1 0 20 0 2 0 00 0 0 0 00 0 0 0 0

Since the dimensions of the null-spaces of U1 and U2 are 3 and 2, we

have g = 5 = n and hence A can be diagonalized.

Example 1

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Linear independent vectors in the null-space of U1 and U2 are11

100

,

01010

,21

001

and

10010

,11

001

.

Then

S =

1 0 2 1 11 1 1 0 1

1 0 0 0 00 1 0 1 00 0 1 0 1

satisfies S1AS =

1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

.

Example 2

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Consider the matrix

A =

1 7 7 8 61 5 5 5 51 0 2 1 10 3 3 3 21 4 5 5 4

.The characteristic polynomial of A is computed with poly(A) to

det(I A) = (+ 1)2( 1)3.The LU-factorization (with command lu) of 1I A and 1I A give

U1 =

1 4 5 5 5

0 7 7 8 60 0 0 0.57 0.570 0 0 0 00 0 0 0 0

, U2 =2 7 7 8 6

0 3.5 0.5 3 20 0 2.57 0.57 0.290 0 0 0.89 0.440 0 0 0 0

.Since the dimensions of the null-spaces of U1 and U2 are 2 and 1, we

have g = 3 < 5 = n and hence A is not diagonizable.

Summary

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Let us be given x = Ax with A Rnn: We can check whether it can be diagonalized as shown on slide 13.

If existing one can compute a coordinate transformation T such that

TAT1 has the eigenvalues of A on its diagonal.

If A can be diagonalized, we can get complete insight into the solutionset of the differential equation as described on slide 5.

With the transformation as on slide 13 the formula on slide 5 reads as

x(t) =(e1tv1 entvn

)S1.

The eigenvalues determine the dynamic characteristics of the solution.

The eigenvectors provide information how these influence the state.

The eigenvalues of A are often called the modes of the system

x = Ax. The corresponding eigenvectors are the mode-shapes.

Matrix Exponential

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If we recall the Taylor series expansion of the function x ex we infer

T1

e1t 0... . . . ...0 ent

T = T1

k=0

1

k!(1t)

k 0...

. . ....

0 k=0

1

k!(nt)

k

T =

=k=0

1

k!

T1k1 0... . . . ...

0 kn

T tk=

k=0

1

k!

T11 0... . . . ...

0 n

Tk tk.

In case that A = T1 diag(1, . . . , n)T we hence conclude that

T1

e1t 0... . . . ...0 ent

T = k=0

1

k!(At)k.

Matrix Exponential

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For any M Rnn recall M0 = I and Mk = MMk1 for k = 1, 2, . . . .

Theorem 4 For any A Rnn and t R define

eAt := limN

Nk=0

1

k!(At)k =

k=0

1

k!(At)k.

The series converges uniformly on [T, T ] for any T > 0. Hencet eAt is a well-defined analytic function on R. eA0 = I and eA(t+) = eAteA and hence eAt = [eAt]1. d

dteAt = AeAt = eAtA.

The third statement follows from

d

dt

k=0

1

k!(At)k =

k=1

Ak1

k!

[d

dttk]

= Ak=1

Ak11

(k 1)!tk1 = AeAt

and the observation that we can also pull A out to the right.

General A

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Theorem 5 If A Rnn the unique solution of x = Ax, x(0) = is given by

x(t) = eAt.

Proof. To show existence we just check that x(t) is a solution:

x(0) = eA0 = and x(t) =d

dteAt = AeAt = Ax(t).

To show uniqueness let y(t) be another solution. Then

d

dt[eAty(t)] = [

d

dteAt]y(t) + eAt[

d

dty(t)] =

= AeAty(t) + eAtAy(t) = eAt[A+ A]y(t) = 0.Since eAty(t) has zero derivative for all t R, it must be constant. Ifwe evaluate at t = 0, we observe that this constant must equal . We

conclude eAty(t) = or y(t) = eAt = x(t) for all t R.

Computation of eAt

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Example. The double integrator q = u has the state-space description

x =

(0 10 0

)x+

(01

)u = Ax+Bu.

Now note that (At)2 = 0. Hence we have

eAt = I + (At) =

(1 00 1

)+

(0 t0 0

)=

(1 t0 1

).

All solutions of x = Ax are thus given by

x(t) =

(1 t0 1

) =

(1 + t22

).

This illustrates the following general structural insight.

All elements of eAt are linear combinations of the terms

e1t, te1t, . . . , tn1e1t, . . . , ept, tept, . . . , tn1ept

if 1, . . . , p are the pairwise different eigenvalues of A.

Jordan Form

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The key tool is the Jordan canonical form of a complex matrix.

Theorem 6 For A Cnn there exists an invertible S Cnn with

S1AS = J =

J1 0... . . . ...0 Jg

, Jl =l 1 0...

. . . . . ....

0 l 10 0 l

.Jl are so-called Jordan blocks. Moreover:

Up to permutation of the Jordan blocks, the Jordan canonicalform diag(J1, . . . , Jg) is uniquely determined by A.

1, . . . , g are the (not necessarily different) eigenvalues of A. There are exactly g linearly independent eigenvectors of A. A is diagonizable iff all Jordan blocks have dimension one.

Example

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Recall the example on slide 16. The command [S, J] = jordan(A) does

indeed return S1AS = J with

S =

3 1 1 5 32 1 1 4 3

0 0 1 1 11 0 1 2 2

2 1 1 3 2

and J =1 1 0 0 0

0 1 0 0 00 0 1 1 00 0 0 1 00 0 0 0 1

. A has three Jordan blocks of dimension 2, 2 and 1. The first, third, fifth column of S are linearly independent eigenvec-

tors of A for the eigenvalues 1, 1, 1. Other columns: Generalized ei-genvectors in the kernel of (AI) for eig(A) and = 2, 3, . . ..

Warning. Computing the Jordan canonical is numerically unrelia-

ble. Do not thoughtlessly use the command jordan in Matlab! The

Schur-decomposition is a much more reliable computational tool.

Computation of eAt

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Once we have computed the Jordan form of A we can determine eAt.

Theorem 7 If A equals SJS1 as in Theorem 6 then

eAt = SeJtS1 = S diag(eJ1t, . . . , eJgt)S1.

Moreover, if Jl has dimension d, then

eJlt =

1 t t2

2! td2

(d2)!td1

(d1)!0 1 t td3

(d3)!td2

(d2)!...

.... . .

......

0 0 0 1 t0 0 0 0 1

elt for l = 1, . . . , g.

This leads to an explicit formula for eAt. In Matlab use expm(A*t).

C. Moler, Ch. Van Loan, Nineteen dubious ways to compute the exponential

of a matrix, twenty-five years later, SIAM Review, 45: 3-49, 2003.

Complex Eigenvalues

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Partition the columns of S and the rows of T = S1 according to J as

S =(C1 Cg

)and T =

R1...Rg

.Then the next formula reveals how the modes of A contribute to eAt:

eAt = C1eJ1tR1 + C2e

J2tR2 + CgeJgtRg

If k is real we can make sure that Ck and Rk are real.

If k is complex and k = l, we can enforce Ck = Cl and Rk = Rl.

For example if 1 = 2 are complex and all other eigenvalues are real,

we then obtain a representation which is analogous to that on slide 9:

eAt = C1eJ1tR1 + C1e

J1tR1 + C3eJ3tR3 + + CgeJgtRg =

= 2Re[C1eJ1tR1] + C3e

J3tR3 + + CgeJgtRg.

Relation to Eigenvectors

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Another way to read the formulas is

A(C1 Cg

)=(C1 Cg

)diag(J1, . . . , Jg)

eAt(C1 Cg

)=(C1 Cg

)diag(eJ1t, . . . , eJgt)

or ACk = CkJk and eAtCk = Cke

Jkt for all k = 1, . . . , g.

If ck denote the first columns of Ck we infer from these two relations:

Ack = ckk = kck. Hence the collection c1, . . . , cg is a (largest) setof linearly independent eigenvectors of A (as on slide 13).

eAtck = ckekt. Hence with k = k + ik we infereAtRe(ck) = Re(cke

kt) = ekt[Re(ck) cos(kt) Im(ck) sin(kt)]eAtIm(ck) = Im(cke

kt) = ekt[Im(ck) cos(kt) + Re(ck) sin(kt)].

Example

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Consider again the matrix on slide 23. We infer that

eAt

3 1 1 5 32 1 1 4 3

0 0 1 1 11 0 1 2 2

2 1 1 3 2

S

=

=

3 1 1 5 32 1 1 4 3

0 0 1 1 11 0 1 2 2

2 1 1 3 2

et tet 0 0 0

0 et 0 0 00 0 et tet 00 0 0 et 00 0 0 0 et

.If is a linear combination of the first two columns of S then eAt

converges to zero for t. This follows fromlimt

et = 0 and limt

tet = 0.

Asymptotic Stability

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The stability analysis of autonomous systems is related to examining the

asymptotic behavior of x(t) if t. For linear systems the explicitrepresentation of the solutions clearly allows to answer such questions.

As an illustration we formulate the most important concept as follows.

Definition 8 The (equilibrium 0 of the) linear system x = Ax is

asymptotically stable if all solutions satisfy limt

x(t) = 0.

Since convergence should hold for all initial conditions, this is often

called global asymptotic stability. The dynamic property of asymptotic

stability can be verified by a simple algebraic test.

Theorem 9 The system x = Ax is asymptotically stable if and only

if all the eigenvalues of A have a negative real part. Matrices A with

this property are called Hurwitz.

Proof

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1) Suppose that all eigenvalues of A have negative real parts. If =

+ i is such an eigenvalue we have < 0. For k N0, t 0 we infer|tket| = |et[tkeit]| = ettk 0 for t.

Since all elements of eAt are linear combinations of such terms we get

limt

eAt = 0 and hence limt

eAt = 0 for all Rn.

2) Conversely suppose that = + i is an eigenvalue of A with 0and with eigenvector v 6= 0. Then eAtv = vet and hence

x(t) = eAtRe(v) = et[Re(v) cos(t) Im(v) sin(t)].Since either Re(v) 6= 0 or Im(v) 6= 0 and since 0, we infer x(t) 6 0for t . Hence we have found a real initial condition (either Re(v)or Im(v)) for which the solution does not converge to zero for t.

Example

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Consider the mechanical system ([AM] pages 142-145) depicted in

and described by

d

dt

q1q2q1q2

=

0 0 1 00 0 0 12k

mkm cm

0km2k

m0 c

m

q1q2q1q2

.The coordinate change z1 =

12(q1 + q2), z2 = z1, z3 =

12(q1 q2),

z4 = z4 leads toz1z2z3z4

=

0 1 0 0 km cm

0 0

0 0 0 10 0 3k

m cm

= ( A1 00 A2)

A

z1z2z3z4

.

Example

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The modes of the system are the eigenvalues of

A1 =

(0 1 km cm

)and A2 =

(0 13k

m cm

).

If all constants m, c, k are positive these are Hurwitz. Therefore the

system on the previous slide is asymptotically stable.

A real 2 2 matrix A is Hurwitz iff trace(A) < 0 and det(A) > 0.

From the definition of the matrix exponential we also infer

eAt =

(eA1t 0

0 eA2t

).

If z3(0) = 0, z4(0) = 0 then z3(t) = 0, z4(t) = 0 as well as(z1(t)z2(t)

)= eA1t

(z1(0)z2(0)

).

Since q1(t) = q2(t) this is the movement in Mode 1 (see figure).

Example

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If there is no damping we have c = 0. With 1 =

km

and 2 =

3km

we then infer

eAt =

(cos(t)

1

sin(t) sin(t) cos(t)

)for = 1, 2.

All system motions are sustained oscillations. We can as well distinguish

the two modes of movement as indicated in the figure:

0 5 101

0.5

0

0.5

1

0 5 101

0.5

0

0.5

1

Systems whose solutions do not decay to zero but only stay bounded

for future times, as in this example, have their own interest.

Lyapunov Stability

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Often we are only interested in analyzing whether the solutions stay

bounded for t and do not explode. This is related to what iscalled Lyapunov stability or neutral stability.

Theorem 10 All solutions of the system x = Ax are bounded for

t iff all eigenvalues of A have a non-positive real part, and allJordan blocks of eigenvalues with real part zero have dimension 1.

Lyapunov stability boils down to eAt being bounded for t .Hence solutions that start close to zero stay close to zero.

Asymptotic stability just means limt

eAt = 0. Note that the decay is

actually exponential.

All other more refined stability questions (decay rates, frequency of os-

cillations) can be obtained by directly analyzing eAt.

Nonlinear Systems

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From now on let us consider

x = f(x) with f C1(G,Rn) and open G Rn.

We denote by (., ) the solution of the ivp with x(0) = G.

Definition 11 The equilibrium xe G of x = f(x) is said to be stable, if for any > 0 there exists > 0 such that G, xe (t, ) xe for all t [0,). unstable if not stable. attractive, if there exists some > 0 such that

G, xe limt

(t, ) = xe.

asymptotically stable, if it is stable and attractive.

Note that these are all local concepts!

Geometric Visualization

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Stability:

Asymptotic Stability:

Lyapunov Functions

36/65

Definition 12 A continuously differentiable function V : G R issaid to be a Lyapunov function for the nonlinear ode x = f(x) if

V (x) := xV (x)f(x) 0 for all x G.

This is an algebraic property which is purely expressed in terms of the

function f and the gradient of V without involving any system trajectory.

As the key feature, this property implies for any state-trajectory of the

nonlinear ode in G, by the chain-rule, that

d

dtV (x(t)) = xV (x(t))x(t) = xV (x(t))f(x(t)) 0.

Therefore along all state-trajectories of x = f(x) in G, the function

t V (x(t)) is monotonically non-increasing for increasing times.

This feature turns V into an potential-function. Intuitively, state-

trajectories should hence converge to points in which V is minimal.

Lyapunov Stability Theorem

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This is made precise in the following Lyapunov theorem.

Theorem 13 Let V be a Lyapunov function for x = f(x) with

equilibrium xe G.1. If V (x) > V (xe) for all x G \ {xe} then xe is stable.2. If V (x) > V (xe) and V (x) < 0 for all x G \ {xe} then xe

is asymptotically stable.

One can assume w.l.o.g. that V (xe) = 0. Functions with V (x) > 0for all x G \ {xe} are called positive definite on G. Guaranteeing stability by searching for a Lyapunov function is the

so-called Direct Method of Lyapunov. There are no general recipes

for doing this. Dont forget: The stability properties are local!

In practice one often reduces G to an open ball around xe, whichfacilitates to fulfill the conditions.

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ProofNotation: Ur := {x Rn | x xe < r} and Ur := {x Rn | x xe r}.Suppose r > 0 is taken with Ur G. If we choose (0, r), we infer that

m := min{V (x) | x Ur, x xe } satisfies m > V (xe).If we hence fix any (V (xe),m) we conclude that

M := {x Rn | x Ur, V (x) } is compact in Rn and M U.Proof: x M x Ur and V (x) < m xxe < x U. Hence Mis bounded. We also infer {x Rn | x Ur, V (x) } = M ; hence M is closed.This implies that M is positively invariant:

M (t, ) M for all t 0.Proof: For M let (t, t+) be the interval of existence of (., ) w.r.t. Ur. Dueto V () we infer V ((t, )) for all t [0, t+) and thus (t, ) M forall t [0, t+). Since this implies (t, ) U for t [0, t+) and since < r, (t, )cannot approach the boundary of Ur which implies t+ =.Since V (xe) < , there exists > 0 with U M U. This implies stability: U M (t, ) M U for t 0.

Example

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Consider the unactuated mass-spring-damper system from Lecture 1:(x1x2

)=

(x2

kmx1 1mc(x2)

)= f(x1, x2) with c(.) C1(R).

Suppose that c(x2) = 0 exactly for x2 = 0 such that xe = (0, 0) is the

unique equilibrium. Motivated by considering the total energy define

V (x1, x2) =1

2kx21 +

1

2mx22.

Note that V (x) > V (0) = 0 for all x 6= 0. Moreover

xV (x)f(x) =(kx1 mx2

)( x2 kmx1 1mc(x2)

)= x2c(x2).

Suppose that c() satisfies, in addition, x2c(x2) 0 for all x2 R.This implies that xe = 0 is Lyapunov stable.

We are not able to conclude asymptotic stability from our result.

Lasalles Invariance Principle

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Theorem 14 Let V be a Lyapunov function for x = f(x) with

equilibrium xe G.1. Let V (x) > V (xe) for all x G \ {xe}.2. Let G and V ((t, )) = 0 for all t (t, t+) imply = xe.

Then xe is asymptotically stable.

Statement 2. trivially implies Statement 2. in Theorem 13.

Example continued

Suppose that c() satisfies x2c(x2) > 0 for all x2 R \ {0}. Thenxe = 0 is asymptotically stable.

Indeed x2(t)c(x2(t)) = 0 for t (t, t+) implies x2(t) = 0 and hencex1(t) = 0 for all t (t, t+) due to the (second) differential equation.

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ProofTake M from the proof of Theorem 13. If M it suffices to show lim

t(t, ) = xe.

Suppose not. Since M is compact, there exists t and x M Ur G withlim(t , ) = x 6= xe.

Since V is continuous on M and hence bounded from below, V ((t, )) is for t 0bounded from below. By monotonicity, the following limit hence exists:

l := limtV ((t, )) R.

Now consider (t, x) for some t (t, t+). Because of the continuous dependenceon initial conditions and the semi-group property we infer

(t, x) = lim(t, (t , )) = lim(t+ t , ).

We thus infer with the continuity of V und and the definition of l:

V ((t, x)) = limV ((t+ t , )) = l.

Here is the crux: Since V ((t, x)) = l for all t (t, t+), we infer ddtV ((t, x)) = 0and thus V ((t, x)) = 0 for all t (t, t+). By the hypothesis in 2. we concludex = xe, a contradiction.

Estimation of Region of Attraction

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Theorem 15 Let V be a Lyapunov function for x = f(x) with

equilibrium xe G.1. For R let M := {x G | V (x) } be compact in Rn.2. Let M and V ((t, )) = 0 for t (t, t+) imply = xe.

For all M we can then conclude limt

(t, ) = xe.

The sublevel-set M comprises points that are attracted by xe. In other

words, M is a subset of the region of attraction of xe, defined as

{ G | limt

(t, ) = xe}.

Note that M can be large and that this is not a local result any more.

Also note that we do not assume/claim stability of xe.

Proof

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M is positively invariant:

M (t, ) M for all t 0.Proof. If M then V () and thus V ((t, )) , i.e. (t, ) M for all t [0, t+).

We claim t+ =: If t+

Global Attractivity

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If the Lyapunov function is a barrier-function for G we can infer global

attractivity of the equilibrium.

Theorem 16 Let V be a Lyapunov function for x = f(x) with

equilibrium xe G.1. For any sequence x G with x x G or x

suppose that V (x) for .2. Let G and V ((t, )) = 0 for t (t, t+) imply = xe.

Then limt

(t, ) = xe for all G.

If G = Rn then 1. reduces to the condition V (x) for x ;such Lyapunov functions are said to be radially unbounded.

Proof

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For G we trivially have M := {x G | V (x) V ()}. Let usshow that M is compact in Rn.

If M is unbounded, there exists a sequence x M with x ; by hypothesis, V (x) such that there exists some mitV (x) > V (); this contradicts x M .

If M is not closed, there exists a sequence x M with x x Rnand x 6M . In case of x G, we could conclude by continuity of Vthat V (x) = lim V (x) V () and thus x M . Therefore, xis not contained in G but in G, which implies x G. By hypothesis,we have V (x), which again contradicts V (x) V ().

Hence M is closed and bounded and thus compact. We can apply Theo-

rem 15 and conclude (t, ) xe for t.

Indirect Method of Lyapunov

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The linearization of x = f(x) at xe was given by

x = Ax with A = xf(xe).

Recall that we hope for x(t) xe + x(t): A solution of the linearsystem leads to a good approximation of that of the nonlinear system.

This hope is reality, at least for the property of local asymptotic stability.

Theorem 17 Suppose that xf(xe) is Hurwitz. Then xe is a locally

asymptotically stable equilibrium of x = f(x).

Colloquially, asymptotic stability of the linearization also implies local

asymptotic stability of the original non-linear system around the point

of linearization. The converse is not true!

This is a major motivation why to study linear systems in such depth!

Example

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Consider again the example on slide 38. The linearization at xe = 0 is

xf(x) =

(0 1 km 1mc(x2)

)and thus A =

(0 1 km 1mc(0)

).

Recall that(0 1

)is Hurwitz iff > 0 and > 0.

Hence the matrix A of the linearization is Hurwitz if and only if c(0) > 0.

xe = 0 is locally asymptotically stable in case that c(0) > 0.

Note that c(0) > 0 implies x2c(x2) > 0 for x2 6= 0 close to zero. The converse is not true. This shows that Lasalles principle is a more

powerful tool to prove asymptotic stability.

Variation-of-Constants-Formula

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Lets come back to linear systems with inputs and outputs described as

x = Ax+Bu and y = Cx+Du

with A Rnn, D Rkm. There exists an explicit formula for thestate- and output-response of such systems.

Theorem 18 For the input u C([a, b],Rm) and the initial condi-tion x() = Rn, [a, b], the unique state-response equals

x(t) = eA(t) + t

eA(ts)Bu(s) ds.

The output-response hence is

y(t) = CeA(t) + t

0

[CeA(ts)B]u(s) ds+Du(t).

Hence the responses are given by a convolution. W.l.o.g. we take = 0.

Proof

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Important Trick: Search for solution x(t) = eAtz(t) with suitable z(t).

Differentiation implies

x(t) = AeAtz(t) + eAtz(t) = Ax(t) + eAtz(t).

This equals Ax(t) +Bu(t) as desired if

eAtz(t) = Bu(t) or z(s) = eAsBu(s).

Hence for any constant vector c the function

z(t) = c+

t

eAsBu(s) ds

does the job. c = leads to satisfaction of x(0) = .

Note that this trick of varying the constants can often be directly

applied for finding responses to explicitly given input signals.

Example

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Consider a constant input: u(t) = ue for all t R. Then

x(t) = eAt +

( t0

eA(ts) ds)Bue = e

At +

( t0

eA d

)Bue.

If A is Hurwitz then it is invertible and t0

eA d =

t0

d

deAA1 d = eAtA1 A1.

The state-response can hence be written with the matrix exponential as

x(t) = eAt[ + A1Bue] A1Bue.For t we conclude x(t) A1Bue =: xe. Therefore the stateconverges to the unique solution of Axe +Bue = 0. Moreover

y(t) = CeAt[ + A1Bue] + [D CA1B]ue.

This nicely displays the transient and the steady-state response. The

matrix D CA1B is called steady-state gain of the system.

Superposition Principle

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Clearly the matrix exponential eAt determines the state-response both to

non-zero initial conditions and to forcing inputs. The response depends

linearly on both and u(). This is called superposition principle.

If the system has multiple inputs (m > 1) we can partition

u(t) =

u1(t)...um(t)

, B = ( B1 Bm ) , D = ( D1 Dm ) .By the superposition principle the full output response is the sum of

CeAt and

t0

CeA(ts)Bkuk(s) ds+Dkuk(t), k = 1, . . . ,m

and each of these contributions can be analyzed separately.

Step and Impulse Response

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Let uk() be equal to the step function s(t) = 1 for t 0. Then t0

CeA(t)Bks() d +Dksk(t) = t

0

CeABk d +Dk

is the step response of the system (in the k-th input).

If uk() is equal to the impulse () at zero then t0

CeA(t)Bk() d +Dk(t) = CeAtBk +Dk(t)

is the impulse response of the system (in the k-th input). Observe

that it is the derivative of the step response.

Definition 19 The step- and impulse-response of the system are t0

[CeAB] d +D and CeAtB +D(t)

and can be obtained by applying m steps/impulses for each input.

Step and Impulse Response

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Steps and impulses should be viewed as test signals that allow to gather

information about the dynamical behavior of a system.

Both responses can be obtained by exiting the system with m well-

defined input signals from initial condition zero. Knowledge of the im-

pulse response

H(t) = CeAtB +D(t)

then allows to determine the response for any other signal by t0

H(t )u() d.

This is a principle path in order to determine a system description

from a finite number of (experimental) responses. This is the topic

in courses on system identification.

Sinusoidal Inputs

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For a complex = + i C and ue Rm consider the inputu(t) = ue e

t = ue et[cos(t) + i sin(t)].

Suppose IA is invertible. As on slide 46 we then have with = t :

y(t) = C

(eAt +

[ t0

eA(t)e d]Bue

)+D(uee

t) =

= C

(eAt + et

[ t0

e(AI) d]Bue

)+D(uee

t) =

= C(eAt + et

[e(AI)t I] (A I)1Bue)+D(ueet).

Reordering leads to the most important formula

y(t) = CeAt[ + (A I)1Bue

]+[C(I A)1B +D] (ueet)

which displays the transient and steady-state response. (Terminology

sensible if A is Hurwitz - then the transient decays to zero for t.)

Sinusoidal Inputs: Summary

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Theorem 20 For exponentially weighted sinusoidal complex input

u(t) = ue et = ue e

t[cos(t) + i sin(t)] ( = + i)

such that I A is invertible, the state-response isx(t) = eAt

[ (I A)1Bue

]+[(I A)1B] (ueet)

and the output-response is

y(t) = CeAt[ (I A)1Bue

]+[C(I A)1B +D] (ueet).

Since A, B, C, D and are real, the responses to the inputs

v(t) = ue et cos(t) and w(t) = ue e

t sin(t)

are just obtained by taking the real and imaginary parts.

Remark. For = 0 (pure sinusoidal inputs) the steady-state respon-

se involves the frequency response matrix C(iI A)1B +D.

The Laplace Transform

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Suppose that f : R C is measurable and of exponential type: Thereexists a constant such that

etf(t) is bounded for t 0.

Definition 21 For all s in the half-plane {s C : Re(s) > }, theone-sided Laplace transform of f is given by

f(s) = L(f)(s) =

0

estf(t) dt.

It is easy to show that f = L(f) is analytic on the above given half-plane(that can often be extended to much larger regions of C).

We rather consider this integral transform as an map which takes f into

f := L(f). L is obviously linear. Moreover L is injective. We only hintat very few more properties of the Laplace transform.

The Laplace Transform

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The following elementary facts suffice for our purposes.

Suppose f is the Laplace transform of f . Then

1. f has the Laplace transform sf(s) f(0).

2. t

0f() d has the Laplace transform 1

sf(s).

3. eptf(t) has the Laplace transform f(s+ p).

By iteratively applying these rules one can find pairs of functions that

are related through the Laplace transformation. An example is

1

(m 1)!tm1ept and

1

(s+ p)m.

Computing Responses with Laplace Transform

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Taking the transforms of all system signals on slide 44 leads to

sx(s) x(0) = Ax(s) +Bu(s), y(s) = Cx(s) +Du(s).

We can hence algebraically solve for x(s) to get with x(0) = that

x(s) = (sI A)1 + (sI A)1Bu(s),y(s) = C(sI A)1 + [C(sI A)1B +D]u(s).

This is the so-called frequency-domain analogue of the time-domain

formulas on slide 44. It is a key observation that the convolution integral

(in the time-domain) is just replaced by a simple multiplication (in the

frequency-domain).

The inverse Laplace transform often allows to determine x(t) and y(t).

Transfer Matrices

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Given the system on slide 44 we have seen that the matrix

G(s) = C(sI A)1B +D

depending the the complex variable s C plays a major role. For s C not being an eigenvalue of A we can compute (sIA)1. The elements of (sI A)1 are rational functions of s, since

(sI A)1 = 1(s)

adj(sI A) with (s) = det(sI A)where adj is the algebraic adjoint. In fact each element can be written

asnij(s)

(s)with a polynomial nij(s) of degree strictly smaller than that

of (s). Such rational functions are called strictly proper.

The elements of C(sIA)1B+D are linear combinations of thoseof (sI A)1 plus a constant matrix D.

Transfer Matrices

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Definition 22 C(sI A)1B + D is the transfer matrix of thesystem with state-space description x = Ax+Bu, y = Cx+Du.

All elements of G(s) = C(sI A)1B + D are rational functions ofs whose numerator degree is not larger than that of the denominator.

Such rational functional are called proper and G(s) is said to be a

proper rational matrix. In Matlab G(s) is computed by tf.

The poles of the transfer matrix G(s) are determined as follows:

Write each element of G(s) in the form nij(s)dij(s)

where the numerator

and denominator polynomials have no common zeros.

The union of all zeros of dij(s) for all elements are the poles of G(s).

Definition 23 G(s) is stable if all its poles have negative real parts.

Example

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Consider the electrical network from [F] p.21

and described by(v1

v2

)=

(1

C1R1+ 1C1R3

1C1R3

1C2R3

1C2R2

+ 1C2R3

)(v1

v2

)+

(1

R1C11

R2C2

)u.

With y = v and C1 = 1, C2 = 2, R1 = 1, R2 = 2, R3 = 3 we have

G(s) =

(s+0.5

s2+1.75s+0.5

0.25s+0.5s2+1.75s+0.5

).

The poles are the zeros of s2 + 1.75s+ 0.5 which are 1.39, 0.36.G(s) is stable. The steady-state gain is G(0) = DCA1B = (1 1)T .

Example

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Lets change C1 in steps from 1 to 4 and consider the step responses

from zero initial conditions:

0 10 20 301

0.5

0

0.5

1v1

0 10 20 301

0.5

0

0.5

1v2

For C1 = 4 the transfer matrix is

G(s) =

(0.25s+0.125

s2+0.75s+0.125

0.25s+0.125s2+0.75s+0.125

)=

(0.25s+0.25

0.25s+0.25

).

We had two different poles for C1 = 1 which reveal themselves in the

different time-constants of the response. For C1 = 4 only one pole

{0.25} remains related to the red response. Why? Well explain later!

Relation to Classical Techniques and Identification

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Lecture 1

First-orderstate-spacesystem

description

System ofhigher orderdifferentialequations

Transferfunction/matrix

description

Laplace & Inverse(Classical)

Laplace(Just Done)

???RealizationTheory

Identification

Experiments

Working with Transfer Matrices: Summary

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In view of slide 54 the transfer matrix is most useful in case that the

initial condition is zero. For a given input the transfer matrix fully

determines the output response.

In addition, if two systems have transfer matrices G1(s) and G2(s) then

the parallel and series interconnections have transfer matrices

G1(s) +G2(s) and G1(s)G2(s)

with the usual matrix operations. Note that the dimensions must be

compatible, and the order of the multiplication is important to be kept!

Moreover if A is Hurwitz and is not an eigenvalue of A then

G(0)ue, G(i)eitue, G()e

tue

are steady-state responses for constant ( = 0), purely sinusoidal ( =

i) and exponentially weighted sinusoidal inputs in the direction ue.

Block-Diagrams

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Block-diagrams for systems translate into block-diagrams for transfer

matrices. Let us stress the similarity to diagrams for linear maps in

linear algebra or functional analysis:

Series interconnection: y(s) = [G1(s)G2(s)]u(s) is depicted as

u(s) y(s)G2(s) G1(s)

u1(s) = y2(s)

Notice the order of the blocks!

Parallel interconnection: y(s) = [G1(s) +G2(s)]u(s) is depicted as

y(s)+u(s)

G2(s)

G1(s)y1(s)

y2(s)

Realizations

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A state-space system (uniquely) determines its transfer matrix by direct

computation. The converse question reads as follows.

Problem: Let us be given a k m-matrix G(s) whose elementsconsist of proper rational functions. Can we compute matrices A Rnn, B Rnm, C Rkn and D Rkm such that

G(s) = C(sI A)1B +D ?

Let us introduce the following important terminology.

Definition 24 If G(s) = C(sI A)1B +D then (A,B,C,D) issaid to be (state-space) realization of G(s).

The above formulated question is called the realization problem. In

Matlab just use ss. The theory will be discussed in detail in Lecture 5.

Coordinate Change

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The matrices (A,B,C,D) of a realization of G(s) are never unique.

One reason (there are many more!) is as follows.

A state-coordinate change modifies the describing matrices of a

state-space system but it does not change the transfer matrix.

More precisely, for the state-space system

x = Ax+Bu, y = Cx+Du

perform the coordinate change z = Tx. This leads to

z = Az+Bu, y = Cz+Du with

(A B

C D

)=

(TAT1 TB

CT1 D

)

while we have C(sI A)1B + D = C(sI A)1B +D.

Proofs

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If we perform for x = Ax + Bu, y = Cx + Du the coordinate change

z = Tx, we infer x = T1z and we can thus conclude

z = T x = TAx+TBu = [TAT1]z+[TB]u and y = [CT1]z+Du.

For A = TAT1, B = TB, C = CT1, D = D this is nothing but

z = Az + Bu, y = Cz + Du.

Invariance of the transfer matrix is shown as follows:

C(sI A)1B + D =

= [CT1](sTT1 [TAT1])1 [TB] +D =

= CT1T (sI A)1 T1TB +D == C(sI A)1B +D.

Covered in Lecture 2

65/65

Analysis of autonomous linear systemscoordinate change, diagonalization, Jordan form, modes, all solutions

matrix exponential

Stability and Lyapunov functionsHurwitz matrices, Lyapunov functions, Lyapunov stability theorems,

indirect method

Analysis of LTI systemsVariation-of-constants formula, computation of responses (steady-

state, transient), Laplace

Transfer matricesdefinition, working with transfer matrices, realizations, behavior under

state coordinate-change