MIT System Theory Solutions

8/10/2019 MIT System Theory Solutions

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MASSACHUSETTSINSTITUTEOFTECHNOLOGYDepartment

of

Electrical

Engineering

and

Computer

Science

6.241: DynamicSystemsSpring2011

Homework

1

Solutions

Exercise1.1 a)GivensquarematricesA1 andA4,weknowthatA issquareaswell:

A

1

A2A=0

A4

I 0 A

1 A2= .0

A4 0 I

Note

that

I

0det

=

det(I)det(A4) =det(A4),0

A4

which

can

be

verified

by

recursively

computing

the

principal

minors.

Also,

by

the

elementary

operationsofrows,wehave

A1 A2 A1 0det

=

=

det

=

det(A1).0

I

0

I

FinallynotethatwhenAandB aresquare,wehavethatdet(AB) =det(A)det(B). Thuswehave

det(A) =det(A1)det(A4).

b) AssumeA11

andA4

1 exist. Then

1

A1 A2 B1 B2 I 0AA =

0

=

0 I

,

A4 B3 B4

whichyieldsfourmatrixequations:

1. A1B1

+A2B3

=I,

2. A1B2

+A2B4

=0,

3. A4B3

=0,

4. A4B4

=I.

From Eqn (4), B 1 1 14 = A

4, with which Eqn (2) yields B2 = A1

A2A4

. Also, from Eqn (3)B3

=0,withwhic 1hfromEqn(1)B1

=A1

. Therefore,

A1 A1A2A1

A1 =

1

1 4

0

A14

.

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Exercise1.2 a)

0

I A1 A2 A A=

3

I 0 A3

A4

4

A1

A2

b)Letusfind

B2B =

B

1

B3 B4

suchthat A1BA =

A20 A A A 14 3 1

A2

Theaboveequation impliesfourequationsforsubmatrices

1. B1A1

+B2A3

=A1,

2.

B1A2+

B2A4 =

A2,

3. B3A1+B4A3 =0,

4. B3A2

+B4A4

=A A3A

14

1

A2.

First two equations yield B1 = I and B2 = 0. Express B3 from the third equation as B =3 4 A fourth.B A3 1

1 and plug it into the After gathering the terms we get B 14 A4

A3A

1A2 =

A4A3A1

1

A2,whichturns intoidentity ifwesetB4 =I. Therefore

I 0B =

1A3A1 I

c)

Using

linear

operations on rows we see that det(B) = 1. Then, det(A) = det(B)det(A) =det

(BA) =

det

(A1)det A4A3A11A2

.

Note

that

A4A3A

1

1

A2

does

not

have

to

be

invertiblefortheproof.

Exercise

1.3

We

have

to

prove

that

det(IAB) =det(IBA).

Proof: SinceI andIBAaresquare, I 0

det(IBA) = detB

IBA

I A I=

detA

= det

B I 0

I

I

A

Idet

A ,

B I 0 I

yet,

from

Exercise

1.1,

we

have

det

I A=

det(I)det(I) = 1.

0 I

Thus, I A

det(IBA) =det .B I

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d

1

d d

(A(t)B(t))

=

A(t)B(t) + t A(t)B(t) + tA(t)

B(t) + h.o.t.

A(t)B(t)

.

oflimits,i.e.d A(t+ t)B(t+ t) A(t)B(t)

(A(t)B(t)) =

lim

dt t0 t

WesubstitutefirstorderTaylorseriesexpansions

d

A(t

+ t) =

A(t) + t A(t) +

o(t)dt

d

B(t

+ t) =

B(t) + t B(t) +

o(t)dt

toobtain

dt t

dt dt

Hereh.o.t. standsfortheterms

d d

h.o.t.

=

A(t) + t A(t)

o(t) +

o(t)

B(t) + t B(t)

o(dt

+ t

2

),dt

amatrix quantity,where limt 0

h.o.t./t

=0(verify). Reducingtheexpressionandtakingthe

limit,weobtaind d d

[A(t)B(t)]= A(t)B(t) +A(t) B(t).dt dt dt

b)

For

this

part

we

write

the

identity

A1(t)A(t) =

I.

Taking

the

derivative

on

both

sides,

we

have

d d dA1(t)A(t)

= A1(t)A(t) +A1(t) A(t) =0

dt dt

dt

3

Now, I A

det

=

det

IAB 0

=

det(IAB).B I B I

Therefore

det(IBA) =det(IAB).

Notethat(IBA) isaqq matrixwhile(IAB) isappmatrix. Thus,whenonewantstocompute

the

determinant

of

(IAB)or(IBA),s/hecancomparepandqtopicktheproduct(AB

or

BA)

with

the

smaller

size.

b)Wehavetoshowthat(IAB)1A=A(IBA)1.Proof: Assumethat(IBA)1 and(IAB)1 exist. Then,

A =

AI=A(IBA)(IBA)1

= (AABA)(IBA)1

= (I

AB)A(I

BA)1

(IAB)1 1A = A(IBA) .

Thiscompletestheproof.

Exercise1.6 a)Thesafestwaytofindthe(element-wise)derivativeisbyitsdefinitioninterms


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RearrangingandmultiplyingontherightbyA1(t),weobtain

dA1(t) =A1

d(t) A(t)A1(t).

dt dt

Exercise1.8Let ( X={g x) = 2 M0+1x+2x + +M

x

|i C}.

a)

We

have

to

show

that

the

set

B

=

{1, x,

, xM}

is

a

basis

for

X.Proof

:

1.

First,

lets

show

that

elements

in

B

are

linearly

independent.

It

is

clear

that

each

element

in

B cannotbewrittenasa linearcombinationofeachother. Moreformally,

c (1)+c (x) + +c (xM1 1 M

) = 0 i ci

= 0.

Thus,elementsofB arelinearly independent.

2.

Then,

lets

show

that

elements

in

B

span

the

space

X.

Every

polynomial

of

order

less

than

or

equal

to

M

looks

like M

p(x) = ixi

i=0

forsomesetof

is.Therefore,

{1, x1, , xM}spanX.

db) T :X X and T(g(x)) = g(x).dx

1. ShowthatT islinear.Proof:

dT

(ag1(x) +bg2(x)) = (ag1(x) +bg2(x))dx

d d= a g1

+b g2dx dx

=

aT

(g1) +bT(g2).

Thus,T islinear.

2.

g(x) =

20+1x+2x + +M

M

x ,

so

T

(g(x))

=

+ 2 x

+

+

M xM

2 M

1

1

.

Thus

it

can

be

written

as

follows:

0 1 0 0 0

0 1

0 0 2 0 0 0 0 0 3

0

. . . . .. . . .

.. . . .

. 0

1

23..

=

22

33...

M M

.

0 0 0 0 M .0 0 0 0 0 M 0

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The big matrix, M, is a matrix representation of T with respect to basis B. The columnvector

in

the

left

is

a

representation

of

g(x)

with

respect

to

B.

The

column

vector

in

the

rightisT(g)withrespecttobasisB.

3.

Since

the

matrix

M

is

upper

triangular

with

zeros

along

diagonal

(in

fact

M

is

Hessenberg),

theeigenvaluesareall0;i = 0i= 1, , M+ 1.

4. OneeigenvectorofM for1 =0mustsatisfyM V1 =1V1 = 0

V1 =

10

...

0

is one eigenvector. Since i

s are not distinct, the eigenvectors are not necessarily independent.

Thus

in

order

to

computer

the

M

others,

ones

uses

the

generalized

eigenvector

formula.

5


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MIT OpenCourseWarehttp://ocw.mit.edu

6.241J / 16.338J Dynamic Systems and ControlSpring2011

For information about citing these materials or our Terms of Use, visit:http://ocw.mit.edu/terms.
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MASSACHUSETTSINSTITUTEOFTECHNOLOGYDepartmentofElectricalEngineeringandComputerScience

6.241: DynamicSystemsFall2007

Homework2Solutions

Exercise1.4 a)Firstdefineallthespaces:

R(A) = {yCm | xCn suchthaty=Axm

}

R(A) = {zC |yz=zy= 0,y

R(A)}

R A ( ) = {pCn | vCm suchthatp=Av}

N(A) = {xCn |Ax= 0}

N(A) = {qCm |Aq= 0}

i)ProvethatR(A) =N(A).Proof: Let

z

R(A) yz= 0y R(A)

xAz = 0xCn

Az= 0z N(A)

R(A) N(A).

Now let

q

N

(A

)

A

q

= 0

xAq= 0xCn

yq= 0y R(A)

q R(A)

N(A) R(A).

ThereforeR(A) =N(A).

ii)ProvethatN(A) =R(A).Proof: From i)weknowthatN(A) =R(A)byswitchingAwithA. Thatimpliesthat

N

(A) =

{R(A)} =

R(A).

b)Showthatrank(A) +rank(B)nrank(AB)min{rank(A),rank(B)}.Proof: i)Showthatrank(AB)min{rank(A),rank(B)}. Itcanbeprovedasfollows:Each column of AB is a combination of the columns of A, which implies that R(AB) R(A).Hence,dim(R(AB))dim(R(A)),orequivalently,rank(AB)rank(A).Each row of AB is a combination of the rows of B rowspace (AB) rowspace (B), but thedimensionofrowspace=dimensionofcolumnspace=rank,sothatrank(AB)rank(B).Therefore,

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rank(AB)min{rank(A),rank(B)}.

ii)Showthatrank(A) +rank(B)nrank(AB).Let

rB = rank(B)

rA = rank(A)

whereACmn , BCnp.Now, let {v1, , vrB} be a basis set of R(B), and add nrB linearly independent vectors{w1, , wn rB}to

thisbasistospanallofCn,{v1, v2, , v , w , , w . n 1 nrB}Let

M = v1| v2 vrB

| w1| wnrB

= V W .

Suppose

x

Cn,

then

x

=

M

for

some

Cn.

1. R(A) =R(AM) =R([AV|AW]).

Proof: i) Let x R(A). Then Ay = xforsomey Cn. But y can be written as a linearcombinationofthebasisvectors ofCn,soy=M forsomeCn.Then,Ay=AM =xx R(AM) R(A) R(AM).

R Cnii) Let x (AM). Then AM y =x for some y . ButM y= zCn Az=xxR(A) R(AM) R(A).Therefore,R(A) =R(AM) =R([AV|AW]).

2. R(AB) =R(AV).Proof: i)Letx R CrB (AV). Then AV y = x for some y . Yet,V y=B forsomeCp

since

the

columns

of

V

and

B

span

the

same

space.

That

implies

that

AV

y

=

AB

=

x

x R(AB) R(AV) R(AB).ii) Let x R(AB). Then (AB)y = x for some y Cp. Yet, again By = V for someCrB ABy=AV=xx R(AV) R(AB) R(AV).Therefore,R(AV) =R(AB).

Usingfact1,weseethatthenumberoflinearlyindependentcolumnsofAislessthanorequaltothenumberoflinearlyindependentcolumnsofAV +thenumberoflinearlyindependentcolumnsofAW,whichmeansthat

rank(A)rank(AV) +rank(AW).

Usingfact2,weseethat

rank(AV) =rank(AB)rank(A)rank(AB) +rank(AW),

yet,therereonlynrB columns inAW. Thus,

rank(AW)nrB

rank(A)rank(AB)rank(AW)nrB

rA(nrB)rAB .

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Thiscompletestheproof.

Exercise2.2 (a)Forthe2nd orderpolynomialp 22(t) =a0 +a1t+a2t , wehave f(ti) =p2(ti) +ei i= 1, . . . ,16, and ti T. We can express the relationship between yi and thepolynomial asfollows;

1

t1 t21

y1 . . .

ea 1 .. . 0. . . . . = . a 1 + .. 2 . 1 t16 t 16 a2

y16

e16

Thecoefficientsa0, a1,anda2 aredeterminedbytheleastsquaressolutiontothis(overconstrained)

problem,a= (A A)1Ay,whereaLS = a0 a1 a2

.

Numerically,thevaluesofthecoefficients

are:

aLS= 0.2061

0.5296

0.375

For the 15th order polynomial, by a similar reasoning

w

e can express the relation between data

pointsyi andthepolynomialasfollows:y1 1 t

2 151 t1 t1 a0 e1

..

. . .

.

. . .=

. .

.

.

.

..

+ .. . . . . . .y 1 t t2 15

16 16 16

t16

a15 e16

This can be rewritten as y = Aa+e e

. Observ that matrix A is invertible for distinct t s.

i So

thecoefficientsa 1i ofthepolynomialareaexact =A y,whereaexact = a0 a1 a15

. The

resulting error in fitting the data is e = 0, thus we have a perfect fit

at these particular

time

instants.

Numerically,thevaluesofthecoefficientsofare:

0.49999998876521 0.39999826604650

0.16013119161635

0.044575313859820

.00699544100513 0.00976690595462

0.02110628552919

0.02986537283027

aexact =

0.03799813521505

0.00337725219202

0.00252507772183

0.00072658523695

0.00021752221402

0.00009045014791

0.00015170733465

0.00001343734075

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The function f(t) as well as the approximating polynomialsp15(t) andp2(t) are plotted inFigure2.2b. Note that whileboth polynomials are agood fit, thefifteenthorderpolynomial is abetterapproximation,asexpected.b)Nowwehavemeasurementsaffectedbysomenoise. Thecorrupteddata is

yi =f(ti) +e(ti) i= 1, . . . ,16 ti T

wherethenoisee(ti) isgeneratedbyacommandrandn inMatlab.Following the reasoning inpart (a), wecanexpress the relation between thenoisydatapoints yiandthepolynomialasfollows:

y =Aa+e

The

solution

procedure

is

the

same

as

in

part

(a),

with

y

replaced

by

y.

Numerically,thevaluesofthecoefficientsare:

0.00001497214861

0.000894425437810

.01844588716755

0 .14764397515270 0.63231582484352

1.

62190727992829

2.61484909708492

2.67459894145774

a

=

exact 1.67594757924772

105

0.56666848864500

0.06211921500456

0.002196227259540.01911248745682

0.01085690854235

0

.00207893294346

0.00010788458590

4

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5Part A Dashed: 2nddegree approximation ; Dashdotted: 15thdegree approximation

Figure2.2a


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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 23

2

1

0

1

2

3

4

5

6

7Part B Dashed: 2nddegree approximation ; Dashdotted: 15thdegree approximation

Figure2.2b

and 1.2239

aLS = 0.10890.3219

The function f(t) as well as the approximating polynomialsp15(t) andp2(t) are plotted inFigure 2.2b. The second order polynomial does much better in this case as the fifteenth orderpolynomial ends up fitting the noise. Overfitting is a common problem encountered when tryingtofitafinitedatasetcorruptedbynoiseusingaclassofmodelsthat istoorich.

Additional

Comments A stochastic derivation shows that the minimum variance unbiased

estimatorforaisa=argminy Aa2 whereW =R1,andRn isthecovariancematrixoftheW n

random

variable

e.

So,

a = (AW A)1AW y.

Roughly speaking, this is saying that measurements with more noise are given less weight in theestimateofa.Inourproblem, Rn =I becausethee

isare independent,zeromeanandhaveunit

variance. Thatis,eachofthemeasurments isequallynoisyortreatedasequallyreliable.

c)p2(t)canbewrittenasp2(t) =a0+a1t+a2t

2.

Inordertominimizetheapproximationerrorinleastsquaresense,theoptimal p2(t)mustbesuchthattheerror,fp2, isorthogonaltothespanof{1,t,t

2}:

< f

p2,

1

>= 0

=< p2,

1

>

< fp2, t>= 0 =< p2, t>

< f

p2, t2 >= 0=< p2, t

2 > .

5


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Figure2.2c

Wehavethatf = 1e0.8t fort[0,2],So,2

2 1 5 = 0

5 e .8tdt= e

8

5

0 2 8

8

2 t

0.8t 15 8 25= e dt= e5 +0 2 32 32

2 t2 85 125=

e0 .8tdt= e5

64 .

0 2 64

And,

8

< p2,

1

>= 2a0+ 2a1+

a

23

8< p2, t>= 2a0+ a

1+ 4a23

28 32

< p2, t >= a0+ 4a1+ a23 5

Thereforetheproblemreducestosolvinganothersetof linearequations:

2 2 8 a 0 3 2 8 4 a

1 = .38 4 32 a 3 5

Numerically,

the

values

of

the coefficien

ts

2

are:

0.5353a

= 0.2032

0.3727

The function f(t) and the approximating

polynomial

p2(t) are plotted in Figure 2.2c. Hereweuseadifferent notion forthe closeness of the approximatingpolynomial,p2(t), totheoriginalfunction, f. Roughly speaking, in parts (a) and (b), the optimal polynomial will be the one for

6

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20.5

1

1.5

2

2.5Part C Dashed: 2nddegree approximation


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whichthereissmallestdiscrepancybetweenf(ti)andp2(ti)forallti,i.e.,thepolynomialthatwillcome closest to passing through all the sample points, f(ti). All that matters is the 16 samplepoints,f(ti). Inthisparthowever,allthepointsoff matter.

y1 C1 e1 S1 0Exercise

2.3

Let

y

= ,

A

= ,

e

=

and

S

= .

y2 C2 e2 0 S2

NotethatAhasfullcolumnrankbecauseC1 hasfullcolumnrank. AlsonotethatS issymmetricpositive definite since both S1 and S2 are symmetric positive definite. Therefore, we know thatx = argmineSeexistsand isuniqueand isgivenbyx= (ASA)1ASy.

Thusbydirectsubstitutionofterms,wehave:

x = (C1

S1C1+C2S2C2)

1(C1

S1y1+C2S2y2)

Recallthatx1 = (C1S1C1)

1C1S1y1 andthatx2 = (C2

S2C2)1C

2S2y2. Hencexcanbere-written

as:

x = (Q1+

Q2)1(Q1x1+

Q2x2)

Exercise 2.8 We can think of the two data sets as sequentially available data sets. x is theleastsquaressolutiontoyAxcorrespondingtominimizingtheeuclideannormofe1 =yAx.

y Ax istheleastsquaressolutionto

z

D x

correspondingtominimizinge1

e1+e2

Se2 where

e2 =zDxandS isasymmetric(hermitian)positivedefinitematrixofweights.Bytherecursionformula,wehave:

x =x + (AA+DSD)1DS(zDx)

Thiscanbere-writtenas:

x =x + (I+ (AA)1DSD)1(AA)1DS(zDx)

=x + (AA)1D(I+SD(AA)1D)1S(zDx)

ThisstepfollowsfromtheresultinProblem1.3(b). Hence

x =x + (AA)1D(SS1+SD(AA)1D)1S(zDx)

=x + (AA)1D(S1+D(AA)1D)1S1S(zDx)

=

x + (A

A)1

D

(S1

+

D(A

A)1

D

)1

(z

Dx)

In order to ensure that the constraint z = Dx is satisfied exactly, we need to penalize thecorresponding error term heavily (S ). Since D has full row rank, we know there exists atleastonevalueofxthatsatisfiesequationz=Dxexactly. Hencetheoptimizationproblemwearesettingupdoes indeed haveasolution. TakingthelimitingcaseasS ,henceasS1 0,wegetthedesiredexpression:

x =x + (AA)1D(D(AA)1D)1(zDx)

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InthetrivialcasewhereD isasquare(hencenon-singular)matrix,thesetofvaluesofxoverwhich we seek to minimize the cost function consists of a single element, D1z. Thus, x in thiscase issimply x =D1z. It is easytoverifythattheexpressionweobtaineddoes in fact reducetothiswhenD isinvertible.

Exercise

3.1

The

first

and

the

third

facts

given

in

the

problem

are

the

keys

to

solve

this

problem,

inadditiontothefactthat:R

U A= .

0

Here note that R is a nonsingular, upper-triangular matrix so that it can be inverted. Now theproblemreducestoshowthat

x =argminyAx2 =argmin(yAx)(yAx)2x x

is indeedequaltox =R1y1.

Letstransformtheproblem intothefamiliarform. We introduceanerroresuchthat

y=Ax+e,

andwewouldliketominimizee2 whichisequivalenttominimizingyAx2. Usingtheproperty

e2 =U e2.

of an orthogonal matrix, we have that

Thus with Ax, we havee=y

2 2 2U e =eUU e= (U(yAx))(U(yAx))=U yU Ax2

2

e =2 2

y1 R = (y1Rx)(y1Rx) +y2

y2.= x0y22

Sincey22 =y

2 y2 isjustaconstant, itdoesnotplayanyrole inthisminimization.2

Thuswewould liktohavey1Rx = 0

andbecauseR isan invertiblematrix,x=R1y1.

8


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of

Electrical

Engineering

and

Computer

Science

6.241: DynamicSystemsFall2007

Homework

3

Solutions

Exercise3.2 i)Wewouldliketominimizethe2-normofu, i.e.,u2 . Sinceyn isgivenas2n

yn = hiun1

we

can

rewrite

this

equality

as

i=1

yn =

h1 h2 hn

un1un

2

.

..

u0

Wewanttofindtheuwiththesmallest2-normsuchthat

y =

Au.

where

we

assume

that

A

has

a

full

rank

(i.e. hi = 0 for some i, 1 in). Then, the solution

reduces

to

the

familiar

form:

u =A(AA)1y.n h2iBynotingthatAA

=

,

we

can

obtain

uj asfollows;i=1

uj =hjyn h2

,i=1 i

for j = 0,1, ,n1.

ii)a)Letsintroduceeasanerrorsuchthatyn =ye. Itcanalsobewrittenasyyn =e. Thennow

the

quantity

we

would

like

to

minimize

can

be

written

as

r(y

yn)2+

u02+

+

un21

whererisapositiveweightingparameter. Theproblembecomestosolvethefollowingminimizationproblem

:

n

u =argmin ui2+re2 =argmin(

u

22+r

e

22),

u u

i=1

from which we see that r is a weight that characterizes the tradeoff between the size of the finalerror,

y

yn,andenergyoftheinputsignal,u.

In order to reduce the problem into the familiar form, i.e, yAx, lets augment re at thebottom

of

u

so

that

a

new

augmented

vector,

u

is u

u =

,

re

1


17/75

Thischoiceofufollowsfromtheobservationthatthisistheuthatwouldhaveu22 =u22+re2,the

quantity

we

aim

to

minimize

.

Now

we

can

writey asfollows

u

y

= A

..

. 1 =

Au =

Au

+

e

=

yn+e.r

re

Now,ucanbeobtainedusingtheaugmentedA,A,as

u =A(AA)1y=A

1

AA+

1y.

rr

By

noting

that

n1

1

AA+ =

hi2+

,

r ri=1

we

can

obtain

uj as

follows

uj = nhjhy2 1 for j = 0, ,n1.i=1 i +rii) b) When r = 0, it can be interpreted that the error can be anything, but we would like tominimizethe inputenergy. Thusweexpectthatthesolutionwillhavealltheui

stobezero. Infact,

the

expression

obtained

in

ii)

a)

will

be

zero

as

r

0.

On

the

other

hand,

the

other

situation

isan interestingcase. Weputaweightoftothefinalstateerror,thentheexpression from ii)a)givesthesameexpressionas ini)asr .

Exercise

3.3

This

problem

is

similar

to

Example

3.4,

except

now

we

require

that

p(T

) = 0.

We

tcanderive,fromx(t) =p(t),thatp(t) =x(t)tu(t) = (t)x()d wheredenotesconvolution

0

and u(t) is the unit step, defined as 1 when t > 0 and 0 when t =

0g(t)f(t)d isan innerproducton

the

space

of

continuous

functions

on

[0, T

],

denoted

by

C[0, T

],

which

we

are

searching

for

x(t).

So,

wehavethaty=p(T) =and0 =p(T) =. Inmatrixform,y

< T

t,

x(t)

>

0 =

=

T

t

1

, x(t)

where .,. denotes the Grammian, as defined in chapter 2. Now, in chapter 3, it was shownthattheminimum lengthsolutiontoy=A,x, isx=AA,A1 y. So,forourproblem,

x =

Tt 1 Tt 1 , Tt 1 1 y0

.

Where,

using

the

definition

of

the

Grammian,

we

have

that:

T

t

1 , Tt 1 < Tt,Tt > < Tt,1> . =

2


18/75

Now,wecanusethedefinition for innerproducttofindthe individualentries,< T t,T t >=T (T

t)2dt

=

T

3

/3,

< T

t,

1

T>= (T

t)dt

=

T

2/2,

and

=

T

.

Plugging

these

in,

one

0 0

cansimplifytheexpressionfor and 12y x obtain x(t) = 12 [ t ]fort[0, T].T 2 TAlternatively,

we

have

that

x(t) =

p(t).

Integrating

both

sides

and

taking

into

account

that

p(0) = 0 and

t tp(0) =

0,

we

hav

tep(t) =

1 x()d

dt =

f(t )dt .

Now,

we

use

the

integrationt

10 0 0 1

u

1

by

parts

formula,

t tdv

=

uv|t0

v

du,

with

u

=

f(t1)

=

1

x()d,

and

dv

= dt ;

hence

du

0 0 1 =

0t

=tdf(t1) =x(t1)dt1 andv t1.Plugging inandsimplifyingwegetthatp(t) = 1 x()d dt 1 =0 0t T

(t

)x()d.Thus,y=p(T) = (T)x()d =< Tt,x(t)> .Inaddition,wehavethat0 0T0 =p(T) = x()d = .That is,weseektofindtheminimumlengthx(t)suchthat

0

y = < Tt,x(t)>0 = .

Recall that the minimum length solution x(t) must be a linear combination of Tt and 1, i.e.,x(t) =

a1(T

t) +

a2.So,

y

=

< T

t,

Ta 21(Tt) +a2 > = a1

(Tt) dt+ Ta2 (T t)dt = a T3 T21 +a20 0 3 2T

2

0 = = (a1(Tt) +

a T2)dt = a1 +a2T.0 2

This isasystemoftwoequationsandtwounknowns,whichwecanrewriteinmatrixform:

T3 T2

y a

1= 3 20 T

2

T

a2

,

2

So,

T3 T2 1 a1 y=

3 2 .

a T2

2 T 02

Exercise4.1NotethatforanyvCm,(showthis!)

v v2 mv . (1)

Therefore,forACmn withxCn

Ax2 AxmAx forx= 0,Ax2 m .x2

x2

But,

from

equation

(1),

we

also

know

that

1

1 .

Thus,x x2

Ax2

mAx m Ax

m

x 2 x 2 A

x ,

(2) Equation(2)mustholdforallx=0,therefore

3


19/75

Axmax =0 2

x = A2 mA .x2

Toprovethelowerbound 1 A

A2,reconsiderequation(1):n

Ax

Ax n AxAx 2 n Ax Ax 2 forx= 0, A 2

x 2

x 2

x 2 2

x 2 nA 2.

(3)

But,fromequation(1)forx nCn,

2

1 . So,x xAx

n Ax

Ax 2

n x2

for

all

x

=

0

Axincluding

x

that

makes

maximum,

so,x

Axmaxx=0 =

A x

nA2,

or

equivalently,

1 An

A2.

Exercise4.5AnymnmatrixA, itcanbeexpressedas

0A

=

U 0 0

V

,

where U and V are unitary matrices. The Moore-Penrose inverse, or pseudo-inverse of A,denoted

by

A+,

is

then

defined

as

the

n

m

matrix

A+ =

V

U.

0 0

1 0

w

a) No we

have

to

show

that

A+A

and

AA+ are

symmetric,

and

that

AA+A

=

A

and

A+AA+ =

+A .Supposethat is adiagonal invertible matrixwith the dimensionofrr. Usingthe givendefinitionsas ellasthefact w thatforaunitarymatrixU,UU =U U =I,wehave

+

0

AA =

U

0 0

0

V

V

1 0U

0 0

1 0

= U I U0 0 0 0

Ir = U r 0

U,0 0

4


20/75

which issymmetric. Similarly, +

1 0 0A A = V U

U V

0 0

0 0

1 0

0

= V I

V0 0

0 0

I 0=

r V r V

0 0

which isagainsymmetric.The

facts

derived

above

can

be

used

to

show

the

other

two.

+ +

Ir r 0AA A

= (AA )A

=

U

UA

0

0 0

Ir r 0

=

U UU

0

0

V

0 0

0=

U V

0 0

= A.

Also,

A+AA+ Ir r 0

= (A+A)A+ =

V

+

V

A

I 0

=

r

V r0 0

1 0

0 0

1 0

V

V

U0 0

=

V

U

0 0= A+.

b)

We

have

to

show

that

when

A

has

full

column

rank

then

A+ = (AA)1A,

and

that

when

A

has full row rank then A+ 1= A(AA) . If A has full column rank, then we know that m n,rank(A) =n,and

nn=U

A V

.

0

Also,

as

shown

in

2,

when

A

has

full

column

rank,

chapter

(AA)1

exists.

Hence

1

1

V

U

(AA) A = V 0 U V

0 U0 1

= VV V 0 U

= V ()1

V

V

0

U

1= V( )

0

U

= V( 1 0 )U

= A+.

5


21/75

Similarly, ifAhasfullrowrank,thennm,rank(A) =m,and

A

=

U

mm 0 V.

ItcanbeprovedthatwhenAhasfullrowrank,(AA)1 exists. Hence,

A(AA)1 =

V

U U

0

V

V

U

1

0 0

= V

U

UU1

0

=

V

UU(1)U

0

1= V U

0

= A+.

c)

Show

that,

of

all

x

that

minimize

y

Ax2,theonewiththesmallest lengthx2 isgivenbyx =A+y. IfA has fullrowrank, we haveshown inchapter3thatthesolutionwiththesmallestlengthisgivenby

x =

A(AA)1y,

andfrompart(b),A(AA)1 =A+. Therefore

x =A+y.

Similary,

it

can

be

shown

that

the

pseudo

inverse

is

the

solution

for

the

case

when

a

matrix

A

hasafullcolumnrank(comparetheresultsinchapter2withtheexpressionyoufoundinpart(b)

for

A+

when

A

has

full

column

rank).

Now, lets consider the case when a matrix A is rank deficient, i.e., rank(A) = r


22/75

solutionwiththeminimumnormcanbeachievedwhenzi =0fori=r+ 1, r+ 2, , n. Thus,wecan

write

this

z

as

z= 1c,

where 1 r

01 = 0 0

and

r is a square matrix with nonzero singular values in its diagonal in decreasing order. Thisvalueofz alsoyieldsthevalueofxofminimal2normsinceV isaunitarymatrix.Thusthesolutiontothisproblemis

x = V z=V1c=V1Uy=A+y.

Itcanbeeasily shownthatthischoiceofA+ satisfiesalltheconditions, ordefinitions, ofpseudoinverse ina).

Exercise 4.6. a) Suppose A Cnm has full column rank. Then QR factorization for A can beeasilyconstructedfromSVD:

nA= U

V

0

wheren isanndiagonalmatrixwithsingularvaluesonthediagonal. LetQ=U andR= nVandwegettheQR factorization. SinceQ isanorthogonalmatrix,wecanrepresentanyY Cmpas

YY

1=

Q

Y2

Next

Y

AX2 Y1 R 2 Y1 RX 2F =Q

Q

XF = Q

Y2

0

Y2

F

Denote

Y2

Y1 RXD=

and note that multiplication by an orthogonal matrix does not change Frobenius norm of thematrix:

QD

2

F

=

tr(DQQD) =

tr(DD) =

2DF

Since

Frobenius

norm

squared

is

equal

to

sum

of

squares

of

all

elements,

square

of

the

Frobenius

normofablockmatrix isequaltosumofthesquaresofFrobeniusnormsoftheblocks: Y

1 RX

2

RXY F

=

2

Y1

2 +

Y 22 F F

SinceY2 blockcannotbeaffectedbychoiceofX matrix,theproblemreducestominimizationofY 21RXF. RecallingthatR is invertible(becauseAhasfullcolumnrank)thesolution is

X=R1Y1

7


23/75

b)EvaluatetheexpressionwiththepseudoinverseusingtherepresentationsofAandY fromparta):

AA

1

AY =

R 0

QQ R

1

R 0

QQ Y1 =R1

R

1

R 0

Y1 =R1Y10 Y2 Y2

From 4.5 b) we know that if a matrix has a full column rank, A+ = (AA)1A, therefore bothexpressionsgivethesamesolutions.c)

Y AY AX2 +ZBX2 2F F = Z B XF

A

Since A has full column rank, also has full column rank, therefore we can apply resultsB

from

parts

a)

and

b)

to

conclude

that

A A 1 A Y 1 X

= =

AA

+

BB AY

+

BZ

B B

B Z

8


24/75





25/75


of

Electrical

Engineering

and

Computer

Science


Homework

4

Solutions

Exercise 4.7 Given a complex square matrix A, the definition of the structured singular valuefunction isasfollows.

1

(A) =min {max() 0 |det(IA) = }

where

is

some

set

of

matrices.

a) If = {I : C}, then det(IA) = det(IA). Here det(IA) = 0 implies thatthereexistsanx=0suchthat(IA)x= 0.Expandingthelefthandsideoftheequationyields

x

=

Ax

1

1

x

=

Ax.

Therefore

is

an

eigenvalue

of

A.

Since

()

=

||, max

1

arg

min{max()|det(IA) = 0}=|| = .

|max(A)

|

Therefore,

(A) =|max(A)|.

b)If={Cnn},thenfollowingasimilarargumentasina),thereexistsanx=0suchthat(I

A)x= 0.That impliesthat

x= Ax x2 =Ax2 2Ax21

Ax

2

max(A)2 x2

1

max().max(A)

Then, weshowthatthe lowerboundcanbeachieved. Since={Cnn},wecanchoosesuch

that 1

=

max(A) 0

V ...

0

U.

where

U

and

V

are

from

the

SVD

of

A,

A

=

UV

.

Note

that

this

choice

results

in

1

0

0IA=IV .

.

.

0

V =V 1 .

V

..

1

which is singular, as required. Also

from the construction

of

, max() =

1 . Therefore,max(A)

(A) =max(A).

1


26/75

c) If={diag(1, , n)|i C}with D {diag(d1, , dn)|di >0}, wefirst note that D1

exists.

Thus:

det(ID1AD) = det(ID1AD)

=

det((D

1

D

1A)D)

= det(D1D1A)det(D)

=

det(D1(IA))det(D)

= det(D1)det(IA)det(D)

= det(IA).

WherethefirstequalityfollowsbecauseandD1 arediagonalandthelastequalityholdsbecausedet(D1) = 1/det(D).

Thus,

(A) =(D1AD).

Now lets show the left side inequality first. Since 1 2 , 1 = {I| C} and 2 =

{diag(1,

, n)},

we

have

that

min{max()|det(IA) = 0} min{max()|det(IA) = 0},

1

2

which

implies

that

1

(A)2

(A).

But

from

part

(a),

1

(A) =

(A),

so,

(A)2

(A).

Now

we

have

to

show

the

right

side

of

inequality.

Note

that

with

3 =

{

C},

we

have

2

3.

Thus

by

following

a

similar

argument

as

above,

we

have

min () det(IA) = 0} min () det(IA) = 0}.

2

{max |

3

{max |

Hence,

2

(A) =2

(D1AD)3

(D1AD) =max(D1AD).

Exercise4.8WearegivenacomplexsquarematrixAwith rank(A)=1. AccordingtotheSVDofAwecanwriteA=uv whereu,varecomplexvectorsofdimensionn. Tosimplifycomputations

we

are

asked

to

minimize

the

Frobenius

norm

of

in

the

definition

of

(A).

So

1(A) =

min{ F |det(IA) = 0}

is

the

set

of

diagonal

matrices

with

complex

entries,

={diag(1, , n)|i C}. Introduce

the column vector = (1, , n)T and the row vector B = (u1v1

, , unvn ), then the original

problemcanbereformulatedaftersomealgebraicmanipulationsas

1(A) =

minCn{ 2 |B= 1}

2


27/75

Toseethis,weusethefactthatA=uv,and(fromexcercise1.3(a))

det(IA) = det(Iuv)

= det(1vu)

= 1

vu

Thusdet(IA)=0 impliesthat1vu= 0.Thenwehave

1 = vu u 1 1

.vn

.

.

.= v1 ..

n

1

n u

=

v1 .

u1

vn

un

= B

.

.

n

Hence,computing(A)reducestoa leastsquareproblem, i.e.,

min{F|det(IA) = 0} 2 s.t.1 = B.

min

Wearedealingwithaunderdeterminedsystemofequationsandweareseekingaminimumnormsolution.

Using

the

oprojection

theorem,

the

optimal

is given

from =

B(BB )1.

Substituting

intheexpressionofthestructuredsingularvaluefunctionweobtain:

n (A) = |uivi

2

i=1

|

In

the

second

part

of

this

exercise

we

define

to

be

the

set

of

diagonal

matrices

with

real

entries,

= {diag(1, , n)|i R}. The idea remains the same, wejust have to alter

the constrain

tRe(B)

equation,

namely

B

=

1 + 0j.

Equivalently

one

can

write

D

=

d

where

D

=

and

d

= Im(B)

1

.

Again

the

optimal

is

obtained

by

use

of

the

projection

theorem

and

o =

D(DDT)1d.

0

Substitutingintheexpressionofthestructuredsingularvaluefunctionweobtain:

1(A) =

dT(DDT)1d

Exercise5.1SupposethatACmn isperturbedby thematrixECmn.

1. Showthat

|max(A+E)max(A)| max(E).

3


28/75

AlsofindanE thatachievestheupperbound.

Note

that

A

=

A

+

E

E

A

=

A

+

E

E A

+

E

+

E A A

+

E E.

Also,

(A

+

E) =

A

+

E A+E A+E A+E A E.

Thus,

putting

the

two

inequalities

above

together,

we

get

that

|A+E A|E.

Note

that

the

norm

can

be

any

matrix

norm,

thus

the

above

inequality

holds

for

the

2-inducednorms

which

gives

us,

|max(A+E)max(A)| max(E).

AmatrixE thatachievestheupperbound is

1 0 0 0.

0 . . 0 0E

=

U

V

=

A,

. .

. .. .

r

0 0

...

0

where

U

and

V

form

the

SVD

of

A.

Here,

A

+

E

=

0,

thus

max(A+E)=0,and

|0 +max(A)|=max(E)

isachieved.

2. Suppose that A has less than full column rank, i.e., the rank(A) < n, but A+E has fullcolumn

rank.

Show

that

min(A

+

E)

max(E).

SinceAdoesnothavefullcolumnrank,thereexistsx=0suchthat

(A+E)x 2 Ex 2Ax= 0(A+E)x=Ex (A+E)x

=

2 =Ex2 E 2 = (E)

x2 max .

x 2

But,(A+E)x

min(A+E)

2

,x2

4


29/75

asshown inchapter4(pleaserefertotheproof inthe lecturenotes!). Thus

min(A+E)max(E).

Finally,amatrixEthatresultsinA+ Ehavingfullcolumnrankandthatachievestheupper

bound

is

0 0 0

0

.

0 . . 0

0

. .. .. 0 r+1 .

0 0

0

r+1

0

E=U

V,

for

1 0 0 0.

0

. . 0 0. .

. .

. 0

r V.A=U .

0

NotethatAhasrankr < n,butthatA+E hasrankn,

A+E=U

1 0 0 0 0.

0 . . 0 0

0

0 0 r 0 00 0 0 r+1 00 0 0

...

r+1

0

V.

Itiseasytoseethatmin(A+E) =r+1,andthatmax(E) =r+1.

The

result

in

part

2,

and

some

extensions

to

it,

give

rise

to

the

following

procedure

(which

is widely used in practice) for estimating the rank of an unknown matrix A from a knownmatrix

A+E,whereE2 isknownaswell. Essentially,theSVDofA+E iscomputed,andthe

rank

of

A

is

then

estimated

to

be

the

number

of

singular

values

of

A

+

E

that

are

larger

thanE2.

5


30/75


31/75

Thus,equation(2)becomes(A

E)z2 r+1.

Finally,(AE)z2 AE2 forallz suchthatz2 = 1.Hence

AE2 r+1

To

show

that

the

lower

bound

can

be

achieved,

choose

=

E U

1 .

.. V. r

0

E hasrankr,

0

...

E=U

0A

r+1 ..

k

V..

0

andAE2 =r+1.

Exercise 6.1 The model is linear one needs to note that the integration operator is a linearoperator. Formallyonewrites

S(u )

(1+ u =

e ts2)(t) (u1(s) +u2(s))ds

0

= e(ts)

u e(ts)1(s) +

u2(s)

0 0

=

(Su1)(t) +(Su2)(t)

Itisnon-causalsincefutureinputsareneededinordertodeterminethecurrentvalueofy. Formallyonewrites

(P Su)(t) = (P S P u)(t) +P (t s)T T T T e u(s)dsT

It

is

not

memoryless

since

the

current

output

depends

on

the

integration

of

past

inputs.

It

is

also

timevaryingsince

0(STu)(t) = (

(t T s)TSu)(t) +

e u(s)ds

T

onecanarguethatiftheonlyvalidinputsignalsarethosewhereu(t) = 0ift


32/75

Exercise6.4(i) linear ,timevarying ,causal ,notmemoryless(ii)

nonlinear

(affine,

tranlated

linear)

time

varying

,

causal

,

not

memoryless

(iii)nonlinear,time invariant ,causal,memoryless(iv)

linear,

time

varying

,

causal,

not

memoryless

(i),(ii)

can

be

called

time

invariant

under

the

additional

requirement

that

u(t) = 0

for

t 0therearetwomoreequilibriumpoints:

0,(Inparticular,this is2

thecasefor inthe interval0< 1).

(b)Linearizingthesystemaround(0,0)wegettheJacobian:

0 1

A =

2 0

Thecharacteristicpolynomialofthesystemisdet(AI) =22. If >0thereisanunstableroot, hence the linearized model is unstable around (0,0). If

0)

we

get

the

Jacobian:

0 1

4 0

The characteristic polynomial for the system is det(AI) = 2 + 4. The complex conjugateroots lieonthej axisandthe linearizedsystem ismarginallystable.

Exercise13.2 a)Noticethatthe input-outputdifferentialequationcanbewrittenas

y =(u

a1y)+(u

a2y

cy2)

andwecanuseobservability-like realizationemployedforadiscrete-timesystemofexercise7.1(c). Thedifferentialequationsforthestatesare

x1 =a1x1+x2+ux2 =a2x1cx12+u

and the output equation is y = x1. You can check that it is indeed a correct realization bydifferentiatingthefirststateequationandplugginginanexpressionforx2 fromthesecondequation.

1


43/75

b)Letusconsiderthesystemwithzero inputanda1 =3,a2 =2andc=2.

x1 = 3x1+x2x2

= 2x1

2x12

Thelinearizedsystemsmatrix is

3 1

A = 2 0

with the characteristic equation 2 + 3+2 = 0, which has the roots 1

= 1 and 2

= 2.Thereforethe linearizedsystem isasymptoticallystablearoundtheorigin,whichalsomeansthattheoriginalnonlinearsystemisa.s. aroundtheorigin(seelecturenotesfortherelevanttheorem).Youcanalsoverifybylinearizationthattheotherequilibriumpoint(1,3)isunstable.c)LetusfindaLyapunovfunctionforthelinearsystem,andthenfindaregionwhereitsderivativeisnegativedefinitealongthetrajectoriesoftheoriginalnonlinearsystem. Sincethe linearsystemisasymptoticallystable,foranysymmetricpositivedefinitematrixQthereexistsauniquepositivedefinitematrixP suchthatA

P +P A=

Q. LetuschooseQ=I. Solvingthesystemof linear

equations

imposed

by

the

matrix

relation

we

obtain

1 1

P =

2

12

12

whichgivesusaquadraticLyapunovfunction

V(x) =xP x=1

x21x1x2+x22 2

TakingaderivativeofV(x)alongthetrajectoryusingthechainruleweget:

V (x) =

x

21

x

22

2x

2(2x2

x1) =

x

2(1

+

2(2x2

x1))

x

2

1 1 2

Thecontour linesofthefoundLyapunovfunctionare

1x1

2x1x2

+x2 =C22 2

forvariousconstantsC.LetusfindsuchCthatifthepoint

x1

x2

iswithintheboundarythen

V (x)


44/75

Exercise14.2 (a)Thesystem isasymptoticallystable ifalltherootsofthecharacteristicpolynomial lie inthe lefthalfofthecomplexplane. Notethatcharacteristicpolynomial formatrixAinacontrolcanonicalformisgivenby

det(AI) =N +a0N1+. . .+aN1

b)

Use

continuity

argument

to

prove

that

destabilizing

perturbation

with

the

smallest

Frobenius

normwillplaceaneigenvalueofA+ontheimaginaryaxis. Supposethattheminimumperturbationis. Assumethatthereisaneigenvalueintherighthalfplane. Consideraperturbationoftheformc, where0c1. Ascchangesfrom0to1atleastoneeigenvaluehadtocrossjaxis,and the resulting perturbation has a smaller Frobenius norm than . This proves contradictionwiththeoriginalassumptionthatA+ hasaneigenvalue intherighthalfplane.c)Thecharacteristicpolynomialfortheperturbedmatrix is

det(AI) =N + (a + )N10

0 +. . .+ (aN1+N )1

Weknowthatthereexistsaroot=j,where isreal. Ifweplugthissolutionin,andassemblethereal and imaginary parts andset themequal to zero, well get two linearequations in with

coefficients

dependent

on

ak and

powers

of

.

For

example,

for

a

4th order

polynomial:

( )4+ ( + )( )3+ ( + )( j a0 0 j a1 1 j)2 + (a2+2)j+a3+3 = 0

results inthefollowingtwoequations:

4( + ) 2 a1

1 +a3

+3

= 0

(a 30

+0) + (a2

+2) = 0

Thisequationcanbewritten inmatrixformasfollows:

003

2

0

10 0

4

1

+

a 2

1

a = 3 032 a a23

orA()=B()

Thereforetheproblemcanbeformulatedasfindingaminimalnormsolutiontoanunderdeterminedsystemofequations:

min A()=B()

By inspection we can see that matrix A has full row rank for any value of unequal to zero. If

=

0

the

solution

is

3 =

a3,

and

the

rest

of

the

k equal

to

zeros.

For

all

other

values

of

the

solutioncanbeexpressedasafunctionof:

() =A()

A( 1

)A()

B()

NotethatthematrixAA isdiagonal,andcanbeeasily

inverted. Byminimizingthenormofthis

expressionoverwecanfindthatcorrespondstotheminimizingperturbation. Thenplugthisinthepreviousequationtocomputeminimizingperturbation()explicitly. Comparethenormsofthesolutionscorrespondingto= and=0(i.e. compare () anda3),andchoosethe

3


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minimumas the solution. Thisway we convertedthe problemtominimizationof a function ofasinglevariable,whichcanbeeasilysolved.d)IncaseN =2thecharacteristicpolynomialoftheperturbedmatrix is

2 + (a0+0)+ (a1+1) = 0

where

=

j .

For

=

0

the

minimizing

solution

is

1 =

a1,

0 =

0.

If

=

0,

plug

in

=

j ,

andtheresultingsystemofequations is

1 = 2a1

0 = a0Thisisapropersystem(numberofequationsisequaltothenumberofunknowns),anditssolutionis given directly by the equations. To minimize the norm of the solution we set =

a1. Note

thatstabilityoforiginalmatrixArequiresthata1

>0, a0

>0(in factpositivityofallcoefficientsisalwaysanecessarycondition,butnotsufficientforN >2- useRouthcriterionforatestinthatcase!). Next, we have to compare |a1| and |a0|, and choose the smallest of them to null with 1or0. Inourproblema0 =a1 =a,thereforethereare2solutions: (0, a)and(a,0)forthesetofs.

Exercise14.5a)Consideranonlineartime-invariantsystem

x =f(x,u)

andits linearizationaroundanequilibriumpoint(x, u) is

x=Ax+Bu.

1)Sincethe feedbacku=Kxasymptoticallystabilizethe linearizedsystem,alltheeigenvaluesofthematrixA+BK are inOLHP.

2)

Without

loss

of

generality,

we

can

take

(x,

u)

=

(0,

0)

where

the

nonlinear

system

is

linearized

around. Then,

fx =f(x,u) =

x

fx+ u+(x,u),(x, u)=(0,0) u

where( theorderof 2x,u) isin x .

(x, u)=(0,0)

Ifu=kx,theaboveequationcanbewrittenasfollows:

f(x,u) =Ax+Bu+(x,kx),

where

(x,

kx) 0asx

x 0.

Thustheoriginalnonlinearsystem is locallystabilizedwiththecontrol lawu=kx.b)ConsiderthesystemS1:

y +y4+y2u+y3 = 0

whereuisthe input.1)Letx1 =y andx2 =y,thenstatespacerepresentationofthesystemS1 is

4


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x1

x

= 2x 32

x 2 4 .1 x2ux2

Thustheuniqueequilibriumpointx isfoundtobex =(0,0),which is independentonu.2)Chooseu =0. Thenthe linearizedsystem is

0 1 0x = x+3 2 2 4 3 2 x x2u x (x,u1 2 )=(0,0) x 2 (x,u)=(0,0)

0 1 0

= x+ u

0 0

0

x = Ax+Bu.

Since the eigenvalues of the matrix A are at 0, and the u term does not enter to the linearizedsystem,the linearizedsystemcannotconcludethe localstabilityofthenonlinearsystemS1.c)Letu=cx32 wherec isafunctionofx1 andx2. Then

x1 = x2

x2 = x31 x22(cx22)x42=x31 cx22.

So, itcanbeconsideredthatthissystemhasanonlinearspring.Now,chooseaLyapunovfunctioncandidate intheV(x) =x4 21+x2. Then,

V(x) =

x4 2x31 2x2

3 2x1 cx2= 4x3x

2 x3x

cx3

1 2 2 1 2

=

x2(2x31

2cx22)

= 2x2(x31 cx22).

InordertomakeV (x)negativedefinite,wewould liketohaveset

23 2 x1

x1cx2 = x2x2

x21+x22+x

31x = 2c .

x32

This

choice

of

c

makes

V (x)

be

V (x) =2x2 212x2


47/75

Exercise14.7 (a)Thelinearizedsystem isgivenby:

x1

=

1 2x2 x1x2 x2 (x1+1)

Hence,thematrixAofthe linearizedsystem

x20

is: 1 0

A=0 1

Ahasrepeatedeigenvaluesat1. Thus,thenonlinearsystemislocallyasymptoticallystableabouttheorigin.

(b)Thelinearizedsystem isgivenby:

x 21

=

3x1

1 x1

x2 1 1 x20

Hence,

the

matrix

A

of

the

linearized

system

is:

0 1A=

1 1

Thecharacteristicpolynomial is: (1+)1=0. Theeigenvaluesarethus1,2=1 2 52 (oneoftheeigenvaluesisintherighthalfplane),andthenonlinearsystemisunstableabouttheorigin.

(c)Thelinearizedsystem isgivenby:

x1

=

x2

x1

1 1

2x1

1

x2

0

Hence,thematrixAofthelinearizedsystemis:

1 1

A=0 1

Ahasrepeatedeigenvaluesat1. Thus,thenonlinearsystemislocallyasymptoticallystableabouttheorigin.

(d)Thelinearizedsystem isgivenby:

x1(k+1)

2 x (k) x= 2 1

(k)

x2(k

+ 1)

1 1

x2(k)

0

Hence,thematrixAofthelinearizedsystemis:

2 0

A=1 1

A has eigenvalues 1 = 2 and 2 = 1. Since one of the eigenvaues is outside the unit disk, thenonlinearsystem isunstableabouttheorigin.

6


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(e)Thelinearizedsystem isgivenby:

x1(k+1) x (k)ex1(k)x=

2(k) 1(k)x2(k)2 x x

1(k)e x1(k)x2(k+ 1)

1 2

x2(k)0

Hence,

the

matrix

A

of

the

linearized

system

is:

0 0

A=1 2

A has eigenvalues 1 = 0 and 2 = 2. Since one of the eigenvalues is outside the unit disk, thenonlinearsystem isunstableabouttheorigin.

7


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6.241J / 16.338J Dynamic Systems and Control

Spring2011

http://ocw.mit.edu/http://ocw.mit.edu/http://ocw.mit.edu/http://ocw.mit.edu/http://ocw.mit.edu/http://ocw.mit.edu/


50/75


of

Electrical

Engineering

and

Computer

Science


Homework

7

Solutions

Exercise15.1(a)Thesystemiscausaliftheimpulseresponseisright-sided. Considerasequenceeatu[t],whereu[t] isaunitstep: u[t] = 1 fort0,andzerootherwise. Laplacetranformofthissequence

converges

if

Re(s)

>a,and isequalto

st 1e eatu[t]dt

= ,

ROC

:

Re(s)

>

a

s+a

Thereforeforasystemrepresentedbyfirst-ordertransferfunctiontobecausaltheROChastobeto

the

right

of

the

pole

(in

fact

this

is

true

for

a

multiple

pole

as

well).

Since

a

rational

function

can

be

represented

by

a

partial

fraction

expansion,

and

region

of

convergence

is

defined

by

the

intersection of individual regions of convergence, the ROC of the system has to lie to the rightof

the

rightmost

pole

for

the

system

to

be

causal.

In

case

of

a

rational

transfer

function

this

is

alsoasufficientcondition. NotethatifanLTIsystemhasarationaltransferfunction, its impulseresponseconsistsofdivergingordecayingexponents(maybemultipliedbypowersoft),thereforeallconcepts

of

p-stability

are

equivalent.

For

BIBO

stability

the

impulse

response

has

to

be

absolutely

integrable,whichisequivalenttoexistenceofFouriertransform. TheFouriertransformisLaplacetransform evaluated at thej axis. Therefore for stability the ROC has to include thej axis.Using

these

two

rules

we

can

see

that

the

system

s

+ 2

G(s) =

(s2)(s+1)

(i)isneithercausalnorstableforROCgivenbyRe(s)


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f(t) = 1for0< t

1:

1

+

2(t) (t2

e e )1

y(t) =

u[t 4 ]f()d = e2 t

1e2 et(e1)

3 3 3

Clearly

this

function

grows

unbounded

and

has

an

infinite

p-norm.

However

the

input

f(t)

has

p-norm equal to 1 for any p including . In case (i) we can use f(t) = 1 for 1 < t

0

there

exists

such

that

O(Cup)up,whichimpliesthatyp (C+)up.Thatconcludesthep-stability,withup < .

c)

When

g(x)

is

a

saturation

function

with

a

scale

of

1,

the

system

is

p-stable

for

p1. Proof:

Againsincethesystemfromutoz isp-stable,thereexistsaconstantC suchthat z p C u p.

So, for all u with u p

, if we take C tobe 1 ,thenwehave:

zP

Cup

1.

Since,

z

g(z) =|z

| 1| | ,

1 |z| 1

for |z| 1wehaveyp =zP Cup 1.

Therefore

this

system

is

p-stable

for

all

p1 in|z|


52/75

Exercise16.1 a)SinceuX,wecanexpressuas

N

u

=

u t where

uR

nie

jii , i R.

i=1

With

N

N u(t) = ejit = uTi e

jit

i=1 i=1

u(t)u(t) = (

N

N N jit e uT)( u ji e

ktj )

i=1 k=1N

=

ej(ki)t uTi uk,i=1 k=1

we

can

computePu

as

follows:

1Pu = lim

L

2L

1

L

u( t

t)u( )dt

LLT

= lim u ej(ji uji)tdt

L2LL i j

lim

u

u

L1 T= ej(j

i)t

L i j dt.

L2i j L

NotethatasL ,becauseoftheorthnormalityofcomplexexponential,

L

0 :

i

=

jlim

ej

(ji)tdt =L

L

1 :

i

=

j

Thus

1

N N

T

P = u u 2u lim

i i(2L) =L2L

i=1

ui 2.

i=1

b)

The

output

of

the

system

can

be

expressed

as

y(t) = H(t)u(t) in time domain or Y(s) =

jit jitH(s)U(s) in frequency domain. For a CT LTI system, we have y = H(ji)uie if u=uie .Thus

N

y(t) =

H(ji)uiejit.

i=1

Followingthesimilarmethodtaken ina),wehave

N N

T

y(t)y(t) = ui H(ji)ejit

i=1

H(jk)u e

jktk

k=1

N

N

=

ej(ki)t Tiu

H(ji)H(jk)uk.i=1 k=1

3


53/75

Thus,Py

canbecomputedasfollows:

= u H

L1 T

P lim (j )H(j )u ej( i)ty i i k kk dt

L2Li k L

1

=

lim

H(j )u

2i i (2L).L2L

i

Py =N

H(ji)u 2

i .i=1

c)Using

the

fact

shown

in

b),

N

Py =

H(ji)ui2

i=1

N

2

max(H(ji))ui

2

i=1

N

max2max(H(ji))i

i=1

u2i

= max2max(H(ji))Pui

Py max2max(H(ji))Pu

i

Py sup2max(H(j))Pu.

sup Py

= H2 .Pu=1

d)Nowwehavetofindan input uXsuchthatPy =H2 P u. ConsideraSVDofH(j0):

| |

1 v1

. .

H(j0) =UV = u1 un . . ..

.

| | n vn

Letu=v ej01 where0 issuchthatH = max(H(j0)),then

P = H(j )v ej0 21

y 2=

0 H(j ) 2

0 v 2 2

1 = 1 u 2

1 2.

2

Py = max(H(j0)).

Indeed,

the

equality

can

be

achieved

by

the

choice

of

u

=

v1ej0.

4


54/75

Exercise16.3Wecanrestrictourattention totheSISOsystemsinceonecanprovetheMIMOcase

with

similar

arguments

and

use

of

the

SVD.

i.)

Inputl

Outputl

this

case

was

treated

in

chapter

16.2

of

the

notes

ii.)

Input

l2 Output

l2this

case

was

treated

in

chapter

16.3

of

the

notes

iii.) InputPowerOutputPowerthiscasewastreated intheExercise16.1. PleasenotethatPy =H

2 Pu,thegivenentry inthetable

corresponds

to

the

rms

values.

iv.) InputPowerOutputl2a

finite

power

input

normally

produces

a

finite

power

output

(unless

the

gain

of

the

system

is

zero

atallfrequencies)and inthatcasethe2-normoftheoutput is infinite.v.) Inputl2 OutputPowerThis

is

now

the

reveresed

situation,

but

with

the

same

reasoning.

A

finite

energy

input

produces

finite

energy

output,

which

has

zero

power.

vi.)

Input

Power

Output

lHere the idea is that a finite power input can produce a finite power output whose -norm isunbounded. Thinkingalong the lines of example 15.2consider the signal u= m=1vm(t)where

vm(t) =m if m < t < m+m3 and

otherwise

0.

This

signal

has

finite

power

and

becomes

unboundedovertime. TakethatsignalastheinputtoanLTIsystemthatisjust

aconstantgain.

vii.)

Input

l2 Outputl

|

|y(t)

=

h(t

)

s)u(s

ds

|y(t)|=|< h(t >

s), u(s) |

h2u2

The

last

step

comes

from

the

application

of

the

Cauchy

Schwartz

inequality.

Taking

the

sup

on

the

lefthandsidegivesy

h2u2;nowtoachievetheboundapplytheinputu(t) =h( t)/ h . 2

viii.)

Input

l Outputl2Applyasinusoidalinputofunitamplitudesuchthatj isnotazeroofthefrequencyresponseofthetransferfunctionH(j).

ix.)

Input

l Output

Power

note

that{u:u

1} isasubsetof{u:Pu

1}. Therefore

sup{Py :u 1} sup{Py :Pu 1};

we use the lower bound from case iii.). Note that the entry in the table corresponds to rms andnottopower.

5


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56/75

MASSACHUSETTSINSTITUTEOFTECHNOLOGYDepartmentofElectricalEngineeringandComputerScience


Homework8Solutions

Exercise 17.4 1) First, in order for the closed loop system to be stable, the transfer functionfrom ( w w )T1 2 to ( y u )

T has to be stable. The transfer function from w1 to y is given by(IP K)1P and iscalledsystemresponsefunction. Thetransfer functionfromw1 tou isgivenby (I KP)1 and is called input sensitivity function. The transfer function from w2 to y is(IP K)1P K and iscalledthecomplementarysensitivity function. Thetransfer function fromw touisgivenby(IKP)12 K. Therefore,wehavethefollowing :

y (I+P K)1P (I+P K)1P K w1=

u

.

(I+KP)1 (I+KP)1

K w2

So, ifK isgivenas

K= Q(IP Q)1 = (IQP)1Q,

then

(I+P K)1P = (I+P Q(IP Q)1)1P

= (((IP Q) +P Q)(IP Q)1)1P

= (IP Q)P

(I+P K)1P K = (I P Q)P Q(IP Q)1

= P(IQP)(IQP)1Q

=

P Q

(I+KP)1 = (I+ (IQP)1QP)1

= ((IQP+QP)(IQP)1)1

= IQP

(I+KP)1K = (IQP)(IQP)1Q

= Q.

Thus,theclosed looptransferfunctioncanbenowwrittenasfollows:

y (I

= P Q)P P Q w1

.

u (IQP) Q w2

In

order

for

the

closed

loop

system

to

be

stable,

then

all

the

transfer

functions

in

the

large

matrix

abovemustbestableaswell.

(IP Q)P IP QP (I+P Q)P

(I+PQ)P P+P2Q

P Q PQ

IQP I+QP I+QP.

1


57/75

SinceP andQarestablefromtheassumptions,weknowthatallthetransferfunctionsarestable.Thereforetheclosed loopsystem isstable ifK=Q(IP Q)1 = (IQP)1Q.

2)From1),wecanexpressQ intermsofP andK inthefollowingmanner.

K

=

Q(I

P Q)1

K(IP Q) = Q

KKP Q = Q

K = (I+KP)Q

Q = (I+KP )1K=K(I+P K)1,

bypushthroughrule.

For some stable Q, the closed loop is stable for a stable P. by the stabilizing controller K =1Q(IP Q) . Yet,notall stableQcanbeusedforthisformulationbecauseofthewell-posedness

of

the

closed

loop.

In

the

state

space

descriptions

of

P

and

Q,

in

order

for

the

interconnectedsystem, in this case K(s) to be well-posed, we have to have the condition ( 17.4 ) in the lecture

note, i.e.,(IDPQ()) is invertible.

3)SupposeP isSISO,w1 isastep,andw2 =0. Then,wehavethe followingclosed looptransferfunction:

Y(s)

(IP Q)P

=

1

,U(s) IQP s

sincetheLaplace transformoftheunitstep is 1 wehaves

1U(s)

=

(1

Q(s)P

(s))

.s

Thenusingthefinalvaluetheorem,inordertohavethesteadystatevalueofu()tobezero,weneed:

1u() = lims(1

s0Q(s)P(s)) =0

s1Q(0)P(0) = 0

Q(0) = 1/P(0).

Therefore,Q(0)mustbenonzeroand isequalto1/P(0). Notethatthiscondition impliesthatP

cannot

have

a

zero

at

s

=

0

because

then

Q

would

have

a

pole

at

s

=

0,

which

contradicts

that

Qisstable.

Exercise17.5 a)Letl(s)bethesignalattheoutputofQ(s),thenwehave

l = Q(r(PP0)l)

(I+Q(PP0))l = Qr

l = (I+Q(PP0))1Qr.

2


58/75

2

Since we can write y =P l, and with P(s) = , P0(s) =1 ,andQ=2,thetransfer function

s1 s1

fromrtoy canbecalculatedasfollows:

Y(s) = P(s)L(s)

= P(I+Q(P

0))

1

P QR(s)

2

1

1

2=

1

+

2

2R(s)

s1

s

1 s

4 s 11

+1

= R(s)s1 s1

4 s 1=

R(s)

s1 s+1Y(s) 4

= .R(s) s+1

b)There isanunstablepole/zerocancellationsothatthesystem isnot internallystable.c)SupposeP(s) =P0(s) =H(s)forsomeH(s). Thenusingapartoftheequation ina),wehave

Y(s) = H(s)(I+Q(s)(H(s)H(s)))1Q(s)R(s)

= H(S)I1Q(s)R(s)

= H(s)Q(s)R(s)

Y(s) = H(s)Q(s).

R(s)

ThereforeinorderforthesystemtobeinternallystableforanyQ(s),H(s)hastobestable.

Exercise19.2Thecharacteristicpolynomfortheclosed loopsystemisgivenby

s(s

+

2)(s

+

a) + 1 = 0

Computing the locus of the closed poles as a function of a can be done numerically. The closedloop system is stable if a0.225. The above bound can also be derived by means of root locustechniques or by evaluating the Routh Hurwitz criterion. Another way of deriving bounds forthe value of a is by casting this parametric uncertainty problem as an additive or multiplicativeperturbationproblem,seealso19.5. Onecanexpectthatthederivedboundsinsuchacasewouldberatherconservative.

Exercise19.4Wecanrepresentanuncertainty infeedbackconfiguration,asshownbelow.Notethattheplant isSISO,andweconsiderblocksandW tobeSISOsystemsaswell,so

wecancommutethem. Thetransferfunctionseenbytheblockcanbederivedasfollows:

z = P0(W wKz )

= (I+P0K)1P0W w

M = (I+P0K)1P0W.

Applythesmallgaintheorem,andobtaintheconditionforstabilityrobustnessoftheclosed loopsystemasfollows:

3


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Figure19.4

W(j)P

0(j)sup

0. Letuscalculatethedeterminant inquestion:

W21K111 1+K11P11 W12W21K1K212W12K22det(IM) = det 1

= 1(1+K11P11) (1+K22P22)1+K22P22

To have a stable perturbed system for arbitrary 1 1 and 2 1 it is necessary andsufficientto impose

W12W21K1K2

(1+K11P11) (1+K22P22)1.

teAt

1=

I+At=

0 1

thus

te

At b=

1

b)thereachabilitymatrix is:

0 1

b Ab =1 0

The

reachability

matrix

has

rank

2,

therefore

the

system

is

reachable.

Now,

we

compute

the

reachabilityGrammianoveran intervalof length1: 1 1 1G= eA (T)bbeA (T)

d = 30

2

1

12

The

system

is

reachable

thus

the

Grammian

is

invertible,

so

given

any

final

statexf wecanalways

findsuchthatxf =G. Inparticular

1

18

=2 10

c)

According

to

23.5

defineFT(t) =eA(1t) b. Thenu(t) =F(t) isacontrol inputthatproduces

atrajectorythatsatisfiestheterminalconstraintxf. Thecontroleffort isgivenas: Tu2 d = G

0

Infactthisinputcorrespondstotheminimumenergyinputrequiredtoreachxf in1second. Thiscanbeverified bysolvingthecorrespondingunderconstrained leastsquares problemby meansofthe

tools

we

learned

at

chapter

3.

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b)NO.Theexplanationisasfollows. Withthecontrolsuggested,theclosedloopdynamicsisnow

x = Ax+ (b+)uTu = f x+v

x = (A+ (b+)fT)x+ (b+)v.

Suppose

that wi was the minimizing eigenvector of unity norm in part a). Then it is also an

eigenvector of matrix A+ (b+)fT since wi is orthogonalto b+. Therefore feedback doesnotimprovereachability.

Exercise24.5a)Thegivensystemingeneralforallt0withu(k) = 0k0hasthefollowingexpressionfortheoutput:

y(t) = CAt k 1Bu(k)k=

1

=

CAt

A

k

1Bu(k)k=1

sincematrixAisstable. NotethatbecauseofstabilityofmatrixAallofitseigenvaluesarestrictlywithin

unit

circle,

and

from

Jordan

decomposition

we

can

see

that

lim Ak 2 = 0k

therefore x() does not influence x(0). Thus the above equation can be used in order to find

x(0)asfollows:

x(0)= Ak1Bu(k).k=1

b) Since the system is reachable, any Rn can be achieved by some choice of an input of theabove

form.

Also,

since

the

system

is

reachable,

the

reachability

matrixRhas fullrow rank. As

aconsequence(RRT)1 exists. Thus, inordertominimizethe inputenergy,wehavetosolvethefollowing

familiar

least

square

problem:

Find minu2

s.t. =

Ak1Bu(k).

k=1

Then

the

solution

can

be

written

in

terms

of

the

reachability

matrix

as

follows:

umin =RT(RRT)1,so

that

its

square

can

be

expressed

as

2u = uTmin minumin

T RRT 1 TRRT RRT= ((( ) ) ( )1= T(RRT)1,

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1

= max{

(t)2 |

u(t)2y 1, u(k) = 0 k 0u

t=0 t=

}

1 =

max{ ()

|

u

s.t.

=

x(0)

and

u(t)22

1

, u(k) = 0

,

k 0

t=}

=

max{2()| min

u 221

}=

max{2()|1()1}.

e)

Now,

using

the

fact

shown

in

d)

and

noting

the

fact

thatP1 =MTM where M isHermitian

squarerootmatrixwhich is invertible,wecancompute:

=

max

{2() |1()1}

{ = max N 2 |M 22 set=M12 1} l

=

max )Tl

{(M1l OTOM1l | l22 1}= max(OM1)= max((M

1 )T

1 T

OTOM1)=

max((M ) QM1)

= max(QM1(M1)T)

=

max(Q P)

Exercise25.2a)Given:

s+f s+f 1H1(s) = = , H2(s) = .

(s+4)3 s3+12s2+48s+64 s2Thusthestate-spacerealizations incontrollercanonicalformforH1(s)andH2(s)are:

12

48

64 1

A1 =

0

1 00

, B1 =

0 1

0

, C1 =

0 1

f

0

, D1

= 0,

and

A2 = 2 , B2 = 1 , C2 = 1 , D2 = 0.

Since

f

is

not

included

in

the

controllability

matrix

for

H1(s)withthisrealization,thecontrollability,whichisequivalenttoreachabilityforCTcases,thecontrollabilityisindependentofthevalueof

f.Thus,checktherankofthecontrollabilitymatrix:

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