8/3/2019 ENEE 660 HW Sol #2

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Electrical and Computer Engineering Department

University of Maryland College Park

ENEE 660

System Theory

Fall 2008

Professor John S. Baras

Solutions to Homework Set #2

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dx1(t)dt =

Ldi(t)dt =

EL

u(t)

RL

u(t)i(t)

1L

v(t)

= EL

u(t) u(t)RLx1(t) x2(t)

dx2(t)dt =

Cdv(t)dt =

i(t)C

= x1(t)

So, the state evolution is described by

d

dt

x1(t)x2(t)

=

0 0

x1(t)x2(t)

+ u(t)

RL 00 0

x1(t)x2(t)

+

EL

0

u(t)

x(to) =

Li(to)Cv(to)

For u(t) = 0

d

dt

x1(t)x2(t)

=

0 0

x1(t)x2(t)

i.e., the system oscillates with frequency .

In the phase plane the trajectories are counterclockwise circles, since ddt

x1(t)x2(t)

T

x1(t)x2(t)

0

x2

x1

For u(t) = 1

dx1(t)dtdx2(t)dt

=

RL

0

x1(t)x2(t)

+

EL

0

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Let

y1(t)y2(t)

=

x1(t)x2(t)

+

z1z2

where

z1z2

=

RL

0

1 EL

0

=

0

EC

Then we see that

dy1(t)dtdy2(t)dt

=

RL

0

y1(t)y2(t)

Since the eigenvalues areR

R24L/C2L , no matter where we start the vector y(t) goes to 0 as

t , in a smooth way (no spirals) if R2 > 4L/C and in a spiral way if R2 < 4L/C. So, inx1, x2 coordinates the x(t) vector goes towards the vector

0

E

C

in a similar fashion. To obtain

some idea how the reachable set looks like, you have to fix an initial condition (it depends on that

heavily). Take, for convenience,

00

as your initial condition. Then, you may have trajectories

such as:

x1

x1

x1

x2

x2

x2

E CE C1/2

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Problem 3

The state variable equations, from the voltage controlled resistors are:

dx1(t)dt = M(v1(t) Vp)x2(t)

dx2(t)dt = M(v2(t) Vp)x1(t)

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or

ddt

x1(t)x2(t)

=

0 M Vp

M Vp 0

x1(t)x2(t)

+ v1(t)

0 M0 0

x1(t)x2(t)

+ v2(t)

0 0

M 0

x1(t)x2(t)

The system is nonlinear since the right-hand side is not simultaneously linear in x(t) and u(t) =v1(t)v2(t)

. The system is time invariant since right-hand side does not depend explicitly on time,

and this implies time invariance by our definition of time invariance.

For the bilinear state space model to be linear it is necessary that B1 = B2 = 0 (as a matrix).

Problem 4

(a) x(t) = Ax(t) with A =

4 00 1

Then eAt =

e4t 0

0 et

and the solution is

x1(t)x2(t)

=

e4txo1

etxo2

.

Clearly, for xo2 = 0 the second component will grow without bound as t increases.

P(t) = eBt =

cos3t sin3t sin3t cos3t

This is continuous with continuous derivative on (, ).

detP(t) = cos2 3t + sin2 3t = 1

So, P(t) is a Liapunov transformation.(b)

z(t) = BeBtx(t) + eBtAx(t)

= BeBteBtz(t) + eBtAeBtz(t)

= (B + eBtAeBt)z(t)= A(t)z(t)

Then

eBtAeBt =

cos3t sin3t

sin3t cos3t

4 00 1

cos3t sin3tsin3t cos3t

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= cos3t sin3t

sin3t cos3t 4cos3t 4sin3tsin3t cos3t

=

4cos2 3t + sin2 3t 4cos3t sin3t + sin 3t cos3t

4cos3t sin3t + sin 3t cos3t 4sin2 3t + cos2 3t

=

1 5cos2 3t 5sin3t cos3t5sin3t cos3t 1 5sin2 3t

So, A(t) =

1 5cos2 3t 5sin3t cos3t + 35sin3t cos3t 3 1 5sin2 3t

The eigenvalues of A(t) are the roots of

det(I A) = det

1 + 5 cos2 3t 5sin3t cos3t 35sin3t cos3t + 3 1 + 5 sin2 3t

= 0

( 1 + 5 cos2 3t)( 1 + 5 sin2 3t) (5sin3t cos3t + 3)(5sin3t cos3t 3 ) = 0 ( 1)2 + 5( 1) + 25sin2 3t cos2 3t 25sin2 3t cos2 3t + 9 = 0

2 + 3 + 5 = 0

1,2 =3 j 11

2

So the eigenvalues of the time varying matrix A(t) are constant and have negative real part for alltime!

Despite this,

z(t) = eBt

x(t) = cos3t sin3t sin3t cos3t

e4txo1etxo2

=

e4t cos3t xo1 + et sin3t xo2e4t sin3t xo1 + et cos3t xo2 can grow without bound as t grows to infinity. For instance, take xo1 = 0, x

o2 = 1. So for time

varying linear systems the sign of the eigenvalues of A(t) has no implications on stability!

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Problem 5

A state x1 is reachable from the zero state at time t = 1 if

x1 =

10

eA(1)bu()d for some u U

Or if

x1 =

1/20

eA(1)bu()d +11/2

eA(1)bu()d =

= eA/21/20

eA(1

2)bu()d +

1/20

eA(1

2)bu(

1

2+ )d

(where we let = 12 + in the second integral)

u(1

2

+ ) = u()

= (I + eA/2) 1/20

eA(1

2)bu()d

Cayley Hamilton = (eA/2 + I)ni=1 Ar1b 1/20 r(1 )u()d=

ni=1 cr(e

A/2 + I)Ar1b

where cr =10 r(1 )u()d.

ThereforeR(1) = Range{(eA/2 + I)[b,Ab, , An1b]}

Problem 6

(a) For A =

1 10 1

, eAt = et

1 t0 1

.

The controllability grammian is

W(0, 1) =

10

et

1 t0 1

t0

t 0

et

1 0t 1

dt

=

10

e2t

t2 00 0

dt =

14(1 5e2) 0

0 0

Then

10

is reachable at time 1 starting from

00

at time 0 iff

e1

1 10 1

10

=

e1

0

is in the range of W(0, 1), which is clearly correct. So the answer is yes.To find the control we solve

14(1 5e2) 0

0 0

=

e1

0

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So,

= 4e5e1

0

So u(t) =

t 0

et

1 t0 1

4e5e1

0

= tet 4

e5e1 .

(b) The observability grammian is

M(0, T) =

T0

et

1 0t 1

1t

1 t

et

1 t0 1

dt

=

T0

e2t

1 2t2t 4t2

dt

=

12(e

2T 1) T e2T 12(e2T 1)

T e2T 12(e2T 1) 2e2TT2 2T e2T+ e2T 1

The question is: is

10

in the null space of M(0, T)? Or is

12(e

2T 1)

T e2T 12(e2T 1)

=

00

true? This is not true unless T = 0. Therefore the answer is: YES;

10

is observable. Any small

T nonzero, will allow observability.

Problem 7

I have followed Arbib from Kalman, Falb and Arbib in the proofs below.Proof of Theorem AAll we have to show is how to exchange hx() of the first commutative diagram by hx() in thesecond diagram.

If we start from the first commutative diagram, since M is reduced, no one state ofM can simulatetwo states of M, and therefore hx is one-to-one. So take X = hx(X); this is a subset of X. Sethx = h

1x on X

.Now if we start from the second commutative diagram, for x X, select hx(x) to be any x suchthat hx(x) = x

. Then the first diagram commutes.

Proof that MSf,if is weakly equivalent to MfMSf,if simulates Mf because of the diagram

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- if - Sf - if - I -u

[u]f

y y

MSf,if

Identity

where if : U Sf via if(u) = [u]f.

Mf simulates MSf,if. Need to find maps hu and hx such that the diagram below commutes:

Sf

U Y

Y

6

-

-

?

fMf, hx(s)

MSf,if,s

hu identity

For every sSf we select hx(s) = uU s.t. s = [u]f. Then set also hx(s) = f(u). Now

fMSf,if,s(s) = if(ss) = f(hu(ss))

= f(hu(s)hu(s)) = fMf,[hu(s)]f(hx(s

))

= fMf,hx(s)(hu(s))

Proof of Theorem B

if Mf simulates Mg the diagram below commutes.

(U)

U Y

Y?

-

-

6

g

f

hu hy

Suppose g = hyf hu.Let Sg = U

/ g and Sf = (U)/ f.Let S =

{[hu(u)]

f: uU

}. Clearly S is a subsemigroup of Sf.

Let Z([hu(u)]f) = [u]g .Z is well defined; indeed [hu(u)]f = [hu(u)]f = [u]g = [u]g, because

g(wuz) = hyf hu(wuz)

= hyf(hu(w)hu(u)hu(z))

(since hu(u) f hu(u)) = hyf(hu(w)hu(u1)hu(z))= hyf hu(wu

1z)

ConverselyIf (Sg, ig) | (Sf, if) the diagram below commutes

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Sg

Sf S Y

Y?

-

-

6

if

ig

hyZ

Given uU, choose hu(u) to be some t(U) s.t. Z([t]f) = [u]g . One exists since Z is onto.Then

hyf hu(u) = hy(if([hu(u)]f)) = ig(Z([hu(u)]f)= ig([u]g) = g(u)

And this completes the proof.

Problem 8 (Optional, Extra Credit)

Look at

Cd

dt(v1(t) + Kv

1(t h)) + (

1

z g)v1(t) K(

1

z+ g)v1(t h) = v

31 (t) Kv

31 (t h)

and suppose that you start at t = 0 and you want to integrate this differential equation for 0 t h.Because of the delayed arguments, you must know v1(t) for

h

t

0 and ddtv

1(t) for

h

t

0

to be able to integrate it. These two functions are therefore the state at time t = 0. Similarlyfor other times. Therefore this system has an infinite dimensional state space, because I need twofunctions on the interval [h, 0] as states.

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