Lund 9904 Sol

8/13/2019 Lund 9904 Sol

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1.

a. (i)-B, (ii)-D, (iii)-A, (iv)-C

b. The equilibria are given by ˙ x1 ˙ x2 0. This gives x2 0 and sin( x1) 0,

hence x (kπ , 0), k 0, ±1, . . . . The linearized equation is given by ˙̃ x A ˜ x

where

A

0 1

(−1)k −1

The eigenvalues are given by λ 2 +λ +(−1)k 0. The system is hence locally

asymptotically stable (a stable focus) for even k and unstable (a saddle) for

odd k.

2.

a. The equilibria are given by ¨ y ˙ y 0. This gives the matrix equation

Ky 0, hence yT

Ky 0, hence y 0 since K is positive definite.

b. Try the function

V 1

2˙ yT M ˙ y +

1

2 yT Ky

as Lyapunov equation. Clearly V ≥ 0 and V 0 gives ˙ yT M ˙ y 0 and

yT Ky 0. Since M and K are positive definite this give y ˙ y 0.

Furthermore V → ∞ as x → ∞. This can be seen from

˙ yM ˙ y+ yT Ky ≥ λ min( M ) ˙ yT ˙ y + λ min( K ) yT y

Differentiation of V gives

V̇ ˙ yT M ¨ y+ yT K ˙ y − ˙ yT D ˙ y ≤ 0

Note that we have V̇ 0 if ˙ y 0. But ˙ y 0 gives ¨ y 0 and hence

Ky 0. Therefore y 0. The only invariant set is hence y ˙ y 0.

By Lyapunov‘s theorem on invariant sets we can therefore conclude global

asymptotic stability.

3.

a. a1 0 (see problem b) and

b1 2

T

2π

0

sinω t( A sinω t)5dt

2 A5

T

2π

0

eiω t − e−iω t

2i

6

dt

2 A5 ⋅ 20

ω T ⋅ 26 ⋅ 2π

5 A5

8

(Use 2π

0 eikxdx 0 for k

0). Therefore

N ( A)

5 A4

8

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b. Put v(t) f ( A sin(ω t)). It is clear that v(T /2 − t) v(t). We therefore have

a1 2

T

3T /4

−T /4

v(t) cos(ω t) dt

4

T T /4

−T /4 v(t) cos(ω t) + v(T /2 − t) cos(ω (T /2 − t)) dt 0

where we have also used that cos(ω (T /2 − t) ) −cosω t.

4.

a. Introduce the states x1 v, x2 y. A state-space description of the system

is obtained from the block diagram:

˙ x1 − 1T t x1 + 1

T tsat( x1)

˙ x2 − x2 + sat( x1)

b. The steady state solution is obtained by setting dxdt

0 in the state-space

model. It can also be obtained directly from the block diagram. Note that

in steady-state, the input to the integrator must equal zero.

0 −1

T i y +

1

T t(u − v)

y G(0)u u

u satv

Substituting y and u in the first equation gives:

v (1 − T t

T i) satv

This has, for T i, T t > 0, the only solution

v 0 y u 0

( A v 0 gives either v < sat v or v and sat v have opposite signs).

c. A round tour in the block diagram gives:

V 1

s

1

T t(U − V ) −

1

T iG p(s)U

s +

1

T t

V

1

T t−

1

T i

1

s+ 1

U

V s + 1 − T t/T i(s + 1)(sT t + 1)

U

Considering the minus sign in the feedback loop gives:

Gtot(s) − s + 1 − T t/T i

(s + 1)(sT t + 1)

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8/13/2019 Lund 9904 Sol


d. We have k1 0 and k2 1 in the circle criterion. The conclusion is that

the system is stable if the Nyquist curve satisfies ReG(iω ) ≥ −1.

5. A linear system is passive if it is stable and its Nyquist plot is in the right

half plane, ReG(iω ) ≥ 0.

G(iω ) iω + b

(iω + a)(iω + c)

Re G(iω ) bac + ω 2(a + c − b)

(ω 2 + a2)(ω 2 + b2)

The system is thus passive if

bac + ω 2(a + c − b) ≥ 0, ∀ω

Since a, b, c > 0, the system is passive precisely when

a + c ≥ b

6. The system is normal so can put n0 1. The Hamiltonian is

H u2 + λ tu

Minimization wrt u gives u −λ (t)t/2. The adjoint equation is

λ̇ − H x

0, λ (T ) 2 x(T ).

This gives u −tx(T ). If this is put into the system equation we get

x(T ) − x(0)

T 0

−t2 x(T ) dt −T 3/3 x(T )

and hence x(T ) x(0)/(1 + T 3/3). The optimal control signal is hence

u − t

1 + T 3/3 x(T ).

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7. The small gain theorem (or the circle crieterion) can be used. The gain of

the linear system is

G maxω

G(iω ) 1

The nonlinear system consists of multiplication with a function f (t) a sinω t. According to the lecture the gain of such a system is

maxt

f (t) a

The SGT says that one has BIBO stability if the loop gain is smaller than 1.

This proves the statement.

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Lund 9904 Sol