Esercizi Fondamenti di automatica

L(s) = -Δ s (-0.5se)

D(s) = 0.5s(s+0.5)(s-1.0) - (1000 - 2 s)

=A₁

=A₂

Siccome μ₁ ed il segno di cambi di coeff. >=

A₁ ≥ 0.30 - 0.25K ( ₋110 ) > 0

A₂ ≥ 0.30 - 0.25( 6 + 220)

K( K ≠ 60)

K

120 - 100

< K < 60

K 60

120

K = 60

(1)

L (sI-A)^-1B Bx(0)+0.05 Ex[0]+0.05K₃[t]

₂(t) -0.05E^{[0.4, 1, 100, 0.05I}

₁x₁[L

x₁[t]+x₂[t]x₂[t]= ={ 0.05 x₂+100 0.05 k(0+1.00)x₃[t]+x₄= 0.004Lx=[500]₃[0] [500]x*

1. x₁ = x₁*^0.957
2. x₂ = x₂*^0.957
3. x₃ = x₃*^0.956

A= -λ +10 0 -λ + (-0.9)0 0 0B= ^{[0][0]… 1.0[0]}

C₂[0 1] D [0]

det [λC^-A]=^-Sλ+1

∞ λ+1 0 0 ^{=[∞-0.5][∞-0.5]}

[λ +1] [λ +0.5 0] λ +0.5

y=^{-λβ[λ+0.5-0.5]=}

1. G(s)=C(sI-A)^-1B=
2. 0 0 0 [0 [0 -1]^λ-0.5

Tr2y

^-sCsub()

-0 λ0 (s) + λ+1λ= [0]A* [s] [10]

1. [λC^-M s[s] -1 = λ+1](0 [λ+0.10 λ-0.5]-0.5)X[100]

[.50][1.100]λ[0 ] 0(*)

B[s][0]-B

Controllo : λ[1]=λ λ[-0.5 λ+0.3]0J[s]=0[R=100] (=) λ-->G₁5 (0)[R(s)-1]=[1] A [0][0-1.0]0 λλ-1

ẋ₁(t) = x₂(t) - (x₁²(t)+ x₂²(t)) x₁ (t) + x₂

ẋ₂(t) = -x₁(t) - (x₁²(t)+ x₂²(t)) x₂ (t) + u (t)

ẋ₁(t)^lin = x₂^lin(t)

ẋ₂(t)^lin = -x₁^lin(t) + u (t)

y (t) = x₂(t) - x₁(t)

{

0 = x₂ - x₁ + x₂ = 0

{ y = x₂ - x₁ = - x₁

{ ẋ₁ = ẋ₂

0 = ẋ₂ - ẋ₁ + ẋ₂ = 0

{ y = ẋ₂ - ẋ₁ = -½ {

ẋ₁ = - ¹⁄₂

ẋ = [^-1⁄₂]

ẋ = [⁵⁄_-½]

{ x₁ = 0

{{ ẋ₂ = x₁ + ẋ₂ = ẋ₁ + ẋ₂

{ x₃ = - ẋ₁ + ẋ₃

{ ẋ₃ = (ẋ₂ + ẋ₁ + ẋ₂ = ẋ₃) + ẋ₂

{ ẋ₂ = - ẋ₂ - ẋ₃

{ ẋ₂ = - x₂ + ẋ₂

{ λ₃ = ẋ - ẋ₂

{ - x₂ = ẋ - ¹⁄₂ẋ

{ ẋ = ẋ₃

ẋ = - ẋ₂ + μ { x

A = [ 0   -2   0 ] [ 1   0 ] [ -1   1 ]

B = [ 1 ], C = [ 0   1 ], D = [ ^-1⁄₂ ]

det (λI - A) = | λ + 2 λ - 2 | = (e.g.)

| λ₂ - c λ - 2

|λ + 2 | = | λ - 2 λ - 2 | ^{possible labo real} y = 1

^λ

G(s) = [0  -3 | s - 2|] , |

[s|s - —³ — ], s = 2 [

G_l(s)

1. ⋅ s - 2 - 2
2. s —
3. zero •
4. [ — ² ] ⁄
5. — [0|s - s₂

7. [ ẋ ]
9. [ s]-
10. ^{- ²} s ⁄
11. [ 5^{-1 ³}]
13. — — ^{- s1/3}
14. — s - ¹
15. [ s ][1]
16. — [0][s - ]
17. ẋ ][1|s_|-
18. — ^—⁄
19. -1
21. end
22. -----
23. ― ⋅―
25. non ẋ non stesso poter i ićeviveren e
27. 15.0

det [(λ-1) * ▢]

A = | 1 4 0 |

 | 0 1 1 |

 | 0 0 2 |

B = | 3 0 |

 | 1 0 |

 | 0 1 |

λ = ±3

C = [ -cos(2) cos(4) ▢ ]

D = [ 1 - 2lex(4) l ]

G = si immane -↠ [ -0 ] ↠ G = G₂

b = parte di λ ↠ fus(λ_s) = 4 ↠ G₁

d = parte di λ & Sub Gl.| 5(1⋅5)_c B₂ G₄

e = colle dei quia in uso G₃

f ≠ incarico [l (s)=0 absorvale inqu plie l (s) = 0l

G₃ = ↠ per velso G₃ in l (s)= 2d plie l (s)=0

B = o o imp inven o = 0 ↠ no asc

C = per trato l’inqunate

D = pole pluv a a nis o

ascese ∴ pole curuio + amiuo ↠ colegio

ele & pole reale -------------------------------- alcalel ∴ pole reale ↠ non

ascese

7 point qualcuno parlette

E = xe | tent | inquato ferro ↠ ricovo & lcm

s & Sub i ', -l² + um_№.' + ∴ 5l ' + & l

l oltre di colle callez & calcu con mestro

E = correggo con finger prague vo

( l ^{'l - 6' = 1} & ^'7.5l ≤-is₅ ) ↠ universo

( 2 ^'l-uni ↠ 1 ) l^l ≤ -16

( 1 2um; _wms. ) l2 ↠ partner

→ eugeguore

1. Convoluzione seno - coseno

2. Rotelle a cono a tre c...

3. Pule in..., pul...

4. Angolo a cono... piacere esclusivo

5. Rotella inversa

Esercizio

K₁=k_b

M₁=2.5

suotasuoto

D=2.8

K₂=3.5 k_b

x_c(t)=4/v

FE₂=-K₂(x₂-x₁)

FE_d=-D...

e_r e_f e_e + FE=Me

x_L=k_u

X= [x₁, x₂, x_c, x_L, μ]

ẋ(t) = A x(t) + B u(t)

y(t) = C x(t) + D u(t) = [1 0] [x₁(t)] = x₁(t)

x(t) = 2 x₁(t) + 2 x₂(t)

u(t) = 2 x₁(t) + 2 x₂(t)

A = -1 02 -1

B = [ 0 ][ 1 ]

C = [ 1 2 ]D = [ 0 ]

det[sI - A] = 0

s 0 1 0 [s+1 0]2 s-1 2 1

[s-1] (s)[s] (2)0100

det[sI - A] = 0 → s = 1+1 2 [x₁(t)]→ X(s) = [ 1 2 ]

[sI-A] → 1 [ 1 ] = [ 0 ]

C [ s⁺¹ 2 ] → [ 1/s +1 0 ]

1/s¹ [s+1](1/s₁ - 1) (2)

ancora scriviamoperché ho fatto cancelifficult-il t-+ →t+ | -1 0 [ 0 ] t- | -s-1 ^-t5| /= [ cosI ]→ [ cos(t) ]x(t) = (1 e -1 sin) t e ^{-1 t} 2 cos(t) 1 x¹ eo

_{∞ e}

1 | 2 [ 0 ]t-e | 2 | → 2 o | e | o 0 3

line ↑limlim x (e) = 2t = 1

lim x(e -e ) = 0

μ(t) = μ(t) 0, esu

y(t) - y 0

u(t 0 L)s [-t e^-t +t]^t-1s x | = ₀ et

. 1/2 | [ → 2 ^sec]

5 = μ → | u

x₁(t) = x₁(t-1) + x₄(t-1) + u₁

x₂(t) = x₂(t-1) + x₁(t-1)

x₃(t) = x₃(t-1)

x₄(t) = x₄(t-1) + x₃(t-1) + u₂

y(t) = x₁(t) x₄(t)

C = [1 0] [x₁] y₀ [x₄] y₂

R

[ x₁(t) ] = A [x₁(t-1)] + B u(t)

[ x₂(t) ] [x₂(t-1)][ x₃(t)] [x₃(t-1)][ x₄(t)] [x₄(t-1)]

1. [ x₁ ] = [x₁ + x₁ + x₃ + x₂ ]
2. [ x₂ ] = [x₁ + x₂ + x₃ + x₂ ]
3. [ x₃ ] = [x₁ + x₂ ]
4. x₁(t+1) = x₁ + x₁c₀ x₂ + x₃+u₁

con polo in: [ -2 -1]

1. v [ 1 ] [ 0 ]
2. w [ C

   [ 1 0 ]]

   con pol verify

3. w [ C ]
4. [ 1 0 ]

1. A = [ 2 0 ] → det (λ I - A = 0 λ
2. [ -1 -2 ] λ = 2 una soluz
3. Δ₂ = [ 1 0 ]→ det (λ I - A² = 0, λ = 2 indvbile
4. [ 0 0 ] c=0

1. Raggiungi b/l a a₀ = N_c = [1 2 ] 1 -2 → range = 2
2. Matrici B N_c : [0 1 -1] range A_B = [1 2] → range = 2
3. Inviinibile matricea A A_C N_c [ 2 A_C I
4. range =
5. Y

   11: A_B = [1 2

6. [0 -1] rang A_B = 2 z = 2 → calool raggiun;ibile
7. X_c = [1 2] → range = 2 z = 2 → scalool razog indible
8. [0 0] range = 2

0_y(λcI - A) = [ -2 ]

安 [ 0 ]

nA v = ( - [ 1 1 1

S+1] 0 [ -S+1] [ [1 -1 ] 1 2 + 1

lo ifero

S+2]

C(s) = [s -1 0]

[-1 0] [1] = [1] [0]

S. C(s) = [s-2] [0] 1

C(s)

[-1 1] [s-1]

[s-2]

C(s) = [1 -1] [1] [s-1] 0

1

[0]

s-1 s-2

C(s) SFT =

(s-1) = 1/(s-1) (s-1) = e e forma minima

U(s) 100 xa C(s) = U(s)

Y(s) [C(s) - 1]/[s(s)] U(s) = 100 - A 50 + 0.5 [A + 50 = 0.5 = 0.5 s ]

[s (s)] [s - 2] s+ 1 s -2

[s(s)]

(A + 50 = 0.5-30 (A)

(A 100 =A 50

d-1 [Ya] = 50 Ya(s) = 50 ye (t) + [e-2t] 0 +

[s+2]

0- ya C(s)

lim y(0,t) = 0 t 50 lim ye(0,t) = 10

lim t - 50 lim te a C(s)

UI(t). 100 sa C[s] = 100/5

10

C(s). Ya (s)

Y(s) = - = A - - e e

[5][s] ya C(s) 50 - a 50 C(s) - 50 e^-2t

C(s)

Ya(s) = o o C(s)

[s]

[s+2][s] =

Ya(t) = 50 [e^-t-] - ya C(s)

Y(t) = 50 e^-t+ 1000

lim dice U

lim y H(t) t -> -1

(lim y H(s) t ->0,

se dv spoil sa sólido ex. Velcro, quello di coile bo un piego pin rigido

• a=8
• prob b uno poco più di elev (pero coituito)
• obiere estube polté F, celato forte
• C= lo uno (pero dolore bajo el mono t y so do pina aco 1 no x o.
• e- sempre pole F o no = contornar condestro California o 0 pero = o o
• z=
• b=
• obtienco forza o e
• c- impostar forzarm o zero en osto o áneo e forza 0
• pole cel b en co o coiso o o en mon un 0=2
• c=2

ẋ₁ = x₂

ẋ₂ = -x₁²

ẋ₃ = -x₄ + 1

ẋ₄ = 5x₃

_{(x = 0)}

ẋ₁ = x₂

ẋ₂ = -x₄ + (x₃ + x₄)x₂

ẋ₃ = 5x₃

ẋ₄ = x₁₀x₄ + x₃(5 - x₄)

x = 0

⟦4, 0, 4, 0⟧

-3 - x₁x² = 0

0x₂ - x₂ = 0

λ = 0

A = ⎡ 0 0 0 0 ⎤

     ⎢ 2 0 0 0 ⎥

     ⎢ 0 0 -5 1 ⎥

     ⎣ 0 0 3 λ + 1 ⎦

• λᵢ = 0
• λ₁ = 0 μ = no
• μ = no golden stable
• μᵢ = 0 xₙ = ⟦0 0 0⟧

L(s) =

(s + 10)/(s(s + 100))

P = 4

C(s) = (s + 1) ... da ab

G₄

v_i

v_o

1. s = 10
2. s = 10
3. s = 10

5 s + 10s = 105 + 10

5 s + 10 s + 1 = 105 s + 50s = 10

5 s + 105 s + 10

5 s + 10

5 (s + 1) -1

5 s + 105 (s + 100)s(s + 100)= 100

g(s) (100)=

• g(s) (1000 - 050)
• g(s) (100) (s)

0 00 -360 -180 -90 -30 = 0

g(s₁)(s + 40)

P

(-180)³

(s_+/)

N = (0)

u = 1

P_e = i

[5 -3]

55 -10

L(s) =

s (s + 10)

(s + 1) 6

(s - 10) 8.18

L(s) = s (s + 1)

L(s) _i = ω (j ω)(j ω + 1)

(L (j ω))

(s + 10)

[41]

-3

P₀ = 2

180°

P₁ = 0

ω

0

1

2

L

e

0

-90°

5

T

90° + 5° = 180°

P_D

e_L

3

P₀

P₁

P₀

P₀

P₁ = 0

P₀

P_q

L = 0

e

_BP₃

Centro←

Punto

Punto

Punto

Punto

Punto

Punto

Punto

C

C

P

P₂

L = 0

R = 0

(P₀)²

5

5 + 2

2

Punto

Punto

Punto

Punto

3

2

1 -

0

Punto

Punto

Punto

-1

-P

-P

-0.3

e

0.02

5

3

5

5

1

-0.3

33

-0.3

-1

9

3

3

(s)

P₂ = 0

(C)

(C)

(C)

3

L(s) =

_P2

e

g

9

9.3

[23]

Punto

Punto

3

L₅ =

16

12

L

₅ =

g

g

G(s)

g

G(s)

g

9

8.03

8.18

≪

5

P_D0

P_D

B

-

1

1

-

-

1

0

↑

↓

↑

←

g

c

Punto

(s)²

Q₂ =

(s - 10)

Q₁ =

Q₃ =

→

c_D

A -1

F

V

g g

P

L

P_C

P_D3

P_C

P_Fi

C

P_C1

k

P

(-0.02)

→

→

- (L)

(L)

0

3

5

5

180°

90°

(

Q₁

g

A

Punto

-0.03

-0.098

-0.093

9

L

l

l

C

C

g

C

g

-

-

L

P

C

(s)

3

5

g

Punto

L

(5)

L = P

(s)

1

-

2

-

1

0

{

(x)₂ =

{

(K)⁰ =

(x)₂

{

((K)₂)

x²

-((κ)₁)

- {((K) ₂}→(x)

x₃

{

()

- {((κ)₁)}(x)₂

(

(x)

0

(L(s)

-

Punto

F

a

9

1

9(+3)

-0.02

0.2

-

[B]

C

8 (s)

-81 (+20)

+

^x

<

=

{

[0,2,2]

-

[8,8,0]

()

80

90

-0

180

-0

(s + 5)

Punto

Punto

(5)

(5)

P_M0.9

Punto

-0.18

Punto

8

(8)

(8)

-18

(0)

Punto

Punto

-

8

8.1

2

(s)

8