Esercizi ed esami di Control Systems

State and Output Response: Time Domain Analysis

Example 4.41 (Battelotti)

A = 1 | 0 | 12 | 0 | 00 | 1 | 0

x₀ = 110

u(t) = 0

calculate the zero input to state response

> calculate eigenvalues

det(λI - A) = λ |1 -1 0|| 2 λ 0|| 0 2 λ|

= (t-λ)(t₂-1, t-2)

= (λ-1)[(λ-1)t₂²] - 4

• λ₁ = 0
• λ₂ = 1
• λ₃ = 1

> calculate e vectors

(A-λ_iI)z_i = 0 →| 1 0 1 | |z_i1| = |0|| 2 0 2 | |z_i2| = |0|| 0 1 2 | |z_i3| = |0|

→

• z_i3 = z_i1
• z_i4 = ¼1/√(3)
• z_i5 = 1/√(3)

> equivalently i can write z_i = λ11

(A-λ₂I)z₂ = 0

| 0 0 01 | |z_{j we have asymptotic stability only if a ≥ 0}

=> now we calculate the response:

L{y_forced}(s) = P(s) L{v(t)}(s)                 sI/O transfer function = C(sI - A)⁻¹ B + D => P(s), 1/(s+a)

Lt cos(t)(s) = s/s²+1

=> L{yₓ₂ₜ}(s) = s___p(s) which is a strictly proper function so we can use …

=> p(s) = …

=> comparing we obtain A: - 2/(1+a)², B: 2a/(1+a)²

=> now we go back in time:

-> G(s): 100/s

5

and no more constraints on G (1

could be even worth to

send

the gain k=100, but he won't do that.

Before doing it, must update the Bode plots to the new function P(s):

100/s

s

by the Bode plot of P(s):

100/s

s

log w have at w 0.1 rads -> magnitude: 80 dB

phase: -186 -> -m₆₀

-> have to decrease the magnitude by 80 dB and m_e by 60°

-> increase m°, decrease magnitude (care D) -> combine anticipative action to m_end,

re phase and alternative action to drop re magnitude

otherwise simply by a proper

control action as r;

can do that -> have no constraints

on G_(i)

-> always start with the anticipative action (no work on the phase) and then alternative

- we can choose m_e=16 and w_i=4 rads/s -> obtain m_e=60

and w_t=w₁>> w_i->40

-> G(s): 100/s

1+τ_oS/

m^*

G_α(s)

-> at w^*: 100/s

1-τ_oS/

m^*

gives 12 dB

60°

-> now the starting values are:

80; 12; 92 dB

phase 170; 60 = -120 -> m_e=60°

- the alternative action must be of -92 dB 0.0101

-> G(s): 100/s

1+4τ₀S 0.0101

1-2.5S

-> from Bode plot we do Nyquist plot of P(s).G(s)

-> o o.c. w occurs around -1

n=0

-> the closed loop system is a stable

s = 2 do Bode plots with sum Arg(P(jw)) = N * pi

w= w0

from that Nyquist plot become

exercise 1.5

P(s) = (5s)^2 / s(s-1)(5s+1)(s+3) => n_p = 1, z_1: z_2: -1, p_1: -1, p_2: 2, p_3: -3, p_4: 0

non considerando il polo m s=0 ho P(0) = 1 / (x1)(x2)(x3)

|P(0,i)|: 6.3 dB

Root Locus (module 10)

Exercise 2.1

P(s):

s : n. number of poles=3, p₁=0 p₂=1 p₃=-2m: n. number of zeros=0n - m = n° root locus=3ξ: asymptotic center: {0-1-2} / 3n-m=1

- have to determine singular points, in which more than one curve pass.

R(s,k): 1 + k P(s): 5(s+1)(s+2) / k, : 5 (3s-25 k 5(s+1)(s+2) (s+1)s(s+2)

• 1 / s₂
• 1 / s
• 1 / s₁
• 1 / s
• 1 / s₁₂
• - 5(2)(35+2.5*6) (25-2.5)5-5
• - (35+6)5-2

{5(s+1)(s+2)} s(s+1)(s+2)

• {
  • s(5k) {s35-25}k=0
  • s35-25{3}k=0
  • 35+6s42=0
  s35-25}

s:3/⁵ : s₁= 1 / sq. 3⁵

s₂: -sq 1/⁵³:-0.423

• {k: 3/⁵₃
• {k₂: 3/⁵₃

s₂: -sq 1/⁵/-1577

Exercise 2.2

• P(s): s4↑ -¹...₁=-1 ₁=₂=0 ₃=-3
• n=3 m -1.
• n−m = 2
• s₀-{1¬0 3}- (¹₂)2

Singular points:

• R(s,k): = 1:k P(s): s3/35
• :k5+k (x=s3/2) ks=3 ks k o o-{:} s(5s3/3)=0
• k= s2*35 =25 & # 0/(2-5)
• s(s+1)(s+2) s(s+4)/ s(s)(s₂)
• {1/}
• if -⁵ # 5
• 1/s ₣₃
• -1/s₃:1/2(2/3:3-15)-0.5005
• 1/s₄
• k= (s/³)=³5
• -2(6)=s2x(s)
• (²5s3) k=s4+5

- 2s+6> [5>k]

₃s⁵ - 55s⁴ + 583 - 5s (21g + 12k₂) - 144k₂

V₁² = 5(210 + k₂ - 290) - k₂ + 1/5 (37 + 36k₂)

V² 37 + 36k₂

V² 360k₂

V' = 5 210 + 72k₂

k₂(37 + 36k₂)

V² = 8k₂ - 9 x 8(72 k₂ - 11240k₂ + 79992)

37 + 36k₂

(58/32 - 11240 k₂ + 79992)

37 + 36k₂

5.8

3/36

0.9

N + - -

D - + +

⇒ Ok k₂<0

Now draw the root locus of Et(s):

P₀: 0

P₁: -1

P₂: 7/33

P₃: 7/33

P₄: 1/2

P₅: 9/33

R(s; k) = (3(s+4)(s+5)(s+4) 1. k + 2(s - 2))/s(s+1)(s+5)(s+4)

Js

-4s³ - 165³ - 11.6s - 216 + 12k

So: 4

-1

(2)

-7/3