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Control and Actuating Devices for Mechanical Systems
Prof. Edoardo Sabbioni
CONTROL STRATEGY
MANIPULATED
NON MANIPULATED
OBSERVABLE
NON OBSERVABLE
- Manipulated: I can control them
- Non manipulated: Unknown or unwanted input of my system
- Observable: All the variables I can measure
- Non observable: Non measurables
FEEDFORWARD CONTROL (FFC)
TARGET INPUTS
CONTROL BASED ON A MODEL → DEMAND → ACTUATOR → TORQUE FORCES → SYSTEM → RESPONSE
mẍ + rẋ + kx = βc + βd + βrnd
- βc: Control forces
- βd: Deterministic forces
- βrnd: Random forces
TARGET X ≡ Xr (reference)
The model is this one of the equation of motion
I don't add βrnd because it's unknown
βc = βc(t) = m̂ẍr + r̂ẋr + k̂xr - βd
We substitute this into our equation of motion
mẍ + rẋ + kx = m̂ẍr + r̂ẋr + k̂xr - βd + βd + βrnd
Under the hyp. of
m=m̂
r=r̂
k=k̂
and βrnd negligible
⇨ X ≡ Xr
FEEDBACK CONTROL (FBC)
TARGET
INPUTS
mẍ + rẋ + νx = βc + βsens
Now I don't consider it just for simplicity, it does not add anything
Now
βc = βc(x, ẋ)
βc = νp(Xr− x) + νd (Ẋr− ẋ) for example
mẍ + rẋ + νx = βsens + νp (Xr− x) + νd (Ẋr− ẋ)
mẍ + (r + νd)ẋ + (ν + νp)x = βsens + νp Xr + νd ẋr
X(t) = XM(t) + Xp, rand(t) + Xp, R(t)
- Homogeneous Solution
Xh = A1 eλ1t + A2 eλ2t
λ1,2 = -α ± iω = −r/ 2m± √((r/2m)2 − ν/m)
λ1,2 = −(r + νd)/2m ± √((r + νd)/2m)2 − ν + νp/m
r, ν are now affected by the presence of the control system
LINEAR
- ASYMPTOTIC STABILITY
- UNSTABLE
- STABLE
NON LINEAR
- ASYMPTOTIC STABILITY
- UNSTABLE
- ? WE CAN'T SAY ANYTHING
STABILITY
- EQ. OF MOTION
mẍ + tẋ + kx = β(x,ẋ)
NON LINEAR TERM
- WORKING CONDITIONS
X = X₀ → ẋ = Ẍ = 0
STATIC EQUILIBRIUM CONDITION
kX₀ = β(X₀,0)
WE MAY HAVE MORE THAN 1 SOLUTION
- 1
- 2
- 3
3 STATIC EQ. POSITIONS:
- STABLE
- INDIFFERENT
- UNSTABLE
- LINEARIZATION
∫(x,ẋ) ≈ ∫(X₀,Ẋ₀) + ∂∫/∂x|ₓ₀ (x-X₀) + ∂∫/∂ẋ|ₓ₀ (ẋ-Ẋ₀)
ADDITIONAL STIFFNESS
-Kₑ
ADDITIONAL DAMPER
-Tₑ
so ∂fx/∂y = ∂fy/∂x
so {kxy = ∂fx/∂y kyx = ∂fy/∂x
kxy = kyx in [k] = [ kxx kxy ] [ kyx kyy ]
[k] is symmetric
- Conservative force fields: ∮ δ x δs = 0 and ∂fx/∂y = ∂fy/∂x (kxy = kyx)
- Non conservative force fields: ∮ δ x δs ≠ 0 → ∂fx/∂y ≠ ∂fy/∂x → kxy ≠ kyx
When we consider a work cycle, energy is introduced i.e. friction forces
Y
x
[k] ∮ δ x δs ∂fx/∂y ; ∂fy/∂x kxy kyx Conservative symm. = 0 = Non conservative Non symm. ≠ 0 ≠2 dofs stability
2 dofs: x, y
Y
x
F(x, y)
fx(x, y)
fy(x, y)
Force field:
- Conservative
- Gravitational
- Elastic
- Non conservative
- Aerodynamic
- Friction
- det(N) > 0 , Vxx or Vyy < 0
δ < 0
Δ = VxxVyy - VxyVyx > 0
- det(N) > 0 so Vxx and Vyy must have the same sign, but at least one must be negative because [N] is not definite positive
- Vxx < 0
- Vyy < 0
β = ( Vxx + Vyy/2)2 - Δ > 0
|√β| > |δ|
λ1,22 = -γ ± √β = < α1
&nsbp;α2 both positive
λ1,2 = ±√α1 = ±α1
λ3,4 = ±√α2 = ±α2
STATIC INSTABILITY
NON CONSERVATIVE FORCE FIELD
- Aerodynamic Forces
- Friction Forces
- Tires-road contact
∂Py/∂y + ∂Px/∂x -> V̅xy ≠ V̅yx
λ1,22 = -δ ± √β
β = ( Vxx + Vyy/2 )2 - det(N) = ( Vxx - Vyy/2 )2 + VxyVyx
β ≥ 0 since Vxy ≠ Vyx
β > 0
λ1,22 = -δ = √β
Both real
det(N) < 0 -> |√β| > |δ|
λ1,22 = <-α1
&nsbp; &nsbp;+α2
λ1,2 = ±iω2
λ3,4 = ±α
STATIC INSTABILITY
4) Stability
Ẑ(t) = Ẑ₀ eλt (λ2[M]+λ[R₁+R₂]+[K]) Ẑ₀ eλt = 0
As before, we have solutions only if the matrix within brackets is not singular. Here the equation is of 4th order and not bi-quadratic, so it's very difficult to solve. Luckily we can use this state-space formulation
5) State-Space
{ [M]ẍ + [R]ẋ + [K]Ẑ = 0
Ẑ = ẏ } → X = {ẏ Ẑ} state space vector
[M] [0] [R₃] [K] {ẋ} = {0}
[0] [I] ẋ + [T₃] [0] X = {0} [E₁,E₂]
[E₁]ẋ + [E₂]Ẑ = 0
Ẋ = -[E₁]-1[E₂]Ẑ X = -[A]X
[A] = [ -[M]-1[R] -[M]-1[K] ]
[ [I] [0] ]
- Solution
- Ẋ-[A]Ẑ = 0
- Ẑ = X0eλt (λ[I]-[A])X0eλt=0
det(λ[I]-[A])=0 → eigenvalues ([A])
- If all eig(A) have Re(λi) < 0 → system is stable