Control and Actuating Devices for Mechanical Systems

Name: Control and Actuating Devices for Mechanical Systems
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Raccolta di appunti direttamente dalle lezioni del professore e da slides basati su appunti personali del publisher presi alle lezioni del prof. Sabbioni dell’università degli Studi …

Esame Controlli e azionamenti

Facoltà Ingegneria industriale

Dal corso del Prof. Sabbioni Edoardo

Università Politecnico di Milano

A.A. 2016-2017

109 pagine

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Control and actuating devices for mechanical systems

Prof. Edoardo Sabbioni

Control strategy

Variables:

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 control based on a model:

Demand
Actuator
Torques
Forces

System inputs response:

ϐ_c: Control forces

ϐ_d: Deterministic forces

ϐ_rand: Random forces

mẍ + rẋ + kx = ϐ_c + ϐ_d + ϐ_rand

Target: X ≡ X_r (reference)

X_r FFC ϐ_c SYSTEM X ϐ_d + ϐ_rand

m̂ẍ_r + r̂ẋ_r + k̂x_r = ϐ_c + ϐ_d

ϐ_c - ϐ_c(t) = m̂ẍ_r + r̂ẋ_r + k̂x_r - ϐ_d

mẍ + rẋ + kx = m̂ẍ_r + r̂ẋ_r + k̂x_r - ϐ_d + ϐ_d + ϐ_rand

The model is the one of the equation of motion:

I don't add ϐ_rand because it's unknown. We substitute this into our equation of motion under this hypothesis of m = m̂, r = r̂, and ϐ_rand negligible

X ≡ X_r

In general:

X(t) = X_H(t) + X_P,r(t) + X_P,rnd(t)

Particular solution due to random forces
Particular solution due to reference forces
Homogenous solution

Homogeneous solution

It's the transient, we have when mẍ + rẋ + kx = 0

X_H = A₁ e^λ₁t + A₂ e^λ₂t

λ_1,2 = -σ ± iω

σ = ^r/_2m = hω₀; ω = ω₀√1-h²

ω₀ = √^k/_m; h = ^r/_2mω₀

FFC does not affect transient, it depends only on m, r, k

Particular solution

Random force f_rand

∫f_rand = ∑_k=1^N_c |F_rand,k| cos(kΩ₀t + φ_k) = ∑_k=1^N_c Re (F_rand,k e^ikΩ₀t)

(Ω_k = kΩ₀)

The response to a harmonic function is an harmonic function too

X_P,rnd = ∑_k=1^N_c |X_rnd,k|cos(Ω_kt + Θ_k)

FRF = G(Ω) = X_end / β_end = 1 / (-mΩ²iΩc+k)

Also in this case FFC does not affect FRF

If π↑, the peak lowers, but the resonance region is larger. We have to make a trade-off

2π / Ω

X_c(t) = Σ _n=1^N |X_n| cos(Ω_nt + α_n) = Σ_n=1^N Re(X_n e^iΩ_nt)

Eq. of motion to get particular solution of β_c

Derivatives

mẍ_c+rẋ_c+kX_c = β_c

{ ẋ_n(t) = Σ_n=1^N Re(iΩ_nX_n e^iΩ_nt)

ẍ_n(t) = Σ_n=1^N Re(-Ω_n²X_n e^iΩ_nt)

So I get

β_c(t) = Σ_n=1^N Re[(-mΩ_n²+irΩ_n+k) X_n e^iΩ_nt]

X_r = G^-1(i)_c = - m_n² + i_n + X_p,R(t) = ∑_k=1^N Re (X_R,k e^i_kt) = ∑_k=1^N |X_R,k| cos(_kt + _k)

If we substitute g_w with g_c w G(i) = ^X/ we get an insanity

In practice it’s impossible: We can’t exactly estimate m,, (R = the hardest)

During resonance we get big displacement and low forces

We may consider only the low freq. component of |X_c| ⇒ X^o

It’s not a triangular wave anymore but a sinusoidal

Summary FFC

Transient not affected
Disturbances are not reduced
In theory we have perfectly X ≅ X_r
In practice a lot of issues are present due to approximation

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