Appunti completi di tutto il corso Control and Actuating Devices for Mechanical Systems

Feedforward and feedback control systems

Feedforward control systems

The feedforward output variables measure the behavior of the system inverse dynamics: model or model of the system very knowledge and complete the control force exactly to reach the different reference x_f(t). x_f = p_c (x₁), it isn’t a force feedback because it doesn’t depend on measurement coordinates p_c = Mx_ic_fx_f. Very variable to unknown disturbances. x(t) = x_p(t) + x_p(t).

If we have more than one external force, we will have more than one partial solution; we use superposition of effects. Example: p_c and point → x_p, p_c and x_p point → x(t) = x_p(t) + x_p, p_c(t) + x_point(t) because the system is linear or originally linear.

Feedback control systems

The control force p_c = p_c (x, x₁) is a force based and so it changes the dynamics of the system p_c = k_p (x_ref - x) + k_d (x_ref - x) + k_i s (x_ref - x).

Example of p_c with PID controller. So you can modify the system, changing characteristic times of the system, changing the reaction parameters, or the damping ratio or the nature frequency of the system.

λ_1,2 = -^c/_xm ± √(c/2m)² - k/m = -W_φ + W_φ√2²-1. c̅ = ^c/_2m, W_φ = √k/m.

1. We have modified the dynamics of the system; you can modify the system in this case for x_ref when using point.

Feedforward inverse dynamics

Mix or model of the system deep knowledge, and generate the control force necessary to reach the desired reference x_r(t). f_c = f_c(x,ẋ,ẍ) it isn't a force fixed because it doesn't depend on independent coordinates. f_c = Mx₁ṡ₁c₁x₁c, very variable to unknown disturbances.

If we have more than one external point, we will have more than one revised function; we use superposition of effects e.g. f_c and feedin → x_r, f_c and x_r feedin. x(t) = x_f(t) + x_p(t) + x_c(t) because the system is linear or originally linear.

Feedback advanced or independent coordinates

The control force f_c = f_c(x,ẋ,ẍ) is a force fixed and doesn’t change the dynamics of the system: f_c = k_p(x_ref - x) + k_d(ẋ_ref - ẋ) + k_IS∫(x_ref - x). Example of f_c with PID controller. So you can modify setpoint, leaving characteristics of the system changing are related parameters, as the damping 1/2ζω_n or the natural frequency of the system.

λ_1,2 = -_c/(xm) ± √((c/2m)² - k/m) = - h Wψ + ω_p√2-1. ω_p = √(k/m), ω_n = c/2m, ω_p = √(k/m).

We have modified the dynamic of the system, you can modify stability e.g., this case for f_c. If the system is originally linear, we are very lucky but in real life, this case is very rare. However, we have a non-linear system, and in the solution can't be completed analytically, so we have to use numeric methods. However, if we do linearize it at the locality by response for large oscillations, we can linearize around this point of equilibrium (0, free dry-rate conditions) and can use analytically the result.

Non-linear EQ of motion (find the trouble). Find equilibrium position. Linearize around this equilibrium position and study. Remember, that if we have a feedback system, we will have for the feedback:

Taylor_c = _c(x,x,x) ≈ _c(x_p,φ,δ) + d_c dx → equilibrium position of forced equilibrium. Example: x^p = x_s = x_s = φm_x + c_x + k_x = _c(^f,ϰ,x)_always positive^d + _d + ^dc. This is _{a simple} mechanic system MT(m+k+)∫_dx + unlace entire methods not solved ideally d_x = x - x_p.

We have local mem → _k stability → _c stability. True symmetric stability → steady irreducibility, _c if system is unstable → αf(t) = |=eα_f element of _c, αe(l)x = 0. Dynamic feedback.