Riassunti di Sistemi di Controllo

28/09/20

Studying a physical process it's necessary to find mathematical laws or algorithm which can help understanding the behavior and the characteristics of the kuola.net workflow. In this way, you can also modify changing some parameters of the process, the behavior will vary you want. The variables of a physical process are:

• inputs
• states described on the right
• outputs, which are the variables on which we will design the control system

The controller is the system which acts on the inputs variables to obtain a determined states or something else. Inputs can be changed by the controller in two ways:

• independently by outputs, which are available thanks to measurement or
• dependently of reference variable (for example if you want to maintain a certain pressure in a box)

C -> INPUTS -> physical process P -> OUTPUTS STATES

don't depend on the process, are independent and given, practically

of P you have to know the independent variables to know its behavior and how they characterize P

are variables which completely describe the physical process il paradiso dello studente For example position and velocity of a vehicle which is moving

Imagine we have a car moving with determinate mass, position and velocity, its dynamic is governed by the Newton's law M_S=F and V=_S & a=_S.

So if I use variable S, I have a mathematical model made by a second order differential equation, so I need to know two border conditions -> _S=F/M & s(0), _S(0).

On the other way we can choose two variables S=x₁ and _S=x₂, so I can state -> x₁,_S=x₂ -> with this eq too I have a first order differential system and so even a first order model

• note: _x2=F/M and so I have a first order differential equation.
• -> F=U: control variable
• parameter

-> the general form is _X=AX+BU, where in our case

x = ...

ricaduto il sistema entra ad uno stato stazionario ...

04/10/20

We now want to obtain a linear model from a first order non-linear model because the latter doesn't have an analytical solution and so its control is more thorough. In doing so, we obviously introduce approximations because ... of the linearization of _{f(x, u)} and _{y = h(x, u)}.

• For non-linear system: _{x = f(x, u)}, _{y = h(x, u)}
• For example in the platform: _{x1 = u2cosx3}, _{x2 = u1sinx3 x3 = u2}, _{y = x1}

Linearization consists in linearizing to the first order, numerically, whom the method is using. In this case, we have to know that the approximation becomes higher and higher as the values comes far from the starting and at a certain point our model is not useful (for example with McLaurin Taylor) it can't be far from zero values).

• So it depends.
  • Not good approximation.
  • In this case the linearized model gives good approximation of the non-linear model.

Depending on the starting conditions, you choose the better linearization model.

But how we choose the point around which we linearize? We can choose the one that satisfy the zero solution _{x1= x2 = 0 or x2 and x3= ...} in the case of the equilibrium.

IMPORTANTE

Remembering that we have to linearize around a point that is a solution of our model, for the robotic platform we can choose constant V and w.

06/10/20

We saw that starting for a first-order non linear model:

• ẋ = f(x,u)
• y = h(x,u)

we can linearize it and make its study easier, thanks to the form:

• ẋ = Ax + Bu
• y = Cx + Du

However, the method we saw has a problem: the linearization gives good approximation only around the point we chose to linearize, and so we need a lot of linearization to solve a good amplitude of the dynamic. But other approaches are possible.

Let's start from a linear problem:

ẋ = Ax where x(t) is the solution

x₀ = x₀

of the problem. This linear problem has a solution which depends on the initial condition x₀, of the form:

x(t) = e^At x₀

matrix exponential, which can be defined by Taylor:

• e^x = 1 + x + (x²)/2! + (x³)/3! + ...

applying this to the case in which you have a matrix A instead of x ∈ ℝ:

e^At = ∑_j=0^∞ (At)^j/j! ; A ∈ ℝⁿ_n

which is a convergent series.

Let's see some properties of e^At:

• e^At e^Bt = e^(A+B)t ⇒ e^At e^At = e^A₁t e^A₁t only if A₁, A₂ are defined in the way that A₁A₂ = A₂A₁
• d(e^At)/dt = Ae^At = e^At A (because e^At and A always commute)
• (d(x))/dt = (d)(y))/dt = Ẋ(t) = (d(xt))/(dt) = ẋ(tt)

this passage is always possible

e^At = ∑_j=0^∞(At)^j/j! * A₁ . A₁^j

So we can see that:

• ẋ = Ax = d/dt(e^At x₀) = e^At x₀
• ẋ = Ae^At x₀, and so x(t) = e^At x₀ is a solution

But as the solution must be unique, x(t) = e^At x₀ must be the solution.

...al this point I have a base {q₁ q₂... q_n q_k}=....

the help of algebra, a real new base

{V_i, w_i......

...

which will be our theory base

So in this case the matched modes move on a plane spanned

by V_i and w_i...

whose can be circles....

convergent electrical or divergent electrical...

08/10/20

2) As we are analyzing the second case σ(A): {μ, μ, i, , μ, μ}{...........

and every e values generates an e vector, z_i: {A, I} z_i=0 .......

z_i V_i w_i.. But how can we decompose x(t,x)₀ along its natural modal?...

=> (A - μ I)z_i = {A - a_iA - jω I}(V_i w_i)...

⇒ {A - a_iI}V_i {A - ...

iu_i V_i + w_i..w_i = 0

=> We have a real and an imaginary part, so we obtain two equations:

{A - a_i}V_i = ω_iw_i

(A - a_iI)w^-1 V^-1i

=> (A - a_iI) V_i - w_i(A - a_iI)w_i = ω_i² V_i => (A - a_iI) V_i = - ω_i² V_i

with the i^th

equation

=> moreover, knowing tho

(A - a₁){A - ω}²...(A - ω)³{w_i}

(A - a₂)V_i=(-1){ω_i}²w_i

those are ->

re..cursive

formulas

(A - a_i) V_i = (-1){ω_iω}^-1{w_iw}^1/1 =

⇒ .......xp ....... - uniform vector

z_i norm vector

(A - a₂........)

(A - a₂)prime, 2₃...

...............

.....norm.....cibal....

more we consider base of IR {V_i, w_i..., V, w, V} which is independent and

is the one we will use to represent x(t, x)... as did for case 1...

...

......

x₀={..........., c.......}

....... which is so decomposed along the new base

=> x(t, x₀) e^x₀ e^{A_i{g_i, V_i, h_i,w}}

...........