Linsys lab forberedelser dag 2

Problem 1

Given $x=\left[\begin{array}{c}p\\ \dot{p}\\ \dot{e}\end{array}\right],u=\left[\begin{array}{c}{\tilde{V}}_s\\ V_d\end{array}\right]$ Then

= A\begin{bmatrix} p \\ \dot{p} \\ \dot{e} \end{bmatrix} + B\begin{bmatrix} \tilde{V}_s \\ V_d \end{bmatrix}$$ **Where** $$A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, B= \begin{bmatrix} 0 & 0 \\ 0 & K_1 \\ K_2 & 0 \end{bmatrix} $$ # Problem 2 The controllability matrix is given by: ![[Controllability Matrix]] Which gives us $$\mathcal{C} = \left[ \begin{array}{cc|cc|cc} 0 & 0 & 0 & K_1 & 0 & 0 \\ 0 & K_1 & 0 & 0 & 0& 0 \\ K_2 & 0 & 0 & 0 & 0 & 0 \end{array} \right]

since $rank(C)=n=3$,The system is controllable.

Problem 3

Given

Where $p$ is the pitch angle, and $\dot{e}$ is the elevation rate which we will try to control with the following state-feedback controller with reference-feed-forward $u=Fr-Kx$ Where

$$\left[\begin{array}{ccc}{k}_{11}& k_{12}& k_{13}\\ k_{21}& k_{22}& k_{23}\end{array}\right],r=\left[\begin{array}{c}{p}_c\\ {\dot{e}}_c\end{array}\right]$$

If we insert our expression for $u$ into the State Space Representation equation, we get:

$$\begin{array}{rl}\dot{x}& =Ax+B(Fr-Kx)\\ \dot{x}& =Ax+BFr-BKx\\ t& \to \infty :\\ {\dot{x}}_{\infty }& =(A-BK)x_{\infty }+BFr=0\\ BFr& =(BK-A)x_{\infty }\end{array}$$

Where

BK-A &= \begin{bmatrix} 0 & 0 \\ 0 & K_1 \\ K_2 & 0 \end{bmatrix} \begin{bmatrix} k_{11} & k_{12} & k_{13} \\ k_{21} & k_{22} & k_{23} \end{bmatrix} - \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ BK-A &= \begin{bmatrix} 0 & 0 & 0 \\ K_{1}k_{21} & K_{1}k_{22} & K_{1}k_{23} \\ K_{2}k_{11} & K_{2}k_{12} & K_{2}k_{13} \end{bmatrix} - \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ BK-A &= \begin{bmatrix} 0 & -1 & 0 \\ K_{1}k_{21} & K_{1}k_{22} & K_{1} k_{23} \\ K_{2}k_{11} & K_{2}k_{12} & K_{2}k_{13} \end{bmatrix} \end{align}

And

$$BF=\left[\begin{array}{cc}0& 0\\ 0& K_1\\ K_2& 0\end{array}\right]\left[\begin{array}{cc}{x}_{11}& x_{12}\\ x_{21}& x_{22}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ K_1{x}_{21}& K_1{x}_{22}\\ K_2{x}_{11}& K_2{x}_{12}\end{array}\right]$$

Which gives us

$$BFr=\left[\begin{array}{c}0\\ K_1{x}_{21}{p}_c+K_1{x}_{22}{\dot{e}}_c\\ K_2{x}_{11}{p}_c+K_2{x}_{12}{\dot{e}}_c\end{array}\right]=\left[\begin{array}{c}-\dot{p}\\ K_1{k}_{21}p+K_1{k}_{22}\dot{p}+K_1{k}_{23}\dot{e}\\ K_2{k}_{11}p+K_2{k}_{12}\dot{p}+K_2{k}_{13}\dot{e}\end{array}\right]$$

Since $\dot{p}=0$, and $p_{\infty }=p_c$, ${\dot{e}}_{\infty }=e_c$, gives us

x_{21} &= k_{21} \\ x_{22} &=k_{23} \\ x_{11} &= k_{11} \\ x_{12} &= k_{13} \end{align}

Meaning

$$F=\left[\begin{array}{cc}{k}_{11}& k_{13}\\ k_{21}& k_{23}\end{array}\right]$$

Problem 4

LQR - Linear Quadratic Regulator

Given a continous-time LTI system in state space form, where D = 0

Link to original

The LQR problem consists of finding $u(t)$ that makes the following equation as small as possible

$$\begin{array}{r}{J}_{LQR}:={\int }_0^{\infty }y(t{)}^{T}Qy(t)+u(t{)}^{T}Ru(t)dt\end{array}$$

• Q and R are positive-definite weighting matrices

  Positive-Definite Matrix

  A symmetric matrix with all positive eigenvalues. NOTE: All symmetric matrices have only real eigenvalues

  For a matrix, $M$, to be positive-semidefinite the following has to hold

  Criterion

  $x^{T}Mx>0,\forall x\in {\mathbb{R}}^n\backslash \{0\}$ Given that $M$ is a $n\times n$ symmetric real matrix

  ---

  Can be written generally as

  $$M=\left[\begin{array}{cc}{m}_1& m_2\\ m_2& m_3\end{array}\right]$$Link to original

• $Q,R>0$, $Q,R=Q^T,R^T$
• The term ${\int }_0^{\infty }y(t{)}^{T}Qy(t)dt$ is a measure of the output energy
• The term ${\int }_0^{\infty }u(t{)}^{T}Ru(t)dt$ is a measure of the control signal energy
• PROTIP: WRITE OUT THE EQUATION TO SEE DIRECTLY HOW EACH PART OF Q AND R AFFECTS EACH STATE

LQR seeks to find a controller that minimizes both energies using the weighting matrices Q and R to establish a trade-off between the control signal and output energy.

• Q - How fast the system reacts
• R - How much energy can we allow in the system

Link to original

For Q:

$$x^{T}Qx=\left[\begin{array}{ccc}p& \dot{p}& \dot{e}\end{array}\right]\left[\begin{array}{ccc}{q}_1& 0& 0\\ 0& q_2& 0\\ 0& 0& q_3\end{array}\right]\left[\begin{array}{c}p\\ \dot{p}\\ \dot{e}\end{array}\right]=q_1{p}^2+{\dot{p}}^2{q}_2+{\dot{e}}^2{q}_3$$

For R:

$$u^{T}Ru=x=\left[\begin{array}{cc}{\tilde{V}}_s& V_d\end{array}\right]\left[\begin{array}{cc}{r}_1& 0\\ 0& r_2\end{array}\right]\left[\begin{array}{c}{\tilde{V}}_s\\ V_d\end{array}\right]=r_1{\tilde{V}}_s^2+r_2{V}_d$$

Meaning that $q_1$ and $q_2$ acts as $k_d$ and $k_p$ for the pitch, and $q_3$ acts as $k_d$ for the elevation And $r_1$ and $r_2$ determines how much the input is amplified

TEST PLAN for $Q_{LQR}$ and $R_{LQR}$:

1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \ce{, and } ; R_{LQR} = \begin{bmatrix} 1 & 0 \ 0 & 1 \ \end{bmatrix}$2.$Q_{LQR} = \begin{bmatrix} 50 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \ce{, and} ; R_{LQR} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$3.$Q_{LQR} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 50 & 0 \ 0 & 0 & 1 \end{bmatrix} \ce{, and} ; R_{LQR} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$4.$Q_{LQR} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 50 \end{bmatrix} \ce{, and} ; R_{LQR} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$5.$Q_{LQR} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \ce{, and} ; R_{LQR} = \begin{bmatrix} 50 & 0 \ 0 & 1 \end{bmatrix}$6.$Q_{LQR} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \ce{, and} ; R_{LQR} = \begin{bmatrix} 1 & 0 \ 0 & 50 \end{bmatrix}$$

Problem 5

We want to include an integral effect, where $\zeta$ and $\gamma$ are our new states

\dot{\gamma} &= p_{c}-p \\ \dot{\zeta} &=\dot{e}_{c}-\dot{e} \end{align}

and the new state vector

$$x=\left[\begin{array}{c}p\\ \dot{p}\\ \dot{e}\\ \gamma \\ \zeta \end{array}\right]$$

Gives us the following system

$$\begin{array}{rl}\dot{x}& =Ax+Bu+Gr\\ u& =Fr-Kx\end{array}$$

We now need to find the augmented $A,B,G$ matrices:

$$\begin{array}{r}\dot{x}=\left[\begin{array}{c}\dot{p}\\ \ddot{p}\\ \ddot{e}\\ \dot{\gamma }\\ \dot{\zeta }\end{array}\right]=A\left[\begin{array}{c}p\\ \dot{p}\\ \dot{e}\\ \gamma \\ \zeta \end{array}\right]+Bu+G\left[\begin{array}{c}{p}_c\\ {\dot{e}}_c\end{array}\right]\end{array}$$

Which is

$$A=\left[\begin{array}{ccccc}0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ -1& 0& 0& 0& 0\\ 0& 0& -1& 0& 0\end{array}\right]$$ $$B=\left[\begin{array}{cc}0& 0\\ 0& K_1\\ K_2& 0\\ 0& 0\\ 0& 0\end{array}\right]$$

and

$$G=\left[\begin{array}{cc}0& 0\\ 0& 0\\ 0& 0\\ 1& 0\\ 0& 1\end{array}\right]$$

TEST PLAN for $Q_{LQR}$ and $R_{LQR}$:

1. Test varying values for $Q_{LQR}$ and $R_{LQR}$