Observer and state-space model

Problemè 1

Consider the following state-space model:

$$\begin{array}{rl}\dot{x}& =\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -5& -6& 0\end{array}\right]x+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u\\ y& =\left[\begin{array}{ccc}1& 0& 0\end{array}\right]x\end{array}$$

Design an observer to place the observer poles at -10, -10, -15

Solution

The characteristic equation of the observer is given by:

$$det(sI-(A-LC))=(s+10)(s+10)(s+15)=s^3+35s^2+350s+1500$$

From the coefficients of the characteristic equation we get

$$det(sI-(A-LC))=s^3+(l_1-6)s^2+(l_2-5-6l_1)s+(l_3-5l_2)$$

Solving for the coefficients we get the observer gain matrix:

$$L=\left[\begin{array}{c}4505\\ 601\\ 41\end{array}\right]$$

Thus the observer dynamics are given by:

$$\dot{\hat{x}}=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ -5& -6& 0\end{array}\right]\hat{x}+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]u+\left[\begin{array}{c}4505\\ 601\\ 41\end{array}\right](y-\hat{y})$$

Problemè 2

Given the plant

$$\begin{array}{rl}\dot{x}& =\left[\begin{array}{cc}-1& 1\\ 0& 2\end{array}\right]x+\left[\begin{array}{c}0\\ 1\end{array}\right]u\\ y& =\left[\begin{array}{cc}1& 1\end{array}\right]x\end{array}$$

Design an integral controller to yield a 10% overshoot, 0.5 second settling time and zero steady-state error for a step input.

Solution

The code for the integral controller is given by p2.py.

Add an integrator to the plant to ensure zero steady-state error for a step input. The augmented state-space model becomes:

$${\dot{x}}_a=\left[\begin{array}{ccc}-1& 1& 0\\ 0& 2& 0\\ -1& -1& 0\end{array}\right]x_a+\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]u$$ $$y=\left[\begin{array}{ccc}1& 1& 0\end{array}\right]x_a$$

where

x_a = \begin{bmatrix} x \\ \int e \, dt \\ \end{bmatrix} $$ and $e = r - y$ is the tracking error. Then, design the state feedback gains

K = \begin{bmatrix} k_1 & k_2 & k_i \ \end{bmatrix}

$$Thecharacteristicequationoftheclosed-loopsystemis:$$

\left| sI - (A_a - B_aK) \right| = 0

$$Expandingthisyields:$$

(s + k_1)(s^2 + (1 - k_2)s + k_i) = 0

$$Thecontrollawthengivenby:$$

u = -Kx_a = -k_1x_1 - k_2x_2 - k_i\int{e , dt}

$$Thecodeyields:‘‘‘prologzeta:0.5911550337988974omeg{a}_n:13.53282902556064Desiredpoles:[-8.+10.91501083j-8.-10.91501083j-135.32829026+0.j]Plantmodel:<LinearIOSystem>:sys[2]Inputs(1):{[}^{\prime }u[0{]}^{\prime }]Outputs(1):{[}^{\prime }y[0{]}^{\prime }]States(2):{[}^{\prime }x[0{]}^{\prime }{,}^{\prime }x[1{]}^{\prime }]A=[[-\mathrm{1.1.}][\mathrm{0.2.}]]B=[[0.][1.]]C=[[\mathrm{1.1.}]]D=[[0.]]Augmentedplantmodel:<LinearIOSystem>:sys[3]Inputs(1):{[}^{\prime }u[0{]}^{\prime }]Outputs(1):{[}^{\prime }y[0{]}^{\prime }]States(3):{[}^{\prime }x[0{]}^{\prime }{,}^{\prime }x[1{]}^{\prime }{,}^{\prime }x[2{]}^{\prime }]A=[[-\mathrm{1.1.0.}][\mathrm{0.2.0.}][-1.-\mathrm{1.0.}]]B=[[0.][1.][0.]]C=[[\mathrm{1.1.0.}]]D=[[0.]]Statefeedbackgains:K=[[-\mathrm{10193.77795361152.32829026}-12391.83976888]]Integralcontrollertransferfunction:-1.239e+04----------sOpen-looptransferfunction:<LinearICSystem>:sys[6]Inputs(1):{[}^{\prime }u[0{]}^{\prime }]Outputs(1):{[}^{\prime }y[0{]}^{\prime }]States(3):{[}^{\prime }sys[4{]}_x[0{]}^{\prime }{,}^{\prime }sys[2{]}_x[0{]}^{\prime }{,}^{\prime }sys[2{]}_x[1{]}^{\prime }]A=[[-0.00000000e+000.00000000e+000.00000000e+00][0.00000000e+00-1.00000000e+001.00000000e+00][-1.23918398e+040.00000000e+002.00000000e+00]]B=[[1.][0.][0.]]C=[[\mathrm{0.1.1.}]]D=[[0.]]Closed-looptransferfunction:<LinearICSystem>:sys[9]Inputs(1):{[}^{\prime }u[0{]}^{\prime }]Outputs(1):{[}^{\prime }y[0{]}^{\prime }]States(3):{[}^{\prime }sys[6{]}_{s}ys[4{]}_x[0{]}^{\prime }{,}^{\prime }sys[6{]}_{s}ys[2{]}_x[0{]}^{\prime }{,}^{\prime }sys[6{]}_{s}ys[2{]}_x[1{]}^{\prime }]A=[[0.00000000e+00-1.00000000e+00-1.00000000e+00][0.00000000e+00-1.00000000e+001.00000000e+00][-1.23918398e+040.00000000e+002.00000000e+00]]B=[[1.][0.][0.]]C=[[\mathrm{0.1.1.}]]D=[[0.]]‘‘‘$$