7 Design with SS

Consider the plant given by

$$G(s)=\frac{Y(s)}{U(s)} = \frac{1}{s^2}$$

which represents a pure inertia. With two poles at the origin, this plant is unstable. Suppose we wish to design a control law for this plant so the poles are at (you can change these later using the interactive figure)

Pick a sample period of $T = 0.1$ second, this is about six times the desired natural frequency. Letting the state vector be

$$\bold{x}=\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} y\\ \dot{y} \end{bmatrix}$$

the state equations are

$$\begin{aligned}\dot{x}_1 &= x_2 \\ \dot{x}_2 &= u\end{aligned}$$

and the output is simply $y = u$, hence the continuous state-space matrices are

$$\bold{A}=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, \bold{B} = \begin{bmatrix} 0\\1 \end{bmatrix}, \optbreak{} \bold{C}= \begin{bmatrix} 1 & 0 \end{bmatrix}, D=0 $$

Discretizing the system at $T = 0.1$ second, we obtain

$$\bold{\Phi} = \begin{bmatrix}1 & 0.1 \\ 0 & 1 \end{bmatrix}, \bold{\Gamma} = \begin{bmatrix} 0.005 \\ 0.1 \end{bmatrix}$$

Using state feedback law $u(k) = −\bold{Kx}(k)$, the controlled system state equation is

$$\bold{x}(k+1)=[\bold{\Phi}-\bold{\Gamma K}]\bold{x}(k)$$

where

$$\bold{K}=\begin{bmatrix}K_1 & K_2 \end{bmatrix}$$

The characterstic equation of the controlled system is given by the determinant

$$|z\bold{I}-\bold{\Phi} + \bold{\Gamma K}|=0$$

and after some algebra we obtain the symbolic characterstic equation as

$$z^2 + (0.005K_1 + 0.1K_2 −2)z \optbreak{}+ (0.005K_1 −0.1K_2 + 1) = 0$$

The desired pole locations in the s plane can be mapped into the z plane using $z = e^{sT}$, and we obtain

thus the numerical form of the desired characteristic polynomial is

Equating coefficients of like powers of z of both characteristic polynomials gives the two simultaneous equations

from which we obtain

The feedback law is negative position and velocity feedback, since state variables $x_1$ and $x_2$ are position and velocity...assuming that the output is displacement, that is!

The controlled system dynamics are

$$\bold{x}(k+1)=[\bold{\Phi}-\bold{\Gamma K}]\bold{x}(k)$$

where $\bold{\Phi}-\bold{\Gamma K}$ is

Simulation of the controlled system’s response from initial condition $x_0 = \begin{bmatrix} 1\\1 \end{bmatrix}$ yields the plotshown in Figure 1.

pole location in s plane and IC response
Figure 1: (left) Desired pole location, (right) Response Solid lines are drawn for clarity. They do not represent the intersample response. of controlled inertia plant from initial conditions $\bold{x}(0) = [1 \quad 1]^T$, sample time $T = 0.10$ sec., and gains $K$=[49.33,10.03].

Drag the desired poles (×) on the s-plane in Figure 1 to change the desired pole locations. The value for $\alpha_c$ is automatically updated in the example and the gain $K$ is recalculated using pole placement. The corresponding discrete time response is shown in the figure.

  1. Change the pole locations so that the system has a faster response. What happens to the gain values?
  2. What pole locations result in very small gain values? How does that compare with the open-loop pole location?
  3. The control effort is $u = -\bold{Kx}$. What is the "price" for having faster closed-loop response?