Control Law for a Pendulum Suppose you have a pendulum with frequency ω0 and a state-space description given by [x1 x2 ] = [0 1 −ω^2 0 0 ] [ x1 x2 ] + [ 0 1 ] u. (7.74) Find the control law that places the closed-loop poles of the system so that they are both at −2ω0. In other words, you wish to

Question

Control Law for a Pendulum
Suppose you have a pendulum with frequency [latex] ω_0 [/latex] and a state-space description given by

[latex] \begin{bmatrix} \dot{x}_1 \ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -w^2_0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} + \begin{bmatrix} 0 \ 1 \end{bmatrix} u . [/latex] (7.74)

Find the control law that places the closed-loop poles of the system so that they are both at [latex] −2ω_0. [/latex] In other words, you wish to double the natural frequency and increase the damping ratio [latex] ζ [/latex] from [latex] 0 [/latex] to [latex] 1.[/latex]

Accepted Answer

From Eq. (7.73) we find that

[latex] α_c(s) = (s − s_1)(s − s_2)...(s − s_n) = 0. [/latex] (7.73)

[latex] α_c(s) = (s + 2ω_0)^2 [/latex] (7.75a)

[latex] \ \ \ \ = s^2 + 4ω_0s + 4ω^2_0. [/latex] (7.75b)

Equation (7.72) tells us that

[latex] det [s \pmb I − (\pmb F − \pmb {GK})] = 0. [/latex] (7.72)

[latex] det [s \pmb I - ( \pmb F - \pmb {GK})] = det \left\{\begin{bmatrix} s & 0 \ 0 & s\end{bmatrix} - \left(\begin{bmatrix} 0 & 1 \ -w^2_0 & 0\end{bmatrix} - \begin{bmatrix} 0 \ 1 \end{bmatrix} \begin{bmatrix} K_1 & K_2\end{bmatrix} ight) ight\} , [/latex]

or

[latex] s^2 + K_2s + ω^2_0 + K_1 = 0. [/latex] (7.76)

Equating the coefficients with like powers of s in Eqs. (7.75b) and (7.76) yields the system of equations

[latex] K_2 = 4ω_0, [/latex]

[latex] ω^2_0 + K_1 = 4ω^2_0, [/latex]

and therefore,

[latex] K_1 = 3ω^2_0, \ K_2 = 4ω_0. [/latex]

More concisely, the control law is

[latex] \pmb K = \begin{bmatrix} K_1 & K_2\end{bmatrix} =\begin{bmatrix} 3w^2_0 & 4w_0 \end{bmatrix}. [/latex]

Figure 7.13 shows the response of the closed-loop system to the initial conditions [latex] x_1 = 1.0, \ x_2 = 0.0, ext{and} ω_0 = 1. [/latex] It shows a very well damped response, as would be expected from having two roots at [latex] s = −2. [/latex] The MATLAB command impulse was used to generate the plot.

Question 7.EX.15: Control Law for a Pendulum Suppose you have a pendulum with ...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions