An Estimator Design for a Simple Pendulum
Design an estimator for the simple pendulum. Compute the estimator gain matrix that will place both the estimator error poles at $-10ω_0$ (five times as fast as the controller poles selected in Example 7.15). Verify the result using MATLAB for $ω_0 = 1.$ Evaluate the performance of the estimator.

Question

An Estimator Design for a Simple Pendulum
Design an estimator for the simple pendulum. Compute the estimator gain matrix that will place both the estimator error poles at [latex] −10ω_0 [/latex] (five times as fast as the controller poles selected in Example 7.15). Verify the result using MATLAB for [latex] ω_0 = 1.[/latex] Evaluate the performance of the estimator.

Accepted Answer

The equations of motion are

[latex] \dot{\pmb x } = \begin{bmatrix} 0 & 1 \ -w^2_0 & 0 \end{bmatrix} \pmb x + \begin{bmatrix} 0 \ 1 \end{bmatrix} u , [/latex] (7.138a)

[latex] y= \begin{bmatrix} 1 & 0 \end{bmatrix} \pmb x .[/latex] (7.138b)

We are asked to place the two estimator error poles at [latex] −10ω_0.[/latex] The corresponding characteristic equation is

[latex] α_e(s) = (s + 10ω_0)^2 = s^2 + 20ω_0s + 100ω^2_0.[/latex] (7.139)

From Eq. (7.136), we get

[latex] det[s \pmb I − (\pmb F − \pmb {LH})] = s^2 + l_1s + l_2 + ω^2_0. [/latex] (7.140)

[latex]det[s \pmb I − (\pmb F − \pmb {LH})]= 0. [/latex] (7.136)

Comparing the coefficients in Eqs. (7.139) and (7.140), we find that

[latex] \pmb L = \begin{bmatrix} l_1 \ l_2 \end{bmatrix} = \begin{bmatrix} 20ω_0 \ 99ω^2_0 \end{bmatrix} . [/latex] (7.141)

The result can also be found from MATLAB for [latex] ω_0 = 1, [/latex] using the following MATLAB statements:

Question 7.EX.25: An Estimator Design for a Simple Pendulum Design an estimato...

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