Disk-Drive Servomechanism: Robust Control to Follow a Sinusoid A simple normalized model of a computer disk-drive servomechanism is given by the equations F = [ 0 1 0 −1 ] , G = [ 0 1 ] , G1 = [ 0 1 ] , H = [ 1 0 ], J = 0. Because the data on the disk are not exactly on a centered circle, the

Question

Disk-Drive Servomechanism: Robust Control to Follow a Sinusoid
A simple normalized model of a computer disk-drive servomechanism is given by the equations

[latex] \pmb F = \begin{bmatrix} 0 & 1 \ 0 & -1 \end{bmatrix}, \ \ \ \ \pmb G = \begin{bmatrix} 0 \ 1 \end{bmatrix} , \ \pmb G_1 = \begin{bmatrix} 0 \ 1 \end{bmatrix} , \ \ \ \ \ \pmb H = \begin{bmatrix} 1 & 0 \end{bmatrix} , \ \ \ J = 0 . [/latex]

Because the data on the disk are not exactly on a centered circle, the servo must follow a sinusoid of radian frequency [latex] ω_0 [/latex] determined by the spindle speed.
(a) Give the structure of a controller for this system that will follow the given reference input with zero steady-state error.
(b) Assume [latex] ω_0 = 1 [/latex] and that the desired closed-loop poles are at [latex] −1 ± j \sqrt{3} [/latex] and [latex] − \sqrt{3} ± j1. [/latex]
(c) Demonstrate the tracking and disturbance rejection properties of the system using MATLAB or SIMULINK.

Accepted Answer

(a) The reference input satisfies the differential equation [latex] \ddot{r} = −ω^2_0 r [/latex] so that [latex] α_1 = 0 [/latex] and [latex] α_2 = ω^2_0. [/latex] With these values, the error-state matrices, according to Eq. (7.218), are

[latex] \pmb A = \begin{bmatrix} 0 & 1 & \pmb 0 \ −α_2 & -α_1 & \pmb H \ \pmb 0 & \pmb 0 & \pmb F \end{bmatrix} , \ \ \ \ \pmb B = \begin{bmatrix} 0 \ 0 \ \pmb G \end{bmatrix} . [/latex] (7.218)

[latex] \pmb A = \begin{bmatrix} 0 & 1 & 0 & 0 \ −ω^2_0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & 0 & −1 \end{bmatrix} , \ \ \ \ \pmb B = \begin{bmatrix} 0 \ 0 \ 0 \ 1 \end{bmatrix} . [/latex]

The characteristic equation of [latex] \pmb A − \pmb {BK} [/latex] is

[latex] s^4 + (1 + K_{02})s^3 + (ω^2_0 + K_{01})s^2 + [K_1 + ω^2_0 (1 + K_{02})]s + K_{01}ω^2_0 K_2 = 0, [/latex]

from which the gain may be selected by pole assignment. The compensator implementation from Eq. (7.220) has the structure shown in Fig. 7.60, which clearly shows the presence of the oscillator with frequency [latex] ω_0 [/latex] (known as the internal model of the input generator) in the controller.[latex] ^{16} [/latex]

[latex] (u + \pmb K_0 \pmb x)^{(2)} + \sum\limits_{i=1}^{2}{α_i(u + \pmb K_0 \pmb x)}^{(2−i)} = − \sum\limits_{i=1}^{2}{ K_ie^{(2−i)}}.[/latex] (7.220)

(b) Now assume that [latex] ω_0 = 1 ext{ rad/sec } [/latex] and the desired closed-loop poles are as given:

[latex] ext{pc} = [−1 + j ∗ \sqrt{3}; −1 − j ∗ \sqrt{3}; − \sqrt{3} + j; − \sqrt{3} − j]. [/latex]

Then the feedback gain is

[latex] \pmb K = [K_2 \ K_1 \ : \ \pmb K_0 ] = [2.0718 \ 16.3923 \ : \ 13.9282 \ 4.4641], [/latex]

which results in the controller

[latex] \dot{\pmb x}_c = \pmb A_c \pmb x_c + \pmb B_ce, [/latex]

[latex] u = \pmb C_c \pmb x_c, [/latex]

with

[latex] \pmb A_c = \begin{bmatrix} 0 & 1 \ −1 & 0 \end{bmatrix}, \ \ \ \ \pmb B_c = \begin{bmatrix} −16.3923 \−2.0718 \end{bmatrix}, \ \pmb C_c = \begin{bmatrix} 1 & 0 \end{bmatrix} . [/latex]

The relevant MATLAB statements are

Question 7.EX.36: Disk-Drive Servomechanism: Robust Control to Follow a Sinuso...

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