Full-State Feedback Control Design Consider the DC motor–driven cart discussed in Example 10.13, in which a PD controller was designed to satisfy the requirements: overshoot Mp

Question

Full-State Feedback Control Design

Consider the DC motor–driven cart discussed in Example 10.13, in which a [latex] ext{PD}[/latex] controller was designed to satisfy the requirements: overshoot [latex]M_p < 10\%[/latex] and rise time [latex]t_r < 0.15 s[/latex].

a. Find a full-state feedback controller such that the closed-loop system meets the same requirements.
b. Use MATLAB to find the control gain matrix [latex]\mathbf{K}[/latex].

Accepted Answer

a. The transfer function of the cart is given by

[latex]G(s) = \frac{Y(s)}{U(s)} = \frac{3.778}{s^2 + 16.883s}[/latex]

where:
the output [latex]y[/latex] is the position of the cart
the input [latex]u[/latex] is the voltage applied to the [latex] ext{DC}[/latex] motor

Using the position and the velocity as the state variables, that is, [latex]x_1 = y[/latex] and [latex]x_2 = \dot{y}[/latex], we find the state-space model as

[latex]\dot{\mathbf{x}} = \begin{bmatrix} 0 & 1 \ 0 & -16.883 \end{bmatrix}\mathbf{x} + \begin{bmatrix} 0 \ 3.778 \end{bmatrix}u[/latex]

[latex]y = \begin{bmatrix} 1 & 0 \end{bmatrix}\mathbf{x}[/latex]

For a second-order system, the gain matrix [latex]\mathbf{K}[/latex] is a [latex]1 × 2[/latex] matrix in the form

[latex]\mathbf{K} = \begin{bmatrix} k_1 & k_2 \end{bmatrix}[/latex]

Applying Equation 10.71 yields the theoretical characteristic polynomial

[latex]\left| s\mathbf{I} - \mathbf{(A - BK)} ight| = 0[/latex] (10.71)

[latex]\left| s\mathbf{I} - \mathbf{(A + BK)} ight| = \left| \begin{bmatrix} s & 0 \ 0 & s \end{bmatrix} - \begin{bmatrix} 0 & 1 \ 0 & -16.883 \end{bmatrix} + \begin{bmatrix} 0 \ 3.778 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix} ight|[/latex]

[latex]= \begin{vmatrix} s & -1 \ 3.778k_1 & s+16.883+3.778k_2 \end{vmatrix} [/latex]

=[latex]s^2 + (16.883+3.778k_2)s + 3.778k_1[/latex]

The time-domain specifications, [latex]M_p < 10\%[/latex] and [latex]t_r < 0.15 ext{s}[/latex] indicate that [latex]ζ > 0.59[/latex] and [latex]ω_n > 12.33 ext{rad/s}[/latex]. Choose the same values of [latex]ω_n[/latex] and [latex]ζ[/latex] as in Example 10.13, where [latex]ω_n = 13.5 ext{rad/s}[/latex] and [latex]ζ = 0.65[/latex]. The desired closed-loop poles are then located at

[latex]p_{1,2}= −ζω_n ± ext{j}ω_n \sqrt{1-ζ^2} = −8.775 ± 10.26 ext{j}[/latex]

Applying Equation 10.72 gives the desired characteristic polynomial

[latex](s − s_1)(s − s_2 )…(s − s_n) = 0[/latex] (10.72)

[latex](s − p_1)(s − p_2 ) = s^2 + 2ζω_ns + ω_n^2 = s^2 + 17.55s + 182.25[/latex]

Equating the two characteristic polynomials, we have

[latex]16.883 + 3.778k_2 = 17.55[/latex]

[latex]3.778k_1 = 182.25[/latex]

which gives [latex]\mathbf{K} = \begin{bmatrix} 48.24 & 0.18 \end{bmatrix}[/latex] . So, the full-state feedback controller is

[latex]u = −\mathbf{Kx} = −48.24x_1 − 0.18x_2[/latex]

b. The following MATLAB script computes a full-state feedback control gain matrix [latex]\mathbf{K}[/latex] using pole placement design

Question 10.24: Full-State Feedback Control Design Consider the DC motor–dri...

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