3.7: Chapter References

References

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Solutions to Practice Problems

1. The answer is a).

and

\[y_t\left(t\right)=0,\ \ y_{ss}\left(t\right)=8\left(1-{\mathrm{cos} t\ }\right)\]

where \(\left|G(jω)\right|\) and \(\angle G(jω)\) are defined below.

\[\left|G(jω)\right|=4{\left({\left(4-ω^2\right)}^2+4ω^2\right)}^{-0.5},\ \ \ \ \ \ \ \angle G\left(jω\right)=-{{\mathrm{tan}}^{-1} \left(\dfrac{4-ω^2}{4ω^2}\right)\ }\]

1. The answer is b).
2. The answer is b).

1. These answers are listed below.
2. Marginally stable.
3. Stable.
4. Marginally stable.
5. Unstable.
6. Unstable.

The stable range for \(K_I\) is \(0<K_I<3\). When \(K_I=3\), there is a pair of imaginary poles \(s_{1,2}=\sqrt{2}j\) due to auxiliary polynomial \(3s^2+6=0\) on the \(s^2\) row.

1. Answers are listed below.
2. Stable.
3. Unstable.
4. Stable.
5. Unstable.

Note that \(ω_g\approx 31\) rad/s, \(ω_p=\infty\), gain margin = \(\infty\), and phase margin = \({47}^{{}^\circ }\).

Hence, we have the following solutions:

a) PO is zero (since first order system does not have any overshoot.

b) For the stable system above, the steady-state response for the unit input is \(y_{ss}={\mathop{\mathrm{lim}}_{s\to 0} G(s)\ }=G\left(0\right)=2\).

c) \(t_s\left(2\%\right)=4T=16\) (second) and \(t_s\left(5\%\right)=3T=12\) (second).

\[PO=100e^{\dfrac{-πζ}{\sqrt{1-ζ^2}}}<20\ \ \ \to \ \ \ ζ>\sqrt{\dfrac

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    at (Bookshelves/Introductory_Engineering/Mechatronics:_Fundamentals_Design_Integration_and_Validation_(Zhu)/03:_System_Responses_and_Controls/3.07:_Chapter_References), /content/body/div/p[12]/span, line 1, column 1

The associated feasible region is plotted below, where red-shaded area is the feasible region.

1. See Example 3.13 for solution.
2. See Example 3.14 for solution.

with closed-loop eigenvalues of

\[s_1=-2.7390,\ \ \ s_{2,3}=-0.6185\pm 0.3657j\]

1. \[L^T=\left[ \begin{array}{ccc} 0.9973 & 1.0734 & 1.7458 \\ 1.0734 & 9.2005 & 42.9005 \end{array} \right]\]

    

The associated dynamic controller system matrices defined in \ref{3.100} are

\[A_c=\left[ \begin{array}{ccc} -0.9973 & -0.0734 & 0 \\ -1.0734 & -9.2005 & 1.0000 \\ -3.1600 & -46.4879 & -3.4892 \end{array} \right],\ B_c=\left[ \begin{array}{cc} 0.9973 & 1.0734 \\ 1.0734 & 9.2005 \\ 1.7458 & 42.9005 \end{array} \right],\ C^T_c=\left[ \begin{array}{c} -2.4142 \\ -5.5873 \\ -4.4892 \end{array} \right]\]

The eigenvalues of the closed loop system matrix defined in \ref{3.102} are as follows.

\[s_{1,2}=-4.0981\pm 3.8466j,\ \ s_3=-1.0015\]

\[s_{4,5}=-0.6929\pm 04384j,\ \ s_6=-2.1033\]

Note that the eigenvalues of \(\left(A_p+B_pF\right)\) are \(s_{4,5}\), \(s_6\) and eigenvalues of \(\left(A_p-LM_p\right)\) are \(s_{1,2}\), \(s_3\), which confirms that the closed-loop system poles are these poles of \(\left(A_p+B_pF\right)\) and \(\left(A_p-LM_p\right)\).

1. Find solutions in Example 3.19.
2. These findings should include the gain scheduling PID controller structure and how the machine learning is applied to obtain scheduled gains.