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ECE4132 Control System Design - MUM S2 2025

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Consider the system defined by

$\frac{d\mathbf{x}}{dt} = \begin{bmatrix} 2 & 1\\1 & 0\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix} 1\\0\end{bmatrix}u(t)\quad\textrm{and}\quad y(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t).$ \frac{d\mathbf{x}}{dt} = \begin{bmatrix} 2 & 1\\1 & 0\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix} 1\\0\end{bmatrix}u(t)\quad\textrm{and}\quad y(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t).

Which of the following statements is true?

If output feedback is used, it is not possible to choose observer feedback gains $\mathbf{L}$ \mathbf{L} and state feedback gains $\mathbf{K}$ \mathbf{K} so that $\mathbf{A} - \mathbf{B}\mathbf{K}$ \mathbf{A} - \mathbf{B}\mathbf{K} and $\mathbf{A} - \mathbf{L}\mathbf{C}$ \mathbf{A} - \mathbf{L}\mathbf{C} have eigenvalues in any desired locations.

If output feedback is used, it is possible to choose observer feedback gains $\mathbf{L}$ \mathbf{L} and state feedback gains $\mathbf{K}$ \mathbf{K} so that $\mathbf{A} - \mathbf{B}\mathbf{K}$ \mathbf{A} - \mathbf{B}\mathbf{K} and $\mathbf{A} - \mathbf{L}\mathbf{C}$ \mathbf{A} - \mathbf{L}\mathbf{C} have eigenvalues in any desired locations.

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Consider the continuous-time state equation:

$\frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1 & 1 & 0\\2 & 0 & 1\\-1 & 0 & 0\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix} 1\\2\\2\end{bmatrix} u(t),\qquad y(t) = \begin{bmatrix} 1 & 0 & 0\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1 & 1 & 0\\2 & 0 & 1\\-1 & 0 & 0\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix} 1\\2\\2\end{bmatrix} u(t),\qquad y(t) = \begin{bmatrix} 1 & 0 & 0\end{bmatrix}\mathbf{x}(t)

True or false, the system is observable.

True

False

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Consider a closed-loop system consisting of a first order plant with transfer function P(s), and a PI controller (k_p, $k_i \neq 0$ k_i \neq 0; k_d=0) with an actuator that has an output range between -1 and 1, as shown in the following diagram. Furthermore, you can assume that there are no noise and disturbance inputs injected into the system, and the plant satisfies P(0)=6.

Image failed to load: Closed-loop system with PI controller, a limited-output range actuator and a plant

The following two plots show the signals y and r for references of the form $r(t)=r_0 \mathbf{1} (t)$ r(t)=r_0 \mathbf{1} (t), for two different values of r_0. Match the system response plots to the possible values of reference input magnitudes r_0. (The plots are not drawn to scale.)

Image failed to load: Response A

Image failed to load: Response B

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Given that the root locus plot of P(s) is now as follows:

Image failed to load: Root locus of new plant P(s)

Three controllers (one P controller, one PD controller, and one PID controller) have been implemented as the controller C(s) in the closed-loop system shown at the start of this quiz. A few step responses are listed below.

Response A	Response B	Response C	Response D
Image failed to load: Response A	Image failed to load: Response B	Image failed to load: Response C	Image failed to load: Response D

For each controller, determine the best matching best-case step response of the closed loop system. Each controller should be matched to exactly one response. (The responses are not drawn to the same scale, but are already showing all the salient features.)

P controller

PD controller

PID controller

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Consider another case where the root loci for the plant, one for which the proportional gain k_p=2 and the integral gain k_i is used as the loop gain (on the left), as well as for which k_i=2 and k_p is used as the loop gain (on the right), are both given in figures below.

Image failed to load: Root locus of the plant with PI controller, fixed Kp and tunable Ki.

Image failed to load: Root locus of the plant with PI controller, fixed Ki and tunable Kp.

You are also given these information about the root locus plots:

the branching point for the root locus on the left has the branching point at loop gain of 4.44, and
the branching point for the root locus on the right has the branching point at loop gain of 1.

As an engineer from Monash Control Design Solutions, you are tasked by your client to design the controller based on the following specified requirements. Match the requirements with design choices you are most likely to make in order to satisfy the design goals. Any design choice can only be matched to a requirement at most.

System

to have roughly similar settling time even though one of the controller

gains is changed in one particular direction without limits, while the other controller gain is

fixed at a value of 2.

System to have slightly increased speed at the expense of overshoots and maybe some oscillations, but still settles down eventually with a step reference.

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Consider a root locus of the plant P(s) which satisfies P(0) = -1 shown in the figures below, with the controller C(s) being only a tunable loop gain defined as k_L.

Image failed to load: Root locus overview of the plant P(s)

Image failed to load: Zoomed in root locus of plant P(s)

Figure on the left shows the overall root locus plot of P(s), while the one on the right shows the same root locus zoomed in near the Origin.

The red crosses (x) mark the open loop poles, the blue circle (o) marks the open loop zero, and the red asterisks (*) mark the closed loop poles at a particular loop gain value $k_{L} = k_{crossover}$ k_{L} = k_{crossover}

such that they fall on the same gray dashed line

indicated in the plot. It is given also that the two complex conjugate closed loop poles are located on the imaginary axis when

$k_{L} = k_{A}$ k_{L} = k_{A}

, while the closed loop pole on the blue-coloured branch (connects to the open loop zero) crosses over

into the right half plane when

$k_{L} = k_{B}$ k_{L} = k_{B}. Note that $0 < k_{crossover} < k_{A} < k_{B} < +\infty$ 0 < k_{crossover} < k_{A} < k_{B} < +\infty

Three possible responses from this closed-loop system are given in the plots that follow, each with their respective labels for your easy reference to answer the question. (The plots are not drawn to the same scales.)

Plot A	Plot B	Plot C
Image failed to load: Response A	Image failed to load: Response B	Image failed to load: Response C

Match the response plots above to each case that best describes the system response when the closed-loop pole of interest has a loop gain that is:

$0 < k_{L} < k_{crossover}$ 0 < k_{L} < k_{crossover}

$k_{crossover} < k_{L} < k_{A}$ k_{crossover} < k_{L} < k_{A}

$k_{A} < k_{L} < k_{B}$ k_{A} < k_{L} < k_{B}

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Consider the state space model with state equation

$\frac{d\mathbf{x}}{dt} = \begin{bmatrix}-1 & 1 \\3 & 2\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix} 1\\-1\end{bmatrix}u(t).$ \frac{d\mathbf{x}}{dt} = \begin{bmatrix}-1 & 1 \\3 & 2\end{bmatrix}\mathbf{x}(t) + \begin{bmatrix} 1\\-1\end{bmatrix}u(t).

If state feedback $u(t) = -\begin{bmatrix} K_1 & K_2 \end{bmatrix}\mathbf{x}(t)$ u(t) = -\begin{bmatrix} K_1 & K_2 \end{bmatrix}\mathbf{x}(t) is used (and the reference is zero) then the closed-loop state equation is:

$\frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1+K_1 & 1+K_2\\3-K_1 & 2-K_2\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1+K_1 & 1+K_2\\3-K_1 & 2-K_2\end{bmatrix}\mathbf{x}(t)

$\frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1-K_1 & 1-K_2\\3+K_1 & 2+K_2\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1-K_1 & 1-K_2\\3+K_1 & 2+K_2\end{bmatrix}\mathbf{x}(t)

$\frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1+K_1 & 1-K_1\\3+K_2 & 2-K_2\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}}{dt} = \begin{bmatrix} -1+K_1 & 1-K_1\\3+K_2 & 2-K_2\end{bmatrix}\mathbf{x}(t)

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