ECE4132 Control System Design - MUM S2 2025 Course Materials & Test Answers | learning.monash.edu

The input v(t) and output $\omega(t)$ \omega(t) of a DC motor are related by the differential equation

$0.005\frac{d^2\omega}{dt^2} + 0.06\frac{d\omega}{dt} + 0.1064\omega(t)= v(t).$ 0.005\frac{d^2\omega}{dt^2} + 0.06\frac{d\omega}{dt} + 0.1064\omega(t)= v(t).

A valid state for the system is

$\mathbf{x}(t) = \begin{bmatrix} \omega(t)\\\frac{d\omega}{dt}\end{bmatrix}.$ \mathbf{x}(t) = \begin{bmatrix} \omega(t)\\\frac{d\omega}{dt}\end{bmatrix}.

Which of the following is the state space model corresponding to this choice of state?

$\frac{d\mathbf{x}(t)}{dt}=\begin{bmatrix}0 & 1\\21.28 & 12\end{bmatrix}\mathbf{x}(t)+\begin{bmatrix}0\\200\end{bmatrix}v(t),\qquad\omega(t)=\begin{bmatrix}1 &0\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}(t)}{dt}=\begin{bmatrix}0 & 1\\21.28 & 12\end{bmatrix}\mathbf{x}(t)+\begin{bmatrix}0\\200\end{bmatrix}v(t),\qquad\omega(t)=\begin{bmatrix}1 &0\end{bmatrix}\mathbf{x}(t)

❌

$\frac{d\mathbf{x}(t)}{dt} =\begin{bmatrix} 0 & 1\\-21.28 & -12\end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} 200\\0\end{bmatrix} v(t),\qquad\omega(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}(t)}{dt} =\begin{bmatrix} 0 & 1\\-21.28 & -12\end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} 200\\0\end{bmatrix} v(t),\qquad\omega(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t)

❌

$\frac{d\mathbf{x}(t)}{dt} = \begin{bmatrix} 0 & 1\\-21.28 & -12\end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} 0\\200\end{bmatrix} v(t),\qquad\omega(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}(t)}{dt} = \begin{bmatrix} 0 & 1\\-21.28 & -12\end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} 0\\200\end{bmatrix} v(t),\qquad\omega(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t)

✅

$\frac{d\mathbf{x}(t)}{dt} =\begin{bmatrix} 0 & 1\\-21.28 & -12\end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} 0\\1\end{bmatrix} v(t),\qquad\omega(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t)$ \frac{d\mathbf{x}(t)}{dt} =\begin{bmatrix} 0 & 1\\-21.28 & -12\end{bmatrix} \mathbf{x}(t) + \begin{bmatrix} 0\\1\end{bmatrix} v(t),\qquad\omega(t) = \begin{bmatrix} 1 & 0\end{bmatrix}\mathbf{x}(t)

❌

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ECE4132 Control System Design - MUM S2 2025

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