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Question: I believe I am correct up until $y(t) = Cx(t)$. I was told I did it incorrectly, but I cannot figure out how to grab the position of the antenna using $y(t) = Cx(t)$. Does it look correct, if not, how can I go about finding $y(t)$? I am stumped.

We want to design and simulate a full state feedback LQR control law for a rotating antenna. The motion of the antenna can be described by the linear second order non-homogeneous differential equation given by $$Iθ''(t) + cθ'(t) = τ (t)$$ with initial conditions: $$θ(0) = θ_0$$ $$θ'(0) = θ_1$$ where $I$ is the effective moment of inertia of all the rotating parts including the antenna, $c ≥ 0$ is the coefficient of viscous (i.e. air; the antenna is in low earth atmosphere, not the vacuum of space) friction, $τ (t)$ is the torque applied by the motor that effects the rotation at time $t ≥ 0$, and $θ(t)$ is the angular position of the antenna at time $t ≥ 0$.

(The basis for the model is simply Newton’s second law, $F = ma$, in the context of rotary motion, i.e. torque equals the moment of inertia times the angular acceleration. The friction term is modeled as a torque that is proportional to the velocity so that it is in the direction opposite to the motion of the antenna.)

The motor torque at time $t ≥ 0$ is assumed to be proportional to $u(t)$, the input voltage to the motor, with constant of proportionality $b > 0$.

In your calculations below use the following values: $\alpha = \frac{c}{I}=4.6s^{-1},$ $λ = \frac{b}{I} = 0.787 \frac{rad}{Vs2} $, and $I = 10kgm^2.$

The output variable is the angular position of the antenna.

Rewrite the model as a first order system of the form $$x'(t) = Ax(t) + Bu(t)$$ $$y(t) = Cx(t)$$ $$x(0) = x_0.$$

ATTEMPT: $$I\theta''(t) + c\theta'(t) + 0\theta(t)= \tau(t)$$ $$\theta''(t) = \frac{1}{I}\tau(t)-\frac{c}{I}\theta'(t)-0\theta(t)$$ $$x_1=\theta(t)$$ $$x_2 =\theta'(t)$$ $$x_1'(t) = \theta'(t)$$ $$x_2'(t) = \theta''(t)=\frac{b}{I}u(t)-\frac{c}{I}\theta'(t)-0\theta(t)$$ $$x_1(0) = \theta(0) = \theta_0$$ $$x_2(0) = \theta'(t) = \theta_1$$ $$x_1'(t) = 0x_1(t)+1x_2(t)+0$$ $$x_2'(t) = 0x_1(t) -\frac{c}{I}x_2(t)+\frac{b}{I}u(t)$$ $$x'(t)=\begin{pmatrix}x_1'\\\ x_2' \end{pmatrix}=\begin{pmatrix}0 & 1\\\ 0 & \frac{-c}{I}\end{pmatrix}x(t)+\begin{pmatrix}0\\\ \frac{b}{I}\end{pmatrix}u(t)$$ $$y(t)=\begin{pmatrix}1\\\ 0\end{pmatrix}x(t)$$ $$x(0)=\begin{pmatrix}\theta_0\\\ \theta_1\end{pmatrix}$$

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You came incredibly close!

It seems that you want $y(t)=\theta(t)=x_1(t).$ Unfortunately, we can't multiply two $2\times 1$ matrices together. In general, if $M$ and $N$ are non-scalar matrices (that is, neither has only one row and one column), say with $M$ being $k\times m$ and $N$ being $n\times p,$ we will have $MN$ defined only if $m=n,$ in which case $MN$ will be a $k\times p$ matrix.

Instead, you want $$y(t)=\begin{pmatrix}1 & 0\end{pmatrix}x(t).$$ By multiplying any $1\times 2$ matrix and a $2\times 1$ matrix together, we obtain a $1\times 1$ matrix--a scalar--as desired.

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