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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: $alpha = fraccI=4.6s^-1,$ $λ = fracbI = 0.787 fracradVs2 $, 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:
$$Itheta''(t) + ctheta'(t) + 0theta(t)= tau(t)$$
$$theta''(t) = frac1Itau(t)-fraccItheta'(t)-0theta(t)$$
$$x_1=theta(t)$$
$$x_2 =theta'(t)$$
$$x_1'(t) = theta'(t)$$
$$x_2'(t) = theta''(t)=fracbIu(t)-fraccItheta'(t)-0theta(t)$$
$$x_1(0) = theta(0) = theta_0$$
$$x_2(0) = theta'(t) = theta_1$$
$$x_1'(t) = 0x_1(t)+1x_2(t)+0$$
$$x_2'(t) = 0x_1(t) -fraccIx_2(t)+fracbIu(t)$$
$$x'(t)=beginpmatrixx_1'\ x_2' endpmatrix=beginpmatrix0 & 1\ 0 & frac-cIendpmatrixx(t)+beginpmatrix0\ fracbIendpmatrixu(t)$$
$$y(t)=beginpmatrix1\ 0endpmatrixx(t)$$
$$x(0)=beginpmatrixtheta_0\ theta_1endpmatrix$$

You came incredibly close!

It seems that you want $y(t)=theta(t)=x_1(t).$ Unfortunately, we can't multiply two $2times 1$ 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 $ktimes m$ and $N$ being $ntimes p,$ we will have $MN$ defined only if $m=n,$ in which case $MN$ will be a $ktimes p$ matrix.

Instead, you want $$y(t)=beginpmatrix1 & 0endpmatrixx(t).$$ By multiplying any $1times 2$ matrix and a $2times 1$ matrix together, we obtain a $1times 1$ matrix--a scalar--as desired.