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Inertial Mechanics

State space expressions can be obtained
$\quad \left\lbrack\matrix{\dot{\theta} \cr \ddot{\theta}} \right\rbrack = \left\lbrack\matrix{0 & 1 \cr -\frac{Mgl}{J} & -\frac{b}{J}} \right\rbrack \left\lbrack\matrix{\theta \cr \dot{\theta}} \right\rbrack + \left\lbrack\matrix{0 \cr 1} \right\rbrack \tau $

$\quad y = \left\lbrack\matrix{0 & 1} \right\rbrack \left\lbrack\matrix{\theta \cr \dot{\theta}} \right\rbrack $

1.1.1.2 PD Control of Pendulum with Motor at Origin (See PD.py)

From the Equation of Motion:
$\quad \ddot{\theta} = \tau-\frac{b}{J}\dot{\theta}-\frac{Mgl}{J}\theta$

Add PD Control terms ($\theta$ defined from upward vertical direction)
$\quad \tau = K_{p}(\theta) + K_{d}(\dot{\theta})$
$\quad \ddot{\theta} = (K_{d}-\frac{b}{J}) \dot{\theta}+ (K_{p}-\frac{Mgl}{J})\theta$
$\quad \ddot{\theta} - (K_{d}-\frac{b}{J}) \dot{\theta}- (K_{p}-\frac{Mgl}{J})\theta = 0$

Laplace Transfrom to obtain quadratic relation
$\quad \theta[s] (s^2- (K_{d}-\frac{b}{J}) s-(K_{p}-\frac{Mgl}{J})) = 0$

Identify roots
$\quad s = \frac{-(K_{d}-\frac{b}{J}) \pm \sqrt{(K_{d}-\frac{b}{J})^2 - 4(K_{p}-\frac{Mgl}{J})}}{2}$

To identify $K_{p}$ and $K_{d}$, select roots S and solve

1.1.1.3 Pendulum with Reaction Wheel (See RW.py)

See (Rigatos et. al., CYBER-PHYSICAL SYSTEMS, 2020) for more details

$\quad (M_{R}L_{R,CoM}^2 + M_{W}L_{W,CoM}^2 + J_{R} + J_{W})\ddot{\theta} + (J_{W})\ddot{\phi} - (M_{R}L_{R,CoM} + M_{W}L_{W,CoM})g\sin(\theta) = 0 $
$\quad J_{R}\ddot{\theta} + J_{W}\ddot{\phi} = \tau $

1.1.1.3 Pendulum with Reaction Wheel LQR Control (See RW_LQR.py)

See (Rigatos et. al., CYBER-PHYSICAL SYSTEMS, 2020) for more details

$\quad (M_{R}L_{R,CoM}^2 + M_{W}L_{W,CoM}^2 + J_{R} + J_{W})\ddot{\theta} + (J_{W})\ddot{\phi} - (M_{R}L_{R,CoM} + M_{W}L_{W,CoM})g\sin(\theta) = 0 $
$\quad J_{R}\ddot{\theta} + J_{W}\ddot{\phi} = \tau $