If we assume the system
${x}=A\sin(\omega t)$ the system 'velocity' is $\dot{x}=A\omega\sin(\omega t)$
Consider some points (taking $\omega t$ as a single angle value)
$\omega t$ | $\cos(\omega t)$ | $x$ | $\dot{x}$ |
0 | 1 | 0 | $\omega A$ |
$45^\circ$ ($\pi/4$) | 0.707 ($\frac1{\sqrt{2}}$) | $0.7A$ | 0.7$\omega A$ |
$90^\circ$ ($\pi/2$) | 0 | $A$ | 0 |
$135^\circ$ ($3\pi/4$) | -0.707 ($-\frac1{\sqrt{2}}$) | $0.7A$ | -0.7$\omega A$ |
$180^\circ$ ($\pi$) | -0.707 ($-\frac1{\sqrt{2}}$) | $0.7A$ | -0.7$\omega A$ |
If we now assume that $A=1$ and $\omega=1$ we can draw a graph of $x$ against $\dot{x}$
This is a phase plane graph, in our case we are plotting a variable against its differential, but it could be two variables linked in some other way.
A simple model of biped walking is based on an 'inverted pendulum'. The assumption is that during the period you have one foot on the ground your knee is 'locked' and your dynamics are similar to a pendulum that can swing in two dimensions [kajita01].
By analysing the forces on a pendulum (left figure), or equivalently the accelerations (right figure) we can derive the pendulum equations. Either way the non-linear pendulum equation is then
\[ \ddot\theta=-\frac{g}{l}\sin\theta \]This can also be written in state space form as
\begin{align} \dot\theta&=\omega\\ \dot\omega&=-\frac{g}{l}\sin\theta \end{align}Where we introduce the additional state variable \[\omega=\frac{d\theta}{dt}\]
We can find a numerical solution to the above state-space equation using 'ode' integrators such 'Euler' or 'Runge-Kutta'.
There is also an analytic approach that can be used to draw and understand the phase plane.
This uses a refactoring of a second differential. If we write $\dot{x}=\frac{dx}{dt}=v$
\[ \ddot{x}=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v \]If we replace $x$ and $v$ in the above equation for $\theta$ and $\omega$ we can substitute that into the non-linear pendulum equation so
\[ \omega\frac{d\omega}{d\theta}=-\frac{g}{l}\sin(\theta) \]Rewrite to get
\[ \omega d\omega=-\frac{g}{l}\sin(\theta){d\theta} \]and integrate with the limits of each integral corresponding.
\[ \int_{\omega(0)}^{\omega(t)}\omega d\omega=-\int_{\theta(0)}^{\theta(t)}\frac{g}{l}\sin(\theta){d\theta} \]Once the integration is done and the limits evaluated the equation becomes
\[ \frac12\left(\omega^2(t)-\omega^2(0)\right)= \frac{g}{l}\left(\cos\theta(t)-\cos\theta(0)\right) \]So it is then possible to rearrange to compute $\omega(t)$ from the initial conditions $\omega(0)$ and $\theta(0)$ and the current angle $\theta(t)$ That is
\[ \omega(t)= \pm\sqrt{ \frac{g}{l}\left(\cos(\theta(t))-\cos(\theta(0))\right) +\omega^2(0) } \]Since this is no longer time dependent we can write it as
\begin{equation} \omega= \pm\sqrt{ \frac{g}{l}\left(\cos\theta-\cos\theta_0\right) +\omega^2_0 } \end{equation}Challenge is to plot $\omega$ vs $\theta$ for a variety of angles
where $l$ and $m$ the pendulum length and mass, in this case the constant $B$ is known as damping (inherent energy loss) of the system, and $g$ is gravitational acceleration.
Closed form solution if $B=0$
Assuming $\dot{x}=v$ Can use the relationship
\[ \frac{d^2 x}{dt^2}= \frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= \frac{dv}{dx}{v} \]Now do the equivalent calculation for $\dot\theta=\omega$ and solve equation t.b.d. (eq:pend1)