|
m(1+q^2)\ddot{q}+b\dot{q}+K_0q+K_1q^3=u
</math>
Now gi
Now given the desired state, <math>q_d</math>, and actual state, <math>q</math>, with error, <math>e = q_d - q</math>, define a function <math>r</math> as
:<math>
r=\dot{e}+\alpha e
</math>
A Control-Lyapunov candidate is then
:<math>
V=\frac{1}{2}r^2
</math>
which is positive definite for all <math> q \ne 0</math>, <math>\dot{q} \ne 0</math>.
Now taking the time derivative of <math>V</math>
:<math>
\dot{V}=r\dot{r}
</math>
:<math>
\dot{V}=(\dot{e}+\alpha e)(\ddot{e}+\alpha \dot{e})
</math>
The goal is to get the time derivative to be
:<math>
\dot{V}=-\kappa V
</math>
which is globally exponentially stable if <math>V</math> is globally positive definite (which it is).
Hence we want the rightmost bracket of <math>\dot{V}</math>,
:<math>
(\ddot{e}+\alpha \dot{e})=(\ddot{q}_d-\ddot{q}+\alpha \dot{e})
</math>
to fulfill the requirement
:<math>
(\ddot{q}_d-\ddot{q}+\alpha \dot{e}) = -\frac{\kappa}{2}(\dot{e}+\alpha e)
</math>
which upon substitution of the dynamics, <math>\ddot{q}</math>, gives
:<math>
(\ddot{q}_d-\frac{u-K_0q-K_1q^3-b\dot{q}}{m(1+q^2)}+\alpha \dot{e}) = -\frac{\kappa}{2}(\dot{e}+\alpha e)
</math>
Solving for <math>u</math> yields the control law
:<math>
u= m(1+q^2)(\ddot{q}_d + \alpha \dot{e}+\frac{\kappa}{2}r )+K_0q+K_1q^3+b\dot{q}
</math>
with <math>\kappa</math> and <math>\alpha</math>, both greater than zero, as tunable parameters
This control law will guarantee global exponential stability since upon substitution into the time derivative yields, as expected
:<math>
\dot{V}=-\kappa V
</math>
which is a linear first order differential equation which has solution
:<math>
V=V(0)e^{-\kappa t}
</math>
And hence the error and error rate, remembering that <math>V=\frac{1}{2}(\dot{e}+\alpha e)^2</math>, exponentially decay to zero.
If you wish to tune a particular response from this, it is necessary to substitute back into the solution we derived for <math>V</math> and solve for <math>e</math>. This is left as an exercise for the reader but the first few steps at the solution are:
:<math>
r\dot{r}=-\frac{\kappa}{2}r^2
</math>
:<math>
\dot{r}=-\frac{\kappa}{2}r
</math>
:<math>
r=r(0)e^{-\frac{\kappa}{2} t}
</math>
:<math>
\dot{e}+\alpha e= (\dot{e}(0)+\alpha e(0))e^{-\frac{\kappa}{2} t}
</math>
which can then be solved using any linear differential equation methods.
==Notes==
| 