Given the following nonlinear system,
$\dot{x}_1 = \text{cos}(x_3) (\text{cos}(x_3 - x_6) + 10 ((x_4 - x_1) \text{cos}(x_3) + (x_5 - x_2) \text{sin}(x_3))) \\ \dot{x}_2 = \text{sin}(x_3) (\text{cos}(x_3 - x_6) + 10 ((x_4 - x_1) \text{cos}(x_3) + (x_5 - x_2) \text{sin}(x_3))) \\ \dot{x}_3 = u + \frac{4}{625} ((x_5 - x_2) \text{cos}(x_3) - (x_4 - x_1) \text{sin}(x_3)) - \frac{4}{25} \text{sin}(x_3 - x_6) \\ \dot{x}_4 = \text{cos}(x_6) \\ \dot{x}_5 = \text{sin}(x_6) \\ \dot{x}_6 = u$
$y = x_4 + x_5 - x_1 - x_2$
How can I derive a controller $u = f(x_1,x_2,x_3,x_4,x_5,x_6)$ such that the output of the closed-loop system $-$i.e., the above system with $u$ being replaced by $f(\cdot)-$ is a non-zero constant (e.g. 1)?
Note that the system has no equilibria, i.e., linearizing the system does work.
You want the output of the system to be constant, i.e. $$y=x_4+x_5-x_1-x_2=c$$ Simply differentiate the above equation: $$\dot{x_4}+\dot{x_5}-\dot{x_1}-\dot{x_2}=0$$ So it is sufficient to determine $u$ such that $$\dot{x_1}+\dot{x_2}=\dot{x_4}+\dot{x_5}$$ In other words, find $x_6$ such that: $$(\cos x_3+\sin x_3)(\cos(x_3 - x_6) + 10 ((x_4 - x_1) \cos x_3 + (x_5 - x_2) \sin x_3))\\=\cos x_6+\sin x_6$$ and then put $u=\dot x_6$
Simplifying the equation yields: $$\sin x_6\cos x_3-\cos x_6\sin x_3=10\frac{\cos x_3+\sin x_3}{\cos x_3-\sin x_3}((x_4 - x_1) \cos x_3 + (x_5 - x_2) \sin x_3))$$ and you can extract $x_6$ from the above: $$x_6=x_3+\arcsin\left(10\tan(x_3+\frac{\pi}4)((x_4 - x_1) \cos x_3 + (x_5 - x_2) \sin x_3))\right)$$