Sliding variables and sliding mode control (SISO only)¶

Vivek Yadav, PhD¶

MEC 560: Advanced Control Systems¶

Motivation¶

Transform an original n-order nonlinear control system into a first order sytem.
Simplify the original dynamics so control design becomes easier.
Force the system dynamics onto a manifold such that the resulting dynamics becomes simple.

Motivation,¶

Given a n-order system, $X^{(n)} = f(X) + g(X) u$ , transform it into a first order dynamics using a variable $s$ . $s$ satisfies the following 2 properties,

$\dot{x}$ contains $u$ .
If $s \rightarrow 0$ then $x \rightarrow 0$ .

Choose control such that $s = 0$ once the states reach sliding manifold.

Example Sliding mode:¶

For $n = 2$ , say a system is given by,

$\ddot{x} = f(x) + g(x)u$

We wish to drive $x \rightarrow x_d$ and $\dot{x} \rightarrow \dot{x}_d$ . Choose $s$ such that

$\dot{x}$ contains $u$ .
If $s \rightarrow 0$ then $x \rightarrow 0$ .

Choose

$s = \lambda (x - x_d ) + ( \dot{x} - \dot{x}_d)=\lambda e+ \dot{e}$

For a general $n$ order system, the sliding variable can be selected as,

$s = \left( \frac{d}{dt} + \lambda \right)^{n-1} e$

Note in the expressions above, $e$ can be considered as a filter response to input $s$ , therefore if $s \rightarrow 0$ , then $x \rightarrow 0$ .

Geometric interpretation¶

For 2-dimension case, $s = \dot{e} + \lambda e$ represents a line with slope of $-\lambda$ in the phasephot, and the line contrains origin.
In general, if we plot $x$ vs $\dot{x}$ , then the line will pass through $x_d$ and $\dot{x}_d$ and will have the slope of $-\lambda$ .

For a general n-dimension system, $s= 0$ represents a hyperplane in the $n-1$ that passes through the origin (or desired point) depending on the variable representation.

Boundedness of states¶

Say we are able to bound the sliding variables $s$ such that $|s|\leq \phi$ after some transient dynamics. We are interested in the bounds on the error $e$ when $|s| \leq \phi$ . We will consider a second order system first, in this case, the sliding variable is given by,

$s = \dot{e} + \lambda e$

$\lambda exp(\lambda t) e + exp(\lambda t) \frac{de}{dt} = exp(\lambda t) s$

$\frac{d(exp(\lambda t)e)}{dt} = exp(\lambda t) s$

$d(exp(\lambda t)e) = exp(\lambda t) s dt$

Integrating between $t_f$ and $0$ gives,

$e(T) = e(0) exp(-\lambda T) + \int_0^T exp(-\lambda (T-t)) s dt$

Sliding mode control:¶

Say we are intersted in controlling a second order system given by,

$\ddot{x} = f(x,\dot{x}) + g(x,\dot{x}) u$

Once we design the sliding variable, we can design control based to drive $s \rightarrow 0$ . Define a Lyapunov function $L$ as,

$L = \frac{1}{2}s^2$

Taking derivative gives,

$\dot{L} = s \dot{s}$

Therefore, if we choose $u$ such that $s \dot{s} < 0$ always, then the $s \rightarrow 0$ . The derivative of Lyapunov can we written as,

$\dot{L} = s ( \ddot{x} + \lambda \dot{x}) = s ( f(x,\dot{x}) + g(x,\dot{x}) u + \lambda \dot{x})$

choosing $u = - g(x,\dot{x})^{-1} (f(x,\dot{x}) + \lambda \dot{x} + \eta sign(s))$

$\dot{L} = - \eta |s|$

The resultant control is chosen as $\eta sign(s)$ instead of $-\eta s$ to have a constant 'velocity' towards the sliding surface.
This ensures that we reach the sliding surface in a finite time and not asymptotically.

Example 1 :¶

Consider the example of a system given by,

$\ddot{x} + 1.5 \dot{x}^2cos(3x) = u$

with $1 \leq a(t) \leq 2$ , we would like to convert this to a first order system. Lets define a sliding variable $s$ ,

$s = \dot{x} + \lambda x$

as $x$ is a first order response to $s$ . $s$ satisfies,

$x \rightarrow 0$ as $s \rightarrow 0$
$\dot{s}$ contains $u$

Therefore, $s$ is a valid choice for sliding variable.

Sliding mode control for control under uncertainty:¶

Conside the second order true system system that is given by

$\ddot{x} = f(x,\dot{x}) + g(x,\dot{x}) u$

and say the approximate system we know is given by,

$\ddot{x} = \hat{f}(x,\dot{x}) + g(x,\dot{x}) u$

such that $|f - \hat{f}| \leq F$ . Sliding mode control lends itself very well to such problems. The technique to design control is similar. We first choose a sliding variable, $s = \dot{x} + \lambda x$ . Similar to process above, we define a Lyapunov,

$L = \frac{1}{2} s^2$

Taking derivative gives,

$\dot{L} = s \dot{s} = s(f(x,\dot{x}) + g(x,\dot{x}) u + \lambda \dot{x})$

Now, we do not know $f(x,\dot{x})$ accurately, so we can use estimate of $f$ for designing control instead of the true $f (x,\dot{x})$ . Choosing $u = - g(x,\dot{x})^{-1} ( \ \hat{f}(x,\dot{x}) +F + \lambda \dot{x} + \eta sign(s) )$ gives,

$\dot{L} = s(f(x,\dot{x}) -\hat{f}(x,\dot{x}) -F - \eta sign(s))$

As $|f - \hat{f}| \leq F$ ,

$\dot{L} \leq s( - \eta sign(s))$

Therefore, the choice of control $u = - g(x,\dot{x})^{-1} ( \ \hat{f}(x,\dot{x}) +F + \lambda \dot{x} + \eta sign(s) )$ drives $s \rightarrow 0$ .

Example 2:¶

Consider the example of a system given by,

$\ddot{x} + a(t) \dot{x}^2cos(3x) = u$

with $1 \leq a(t) \leq 2$ , we would like to control this sytem so $x \rightarrow 0$ . As we do not know $a(t)$ precisely, we choose an estimate as, $1.5 \dot{x}^2cos(3x)$ . Therefore,

$f - \hat{f} = (a(t)-1.5) x^2 cos(3x) \leq .5x^2|cos(3x)|$

We will now define a sliding variable, $s$ as,

$s = \dot{x} + \lambda x$

$L = \frac{1}{2} s^2$