I am trying to understand an example from [1]. In detail I do not understand how the equation for the dynamic of the tracking error is chosen. I am not a mathematician so please forgive me if I may not use the most proper jargon. The example below, from [p. 216, 1], refers to a third order system. The nonlinear state equations are:
$\dot{x_1}= sin (x_{2})+ (x_2 + 1) x_3$
$\dot{x_2}={x_1}^5 + x_3$
$\dot{x_3}={x_1}^2 + u$
$y=x_1$
where $x_1$, $x_2$, $x_3$ are the states, $u$ is the control input and $y$ the output. Now, the objective of the example is to find a direct equation linking the output to the control input $u$.
Differentiating $y$ yields
$\dot{y} = sin (x_{2}) + (x_2 + 1) x_3 $,
and differentiating the above gives:
$\ddot{y}=(x_2 + 1)u + f(x)$
showing a direct relationship between the output $y$ and the control $u$, also
$f(x)=({x_{1}}^5 + x_3) x3 + cos(x_2) + (x_2+1){x_1}^2$.
If the control input is chosen as
$u=\frac{1}{x_2 + 1}(v-f(x_1))$
with $v$ a new 'auxiliary' control input, then the nonlinearity in the equation for $\ddot{y}$ is canceled, and the simple linear relation $\ddot{y}=v$ is obtained. Now, [p. 217, 1] says that designing a tracking controller for this double integrator is simple. Letting the tracking error be $e=y-y_d$, where $y_d$ is the reference, and choosing the new input as
$v=\ddot{y_d} – k_{1} e – k_{2} \dot{e}$
with $k_1, k_2 >0$, the tracking error dynamic is:
$\ddot{e}+ k_{1} e + k_{2} \dot{e} =0$
which represents a stable error dynamic.
My question is
1) how does one come up with the choice $v=\ddot{y} – k_{1} e – k_{2} \dot{e}$? Is this done in order to have $\ddot{e}+ k_{1} e + k_{2} \dot{e} =0$.
2) Why is the choice $\ddot{e}+ k_{1} e + k_{2} \dot{e} =0$ made? Rather than for instance simply choosing $k_{1} e + k_{2} \dot{e} =0$?
Any (simple) explanation would help a lot!
Thank you
References:
[1] Slotine and Li, Applied Nonlinear Control, Prentice-Hall, New Jersey, 1991
Best Answer
I assume that you made a typo and the choice for the input should be
$$ v = \ddot{y}_d - k_1\,e - k_2\,\dot{e}, $$
so using $\ddot{y}_d$ instead of $\ddot{y}$, which acts as a feedforward term. Such feedforward term makes sure that when the error is zero it will remain zero for any $y_d$ which is at least twice differentiable.
The choice for the input-output linearization gives $\ddot{y} = v$. So if we now look at the error dynamics one gets
\begin{align} \ddot{e} &= \ddot{y} - \ddot{y}_d \\ &= v - \ddot{y}_d \\ &= \ddot{y}_d - k_1\,e - k_2\,\dot{e} - \ddot{y}_d \\ &= - k_1\,e - k_2\,\dot{e} \end{align}
Coming up with this $v$ is common choice in linear control. Namely as stated before $\ddot{y}_d$ acts as feedforward. The remaining two terms act as state feedback, for example after applying the feedforward, so define $v = \ddot{y}_d + w$, the error dynamics can also be written as
$$ \dot{z} = \underbrace{ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} }_A z + \underbrace{ \begin{bmatrix} 0 \\ 1 \end{bmatrix} }_B w, $$
with $z = \begin{bmatrix}e & \dot{e}\end{bmatrix}^\top$. Now by using state feedback $w = -K\,z$ we get the closed loop dynamics $\dot{z} = (A - B\,K)\,z$, which decays exponentially to zero if $A - B\,K$ is Hurwitz. Choosing a $K$ can be done with things like pole placement or LQR, but it can be shown that $A - B\,K$ is always Hurwitz when $k_1,k_2>0$, with $K = \begin{bmatrix}k_1 & k_2\end{bmatrix}$.
Lets say one would choose $v = \ddot{y} - k_1\,e - k_2\,\dot{e}$, plugging this into $\ddot{y}$ would give $k_1\,e + k_2\,\dot{e} = 0$. However, $v$ is chosen to be equal to $\ddot{y}$ so solving for $v$ would imply solving $v = v - k_1\,e - k_2\,\dot{e}$. This equation is only true when $k_1\,e + k_2\,\dot{e} = 0$ but you are not able to choose what $e$ and $\dot{e}$ are at any given moment. And if that equation would be true, then all values for $v$ would satisfy it. So either way it does not lead to a very sensible result.