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:

$\dot{x_1}= sin (x_{2})+ (x_2 + 1) x_3$

$\dot{x_2}={x_1}^5 + x_3$

$\dot{x_3}={x_1}^2 + u$


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


2 Answers 2


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.

  • $\begingroup$ thank you for your effort and detailed answer. Yes, there was a typo before as you have indicated (it is corrected now). I was also wondering that, since $\ddot{y}=v$, and $e=y-y_d$, , then it follows $\ddot{e}=\ddot{y}-\ddot{y_d}$ and $\ddot{e}=v-\ddot{y_d}$. Now, I don't understand why/how it is chosen $\ddot{e} = - k_1 e - k_2 \dot{e}$ ? It seems to be a straightforward choice, but not sure why. Maybe is because the system of the second order? I feel I am missing some simple detail here $\endgroup$
    – Lello
    Feb 1, 2019 at 7:37
  • $\begingroup$ @Lello Indeed, namely if the relative degree would have been $n$, so after input-output linearization you would get $$\frac{d^n y}{dt^n}=v,$$ then the natural choice for $v$ would be $$v=\frac{d^n y_d}{dt^n} + \sum_{i=0}^{n-1} k_i\,\frac{d^i e}{dt^i}.$$ $\endgroup$ Feb 1, 2019 at 7:47
  • $\begingroup$ So this is essentially feedforward plus (error-) state feedback with the characteristic polynomial equal to $$\lambda^n - \sum_{i=0}^{n-1} k_i\,\lambda^i,$$ so if its roots all lie inside the left half plane (which is equivalent to $A-B\,K$ being Hurwitz) then the error will decay exponentially to zero. $\endgroup$ Feb 1, 2019 at 7:57

@Kwin thank you very much again. Maybe I get the algebra right this time... we have $\ddot{y}=v$ and writing this equation in terms error $e$ yields $\ddot{e}+\ddot{y_d}=v$. If the choice $\ddot{e}=-k_1 e - k_2 \dot{e}$ is made, then $v=\ddot{y_d}-k_1 e - k_2 \dot{e}$. Initially we had $\ddot{y}=v$, hence we have $\ddot{y}=\ddot{y_d}-k_1 e - k_2 \dot{e}$, moving $\ddot{y_d}$ on the left hand side gives $\ddot{y}-\ddot{y_d}=\ddot{e}=-k_1 e - k_2 \dot{e}$. The equation $\ddot{e}=-k_1 e - k_2 \dot{e}$ is written as $\ddot{e} +k_1 e + k_2 \dot{e}=0$, whose characteristic equation is $\lambda^2 + k_1 + k_2 \lambda =0$, which is stable if $k_1>0$ and $k_2>0$ (it's Hurwitz), i.e. the solutions will be converging to zero as $t\rightarrow \infty$, since they will be $e^{-\lambda_1 t} + e^{-\lambda_2 t}$, with $\lambda_1, \lambda_2>0$.


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