To illustrate the result in and compare it with the control law resulting from standard backstepping, consider the two-dimensional system
$$
\begin{aligned}
&\dot{x}{1}=-x{1}+\lambda x_{1}^{3} x_{2}, \
&\dot{x}{2}=u . \end{aligned} $$ A backstepping-based stabilising control law is $$ u=-\lambda x{1}^{4}-x_{2},
$$
whereas a direct application of the procedure described in the proof of Proposition $2.1$ shows that the I\&I stabilising control law is
$$
u=-\left(2+x_{1}^{8}\right) x_{2} .
$$
The latter does not require the knowledge of the parameter $\lambda$, however, it is in general more aggressive because of the higher power in $x_{1}$.
As a second example consider the system
$$
\begin{aligned}
&\dot{x}{1}=A(\theta) x{1}+F\left(x_{1}, x_{2}\right) x_{2}, \
&\dot{x}_{2}=u
\end{aligned}
$$
MATH3063 COURSE NOTES ：
and the problem of constructing a (reduced-order) observer for the unmeasured state $\eta$. In an attempt to apply the methodology in [12], rewrite the system in the form
$$
\begin{aligned}
&\dot{\eta}=A_{1} \eta+G_{1} \gamma(\eta)+u, \
&\dot{y}=A_{2} \eta+G_{2} \gamma(\eta)-y
\end{aligned}
$$
where $\gamma(\eta)=\left[\eta^{3}, \eta+\eta^{2}+\eta^{3}\right]^{\top}$ satisfies the monotonicity condition, and $A_{1}=-1, G_{1}=[-1,1], A_{2}=1$, and $G_{2}=[1,0]$. Defining the error $z=\xi+B y-\eta$ and the observer dynamics
$$
\dot{\xi}=\left(A_{1}-B A_{2}\right)(\xi+B y)+\left(G_{1}-B G_{2}\right) \gamma(\xi+B y)+u+B y
$$
yields the error system
$$
\dot{z}=\left(A_{1}-B A_{2}\right) z+\left(G_{1}-B G_{2}\right)(\gamma(\eta+z)-\gamma(\eta))
$$
which has an asymptotically stable equilibrium at zero if there exist constants $\nu_{1}>0, \nu_{2}>0$ and $B$ such that ${ }^{11}$
$$
A_{1}-B A_{2} \leq 0, \quad G_{1}-B G_{2}=\left[-\nu_{1},-\nu_{2}\right] .
$$
It can be readily seen that the last condition cannot be satisfied, since it requires $\nu_{2}=-1$.
