$8\mathrm{.}2$ Exercises

Linearize,

$x^{}(t)={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(x,u)=-x^2(l)+9u(l)$

Equilibrium conditions are $x^{}(t)=0$ and $u_p=1.$

Linearize ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(x)=5cosx$ about $x=\frac{}{2}$

Linearize $8\mathrm{.}12$ about $x=\frac{}{4}$

$\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+cosx=0$

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Consider the nonlinear equation

$y^{}+y^{}+y^{}y=u$

Linearize about $y_o={\left[\begin{array}{cc}1& 0\end{array}\right]}^T$ and $u_o=1$ and form into a state$-$space equation.

Given the nonlinear equation of a pendulum, linearize this about the operating point, ${}_o=0.$

$\frac{L}{g}\frac{d^2(t)}{d{t}^2}=-sin(t)$

Linearize $8\mathrm{.}15$ and form into a state$-$space equation where $i(l)$ is the input and $h(t)$ is the output.

$($Ignore units for simplicity.$)$

$\frac{m}{b}ar(h)(t)=mg-\frac{i^2(l)}{h^2(t)}-h^{}(t)$

give that at operating point $i(t)=i_p=0\mathrm{.}12,$

and $m=0\mathrm{.}1,=0\mathrm{.}9,=0\mathrm{.}95.$

In a paint mixing plant, two tanks supply fluids to a mixing cistern. the height of the fluid in the

cistern is dependent upon the difference between the input mass flow rate, $q,$ and the output flow

rate, $q_e.$ A nonlinear differential equation describing this dependency is given by

$\frac{dh}{dt}+\frac{A_c}{}\sqrt[\phantom{2}]{2gh}=\frac{q}{}$

where $A=$ cross$-$sectional area of the cistern, $A_e=$ cross$-$sectional area of the exit pipe, $g=$

acceleration due to gravity, and $=$ liquid density.

Linearize the nonlinear equation about the equilibrium point $(h_{},q_o).$