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��Chapter 4�Equilibria and Linearization

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Course Overview

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Motivation

In the real world, it is common for equations of motion describing real-world behavior to have nonlinear terms
To use powerful linear control design techniques (taught in this class) requires the system models to be linear

In the words of Arthur Bryson, sometimes referred to as the father of modern control theory, “Be wise, linearize!”

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What is a Taylor Series Linearization?

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Example

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What does that look like in two dimensions?

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Equilibrium

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Example of Equilibrium

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Example: Linearize About Equilibrium

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\[

\ddot{y} + a\dot{y} + by + g(y,\dot{y}) = u,

\]

\noindent Define $\tilde{y}\defeq y-y_e$ and $\tilde{u}=u-u_e$ and linearize each term about equilibrium:

\begin{align*}

\ddot{y} &\approx \ddot{y}_e + \frac{\partial \ddot{y}}{\partial \ddot{y}}\Big|_e (\ddot{y}-\ddot{y}_e) = \ddot{\tilde{y}} \\

a\dot{y} &\approx a\dot{y}_e + a\frac{\partial \dot{y}}{\partial \dot{y}}\Big|_e (\dot{y}-\dot{y}_e) = a\dot{\tilde{y}} \\

by &\approx by_e + b\frac{\partial y}{\partial y}\Big|_e (y-y_e) = by_e + b\tilde{y} \\

g(y,\dot{y}) &\approx g(y_e,\dot{y}_e) + \frac{\partial g}{\partial y}\Big|_e(y-y_e) + \frac{\partial g}{\partial \dot{y}}\Big|_e(\dot{y}-\dot{y}_e) \\

&= g(y_e,0) + \frac{\partial g}{\partial y}\Big|_e\tilde{y} + \frac{\partial g}{\partial \dot{y}}\Big|_e\dot{\tilde{y}} \\

u &\approx u_e + \frac{\partial u}{\partial u}\Big|_e (u-u_e) = u_e + \tilde{u}.

\end{align*}

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Example: Linearized Equations

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Example: Block Diagram – Jacobian Linearization

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Example: Feedback Linearization

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Jacobian vs Feedback Linearization

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Jacobian Linearization

Feedback Linearization

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Jacobian vs Feedback Linearization

Jacobian Linearization

More general–can always be done.
Linearization is about an equilibrium
Region of attraction can be small
Not as sensitive to model parameters

Feedback Linearization

Not always possible.
Linearization is NOT about an equilibrium
Region of attraction can be large
Sensitive to model parameters for exact cancellation

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Design Study A

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Design Study A

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\noindent{\bf Linearize Each Term About Equilibrium}

\[

\frac{m\ell^2}{3}\ddot{\theta} + \frac{mg\ell}{2}\cos\theta = \tau - b\dot{\theta}.

\]

Define $\tilde{\theta}\defeq\theta-\theta_e$, $\dot{\tilde{\theta}}\defeq\dot{\theta}-\dot{\theta}_e=\dot{\theta}$, $\ddot{\tilde{\theta}}=\ddot{\theta}-\ddot{\theta}_e=\ddot{\theta}$, and $\tilde{\tau}=\tau-\tau_e$,

\begin{align*}

\frac{m\ell^2}{3}\ddot{\theta} &= \frac{m\ell^2}{3}\ddot{\theta}_e + \frac{m\ell^2}{3}\left.\frac{\partial \ddot{\theta}}{\partial \ddot{\theta}}\right|_e (\ddot{\theta}-\ddot{\theta}_e) = \frac{m\ell^2}{3} \ddot{\tilde{\theta}}, \\

\frac{mg\ell}{2}\cos\theta &\approx \frac{mg\ell}{2} \cos\theta_e + \frac{mg\ell}{2}\frac{\partial}{\partial \theta}

(\cos\theta)\Big|_{\theta_e}

(\theta-\theta_e)

= \frac{mg\ell}{2}\cos\theta_e - \frac{mg\ell}{2}\sin\theta_e \tilde{\theta} \\

\tau &= \tau_e + \frac{\partial \tau}{\partial \tau}\Big|_e (\tau-\tau_e)

= \tau_e + \tilde{\tau} \\

b\dot{\theta} &= b\dot{\theta}_e + b\frac{\partial \dot{\theta}}{\partial \dot{\theta}}\Big|_e (\dot{\theta}-\dot{\theta}_e) = b\dot{\tilde{\theta}}.

\end{align*} \tau_e = \frac{mg\ell}{2}\cos\theta_e

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Design Study A

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Design Study A

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Design Study B

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Design Study B

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\noindent{\bf Linearize Each Term About Equilibrium}

\\

Define $\tilde{\theta}\defeq\theta-\theta_e$, $\tilde{z}\defeq z-z_e$, $\tilde{F}\defeq F-F_e$.

\begin{align*}

\ddot{\theta}\cos\theta &\approx

\ddot{\theta}_e\cos\theta_e

+ \frac{\partial}{\partial\theta}(\ddot{\theta}\cos\theta)\Big|_{(\ddot{\theta}_e,\theta_e)}(\theta-\theta_e)

% \\ & \qquad

+ \frac{\partial}{\partial\ddot{\theta}}(\ddot{\theta}\cos\theta)\Big|_{(\ddot{\theta}_e,\theta_e)}(\ddot{\theta}-\ddot{\theta}_e) \\

&= \ddot{\theta}_e\cos\theta_e

- (\ddot{\theta}_e\sin\theta_e)\tilde{\theta}

+ (\cos\theta_e)\ddot{\tilde{\theta}}

= \ddot{\tilde{\theta}} \\

%

\dot{\theta}^2\sin\theta &\approx \dot{\theta}_e^2\sin\theta_e + \frac{\partial}{\partial\theta}(\dot{\theta}^2\sin\theta)\Big|_{(\theta_e,\dot{\theta}_e)}(\theta-\theta_e)

% \\ & \qquad

+ \frac{\partial}{\partial\dot{\theta}}(\dot{\theta^2}\sin\theta)\Big|_{(\theta_e,\dot{\theta}_e)}(\dot{\theta}-\dot{\theta}_e) \\

&= \dot{\theta}_e^2\sin\theta_e + \dot{\theta}_e^2\cos\theta_e\tilde{\theta} + 2\dot{\theta}_e\sin\theta_e\dot{\tilde{\theta}}

= 0, \\

%

\ddot{z}\cos\theta &\approx

\ddot{z}_e\cos\theta_e

+ \frac{\partial}{\partial\theta}(\ddot{z}\cos\theta)\Big|_{(\ddot{z}_e,\theta_e)}(\theta-\theta_e)

% \\ & \qquad

+ \frac{\partial}{\partial\ddot{z}}(\ddot{z}\cos\theta)\Big|_{(\ddot{z}_e,\theta_e)}(\ddot{z}-\ddot{z}_e) \\

&= \ddot{z}_e\cos\theta_e

- (\ddot{z}_e\sin\theta_e)\tilde{\theta}

+ (\cos\theta_e)\ddot{\tilde{z}}

= \ddot{\tilde{z}} \\

%

\sin\theta &\approx \sin\theta_e + \frac{\partial}{\partial \theta}

(\sin\theta)\Big|_{\theta_e}

(\theta-\theta_e)

= \sin\theta_e + (\cos\theta_e) \tilde{\theta}

= \tilde{\theta},

\end{align*}

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Design Study B

Introduction to Feedback Control, Beard, McLain, Peterson, Killpack

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\noindent{\bf Find Linearized Equations using Jacobian Linearization}

\begin{align*}

(m_1+m_2)\ddot{z} + m_1 \frac{\ell}{2} \ddot{\theta} \cos\theta &=

m_1 \frac{\ell}{2} \dot{\theta}^2\sin\theta -b\dot{z} + F

\\

m_1 \frac{\ell}{2} \ddot{z} \cos\theta + m_1 \frac{\ell^2}{3} \ddot{\theta} &=

m_1 g \frac{\ell}{2} \sin\theta

\end{align*}

Substituting linearized terms gives gives

\begin{align*}

(m_1+m_2)[\ddot{z}_e+\ddot{\tilde{z}}] + m_1 \frac{\ell}{2} [\ddot{\tilde{\theta}}] &=

m_1 \frac{\ell}{2} [ 0 ] -b[\dot{z}_e+\dot{\tilde{z}}] + [F_e+\tilde{F}]

\\

m_1 \frac{\ell}{2} [ \ddot{\tilde{z}}] + m_1 \frac{\ell^2}{3} [\ddot{\theta}_e+\ddot{\tilde{\theta}}] &=

m_1 g \frac{\ell}{2} [\tilde{\theta}],

\end{align*}

Simplify using $F_e = 0$:

\[

\begin{pmatrix} (m_1+m_2) & m_1 \frac{\ell}{2} \\

m_1 \frac{\ell}{2} & m_1 \frac{\ell^2}{3} \end{pmatrix}\begin{pmatrix}

\ddot{\tilde{z}} \\ \ddot{\tilde{\theta}} \end{pmatrix} =

\begin{pmatrix} -b\dot{\tilde{z}} + \tilde{F}

\\

m_1 g \frac{\ell}{2} \tilde{\theta} \end{pmatrix}

\]

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Design Study B

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Note that direct feedback linearization is not possible.

\noindent{\bf Find Linerized Equations using Jacobian Linearization}

\begin{align*}

(m_1+m_2)\ddot{z} + m_1 \frac{\ell}{2} \ddot{\theta} \cos\theta &=

m_1 \frac{\ell}{2} \dot{\theta}^2\sin\theta -b\dot{z} + F

\\

m_1 \frac{\ell}{2} \ddot{z} \cos\theta + m_1 \frac{\ell^2}{3} \ddot{\theta} &=

m_1 g \frac{\ell}{2} \sin\theta

\end{align*}

Substituting linearized terms gives gives

\begin{align*}

(m_1+m_2)[\ddot{z}_e+\ddot{\tilde{z}}] + m_1 \frac{\ell}{2} [\ddot{\tilde{\theta}}] &=

m_1 \frac{\ell}{2} [ 0 ] -b[\dot{z}_e+\dot{\tilde{z}}] + [F_e+\tilde{F}]

\\

m_1 \frac{\ell}{2} [ \ddot{\tilde{z}}] + m_1 \frac{\ell^2}{3} [\ddot{\theta}_e+\ddot{\tilde{\theta}}] &=

m_1 g \frac{\ell}{2} [\tilde{\theta}],

\end{align*}

Simplify using $F_e = 0$:

\[

\begin{pmatrix} (m_1+m_2) & m_1 \frac{\ell}{2} \\

m_1 \frac{\ell}{2} & m_1 \frac{\ell^2}{3} \end{pmatrix}\begin{pmatrix}

\ddot{\tilde{z}} \\ \ddot{\tilde{\theta}} \end{pmatrix} =

\begin{pmatrix} -b\dot{\tilde{z}} + \tilde{F}

\\

m_1 g \frac{\ell}{2} \tilde{\theta} \end{pmatrix}

\]

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Design Study C

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Design Study C

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\noindent{\bf Find Equilibrium}

\begin{align*}

J_s\ddot{\theta} + b(\dot{\theta}-\dot{\phi}) + k(\theta-\phi) &= \tau \\

J_p\ddot{\phi} + b(\dot{\phi}-\dot{\theta}) + k(\phi-\theta) &= 0

\end{align*}

Defining $x=(\theta, \phi, \dot{\theta}, \dot{\phi})^\top$, and $u=\tau$ we get

\begin{align*}

\dot{x} = \begin{pmatrix} \dot{\theta} \\ \dot{\phi} \\ \ddot{\theta} \\ \ddot{\phi}\end{pmatrix}

= \begin{pmatrix} \dot{\theta} \\ \dot{\phi} \\ -\frac{b}{J_s}(\dot{\theta}-\dot{\phi}) -\frac{k}{J_s}(\theta-\phi)+\frac{1}{J_s}\tau \\ - \frac{b}{J_p}(\dot{\phi}-\dot{\theta})-\frac{k}{J_p}(\phi-\theta) \end{pmatrix} \defeq f(x,u).

\end{align*}

The equilibrium is when $f(x_e,u_e)=0$, or in other words when

\[

\theta_e = \phi_e = \text{~anything~}, \qquad \dot{\theta}_e = \dot{\phi}_e =0, \qquad \tau_e = 0

\]

NOTE: The equations of motion are already linear and therefore do not need to be linearized.