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��Chapter 6�State Space Models

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Course Overview

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

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Equations of motion: Big picture

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Mechanical

Newton’s Laws

Conservation of Momentum

Lagrange’s Equations

Etc.

Electrical

Kirchhoff’s Laws

Etc.

Fluid

Continuity Equation

Etc.

Fluid-Mechanical

Electromechanical

Nonlinear ODE’s

Linear ODE’s

State-variable

form

State-space

form

Transfer function

Design models

Simulation models

Linearization

Laplace x-form

Linearization

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State Space Equations

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State Space Equations

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\par\noindent{\bf Any nonlinear $n^{th}$ order differential equation can be transformed into state space form.}

�

\hspace{1cm}

\par\noindent{\bf Example 1:}

\[

\dddot{y} + y^2\ddot{y} + \sin(y\dot{y}) + e^y = u, \qquad \text{$y$ is the output, $u$ is the input}

\]

Define the state

\[

x=\begin{pmatrix}x_1 \\ x_2 \\ x_3 \end{pmatrix}\defeq\begin{pmatrix} y \\ \dot{y} \\ \ddot{y} \end{pmatrix}.

\]

Then the state evolution equation is

\begin{align*}

\dot{x} &= \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{pmatrix} = \begin{pmatrix} \dot{y} \\ \ddot{y} \\ \dddot{y} \end{pmatrix}

= \begin{pmatrix} \dot{y} \\ \ddot{y} \\ -y^2\ddot{y} - \sin(y\dot{y}) - e^y + u \end{pmatrix}

= \begin{pmatrix} x_2 \\ x_3 \\ -x_1^2x_3 - \sin(x_1x_2) - e^{x_1} + u \end{pmatrix}

\defeq f(x,u),

\end{align*}

and the output equation is

\[

y = x_1 \defeq h(x,u).

\]

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State Space Equations

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\par\noindent{\bf Example 2:}

\begin{align*}

\ddot{y}_1 + \dot{y}_1y_2 + y_2 &=u \\

\ddot{y}_2 + \dot{y}_2\cos(y_1)+\dot{y_1} &=0.

\end{align*}

Defining the output as $y=(y_1, y_2)^{\top}$ and the state as

\[

x=\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \end{pmatrix}^\top \defeq \begin{pmatrix} y_1 & y_2 & \dot{y}_1 & \dot{y}_2 \end{pmatrix}^\top,

\]

the state evolution equation is

\begin{align*}

\dot{x} &= \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \\ \dot{x}_4 \end{pmatrix} = \begin{pmatrix} \dot{y}_1 \\ \dot{y}_2 \\ \ddot{y}_1 \\ \ddot{y}_2 \end{pmatrix}

= \begin{pmatrix} \dot{y}_1 \\ \dot{y}_2 \\ -\dot{y}_1y_2 - y_2 +u \\ -\dot{y}_2\cos(y_1)-\dot{y_1} \end{pmatrix}

= \begin{pmatrix} x_3 \\ x_4 \\ -x_3x_2 - x_2 +u \\ -x_4\cos(x_1)-x_3 \end{pmatrix}

\defeq f(x,u),

\end{align*}

and the output equation is given by

\[

y = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \defeq h(x,u).

\]

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Jacobian Linearization

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Linearization using the state-space equations is relatively straightforward.

\begin{align*}

\dot{x} &= f(x,u) \qquad \qquad \text{Equilibrium state } (x_e,u_e)\\

y &= h(x,u)

\end{align*}

Taylor series expansion about $(x_e,u_e)$:

\begin{align*}

f(x,u) &\approx f(x_e,u_e) + \left.\frac{\partial f}{\partial x}\right|_e (x-x_e) + \left.\frac{\partial f}{\partial u}\right|_e(u-u_e) + H.O.T.

\\

h(x,u) &\approx h(x_e,u_e) + \left.\frac{\partial h}{\partial x}\right|_e (x-x_e) + \left.\frac{\partial h}{\partial u}\right|_e(u-u_e) + H.O.T.,

\end{align*}

where

\begin{alignat*}{3}

& \frac{\partial f}{\partial x} &\defeq

\begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \dots & \frac{\partial f_1}{\partial x_n} \\

\vdots & & \vdots \\

\frac{\partial f_n}{\partial x_1} & \dots & \frac{\partial f_n}{\partial x_n}

\end{pmatrix},

&\qquad\qquad&

\frac{\partial f}{\partial u} &\defeq

\begin{pmatrix} \frac{\partial f_1}{\partial u_1} & \dots & \frac{\partial f_1}{\partial u_m} \\

\vdots & & \vdots \\

\frac{\partial f_n}{\partial u_1} & \dots & \frac{\partial f_n}{\partial u_m}

\end{pmatrix} \\

& \frac{\partial h}{\partial x} &\defeq

\begin{pmatrix} \frac{\partial h_1}{\partial x_1} & \dots & \frac{\partial h_1}{\partial x_n} \\

\vdots & & \vdots \\

\frac{\partial h_p}{\partial x_1} & \dots & \frac{\partial h_p}{\partial x_n}

\end{pmatrix},

&\qquad\qquad&

\frac{\partial h}{\partial u} &\defeq

\begin{pmatrix} \frac{\partial h_1}{\partial u_1} & \dots & \frac{\partial h_1}{\partial u_m} \\

\vdots & & \vdots \\

\frac{\partial h_p}{\partial u_1} & \dots & \frac{\partial h_p}{\partial u_m}

\end{pmatrix}.

\end{alignat*}

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Jacobian Linearization

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Converting State Space to Transfer Function

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Example

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

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\begin{align*}

\dot{x} &= \begin{pmatrix} 0 & 1 \\ -4 & -3 \end{pmatrix} x + \begin{pmatrix} 0 \\ 2 \end{pmatrix} u \\

y &= \begin{pmatrix} 1 & 0 \end{pmatrix} x + (0)u

\end{align*}

The transfer function from $u$ to $y$ is

\begin{align*}

H(s) &= C(sI-A)^{-1}B + D \\

&= \begin{pmatrix} 1 & 0\end{pmatrix}\left(s\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} - \begin{pmatrix} 0 & 1 \\ -4 & -3 \end{pmatrix}\right)^{-1}\begin{pmatrix} 0 \\ 2\end{pmatrix} + 0 \\

&= \begin{pmatrix} 1 & 0\end{pmatrix}\begin{pmatrix} s & -1 \\ 4 & s+3\end{pmatrix}^{-1}\begin{pmatrix} 0 \\ 2\end{pmatrix}\\

&= \begin{pmatrix} 1 & 0\end{pmatrix}\frac{\begin{pmatrix} s+3 & 1 \\ -4 & s\end{pmatrix}}{s^2+3s+4}\begin{pmatrix} 0 \\ 2\end{pmatrix}\\

&= \frac{\begin{pmatrix} s+3 & 1 \end{pmatrix}}{s^2+3s+4}\begin{pmatrix} 0 \\ 2\end{pmatrix}\\

&= \frac{2}{s^2+3s+4}.

\end{align*}

--

{\color{blue}

\[

\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}}{ad-bc}

\]

}

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Poles vs. Eigenvalues

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Design Study A

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Given the linear ODE

\[

\ddot{\theta} = \frac{3}{m\ell^2}\tilde{\tau} - \frac{3b}{m\ell^2}\dot{\theta}

\]

{\color{blue} Find the state space model.}

�

\hspace{1cm}

�

Define

\[

x=(x_1, x_2)^\top = (\theta, \dot\theta)^\top, \qquad u=\tilde{\tau}, \qquad y = \theta

\]

\begin{align*}

\dot{x} &\defeq \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix}

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

= \begin{pmatrix} \dot{\theta} \\ \frac{3}{m\ell^2}\tilde{\tau} - \frac{3b}{m\ell^2}\dot{\theta} \end{pmatrix}

=\begin{pmatrix} x_2 \\ \frac{3}{m\ell^2}\tilde{u} - \frac{3b}{m\ell^2}x_2 \end{pmatrix}

= \begin{pmatrix} 0 & 1 \\ 0 & -\frac{3b}{m\ell^2} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{3}{m\ell^2} \end{pmatrix} \tilde{u}.

\end{align*}

Also

\[

y = \theta = x_1 = \begin{pmatrix} 1 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix} + (0)\tilde{u}.

\]

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Design Study A

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Design Study A

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\par\noindent{\color{blue} Jacobian Linearization:}

\begin{align*}

\dot{x} &\defeq \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix}

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

= \begin{pmatrix} \dot{\theta} \\ \frac{3}{m\ell^2}\tau - \frac{3b}{m\ell^2}\dot{\theta} - \frac{3g}{2\ell}\cos\theta \end{pmatrix} \defeq f(x,u).

\end{align*}

Evaluating the Jacobians at the equilibrium $(\theta_e, \dot{\theta}_e, \tau_e) = (0, 0, \frac{mg\ell}{2})$ gives

\begin{align*}

A &= \frac{\partial f}{\partial x}\Big|_e = \begin{pmatrix} \frac{\partial f_1}{\partial \theta} & \frac{\partial f_1}{\partial \dot{\theta}} \\ \frac{\partial f_2}{\partial \theta} & \frac{\partial f_2}{\partial \dot{\theta}} \end{pmatrix}\Big|_e

= \begin{pmatrix} 0 & 1 \\ \frac{3g}{2\ell}\sin\theta & -\frac{3b}{m\ell^2} \end{pmatrix}\Big|_e

= \begin{pmatrix} 0 & 1 \\ 0 & -\frac{3b}{m\ell^2} \end{pmatrix} \\

B &= \frac{\partial f}{\partial u}\Big|_e = \begin{pmatrix} \frac{\partial f_1}{\partial \tau} \\ \frac{\partial f_2}{\partial \tau} \end{pmatrix}\Big|_e

= \begin{pmatrix} 0 \\ \frac{3}{m\ell^2} \end{pmatrix}\Big|_e

= \begin{pmatrix} 0 \\ \frac{3}{m\ell^2} \end{pmatrix} \\

C &= \frac{\partial h}{\partial x}\Big|_e = \begin{pmatrix} \frac{\partial h_1}{\partial \theta} & \frac{\partial h_1}{\partial \dot{\theta}} \end{pmatrix}\Big|_e

= \begin{pmatrix} 1 & 0 \end{pmatrix}\Big|_e

= \begin{pmatrix} 1 & 0 \end{pmatrix} \\

D &= \frac{\partial h}{\partial u}\Big|_e = 0.

\end{align*}

�

\par\noindent{\color{blue} Same as before.}

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Design Study A

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\par\noindent Given

\begin{align*}

\dot{\tilde{x}} &= \begin{pmatrix} 0 & 1 \\ \frac{3g\sin\theta_e}{2\ell} & -\frac{3b}{m\ell^2} \end{pmatrix} \tilde{x} + \begin{pmatrix} 0 \\ \frac{3}{m\ell^2} \end{pmatrix} \tilde{u}

\notag \\

\tilde{y} &= \begin{pmatrix} 1 & 0\end{pmatrix} \tilde{x},

\end{align*}

the transfer function from $U(s)$ to $Y(s)$ is

\begin{align*}

H(s) &= C(sI-A)^{-1}B+D \\

&= \begin{pmatrix} 1 & 0 \end{pmatrix} \left(s\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} -\begin{pmatrix} 0 & 1 \\ 0 & -\frac{3b}{m\ell^2}\end{pmatrix} \right)^{-1} \begin{pmatrix} 0 \\ \frac{3}{m\ell^2} \end{pmatrix}

= \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} s & -1 \\ 0 & s+\frac{3b}{m\ell^2}\end{pmatrix}^{-1} \begin{pmatrix} 0 \\ \frac{3}{m\ell^2} \end{pmatrix} \\

&= \frac{\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} s+\frac{3b}{m\ell^2} & 1 \\ 0 & s \end{pmatrix} \begin{pmatrix} 0 \\ \frac{3}{m\ell^2}\end{pmatrix}}{s^2 + \frac{3b}{m\ell^2}s}

= \frac{\begin{pmatrix} s+\frac{3b}{m\ell^2} & 1 \end{pmatrix} \begin{pmatrix} 0 \\ \frac{3}{m\ell^2}\end{pmatrix}}{s^2 + \frac{3b}{m\ell^2}s} \\

&= \frac{\frac{3}{m\ell^2}}{s^2 + \frac{3b}{m\ell^2}s}.

\end{align*}

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Design Study B

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\par\noindent Starting with

\[

\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},

\]

{\color{blue}Find the state space equations}

�

\hspace{1cm}

�

\par\noindent Invert matrix and simplify

\begin{align*}

\begin{pmatrix} \ddot{\tilde{z}} \\ \ddot{\tilde{\theta}} \end{pmatrix} &= \frac{\begin{pmatrix} m_1\frac{\ell^2}{3} & -m_1\frac{\ell}2 \\ -m_1\frac{\ell}{2} & (m_1+m_2)\end{pmatrix}}{m_1^2\frac{\ell^2}{3}+m_1m_2\frac{\ell^2}{3}-m_1^2\frac{\ell^2}{4}}\begin{pmatrix} -b\dot{\tilde{z}}+\tilde{F} \\ m_1g\frac{\ell}{2}\tilde{\theta} \end{pmatrix} \\

&= \begin{pmatrix} -\frac{b}{\frac{1}{4}m_1 + m_2}\dot{\tilde{z}} + \frac{1}{\frac{1}{4}m_1 + m_2}\tilde{F} - \frac{\frac{3}{4} m_1g}{\frac{1}{4}m_1 + m_2}\tilde{\theta} \\ \frac{3b}{2(\frac{1}{4}m_1 + m_2)\ell}\dot{\tilde{z}} - \frac{3}{2( \frac{1}{4}m_1 + m_2)\ell}\tilde{F} + \frac{3 (m_1+m_2)g}{2 (\frac{1}{4}m_1 + m_2)\ell}\tilde{\theta} \end{pmatrix}.

\end{align*}

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Design Study B

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\begin{align*}

\begin{pmatrix} \ddot{\tilde{z}} \\ \ddot{\tilde{\theta}} \end{pmatrix} = \begin{pmatrix} -\frac{b}{\frac{1}{4}m_1 + m_2}\dot{\tilde{z}} + \frac{1}{\frac{1}{4}m_1 + m_2}\tilde{F} - \frac{\frac{3}{4} m_1g}{\frac{1}{4}m_1 + m_2}\tilde{\theta} \\ \frac{3b}{2(\frac{1}{4}m_1 + m_2)\ell}\dot{\tilde{z}} - \frac{3}{2( \frac{1}{4}m_1 + m_2)\ell}\tilde{F} + \frac{3 (m_1+m_2)g}{2 (\frac{1}{4}m_1 + m_2)\ell}\tilde{\theta} \end{pmatrix}.

\end{align*}

Define

\[

x = (x_1, x_2, x_3, x_4)^\top \defeq (z, \theta, \dot{z}, \dot{\theta})^\top, \qquad u\defeq F, \qquad y\defeq (z, \theta)^\top

\]

Then

\begin{align*}

\dot{\tilde{x}} &\defeq \begin{pmatrix} \dot{\tilde{x}}_1 \\ \dot{\tilde{x}}_2 \\ \dot{\tilde{x}}_3 \\ \dot{\tilde{x}}_4\end{pmatrix} = \begin{pmatrix}\dot{\tilde{z}} \\ \dot{\tilde{\theta}} \\ \ddot{\tilde{z}} \\ \ddot{\tilde{\theta}}\end{pmatrix}

= \begin{pmatrix} \tilde{x}_3 \\ \tilde{x}_4 \\

-\frac{b}{\frac{1}{4}m_1 + m_2}\tilde{x}_3 + \frac{1}{\frac{1}{4}m_1 + m_2}\tilde{u} - \frac{\frac{3}{4} m_1g}{\frac{1}{4}m_1 + m_2}\tilde{x}_2 \\

\frac{3b}{2(\frac{1}{4}m_1 + m_2)\ell}\tilde{x}_3 - \frac{3}{2(\frac{1}{4}m_1 + m_2)\ell}\tilde{u} + \frac{3 (m_1+m_2)g}{2 (\frac{1}{4}m_1 + m_2)\ell} \tilde{x}_2 \end{pmatrix} \\

&= \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & -\frac{\frac{3}{4} m_1g}{\frac{1}{4}m_1 + m_2} & -\frac{b}{\frac{1}{4}m_1 + m_2} & 0 \\ 0 & \frac{3 (m_1+m_2)g}{2 (\frac{1}{4}m_1 + m_2)\ell} & \frac{3b}{2(\frac{1}{4}m_1 + m_2)\ell} & 0 \end{pmatrix} \begin{pmatrix} \tilde{x}_1 \\ \tilde{x}_2 \\ \tilde{x}_3 \\ \tilde{x}_4 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ \frac{1}{\frac{1}{4}m_1 + m_2} \\ \frac{-3}{2(\frac{1}{4}m_1 + m_2)\ell} \end{pmatrix} \tilde{u}.

\end{align*}

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Design Study B

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Design Study B

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Design Study C

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

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Design Study C

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