1 of 30

��Chapter 3�Euler-Lagrange Equations

2 of 30

Euler-Lagrange Equations

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

2

3 of 30

Potential Energy

Two types in this class, 1) gravity field, 2) compliance or springs
Gravity (typically zero when y, or other configuration variable is zero)

Springs (works for rotary and linear springs)

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

3

4 of 30

Generalized Coordinates and Forces

Generalized Coordinates – Minimum set of variables used to uniquely define the configuration of a mechanical system

Examples:

Robot arm – one degree of freedom or variable (theta)
Pendulum on cart – two degrees of freedom (theta and z)
Satellite – two degrees of freedom (theta and phi)

Generalized Forces – nonconservative forces and torques that act in the direction of each generalized coordinate.

E.g. the applied forces, either by actuators or by friction and other external forces.

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

4

5 of 30

Friction and Any Other External Force

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

5

Book presents one model of what is called “viscous friction” (where a constant coefficient multiplies the derivative of our generalized coordinates), but there are many other models.
All of the nonconservative forces (including other friction models) would be placed on the right-hand side!!!

6 of 30

Friction and Any Other External Force

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

6

7 of 30

Friction and Any Other External Force

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

7

8 of 30

Design Study A

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

8

9 of 30

Design Study A

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

9

10 of 30

Design Study B

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

10

11 of 30

Design Study B

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

11

Lagrangian: \quad

$

L(\mathbf{q},\dot{\mathbf{q}}) = \frac{1}{2}(m_1+m_2)\dot{q}_1^2 + \frac{1}{2}m_1\frac{\ell^2}{3}\dot{q}_2^2 + m_1\frac{\ell}{2}\dot{q}_1\dot{q}_2\cos q_2- P_0 - m_1 g \frac{\ell}{2}(\cos q_2-1)

$

\begin{align*}

\frac{\partial L}{\partial\dot{\mathbf{q}}} &= \begin{pmatrix}

(m_1+m_2)\dot{z} + m_1\frac{\ell}{2}\dot{\theta}\cos\theta \\

m_1\frac{\ell^2}{3}\dot{\theta} + m_1\frac{\ell}{2}\dot{z}\cos\theta

\end{pmatrix} \\

\frac{\partial L}{\partial\mathbf{q}} &= \begin{pmatrix}

0 \\ -m_1\frac{\ell}{2}\dot{z}\dot{\theta}\sin\theta + m_1 g \frac{\ell}{2} \sin\theta

\end{pmatrix}.

\end{align*}

Differentiating $\frac{\partial L}{\partial \dot{\mathbf{q}}}$ with respect to time gives

\[

\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\mathbf{q}}}\right) = \begin{pmatrix}

(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 \\

m_1 \frac{\ell^2}{3} \ddot{\theta} + m_1 \frac{\ell}{2} \ddot{z}\cos\theta - m_1 \frac{\ell}{2} \dot{z}\dot{\theta}\sin\theta

\end{pmatrix}.

\]

12 of 30

Design Study B

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

12

\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 &= F - b\dot{z} \\

m_1 \frac{\ell^2}{3} \ddot{\theta} + m_1 \frac{\ell}{2} \ddot{z}\cos\theta - m_1 \frac{\ell}{2} \dot{z}\dot{\theta}\sin\theta + m_1\frac{\ell}{2}\dot{z}\dot{\theta}\sin\theta - m_1 g \frac{\ell}{2} \sin\theta &= 0.

\end{align*}

Simplifying and moving all second order derivatives to the left-hand side, and all other terms to the right-hand side gives

\[

\begin{pmatrix}

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

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

\end{pmatrix}

= \begin{pmatrix} m_1 \frac{\ell}{2} \dot{\theta}^2\sin\theta + F -b\dot{z} \\ m_1 g \frac{\ell}{2} \sin\theta \end{pmatrix}.

\]

Using matrix notation, this equation can be rearranged to isolate the second order derivatives on the left-hand side as

\[

\begin{pmatrix}

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

m_1 \frac{\ell}{2} \cos\theta & m_1 \frac{\ell^2}{3}

\end{pmatrix} \begin{pmatrix}\ddot{z} \\ \ddot{\theta} \end{pmatrix}

= \begin{pmatrix} m_1 \frac{\ell}{2} \dot{\theta}^2\sin\theta + F -b\dot{z} \\ m_1 g \frac{\ell}{2} \sin\theta \end{pmatrix}.

\]

13 of 30

Design Study C

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

13

14 of 30

Design Study C

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

14

15 of 30

Design Study C

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

15

\noindent Therefore the Euler-Lagrange equation gives

\[

\begin{pmatrix}

J_s\ddot{\theta} - k(\phi-\theta) \\

J_p\ddot{\phi} + k(\phi-\theta)

\end{pmatrix}

= \begin{pmatrix} \tau \\ 0 \end{pmatrix} + \begin{pmatrix} -b(\dot{\theta}-\dot{\phi}) \\ -b(\dot{\phi}-\dot{\theta}) \end{pmatrix}.

\]

Simplifying and moving all second order derivatives to the left-hand side, and all other terms to the right-hand side gives

\[

\begin{pmatrix}

J_s\ddot{\theta} \\

J_p\ddot{\phi}

\end{pmatrix}

= \begin{pmatrix} \tau -b(\dot{\theta}-\dot{\phi})-k(\theta-\phi) \\ -b(\dot{\phi}-\dot{\theta}) - k(\phi-\theta) \end{pmatrix}.

\]

Using matrix notation:

\[

\begin{pmatrix}

J_s & 0 \\

0 & J_p

\end{pmatrix} \begin{pmatrix}\ddot{\theta} \\ \ddot{\phi} \end{pmatrix}

= \begin{pmatrix} \tau-b(\dot{\theta}-\dot{\phi})-k(\theta-\phi) \\ -b(\dot{\phi}-\dot{\theta}) - k(\phi-\theta) \end{pmatrix}.

\]

16 of 30

Appendix P.4�Numerical Solutions to Differential Equations

16

17 of 30

Numerically Solving an ODE

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

17

18 of 30

Numerically Solving an ODE

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

18

19 of 30

Euler Method / Runge-Kutta 1 (RK1)

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

19

20 of 30

Euler Method / Runge-Kutta 1 (RK1)

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

20

import numpy as np

class systemDynamics:

# Model the system yddot + a1*ydot + a2*sin(y) = b*u

def __init__(self, Ts=0.01):

self.Ts = Ts # sample rate

# Specify the initial condition

y0 = 0.0 # y at time zero

ydot0 = 0.0 # ydot at time zero

self.state = np.array([[y0], [ydot0]]) # state at time zero

self.a1 = 2.0

self.a2 = 2.0

self.b = 4.0

def update(self, u):

# This is the external method that takes the input u and

# returns the output y.

self.rk1_step(u) # propagate the state by one time sample

y = self.h() # return the corresponding output

return y

def rk1_step(self, u):

# Integrate ODE using Runge-Kutta RK1 algorithm

F1 = self.f(self.state, u)

self.state = self.state + self.Ts * F1

21 of 30

Runge-Kutta 2 (RK2)

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

21

22 of 30

Runge-Kutta 2 (RK2)

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

22

import numpy as np

�

class systemDynamics:

# Model the system yddot + a1*ydot + a2*sin(y) = b*u

def __init__(self, Ts=0.01):

self.Ts = Ts # sample rate

# Specify the initial condition

y0 = 0.0 # y at time zero

ydot0 = 0.0 # ydot at time zero

self.state = np.array([[y0],[ydot0]]) # state at time zero

self.a1 = 2.0

self.a2 = 2.0

self.b = 4.0�

def update(self, u):

# This is the external method that takes the input u and

# returns the output y.

self.rk2_step(u) # propagate the state by one time sample

y = self.h() # return the corresponding output

return y�

def rk2_step(self, u):

# Integrate ODE using Runge-Kutta RK2 algorithm

F1 = self.f(self.state, u)

F2 = self.f(self.state + self.Ts / 2 * F1, u)

self.state += self.Ts / 2 * (F1 + F2)

23 of 30

Runge-Kutta 4 (RK4)

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

23

24 of 30

Runge-Kutta 4 (RK4)

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

24

import numpy as np

�

class systemDynamics:

# Model the system yddot + a1*ydot + a2*sin(y) = b*u

def __init__(self, Ts=0.01):

self.Ts = Ts # sample rate

# Specify the initial condition

y0 = 0.0 # y at time zero

ydot0 = 0.0 # ydot at time zero

self.state = np.array([[y0],[ydot0]]) # state at time zero

self.a1 = 2.0

self.a2 = 2.0

self.b = 4.0�

def update(self, u):

# This is the external method that takes the input u and

# returns the output y.

self.rk4_step(u) # propagate the state by one time sample

y = self.h() # return the corresponding output

return y�

def rk4_step(self, u):

# Integrate ODE using Runge-Kutta RK4 algorithm

F1 = self.f(self.state, u)

F2 = self.f(self.state + self.Ts / 2 * F1, u)

F3 = self.f(self.state + self.Ts / 2 * F2, u)

F4 = self.f(self.state + self.Ts * F3, u)

self.state += self.Ts / 6 * (F1 + 2 * F2 + 2 * F3 + F4)

25 of 30

Design Study A

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

25

26 of 30

Design Study A

Discuss python implementation

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

26

27 of 30

Design Study B

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

27

\begin{align*}

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

\implies & \ddot{\theta} = \frac{3}{m\ell^2} \tau - \frac{3b}{m\ell^2} \dot{\theta} - \frac{3g}{2\ell}\cos\theta

\end{align*}

Define

\[

\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \defeq \begin{pmatrix} \theta \\ \dot{\theta} \end{pmatrix}

\]

then

\begin{align*}

\dot{\mathbf{x}} &= \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} \\

&= \begin{pmatrix} x_2 \\ \frac{3}{m\ell^2} u - \frac{3b}{m\ell^2} x_2 - \frac{3g}{2\ell}\cos x_1 \end{pmatrix} \\

&\defeq f(\mathbf{x}, u)

\end{align*}

--

Then

\[

\begin{pmatrix} \ddot{z} \\ \ddot{\theta} \end{pmatrix} = M^{-1}(\mathbf{x}) \mathbf{c}(\mathbf{x},u)

\]

and

\begin{align*}

\dot{\mathbf{x}} &= \begin{pmatrix} \dot{z} \\ \dot{\theta} \\ \ddot{z} \\ \ddot{\theta} \end{pmatrix}

= \begin{pmatrix} \dot{z} \\ \dot{\theta} \\ M^{-1}(\mathbf{x}) \mathbf{c}(\mathbf{x}, u) \end{pmatrix} \\

&\defeq f(\mathbf{x}, u)

\end{align*}

28 of 30

Design Study B

Discuss python implementation

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

28

29 of 30

Design Study C

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

29

\[

\begin{pmatrix}

J_s & 0 \\

0 & J_p

\end{pmatrix} \begin{pmatrix}\ddot{\theta} \\ \ddot{\phi} \end{pmatrix}

= \begin{pmatrix} \tau-b(\dot{\theta}-\dot{\phi})-k(\theta-\phi) \\ -b(\dot{\phi}-\dot{\theta}) - k(\phi-\theta) \end{pmatrix}.

\]

�

Define

\begin{align*}

\mathbf{x} &\defeq \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} z \\\theta \\ \dot{z} \\ \dot{\theta} \end{pmatrix} \\

M &\defeq \begin{pmatrix}

J_s & 0 \\

0 & J_p

\end{pmatrix} \\

\mathbf{c}(\mathbf{x}, u) &\defeq \begin{pmatrix} \tau-b(\dot{\theta}-\dot{\phi})-k(\theta-\phi) \\ -b(\dot{\phi}-\dot{\theta}) - k(\phi-\theta) \end{pmatrix}

\end{align*}

--

Then

\[

\begin{pmatrix} \ddot{\theta} \\ \ddot{\phi} \end{pmatrix} = M^{-1} \mathbf{c}(\mathbf{x},u)

\]

and

\begin{align*}

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

= \begin{pmatrix} x_3 \\ x_4 \\ M^{-1} \mathbf{c}(\mathbf{x}, u) \end{pmatrix} \\

&\defeq f(\mathbf{x}, u)

\end{align*}

30 of 30

Design Study C

Discuss python implementation

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

30