pyrobolearn/tutorials/robotics/mass-spring-damper.ipynb

Mass-spring-damper

In this tutorial, we will describe the mechanics and control of the one degree of freedom translational mass-spring-damper system subject to a control input force. We will first derive the dynamic equations by hand. Then, we will derive them using the sympy.mechanics python package.

The system on which we will work is depicted below:

Note that in what follows, we use the notation u(t) = F.

1. Mechanics

Deriving the dynamical equations by hand

1.1 By using Newton equations

Using Newton's law, we have:

\begin{align} m \ddot{x}(t) &= \sum F_{ext} \ &= - b \dot{x}(t) - k x(t) + u(t) \end{align}

1.2 By using the Lagrange Method

Let's first derive the kinematic and potential energies.

\begin{equation} T = \frac{1}{2} m \dot{x} \ V = - \int \vec{F} . \vec{dl} = - \int (-kx \vec{1_x}) . dx \vec{1_x} = \frac{k x^2}{2} \end{equation}

The Lagrangian is then given by: \begin{equation} \mathcal{L} = T - V = \frac{1}{2} m \dot{x} - \frac{k x^2}{2} \end{equation}

Using the Lagrange's equations we can derive the dynamics of the system:

\begin{equation} \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}} - \frac{\partial \mathcal{L}}{\partial q} = Q \end{equation}

where q are the generalized coordinates (in this case x), and Q represents the non-conservative forces (input force, dragging or friction forces, etc).

• \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} = \frac{d}{dt} m \dot{x}(t) = m \ddot{x}(t)
• \frac{\partial \mathcal{L}}{\partial x} = - k x(t)
• Q = - b \dot{x}(t) + u(t)

which when putting everything back together gives us:

\begin{equation} m \ddot{x}(t) + b \dot{x}(t) + k x(t) = u(t) \end{equation}

Deriving the dynamical equations using sympy

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import sympy
import sympy.physics.mechanics as mechanics
from sympy import init_printing
init_printing(use_latex='mathjax')
from sympy import pprint

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# define variables 
q = mechanics.dynamicsymbols('q')
dq = mechanics.dynamicsymbols('q', 1)
u = mechanics.dynamicsymbols('u')

# define constants
m, k, b = sympy.symbols('m k b')

# define the inertial frame
N = mechanics.ReferenceFrame('N')

# define a particle for the mass
P = mechanics.Point('P')
P.set_vel(N, dq * N.x) # go in the x direction
Pa = mechanics.Particle('Pa', P, m)

# define the potential energy for the particle (the kinematic one is derived automatically)
Pa.potential_energy = k * q**2 / 2.0

# define the Lagrangian and the non-conservative force applied on the point P
L = mechanics.Lagrangian(N, Pa)
force = [(P, -b * dq * N.x + u * N.x)]

# Lagrange equations 
lagrange = mechanics.LagrangesMethod(L, [q], forcelist = force, frame = N)
pprint(lagrange.form_lagranges_equations())

⎡                              2             ⎤
⎢  d                          d              ⎥
⎢b⋅──(q(t)) + 1.0⋅k⋅q(t) + m⋅───(q(t)) - u(t)⎥
⎢  dt                          2             ⎥
⎣                            dt              ⎦

2. Laplace transform and transfer function

Applying the Laplace transform on the dynamic equation:

\begin{equation} m \ddot{x}(t) + b \dot{x}(t) + k x(t) = u(t) \stackrel{L}{\rightarrow} m s^2 X(s) + b s X(s) + k X(s) = U(s) \end{equation}

The transfer equation is given by:

\begin{equation} H(s) = \frac{X(s)}{U(s)} = \frac{1}{m s^2 + b s + k} \end{equation}

By calculating the pole:

\begin{equation} m s^2 + b s + k = 0 \Leftrightarrow s = \frac{-b}{2m} \pm \sqrt{\left(\frac{b}{2m}\right)^2 - \frac{k}{m}} \end{equation}

Note that b, k, m > 0 because they represent real physical quantities.

LTI system

We can rewrite the above equation as a first-order system of equations. Let's first define the state vector \pmb{x} = \left[ \begin{array}{c} x(t) \\ \dot{x}(t) \end{array} \right] and the control vector \pmb{u} = \left[ \begin{array}{c} u(t) \end{array} \right], then we can rewrite the above equation in the form \pmb{\dot{x}} = \pmb{Ax} + \pmb{Bu}, as below:

\begin{equation} \left[ \begin{array}{c} \dot{x}(t) \ \ddot{x}(t) \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \ -\frac{k}{m} & -\frac{b}{m} \end{array} \right] \left[ \begin{array}{c} x(t) \ \dot{x}(t) \end{array} \right] + \left[ \begin{array}{c} 0 \ \frac{1}{m} \end{array} \right] \left[ \begin{array}{c} u(t) \end{array} \right] \end{equation}

If there is no u(t), i.e. u(t) = 0 \; \forall t, then we have \pmb{\dot{x}} = \pmb{Ax}. The solution to this system of equation is \pmb{x}(t) = e^{\pmb{A}t} \pmb{x}_0.

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