Code covered by the BSD License  

Highlights from
Teaching System Dynamics with MATLAB & Simulink

image thumbnail

Teaching System Dynamics with MATLAB & Simulink

by

 

MATLAB, Simulink, MuPAD, & Simscape models of 2nd order systems used in “Teaching System Dynamics.”

publish_spring_mass_damper_free_response.m
%% Transient Behavior and Location of Characteristic Roots in the Complex Plane
% Copyright 2011 The MathWorks, Inc.
%% Initial Value Problem
%
% Consider the system described by the following second-order, 
% ordinary differential equation and associated initial conditions:
%
% $$\frac{d^2x}{dt^2} + 2\zeta\omega_n\frac{dx}{dt} + \omega_n^2x = 0; x(0)=x0,\frac{dx}{dt}(0)=0 $$
%
% We would like to examine the behavior of the free response of the 
% system as the damping ratio changes.  This will be done first by solving
% the initial value problem for various damping ratios and plotting the
% results.  Then, we will explore changes in the damping ratio affect the
% characteristic roots.  (Recall that the characteristic roots dictate the
% free-response behavior.)

%% Set physical and derived parameters.
%
% To enable numerical solution of the differential equation, we will set
% specific values for the mass and (linear) spring rate.  We will then use
% these to derive the (circular) natural frequency.
%
param.m = 1;                            % Mass [kg]
param.k = 1;                            % Spring rate [N/m]
param.wn = sqrt(param.k/param.m);       % (Circular) natural frequency [rad/s]


%% Set initial conditions.
%
% To solve our 2nd order differential equation, we need two initial
% conditions: one on position, the other on velocity.  Moreover, recall 
% from the accompanying PowerPoint slides that we have represented our
% 2nd order linear ODE into a system of 1st order ODE's amenable to
% solution using MATLAB. Having defined position and velocity as our
% system states, we must define an initial state vector.

x0 = 1;              % Position [m]
x0_dot = 0;          % Velocity [m/s]
z0 = [x0 ; x0_dot];  % Initial condition on state vector


%% Set simulation parameters.
%
% We will define the duration of our simulation to be approximately 4
% periods.  To accommodate changes in physical parameters, we will
% parameterize the duration by the natural frequency.  We then define the
% associated time step and vector of simulation time.  Note that the time
% step dt serves to define the times at which output will be provided, not
% the solver time step.

tend = 4*round((2*pi)/param.wn);    % Simulate for approximately 4 periods [s]
dt = round((2*pi)/param.wn)/1000;   % Time step [s]
tsim = 0:dt:tend;                   % Vector of simulation times [s]


%% Undamped system.
%
% We first consider an undamped system, characterized by a damping ratio of
% zero.  Physically, this represents a system with zero dissipation of
% energy.  We will plot the free response to confirm the expected behavior.
%
% To solve the system of 1st order ODE's, we will use the baseline ode45
% solver.  The solver requires a vector of state derivatives; in this
% case, we call upon a function to define them.

param.zeta = 0;     % Damping ratio


[t,z] = ode45(@(t,z) odesys( t,z,param ),tsim,z0);
x(:,1) = z(:,1);    % Save solution for future plotting.


plot(t,x(:,1),'linewidth',2);grid on;hold all


    ylim([-1 1])
    xlabel('Time [s]');
    ylabel('Position [m]')
    title('Free Response: Undamped 2nd Order System');


%% Consider several damping ratios.
%
% Now we will consider four distinct values of the damping ratio
% corresponding to four distinct types of system behavior: undamped,
% underdamped, critically damped, and overdamped.  These terms refer to the
% behavior of the system as it returns to its equilibrium state.  We will
% plot and compare the free responses for these cases.

zetavals = [0 1/sqrt(2) 1 1.5]; % Damping ratios to consider.

case1 = 'Undamped: zeta = 0' ;
case2 = 'Underdamped: zeta = 0.707' ;
case3 = 'Critically Damped: zeta = 1' ;
case4 = 'Overdamped: zeta = 1.5' ;

for i = 2: length(zetavals)
    
    param.zeta = zetavals(i); % Set current damping ratio.
    [t,z] = ode45(@(t,z) odesys( t,z,param ),tsim,z0); % Compute free response.
    x(:,i) = z(:,1); % Save solution for future plotting.

end;


figure      % Plot results

    plot(t,x,'linewidth',2);grid on
    xlabel('Time [s]');
    ylabel('Position [m]')
    title('Free Response of 2nd Order System for Various Damping Ratios');
    legend(case1,case2,case3,case4,0)
    
    
%% Track characteristic roots as damping ratio increases.
%
% Recalling that the free response of a system is governed by its
% characteristic roots, we will examine the changes in these roots as the
% damping ratio changes over a more finely spaced set of values.  Since the
% roots may have real and imaginary parts, we will plot them in the complex 
% plane.  For each value of the damping ratio, we will compute the
% characteristic roots as the eigenvalues of the associated state matrix.
 
zeta_range = 0:0.01:1.5;          % Damping ratio: undamped to overdamped.
S = zeros(2,length(zeta_range));  % Initialize characteristic roots array.

for i = 1:length(zeta_range)
    
    S(:,i) = eig([0 1; -param.wn^2 -2*zeta_range(i)*param.wn]); % Characteristic roots.

end;

figure      % Plot characteristic roots in complex plane.

    plot(S,'bx','linewidth',2,'markersize',5); grid on;
    set(gca,'YAxisLocation','right')
    xlabel('Real Axis','Position',[-1.5 -0.1 1])
    ylabel('Imaginary Axis')
    title('Loci of Characteristic Roots in Complex Plane for Various Damping Ratios')

    
%% Show the "odesys" function.
%
% Here we record the function used to define the state matrix
% corresponding to our system of ordinary differential equations.

type odesys.m

Contact us