May 30, 2012 Compiled on September 9, 2023 at 7:36pm

1 download examples source code

2 description

3 Simulation

4 Using ode45 with piecewise function

5 Listing of source code

2 description

3 Simulation

4 Using ode45 with piecewise function

5 Listing of source code

- first_order_ode.m.txt
- second_order_ode.m.txt
- engr80_august_14_2006_2.m.txt
- engr80_august_14_2006.m.txt
- ode45_with_piecwise.m.txt

This shows how to use Matlab to solve standard engineering problems which involves solving a standard second order ODE. (constant coeﬃcients with initial conditions and nonhomogeneous).

A numerical ODE solver is used as the main tool to solve the ODE’s. The matlab function ode45 will be used. The important thing to remember is that ode45 can only solve a ﬁrst order ODE. Therefore to solve a higher order ODE, the ODE has to be ﬁrst converted to a set of ﬁrst order ODE’s. This is possible since an \(n\) order ODE can be converted to a set of \(n\) ﬁrst order ODE’s.

Gives a ﬁrst order ODE \[ \frac {dx}{dt}=f(x,t) \]

An example of the above is \(\frac {dx}{dt}=3 e^{-t}\) with an initial condition \(x(0)=0\). Here is the result of solving this ODE in Matlab. Source code is first_order_ode.m.txt

To solve a second order ODE, using this as an example. \[ \frac {d^{2} x}{dt^{2}}+5 \frac {dx}{dt}- 4 x(t) = \sin (10\ t) \]

Since ode45 can only solve a ﬁrst order ode, the above has to be converted to two ﬁrst order ODE’s as follows. Introduce 2 new state variables \(x_{1},x_{2}\) and carry the following derivation

\[ \left . \begin {array} [c]{c}x_{1}=x\\ x_{2}=x^{\prime }\end {array} \right \} \overset {\text {take derivative}}{\rightarrow }\left . \begin {array} [c]{c}x_{1}^{\prime }=x^{\prime }\\ x_{2}^{\prime }=x^{\prime \prime }\end {array} \right \} \overset {\text {do replacement}}{\rightarrow }\left . \begin {array} [c]{l}x_{1}^{\prime }=x_{2}\\ x_{2}^{\prime }=-5x^{\prime }+4x+\sin \left ( 10t\right ) \end {array} \right \} \rightarrow \left . \begin {array} [c]{l}x_{1}^{\prime }=x_{2}\\ x_{2}^{\prime }=-5x_{2}+4x_{1}+\sin \left ( 10t\right ) \end {array} \right \} \]

The above gives 2 new ﬁrst order ODE’s. These are

\[\begin {array} [c]{l}x_{1}^{\prime }=x_{2}\\ x_{2}^{\prime }=-5x_{2}+4x_{1}+\sin \left ( 10t\right ) \end {array} \]

Now ode45 can be used to solve the above in the same way as was done with the ﬁrst example. The only diﬀerence is that now a vector is used instead of a scalar.

This is the result of solving this in Matlab. The source code is second_order_ode.m.txt

Now ode45 is used to perform simulation by showing the solution as it changes in time.

Given a single degree of freedom system. This represents any engineering system whose response can move in only one direction. A typical SDOF (single degree of freedom) is the following mass/spring/damper system.

The ﬁrst step is to obtain the equation of motion, which will be the second order ODE. Drawing the free body diagram and from Newton’s second laws the equation of motion is found to be

\[ m x'' + c x' + k x = f( \omega _f t) \]

In the above, \(\omega _f\) is the forcing frequency of the force on the system in rad/sec.

The response of the system (the solution of the system, or \(x(t)\)) is simulated for diﬀerent parameters.

For example, the damping \(c\) can be changed, or the spring constant (the spring stiﬀness) to see how \(x(t)\) changes. The forcing function frequency \(\omega _f\) can also be changed.

The following deﬁnitions are used in the Matlab code.

Natural frequency of the system \(\omega =\sqrt {\frac {k}{m}-\left (\frac {c}{2m}\right )^2}\)

Damping ratio \(\varsigma =\frac {c}{c_r}\) where \(c\) is the damping coeﬃcient and \(c_r\) is the critical damping. \[ c_{r} = 2 \sqrt {k m} \]

When \(c>c_{r}\) the system is called over damped. When \(c<c_r\) the system is called underdamped

The following example runs a simulation showing the eﬀect of changing the damping when the forcing function is a step function. The response to a step function is a standard method used to analyze systems.

ode45 can be used with piecewise function deﬁned for the RHS. For example, given \(x''(t)-x(t)=c\) where \(c=1\) for \(0<=t<1\) and \(c=20\) for \(1<=t<2\) and \(c=3\) for \(2<=t<=3\), the following code example shows one way to implement the above.

function first_oder_ode % SOLVE dx/dt = -3 exp(-t). % initial conditions: x(0) = 0 t=0:0.001:5; % time scalex initial_x=0; [t,x]=ode45( @rhs, t, initial_x); plot(t,x); xlabel('t'); ylabel('x'); function dxdt=rhs(t,x) dxdt = 3*exp(-t); end end

function second_oder_ode % SOLVE d2x/dt2+5 dx/dt - 4 x = sin(10 t) % initial conditions: x(0) = 0, x'(0)=0 t=0:0.001:3; % time scale initial_x = 0; initial_dxdt = 0; [t,x]=ode45( @rhs, t, [initial_x initial_dxdt] ); plot(t,x(:,1)); xlabel('t'); ylabel('x'); function dxdt=rhs(t,x) dxdt_1 = x(2); dxdt_2 = -5*x(2) + 4*x(1) + sin(10*t); dxdt=[dxdt_1; dxdt_2]; end end

function engr80_august_14_2006_2() % shows how to use Matlab to animation response of one degree of % freedom system. % show the effect of changing the damping of the system on the response. % by Nasser Abbasi, UCI. clear all; close all; t_start = 0; t_end = 6; %final time in seconds. time_span =t_start:0.001:t_end; k = 40; % spring stiffness. N/m m = 5; % mass, kg cr = 2*sqrt(k*m); %critical damping fprintf('critical damping coef. of system is %f\n',cr); initial_position = 0; initial_speed = 0; x0 = [initial_position initial_speed]; % Now start the simulation, change damping. for c = 0: .5 : cr+.1*cr [t,x]=ode45(@rhs,time_span,x0); plot(t,x(:,1)); title(sprintf('Critical damping=%4.1f, current damping coeff. =%4.1f',cr,c)); ylim([-.1 .5]); drawnow; pause(.1); end grid %************************************** % solves m x''+ c x' + k x = f(t) %************************************** function xdot=rhs(t,x) xdot_1 = x(2); xdot_2 = -(c/m)*x(2) - (k/m)*x(1) + force(t)/m; xdot = [xdot_1 ; xdot_2 ]; end %******************** % The forcing function, edit to change as needed. %******************** function f=force(t) P = 100; % force amplitude %f=P*sin(omega*t); f=10; %unit step %if t<eps %impulse % f=1 %else % f=0; %end %f=P*t; %ramp input end end

function engr80_august_14_2006() % shows how to use Matlab to animation response of one degree of % freedom system. % by Nasser Abbasi, UCI. clear all; close all; t_start = 0; t_end = 6; %final time in seconds. time_span =[ t_start t_end]; time_span =t_start:0.001:t_end; k = 100; % spring stiffness. N/m c = 20; % damping coeff. N-s/m m = 5; % mass, kg natural_damped_omega = sqrt( k/m - (c/(2*m))^2 ); fprintf('Natural damped frequency of system is %f\n',natural_damped_omega); initial_position = 0; initial_speed = 0; x0 = [initial_position initial_speed]; for omega=0:0.1:natural_damped_omega+0.1 [t,x]=ode45(@rhs,time_span,x0); plot(t,x(:,1)); title(sprintf('forcing freq=%4.1f',omega)); % ylim([-.1 2]); drawnow; % pause(.1); end grid %************************************** % solves m x''+ c x' + k x = f(t) %************************************** function xdot=rhs(t,x) xdot_1 = x(2); xdot_2 = -(c/m)*x(2) - (k/m)*x(1) + force(t)/m; xdot = [xdot_1 ; xdot_2 ]; end %******************** % %******************** function f=force(t) P = 100; % force amplitude f=P*sin(omega*t); %f=10; %unit step %if t<eps %impulse % f=1 %else % f=0; %end %f=P*t; %ramp input end end

%--------------------------------- % %Example solve x''- x = c % %where c=1 for 0<=t<1 % c=20 for 1<=t<2 % c=3 for 2<=t<=3 % %IC x = 0, t = 0 % x'= 1, t = 0 % %-------------------- function ode45_with_piecwise() t = 0:0.1:3; % time scale initial_x = 0; initial_dxdt = 1; [t,x] = ode45( @rhs, t, [initial_x initial_dxdt] ); plot(t,x(:,1)); xlabel('t'); ylabel('x'); end %-------------------- %ode45 rhs function dxdt=rhs(t,x) dxdt_1 = x(2); dxdt_2 = x(1)+ 1*((0<=t)&(t<1)) + 20*((1<=t)&(t<=2))+ 3*((2<=t)&(t<=3)); dxdt = [dxdt_1; dxdt_2]; end