Symbolic package: Difference between revisions
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[-0.777688831121563, -0.7776888311215626] | [-0.777688831121563, -0.7776888311215626] | ||
[0.22911205809043574, 0.2291120580904359] | [0.22911205809043574, 0.2291120580904359] | ||
</source> | |||
* '''Demo of ODE with a step input and initial conditions.''' | |||
<source lang="octave"> | |||
## This is a demo of a second order transfer function and a unit step input. | |||
## in laplace we would have 1 1 | |||
## _______________ * _____ | |||
## s^2 + sqrt(2)*s +1 s | |||
## | |||
## So the denominator is s^3 + sqrt(2) * s^2 + s | |||
# and for laplace initial conditions area | |||
## t(0)=0 t'(0) =0 and the step has initial condition of 1 | |||
## so we set t''(0)=1 | |||
## In the code we use diff(y,1)(0) == 0 to do t'(0)=0 | |||
## | |||
## I know that all this can be done using the control pkg | |||
## But I used this to verify that this solution is the | |||
## same as if I used the control pkg. | |||
## With this damping ratio we should have a 4.321% overshoot. | |||
## | |||
syms y(x) | |||
de =diff(y, 3 ) +sqrt(2)*diff(y,2) + diff(y) == 0; | |||
f = dsolve(de, y(0) == 0, diff(y,1)(0) == 0 , diff(y,2)(0) == 1) | |||
ff=function_handle(rhs(f)) | |||
x1=0:.01:10; | |||
y=ff(x1); | |||
plot(x1,y) | |||
grid minor on | |||
</source> | </source> |
Revision as of 12:47, 6 July 2015
The symbolic package is part of the octave-forge project.
Demos and usage examples
- Demo of Anonymous function to symbolic function and back to anonymous function and then the use of the interval pkg.
% this is just a formula to start with,
% have fun and change it if you want to.
f=@(x) x.^2 +3*x-1 + 5*x.*sin(x);
% the next 2 line take the Anonymous function into a symbolic formula
syms x;
ff=formula(f(x));
% now calculate the derivative of the function
ffd=diff(ff);
% and convert it back to an Anonymous function
df=function_handle(ffd)
% this uses the interval pkg. to find all the roots between -15 an 10
fzero (f, infsup (-15, 10), df)
ans ⊂ 4×1 interval vector
[-5.743488743719015, -5.743488743719013]
[-3.0962279604822407, -3.09622796048224]
[-0.777688831121563, -0.7776888311215626]
[0.22911205809043574, 0.2291120580904359]
- Demo of ODE with a step input and initial conditions.
## This is a demo of a second order transfer function and a unit step input.
## in laplace we would have 1 1
## _______________ * _____
## s^2 + sqrt(2)*s +1 s
##
## So the denominator is s^3 + sqrt(2) * s^2 + s
# and for laplace initial conditions area
## t(0)=0 t'(0) =0 and the step has initial condition of 1
## so we set t''(0)=1
## In the code we use diff(y,1)(0) == 0 to do t'(0)=0
##
## I know that all this can be done using the control pkg
## But I used this to verify that this solution is the
## same as if I used the control pkg.
## With this damping ratio we should have a 4.321% overshoot.
##
syms y(x)
de =diff(y, 3 ) +sqrt(2)*diff(y,2) + diff(y) == 0;
f = dsolve(de, y(0) == 0, diff(y,1)(0) == 0 , diff(y,2)(0) == 1)
ff=function_handle(rhs(f))
x1=0:.01:10;
y=ff(x1);
plot(x1,y)
grid minor on