LaPlace Transform with initial conditions
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Kyle Langford
el 27 de Mzo. de 2022
I am having a hard time using MATLAB to solve LaPlace transforms.
I need to solve ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/943164/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/943164/image.png)
I don't really understand how to write the derivatives in MATLAB, or set the initial conditions using built ins. Or are there built ins?
2022a
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Star Strider
el 27 de Mzo. de 2022
The procedure is not exactly straightforward, so I will demonstrate a working approach —
syms s t y(t) Y(s)
D1y = diff(y);
D2y = diff(D1y);
Eqn = D2y + 7*D1y + 12*y == 14*heaviside(2)
LEqn = laplace(Eqn)
LEqn = subs(LEqn, {laplace(y(t),t,s), y(0), D1y(0)}, {Y(s), 2, 0})
LEqn = isolate(LEqn, Y(s))
y(t) = ilaplace(rhs(LEqn))
figure
fplot(y, [0 2.5])
grid
Experiment with my code to understand how it works.
.
12 comentarios
Star Strider
el 29 de Mzo. de 2022
Thank you!
It’s much more useful now than when it was introduced.
Paul
el 29 de Mzo. de 2022
Editada: Paul
el 29 de Mzo. de 2022
The posted solution in this comment is a peculiar approach. The first equation for Y(s) seems o.k. with the u(t) = 2 being expressed more formally as u(t) = 2*heaviside(t) and U(s) = 2/s, which is why we see the 2 in the numerator and s in the denominator on the first term on the RHS. But in the next step it looks like the 2 in the numerator is replaced with a symbolic constant, also called u (with the expectation that eventually u will be replaced with 2 as stated in the parenthetical). So in the end, the posted solution is the response of the system with the initial conditions y(0) = y0, ydot(0) = 0, and input u(t) = u*heaviside(t).
syms s
syms t y0 real
syms y(t) Y(s) u(t) U(s)
d1y=diff(y);
d2y=diff(d1y);
eq1=d2y+7*d1y+12*y==14*u(t)
Leq1=laplace(eq1);
Leq1=subs(Leq1,{laplace(y(t),t,s),y(0),d1y(0), laplace(u(t),t,s)},{Y(s),y0,0,U(s)})
Leq1=isolate(Leq1,Y(s))
syms u real % reusing the same variable name, unfortunately
Leq1 = subs(Leq1,U(s),laplace(u*heaviside(t)))
Leq1 = partfrac(Leq1,s) % same as shown in the posted solution
y(t) = ilaplace(rhs(Leq1))
Which is the same as the posted solution. I would write it as
y(t) = ilaplace(rhs(Leq1))*heaviside(t)
The steps in the posted solution are very strange, IMO, to first use the explicit expression u(t) = 2*heaviside(t), based on the problem statement, and then subsequently switch to the general u(t) = u*heaviside(t), all the while only ever using the symbolic constant for the initial condition.
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