LaPlace Transform with initial conditions

I am having a hard time using MATLAB to solve LaPlace transforms.

I need to solve

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

Answer 1

[Star Strider]

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)

Eqn(t) = 

LEqn = laplace(Eqn)

LEqn = 

LEqn = subs(LEqn, {laplace(y(t),t,s), y(0), D1y(0)}, {Y(s), 2, 0})

LEqn = 

LEqn = isolate(LEqn, Y(s))

LEqn = 

y(t) = ilaplace(rhs(LEqn))

y(t) = 

figure
fplot(y, [0 2.5])
grid

Experiment with my code to understand how it works.

.

[Paul]

Shoud the RHS of Eqn be: 14*2*heavisde(t)?

[Walter Roberson]

The u function involved is some constant function, not heaviside. The initial conditions say that u(t)=2 not u(0)=2. Heaviside does not have a strict definition at 0, with u(0)=0 and u(0)=1 and u(0)=1/2 all having their uses, so it would be pretty unusual but not strictly wrong to say u(0)=2. But we are not told u(0)=2, we are told u(t)=2, which includes u(-1) and u(0) and u(1) all = 2. Not heaviside.

[Paul]

The problem statement says that "u(t) = 2." The problem statement also says to solve the equation via the Laplace transform, which typically is the one-sided transform, and certainly is in Matlab's laplace() function, which implies the input is zero for t < 0-. All in all within the context of the problem statement, I'd say it's safe to assume that the statement "u(t) = 2" really means u(t) = 2*heavisde(t). Only the OP will know for sure.

[Kyle Langford]

You guys are awesome. I haven't played any laplace stuff in MATLAB like heaviside or laplace/ilaplace. I am still a little shakey on how to correctly use subs, especially when subbing multiple things.

I need to do some more MATLAB classes haha.

I kept trying it by making an equation with a @(t) handle, which was not working.

This is extremely helpful. Though, I am not 100% confident I could repeat this on my own, I will play with this and apply it to some other problems.

[Kyle Langford]

Would this be an acceptable way of solving this? By not using heaviside?

I know this is more of a math question vs MATLAB.

clear;clc;
disp('Problem 1')

Problem 1

syms s t y(t) Y(s) u(t) U(s)
d1y=diff(y);
d2y=diff(d1y);
eq1=d2y+7*d1y+12*y==14*u

eq1(t) = 

Leq1=laplace(eq1)

Leq1 = 

Leq1=subs(Leq1,{laplace(y(t),t,s),y(0),d1y(0), laplace(u(t),t,s)},{Y(s),2,0,2})

Leq1 = 

Leq1=isolate(Leq1,Y(s))

Leq1 = 

y(t)=ilaplace(rhs(Leq1))

y(t) = 

fplot(y,[0 2.5])
grid on

[Star Strider]

Yes!

(I usually interpret as the unit step or Heaviside function, however here that appears not to be correct.)

.

[Paul]

What is the justification for the substitution U(s) = 2?

[Kyle Langford]

@Paul I guess you are right. u(t)=2, not U(s)=2

This seems to be the (or a) solution to this problem.

I have been trying to maniupulate MATLAB to give a similar solution, but I cannot get there.

[Star Strider]

The original Laplace transform:

does not appear to contain u.

I don’t understand where it came from.

If it’s just a constant:

syms s t u y(t) Y(s) 
D1y = diff(y);
D2y = diff(D1y);
Eqn = D2y + 7*D1y + 12*y == 14*2*u;
LEqn = laplace(Eqn);
LEqn = subs(LEqn, {laplace(y(t),t,s), y(0), D1y(0)}, {Y(s), 2, 0});
LEqn = isolate(LEqn, Y(s))

LEqn = 

y(t) = ilaplace(rhs(LEqn))

y(t) = 

y(t) = expand(y(t))

y(t) = 

That’s likely as close as it’s possible to get to the intended result.

.

[Kyle Langford]

Looks good enough to me. I appreciate the knowledge of laplace in MATLAB!

[Star Strider]

Thank you!

It’s much more useful now than when it was introduced.

[Paul]

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)

eq1(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 = 

Leq1=isolate(Leq1,Y(s))

Leq1 = 

syms u real % reusing the same variable name, unfortunately
Leq1 = subs(Leq1,U(s),laplace(u*heaviside(t)))

Leq1 = 

Leq1 = partfrac(Leq1,s)  % same as shown in the posted solution

Leq1 = 

y(t) = ilaplace(rhs(Leq1))

y(t) = 

Which is the same as the posted solution. I would write it as

y(t) = ilaplace(rhs(Leq1))*heaviside(t)

y(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.