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Trying to do a Laplace transform on a discontinuous function

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Emily Friedman
Emily Friedman 2023-9-5
编辑： Paul 2023-9-5
Ran in:
I am trying to write code to solve g(t).
I rewrote the function as g(t) = 4 + 5*(t-2)*e^(t-2)*u(t-2).
In MATLAB,
syms t
oldVal = sympref("HeavisideAtOrigin",4);
eqn = 4 + 5*(t-2)*exp(t-2)*heaviside(t)
eqn = 
L = laplace(eqn)
L = 
Though this runs, it doesn't seem right to me.
What am I doing wrong?
  1 个评论
Paul
Paul 2023-9-5
编辑：Paul 2023-9-5
Recheck the code for eqn. It doesn't match how g(t) was rewritten.
Also, the sympref isn't really necessary. Try different values of HeavisideAtOrigin and see if you get different results for L.

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回答（1 个）

Star Strider
Star Strider 2023-9-5
Ran in:
I initially wanted to see if piecewise would work. It didn’t.
This is the result I get using heaviside to define the areas of interest, and then combine them into one expression —
syms s t
g_1(t) = 4*(heaviside(t)-heaviside(sym(t-2)))
g_1(t) = 
g_2(t) = heaviside(sym(t-2))*(4+5*(t-2)*exp(t-2))
g_2(t) = 
G(s) = laplace(g_1) + laplace(g_2)
G(s) = 
G(s) = simplify(G,500)
G(s) = 
figure
fplot(g_1, [-1 5])
ylim([0 50])
title('g_1(t)')
figure
fplot(g_2, [-1 5])
ylim([0 50])
title('g_2(t)')
figure
fplot(g_1+g_2, [-1 5])
ylim([0 50])
title('g_1(t)+g_2(t)')
The time-domain function appears to be reasonable, so I assume the Laplace transform is as well.
.

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