I want to perform inverse Laplace transform but my code does not seem to work

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Carlos Puente 2023-10-27

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https://ww2.mathworks.cn/matlabcentral/answers/2039326-i-want-to-perform-inverse-laplace-transform-but-my-code-does-not-seem-to-work

评论： Paul 2023-10-27

I am trying to perform inverse Laplace transform the following way, but I am getting an answer that is not satisfying.

syms s K tau a
fun2 = ((60*K*(exp(-a*s)))/tau)/(s*(s + 1/tau));
y2 = ilaplace(fun2);
pretty(y2)
          /   exp(-a s)         \
K ilaplace| -------------, s, t | 60
          |   /      1  \       |
          | s | s + --- |       |
          \   \     tau /       /
------------------------------------
                 tau

I am getting the answer that code shows, but it is not helpful as it does not perform the Laplace inverse transform. I do not know how to solve the problem.

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Answer 1

Paul 2023-10-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2039326-i-want-to-perform-inverse-laplace-transform-but-my-code-does-not-seem-to-work#answer_1341611

在 MATLAB Online 中打开

Hi Carlos,

ilaplace assumes that all signals are causal, which wouldn't be true in this case if a < 0.

Add the appropriate assumption, and ilaplace returns a closed-form expression.

syms s K tau a

assume(a >= 0)

fun2 = ((60*K*(exp(-a*s)))/tau)/(s*(s + 1/tau))

fun2 =

y2 = ilaplace(fun2)

y2 =

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Paul 2023-10-27

@Carlos Puente

You may be more familiar with the term unit step, or step, function, rather than Heaviside function.

It shows up in y2 as a result of the time shift property of the unilateral Laplace transform:

f(t-a)*u(t-a) -> exp(-a*s)*F(s) , a > = 0

u(t) is the same as heaviside(t)

@Walter Roberson

y2(a) = 0 regardless of the value of heaviside(0).

Paul 2023-10-27

在 MATLAB Online 中打开

Also, you have to be careful with Matlab if you want to apply the time shift property by hand because ilaplace does not always tack on the heaviside to the result like it should (others may have different opinion).

For example

syms s K tau t a

assume(a >= 0)

fun2(s) = ((60*K)/tau)/(s*(s + 1/tau))

fun2(s) =

y2(t) = ilaplace(fun2)

y2(t) =

Now, if we want to find the ilapalce of exp(-a*s)*fun2(s), it's tempting to just shift y2(t) to the right by a

y2(t-a)

ans =

whch is incorrect. Instead, we have to tack on the heaviside to the y2(t) first and then shfit

y2(t) = y2(t)*heaviside(t);

y2(t-a)

ans =

which is the same result as above (after canceling tau in result above).

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