ilaplace returns unexpected answer

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Abdulla Al Blooshi
Abdulla Al Blooshi 2020-11-6
评论： David Goodmanson 2020-11-17
I am having trouble understanding MATLAB's unability to find the inverse Laplace of delayed step response ( heaviside()).
For example, notice in the below code, the coefficient of "s" is "-3".
syms t s ;
f3=(exp(-6*s))/(s-4)
F=ilaplace(f3,s,t);
pretty(F);
which returns a fine answer of
heaviside(t - 6) exp(4 t - 24)
But, when I change the coefficient of “s” to “3” ( positive 3), I get the below answer.
/ exp(6 s) \
ilaplace| --------, s, t |
\ s - 4 /
what could be the problem! the hand written answer is heaviside(t + 6) exp(4 t + 24)
Appreicate the help.

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David Goodmanson
David Goodmanson 2020-11-10
编辑：David Goodmanson 2020-11-10
Hi Abdulla,
You mention +- 3 as the coefficient, but are using +-6 in the actual example so I will stay with that. Looking at things the other way around, suppose s real for convenence, and replace 4 by n for generality. Then
Integral{0,inf} heaviside(t-A)*exp(-s*t)*exp(n*t) dt
= Integral{A,inf} exp(-s*t)*exp(n*t) dt
= (1/(s-n))*exp(-s*A)*exp(n*A) s > n A >= 0
The function exp(-6*t) has A = 6, giving heaviside(t-6). The integral is from 6 to inf, which works.
On the other hand, exp(6*t) has A = -6, giving heaviside(t+6) as you noted. But this implies
Integral{-6,inf} exp(-s*t)*exp(n*t) dt
which can't be, because the lower limit for t is 0. ilaplace doesn't know what to do, and punts.
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Paul
Paul 2020-11-14
Setting aside what ilaplace does and just continuing with a math discussion ....
To derive your result I start with the one-sided Laplace transform, which is defined with limits of integration from t = 0 to t = inf (or from t = 0-epsilon to t inf if we care about Dircal delta functions). To evaluate that integral we use a change of variables that leads to your expression:
Integral{A,inf} (etc.). (with A < 0 for the case of interest)
But I started with the correct definition of the one-sided Laplace transform, so there must be some reason other than the defintion of the one-sided Laplace transform as to why that resulting integral can't be evaluated.
One possibility is that the one-sided Laplace transform is only defined for functions that themsevles are only defined over the domain t >= 0. In this case, the integrand is undefined (for A < 0).
However, is that limitation of the domain fundamentally required? Can the one-sided Laplace transform apply to functions that are defined over the domain -inf < t < inf, but limited to functions that must satisfy f(t) = 0 for t < 0? If this were allowed, would the integrand be well-defined?
A few references I've checked all state that f(t) is only defined for t>=0. But I don't know if that's just because those are the class of functionst that we typically care about, or if the whole process breaks down if we extend the domain to t<0 with the constraint f(t<0) = 0.
David Goodmanson
David Goodmanson 2020-11-17
Hi Paul,
For another point of view It makes sense to calculate the inverse laplace transform, which involves an integral in the complex plane.
In what follows we have the lemma
h(t-c)*h(t-d) = h(t-max(c,d)) a)
With the one sided laplace transform, you start with the agreement that the integral in the time domain is 0 to inf. That's the same as saying that the function being transformed is h(t)*f(t). The function h(t)*f(t) can be zero for awhile after t = 0, but if can't spill into t <0.
For A > 0, the laplace transform of
h(t-A)*exp(n*t)
is
(1/(s-n))*exp(-s*A)*exp(n*A) s > n A >= 0
and doing the explicit inverse laplace transform goes right back to
h(t-A)*exp(n*t) = h(t)*h(t-A)*exp(n*t) % from a)
This is all in accord with everything happening for t>0.
For A < 0 the inverse laplace procedure applied to
(1/(s-n))*exp(-s*A)*exp(n*A)
still produces
h(t-A)*exp(n*t) A < 0
as you might expect. But this function has nonzero values for t < 0, and the inverse laplace transform does not meet the requirements. Also, saying
h(t-A)*exp(n*t) = h(t)*h(t-A)*exp(n*t) A < 0
does not work since
h(t)*h(t-A) = h(t) A < 0 % from a)

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