ilaplace returns unexpected answer

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.

Answer 1

[David Goodmanson]

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.

[Abdulla Al Blooshi]

I appreciate your simple and systematic answer. now I understand the issue :D

Regards,

[Paul]

David,

I don't understand what you're saying "can't be." What exactly are you saying is a contradiction?

[David Goodmanson]

Hi Paul,

Yes, I might have said that better. The integral in the second case is from t = -6 to inf, but by the definition of the laplace transform the region of integration is t >= 0. So the section of the integration from -6 to 0 is impossible. The same objection doesn't arise in the first case, where integration from 6 to inf does not conflict with t >= 0.

[Paul]

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]

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)