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QUESTION: Hello again.

My question this time isn't a direct mathematics question, but I feel it's most suited to be answered by a mathematician.

As you know the Laplace transform require the function to be exponentially bounded to exist. I am trying to understand the implication behind it in terms of frequency domain.

I mean physically I haven't seen any exponentially unbounded signals in the electrical engineering applications, event in unstable systems it would be the poles of the transfer function on the right plane, but I am trying to get an idea of the frequency behavior of exponentially unbounded signals.

What does it mean that the laplace transform does not exist. I mean what will physically happen if sent those signals and studied magnitude vs w the frequency ?

The issue I see is the output signals will always be unstable even in a stable system but stability isn't really a mathematical definition so there should be a mathematical interpretation right?

let's just take the basic system like

Y(s)/X(S) =K/(s)

Y(s) = K X(s)/(s)

now if X(s) is like exp(t^2)

y(t) would be integral from 0 to t exp(t^2) which isn't an analytically function but it exists however what about

Y(s) ?

will there be any mathematical indication of *something going wrong* ?

I mean so far to what I have studied such thing isn't physically realizable, so I am curious.

ANSWER: I am not sure if I understand the phrasing of your question, so I'll try to answer it and explain what I think you are asking along the way. You should follow up if you need to clarify the question at all.

First, the Laplace transform is defined as:

L(s) = lim [R→∞] ∫ [0 to R] f(t) e^(-st) dt

For example, L(0) would give the improper integral ∫ [0 to ∞] f(t) dt.

Now, you are talking about "requiring the function to be exponentially bounded." What you are talking about (I think) is the question: for which values of s (if any) does L(s) exist? We typically consider the domain for s to be the complex numbers, so there is some set of values of s where L(s) is a convergent limit (i.e. the integral converges at infinity).

Generally, the first criterion for f to have any reasonable Laplace transform is that it is locally integrable -- that it can be integrated at all on a finite interval. If not, then the notion of integrating f (whether or not R goes to ∞) is not necessarily useful.

It also appears to be true that f(t) e^(-st) → 0 as t → ∞ . Generally this means that f(t) grows (worst case) slower than e^(st).

So the reason, to be blunt, that a function must be exponentially bounded to take the Laplace transform is that -- by the definition of the L.T. -- L(s) would be infinite for all values of s if f(t) grew super-exponentially.

I do not believe stability is relevant to the discussion, although I feel inclined to point out that stability has several concrete, rigorous, meaningful definitions in mathematics: https://en.wikipedia.org/wiki/Stability

Are there electrical signals that correspond to functions that are not integrable / transformable by the Laplace transform? I don't know. That sounds almost like a philosophy question, because it assumes that an electrical signal f(t) could be arbitrarily large and/or go for a infinite length of time -- neither of which might be physically possible, depending on how you feel about some philosophical or cosmological questions.

Mathematically speaking, nothing has "gone wrong."

A "transform" is, in mathematical terms, a function that takes one function (in our case, f(t)) and gives you another function (for us, L(s)).

The "domain" of the L.T. will be a set of functions -- much like the domain of any other scalar or vector function. In fact, the L.T. is a linear transformation on a set of vectors (functions can comprise vector spaces). But, like the square root function (or many other functions), the natural domain of the L.T. is not "all possible functions" but a subset of functions -- the ones that are exponentially bounded. (You can prove pretty easily this is a subspace of the set of all functions.)

So what this means, practically speaking, is that the domain of the Laplace transform only includes those functions that we care about anyway. And you can also start talking about L.T. functions that converge only in certain regions in the complex plane, etc. etc., which gives more restrictions on the possible functions f(t). You can see more discussion of this on Wikipedia: https://en.wikipedia.org/wiki/Laplace_transform#Region_of_convergence

But nothing has "gone wrong." We just have some limits that don't converge.

---------- FOLLOW-UP ----------

QUESTION: That sounds almost like a philosophy question, because it assumes that an electrical signal f(t) could be arbitrarily large and/or go for a infinite length of time -- neither of which might be physically possible, depending on how you feel about some philosophical or cosmological questions.

Well Yes I think I understand my question better now. what I am trying to ask is if such signals did exist, how would mathematics expect them to operate with respect to frequency, and would that be how they would actually physically operate.

I mean the difference between laplace and square root is laplace also models how the function acts with changing frequency which is a physical concept right?

But since as you said they cannot exist in our universe we cannot test it right? that's what you are saying?

The restriction on functions for a Laplace transform requires integrating on an infinite interval of "t" which is time in this case. Even the presumption that a real-life quantity could be unbounded for all time is physically meaningless from most scientific perspectives.

If such a function could truly exist, or if somehow such a function required analysis (regardless of whether it exists), the Laplace transform would not be defined at any values of s. What it means for the frequency domain is that for this function f(t), L(s) has empty domain (no valid input). What that means for physics is anybody's guess.

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Comment | Thank you very much~ I asked at my university our math instructors, but they weren't even interested in thinking about the question :/ |

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