# question for the possible application of the laplace transform

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I have a question in regards to the usage of the Laplace transform.

( Please note that I only have a single class on the understanding of the Laplace transform so I have surface level confidence in its understanding. )

Would it be possible, hypothetically, to store information using the Laplace transform to change signal data into a signal representation and then stacking signals over and over again until you have a larger and more "complex", using the term complex as a way to say that it would be a very large amount of signals stacked on top of each other, signal. Which you can then use the inverse Laplace transform to select certain harmonic levels to retrieve original signal data?

Below is a picture to visually represent what I am thinking of.

Thank you for taking the time out of your day to assist me in this question

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Posted (edited)
2 hours ago, ALine said:

Would it be possible, hypothetically, to store information using the Laplace transform to change signal data into a signal representation and then stacking signals over and over again until you have a larger and more "complex", using the term complex as a way to say that it would be a very large amount of signals stacked on top of each other, signal. Which you can then use the inverse Laplace transform to select certain harmonic levels to retrieve original signal data?

Below is a picture to visually represent what I am thinking of.

Yes indeed the Laplace and Fourier transforms are related and very similar types of integral transform.

You are effecteively asking if the 'composed time varying signal' in your diagram can be processed or decomposed into simpler signals.
This is the question also answered  (task accomplished) by the Fourier Transform which decomposes into simusoids.

Note these simpler signals are periodic in nature.
Non periodic signals may well not converge when the integral is taken.
So if there periodic signals embodied in your 'composed signal' the integral proabably will not converge ie the method will not work.

.
The Laplace transform is actually more powerful in this respect since you can take (and untake) the Laplace tranform of allsorts of peridoc and other functions such as square waves, step functions, sawtooth waves and staircase functions.

The 'formula' for the Laplace transform of a periodic function f is

$L\left( f \right) = \frac{{\int_0^p {{e^{ - st}}} f\left( t \right)dt}}{{1 - {e^{ - ps}}}}$

so for the square wave shown

$L\left( f \right) = \frac{{\int_0^2 {{e^{ - st}}} f\left( t \right)dt}}{{1 - {e^{ - 2s}}}}$

$= \frac{{\int_0^1 {{e^{ - st}}} dt}}{{1 - {e^{ - 2s}}}}$

$= \frac{{1 - {e^{ - s}}}}{{s\left( {1 - {e^{ - 2s}}} \right)}}$

$= \frac{1}{{s\left( {1 + {e^{ - s}}} \right)}}$

Which may be useful if your input data is digital.

Does this help?

Edited by studiot

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yep, I do not quite fully understand it, however I kind of get the gist of it.

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11 hours ago, ALine said:

yep, I do not quite fully understand it, however I kind of get the gist of it.

You might find this discussion interesting and useful

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So the Fourier transform is like a subset of the Laplace transform, in where it only deals with sinusoidal periodic functions and where Laplace transforms deal with all kinds of continuous functions? Am I getting that correct?

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