# Physical Interpretation of Laplace's Transform

1. Aug 22, 2013

### muzialis

Hi all,
I am trying to hard to understand integral's transform.
While an interpretation of Fourier transform is relatively easy to furnish in terms of signal decomposition and harmonics, it seems the "meaning" of Laplace transfrom is more subtle (in spite of the similarities between the two).
I have then been through the derivations.
In Fourier series a function of period 2π is expressed as a series of sines and cosines
$$f(x) = \sum_{k=0} ^{\infty} a_n cos (nx) + b_n sin (nx)$$
The coefficient a_n and b_n are then found using the orthogonality of functions such as sin(nx), cos(nx).
Considering an aperiodic function one considers the limit of the period to infinity and demonstrates the series converges to an integral.
All this makes a physical meaning very clear: the functions is decomposed onto a basis of harmonic functions.
Why is not the same approach followed for Laplace transforms?
One could start by
$$f(x) = \sum_{k=0} ^{\infty} a_n e^{b_n x}$$
and try to follow the same approach.
One really would need to have some information on a_n in order to write in the limit the proper integral, for I am not so sure all I am saying make any sense.. If it did, the Laplace transform could be seen as a decomposition on an expnentail basis (by the way, I remember reading somehwere that the exponentials form a complete basis for continous functions, is this true?)
Am I missing something here? Can one follow this approach, and conclude the Laplace transform is the integral representation of a series of exponentials?
Many thanks

2. Aug 22, 2013

### saminator910

An important identity is $e^{i\theta} = cos(\theta)+i sin(\theta)$
As you know the Laplace transform changes a function of $t$ into a function of $s$,which is a complex variable. Essentially, as I see the Laplace transform as the Fourier transform with a real component.

Fourier Transform $\hat f = \displaystyle\int_ℝ f(x)e^{-i\omega x}\,dx$

Laplace Transform $F(s) = \displaystyle\int_{0}^{∞} f(t)e^{(-σ-i\omega)t}\,dt$ where $s=\sigma+i\omega$

Last edited: Aug 22, 2013
3. Aug 23, 2013

### muzialis

Saminator910, thanks for your post. What you say makes perfect sense to me, I am trying to work out the details on how this relates to my original questions. thanks

4. Sep 5, 2013

### bolbteppa

The question of interpreting Fourier & Laplace transforms is something I'm struggling with as well.

This video:

makes a great point. If we interpret the $f$ in $f(x)$ as a function of time, then the fourier transform of $f$ takes the representation of this function in terms of time to a new representation of $f$ simply in terms of frequency, & this makes sense intuitively since a function can be represented, via Fourier series, as a http://canopus.mogi.bme.hu/letoltes/SYSTEMTECHNIK/lc-trig.pdf [Broken], & because any sine function is completely determined by it's amplitude, frequency & phase we should be able to completely characterize the function represented in terms of time by a new function in terms of frequency - if we know the amplitude & phase at every frequency in our sum of sines (& this is found by integrating over all time, across the entire spectrum, in the time representation). From this perspective the utility of complex numbers leaps out since amplitude & phase can be represented in a single complex number...

In other words, take a function, represent it as it's fourier series, decompose the coefficients into of the sines & cosines into their integral representations, do the algebra to reduce this to the Fourier transform representation, then if you wish represent this in terms of complex numbers - this decomposes a function represented in terms of time to a new representation in terms of frequency. This whole process can also be achieved by "pre-multiplying" the function by a complex exponential with a purely imaginary argument and integrating.

From this perspective the Laplace transform is merely just the Fourier transform where the complex exponential also has a real argument:

But I just do not see how that extra step falls out of the development I've written above, I don't see where the real argument rears it's ugly head.

The scary thing is that there's a completely different approach to the Laplace transform, thinking of it as the continuous analogue of a power series:

From this perspective, think of the function $a : \mathbb{N} \rightarrow \mathbb{R}$ as a function giving the coefficients of the taylor series of a function $f$ in $f(x) = \sum _{i=0}^\infty a(n)x^n$. In other words the 'sequence' $a$ is a discrete representation of the function $f$. What is the continuous analogue? Can we use this to find a continuous representation for the function $f$? We basically want:

$$f(x) = \sum _{i=0}^\infty a(n)x^n \rightarrow f(x) = \smallint_0^\infty a(t) x^t dt$$

Then applying 'cosmetic' adjustments like $x = e^{\ln(x)}$, $x^t = e^{t\ln(x)}$ & $\ln(x) = - s$ imply

$$f(x) = \sum _{i=0}^\infty a(n)x^n \rightarrow f(x) = \smallint_0^\infty a(t) e^{-st} dt = F(s)$$

So every time you apply the Laplace transform to a function, you can interpret it as saying that we are saying that this function, restricted to lying over $\mathbb{N}$, represents the Taylor series of some other function, & the continuous representation we've been given is merely a tool we're using in finding a continuous representation of that other function via the Laplace transform, though why you can just magically go from a discrete function $a(n)$ to a continuous function $a(t)$ is beyond me.

That's where I am, a mass of confusion & half-truths, how would you guys push what I've written above? I can't merge it all together in a cogent representation... I think there's something fundamental unifying all of this but I don't see it yet.

Note the Laplace transform video makes some comments relating to the op's sentence

"Can one follow this approach, and conclude the Laplace transform is the integral representation of a series of exponentials?"

However I've shown that the Laplace transform is like a shift in the 'real-direction' of the Fourier integral representation of a function (which is a series of exponentials), so I think it's more complicated.

Last edited by a moderator: May 6, 2017