# What is the physical meaning of a Laplace Transform?

Today I have learned what a Laplace Transform is and how it is defined. I have heard that it has wide applications in Physics/Engineering and so I started to wonder. I started with probably the simplest function in all of physics: distance formula in a straight line movement with no acceleration:

$$d(t) = vt$$ Where $d$ stands for distance, $v$ for velocity, and $t$ for time.

Taking Laplace Transform of this function leads to:

$$\mathscr{L}\{d(t)\}(s) = \frac{v}{s^2}$$

Now as the $s$ shows up in the exponential in the definition of the Laplace Transform along with $t$ it seems to me that it has to have dimensions of $1/\text{time}$ for it to make any physical sense. This would suggest that $s$ is some sort of frequency, but I cannot understand what this new function tells us about the movement.

I understand that perhaps I'm using this transform in a wrong way/place and it has no real physical in that particular case. Therefore I would like to ask if there is any physical meaning in what I have done? If yes, what meaning? If no, why?

Frequency Space and Fourier Transforms

The Laplace transform of a function $$f(t)$$ for $$t>0$$ is defined as,

$$F(s) = \mathcal L\{f(t)\}(s) = \int_0^\infty e^{-st}f(t) \, dt$$

for a complex variable $$s$$ with - as you correctly stated - the interpretation of a frequency. This in turn is related to the Fourier transform, if integrated over $$(-\infty, \infty)$$, for $$s = i\omega$$. However, note that generally substituting $$F(i\omega)$$ will not yield the Fourier transform $$\hat f(\omega)$$ as $$i\omega$$ is often a pole, reflecting the fact that the Fourier transform may have a delta function.

The Fourier transform of some signal $$f(t)$$ gives us a prescription on how to weigh different sinusoids of varying frequencies that make up the signal $$f(t)$$. Thus, the Laplace transform has the same interpretation, but instead we are interested in weights for representing it as a sum of exponentials.

Continuous Analogue of a Power Series

There is a mathematical motivation of the Laplace transform which nicely demystifies it further. To be specific, consider a function $$f(x)$$ with a power series representation,

$$A(x) = \sum_{n= 0 }^\infty a_n x^n.$$

We could ask the question: what is the relation between $$a_n$$ and $$A(x)$$? The answer is you perform the discrete sum. For example, $$a_n = \frac{1}{n!}$$ corresponds to $$A(x) = e^x$$.

We now ask, what is the continuous analogue of this prescription? Let us now consider instead of a discrete $$a_n$$, function $$a(t)$$ with $$t\in [0,\infty)$$ and instead of a sum we get an integral,

$$A(x) = \int_0^\infty a(t) x^t \, dt.$$

Thus to the continuous 'set of coefficients' $$a(t)$$, we can associate an $$A(x)$$. This integral has the best chance of converging if $$x < 1$$, as we are essentially taking higher powers as we sum. This motivates the substitution, $$x := e^{-s}$$ and we recover the Laplace transform,

$$A(s) = \int_0^\infty a(t)e^{-st} \, dt.$$

To convince yourself of this, take the Laplace transform,

$$\mathcal{L}\{\sin t\} = \frac{1}{1+s^2}.$$

$$\int_0^\infty (\sin t )x^t \, \mathrm dt = \frac{1}{1+\log^2 x }$$

which is defined for $$x < 1$$, substituting $$x= e^{-s}$$ would recover the Laplace transform.

• Is there any reason why the two transforms have two different integration intervals? The Laplace transform would not converge for $t\to - \infty$ but the Fourier transform from $0$ to $\infty$ should be well defined. – FrodCube Dec 22 '16 at 18:14
• Thanks for the answer as it nicely fits in with JMLCarter's. The motivation behind the series as a 'continous power series' is extremely helpful as the Laplace Transform came completely out of blue as a kind of black box. Now it seems much more logical. – MeyCJey Dec 22 '16 at 22:57
• @MeyCJey Well yes, JMLCarter's answer fits with mine as his explanation of it as weighed decompositions in terms of exponentials is basically my first paragraph. And thanks, I'm glad the power series interpretation helped you. – JamalS Dec 22 '16 at 23:20
• yes the answers agree – JMLCarter Dec 23 '16 at 0:12

My experience was that Laplace is typically taught as a toolkit, not explained. Which is a massive loss.

Laplace is a transform, allowing a function to be mapped to another like the usually more familiar Fourier transform. Whereas Fourier maps amplitude(time) into amplitude(frequency), using sin/cosine functions; Laplace is targeted at a function with a special property in calculus, $$f(x) = e^x$$

The special property is that "it is its own derivative", i.e.

$$de^x/dx = e^x$$

That is the rate of change/gradient at any value x is the value f(x).

Just as in Fourier the transformed domain is the input of the transforming function (the frequencies of a sine function),

• the collection of all values of s define the set of exponential functions that could be present as components in the original function,
• and the transformed function describes the actual magnitudes with which they are present in the original function.

The self-derivative property means that if you can transform a function using Laplace and the resultant function can usually be manipulated easily by calculus. Then it can be transformed back again to get the derivative or integral of the original function. It was a new toolkit for solving differential equations at the time.

It's rather like the much simpler log functions, if you take the $log_{10}$ of a number to multiply it by 10 you just need to add one; because the log function has special properties with respect to base 10. Only for logs the element undergoing "transformation" is a single number not a function.

• I see. So a Laplace Transform decomposes the function into exponentials, just as a Fourier decomposes into sine/cosine waves. That helped, thanks. – MeyCJey Dec 22 '16 at 22:45
• @MeyCJey ... and it will help even more if you realize the connection between "exponential" and "sine/cosine waves" for complex variables: $e^{iz} = \cos z + i \sin z$. Note, this is true when $z$ is any complex number, not just when it is real! In fact the Fourier transform is just the special case of the Laplace transform when the real part of $s$ is zero, with the notation changed to get rid of a few $i$'s. – alephzero Dec 23 '16 at 2:21
• As long as you don't start thinking that it's only exponential functions that can be used to create transforms. That part of the technique is quite general. – JMLCarter Dec 23 '16 at 2:44
• This answer implies that the Fourier transform does not take advantage of the fact that $e^x$ is its own derivative, which is false. – DanielSank Feb 5 at 2:53