# How does the Laplace Transform work for circuit analysis? [closed]

I would like to understand how signals transformed from the time domain to the frequency domain for algebraic manipulation, can be transformed back to give solutions in the time domain. Knowing how to do it is not enough. I would like to know why it works.

Is it that when multiplying a function by exp(-st) that the area captured beneath the curve during integration in the time domain, gives a transform of the function to a unique function in another linear vector space, the frequency domain, and that combinations in that other space uniquely transform back to functions in the time domain?

John

-

## closed as not a real question by Andreas Blass, Yemon Choi, Suvrit, Qiaochu Yuan, Andy PutmanMar 8 '12 at 20:16

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

$\exp(-st)$ is an eigenfunction of $d/dt$. – Terry Tao Mar 7 '12 at 18:17
As a cartoon, you can view the Laplace transform as a mapping between time and energy representations (this is what the Gibbs distribution is about). Meanwhile, the Fourier transform is a mapping between time and complex frequency representations. – Steve Huntsman Mar 7 '12 at 18:28
"why the Laplace transform works" - do you mean "why are the usual formulas for Laplace transforms correct"? or "what is the conceptual reason why Laplace transforms turn convolutions into products"? – Yemon Choi Mar 7 '12 at 18:46
maybe it's also worth mentioning Stieltjes integrals at this point? – Suvrit Mar 8 '12 at 0:38
The main question is not well posed. Works for doing what? – Michael Bächtold Mar 8 '12 at 9:20

## 2 Answers

As other posters indicated, the Laplace transform is closely related to the Fourier transform. It is easier to explain the versatility of the Fourier transform.

If you are interested in differential equations, you wish that all functions were linear combinations of exponentials

$$e_\xi(x)= e^{ i \xi x}.$$

The reason is the following simple identity

$$\frac{d}{dx}\left(\sum_k A(\xi_k) e_{\xi_k}(x)\right) = i\sum_k \xi_k A(\xi_k) e_{\xi_k}(x)$$

which shows that for linear combinations of exponentials the transcendental operation of derivation is replaced with a much simpler algebraic operation. It is natural to ask if any function f(x) can be described as a linear combination of exponentials

$$f(x) = \sum_\xi A(\xi) e_\xi(x).$$

The answer is yes, if we allow for infinte superpositions of exponentials

$$f(x) "= " \sum_{\xi\in\mathbb{R}} A(\xi) e_\xi(x) :=\int_{\mathbb{R}} A(\xi) e_\xi(x) d\xi.$$

More precisely, the above function $\xi\mapsto A(\xi)$ is the Fourier transfrom of $f(x)$

$$A(\xi)=\frac{1}{2\pi} \int_{\mathbb{R}} f(x) e_{-\xi}(x) dx.$$

-
I like this a lot. – Jeff Strom Mar 8 '12 at 15:32
I feel that this comment, and the eigenfunction comment, are most likely to lead me to a better understanding of why circuit problems can be represented and solved when transformed to another space. Thanks to all. – John Mar 9 '12 at 18:19

Consider these relations for the Fourier, Mellin, and Laplace transforms:

$\int^{\infty}_{-\infty}{exp(2 \pi ifx)exp(-2 \pi ify)df} = \delta(x-y)$

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(ln(x)-ln(y))= y \delta(x-y)$

$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y)$.

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups), and the eqns. encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g., Plancherel, convolution, Poisson summation) and illustrate the transformations from one transform to another.

(Tried this as a comment initially, but had formatting problems.)

-
Per Fourier's original motivation, apply F(d/dy) to the LPT/FT relations to obtain an operational calculus. For the Mellin relation, apply G(xd/dx). A good part of operational calculus deals with the issue of for what kinds of functions this is valid. – Tom Copeland Mar 8 '12 at 23:06