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

Question

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

Answer 1

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 $f(x)$

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

Answer 2

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.)