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



How does the Laplace Transform work for circuit analysis?
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 $f(x)$


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


