Yes, absolutely. Think of the Fourier transform as plot of how much of each frequency is "inside" a signal. If you added up all the frequencies and weighted them according to the Fourier transform, you would get your signal back.

You can also see a Fourier transform at the focal spot of a lens. Create a shaped hole and place it immediately in front of a lens. At the focal spot, the image will be the Fourier transform of that hole. Even neater, all the planes between the lens and the focal point are different instances of a fractional Fourier transform.

A Fourier transform is a special case of a Laplace transform. That is, the Laplace transform is a generalization to the Fourier transform to include damping, or signals that decay with time.

In addition to Raymond Rumpf´s explanation I like to mention that the FT can be used to calculate the bandwidth that is occupied by a certain signal waveform.

As far as the Laplace transform is concerned (yes - it is a generalization of the FT) I think there is NO physical representation because there are no steady state complex frequencies.

Hi Srikanth,

May be this brief note can help you.

http://www.cambridge.org/us/features/chau/webnotes/chap2laplace.pdf

Hi Srikanth,,

following Raymond's explanation and considering the spatial domain, let's consider a scene and fix a direction across that scene. If you measure the light coming from each point along that direction (e.g. the intensity distribution I(x,y) ), you can describe that light in terms of the inverse Fourier transform. Moving to a different direction and defining a new I(x,y, theta) this example shows that you can desribe the light distribution across a whole image in terms of Fourier components.

The above example is analogous to the classical approach used to move from the time domain to the frequency domain (visit http://fourier.eng.hmc.edu/e101/lectures/handout2_tex/node2.html ).

If the Fourier transform exists then it can be got  analytically from the  two-sided Laplace transform (which is a more general Laplace transform than the standard one ) which is got with the defining integral taken from minus infinity to plus infinitty.

However, Laplace transform exists under more general conditions than Fourier transform since there is a decreasing exponential inside the integral and the anti-Laplace transform is informative also of the (time ) transient behavior, not only of the stationary one.

The last paragraph of R. Rumpf response is very illustrative for physicalk interpretations.

Dear Srikanth Kotra, 

If we put the exp(-sigma t)  to multiply the Fourier transform, then it becomes the Laplace transform. We can easily see that what is written in the Fourier transform as exp(-jwt), now becomes exp(-st) where s = sigma + jw. The sigma is the real or called as the exponential part, where the jw is the imaginary or called sinusoidal part. The Fourier transform does not really care on the changing magnitudes of a signal, whereas the Laplace transform 'care' both the changing magnitudes (exponential) and the oscillation (sinusoidal) parts. We can say that Fourier transform is a subset of Laplace transform.

The Laplace transform is essentially helpful for solving differential equations, since most of any differential equation's solution will contain exponential and sinusoidal parts. The solution can be more easily express and understand in the s domain.

Please have a look the the following video.

https://www.youtube.com/watch?v=ZGPtPkTft8g

I hope this would help. 

Using this Laplace,  Differential equations can be changed in the form of algebraic equations.  

Dear Friends,

I have come across this video which elaborates the intuitive understanding of Fourier Transform. I hope you will find it interesting.

https://www.youtube.com/watch?v=1JnayXHhjlg&index=6&list=PLUMWjy5jgHK1NC52DXXrriwihVrYZKqjk

You may want to take a look at videos I have. Here is the link.

https://www.youtube.com/watch?v=ZKNzMyS9Z6s&t=389s

They come from my book, "Intuitive Guide to Fourier analysis and Spectral Estimation".