How to solve the Fast Fourier transform?

Why do some functions, which have Fourier transform, don't have Laplace transform?

• For eg: signum(t), sin(t) and cos(t) (When I say sin(t) and cos(t) I am not talking about the one sided sin(t) *u(t) and cos(t) *u(t) ). I have always thought that laplace is a generalized transform and fourier is a special case of laplace transform. So Laplace transform should exist even for functions which don't have fourier transform(Which actually happens in case of u(t)). But why does the opposite happens in case of the functions mentioned above?

• Answer:

  It is definitely true that Fourier is a special case of Bilateral Laplace (where in most nonperiodic cases you can put s = jw in L to get F). But this does not stop a certain class of functions from having either Laplace or Fourier transforms - it just means that the end result of the transform is different for different transforms. As long as a time function satisfies the existence conditions, it will have a transform. I think you are conflating between the unilateral and bilateral laplace transforms in your question, though. To answer this question, we have to begin at the Dirichlet's first condition. In simple terms, this means that the integral of the |x(t)| must absolutely converge. ∫∞−∞|x(t)|dt<∞∫−∞∞|x(t)|dt<∞\int_{-\infty}^{\infty} |x(t)| dt < \infty This condition is absolutely necessary for the Fourier transform to exist. When we have to find the Fourier transform of some function, say like u(t), whose integral does not converge, the simplest way is to multiply it with an exponentially decaying term like e^{-\sigma |t}}e^{-\sigma |t}}e^{-\sigma |t}}. You can calculate the transform because now the Dirichlet's condition is satisfied, and put σ=0σ=0\sigma = 0 later. Strictly speaking, this means that signum t, sint and cost DO have Fourier Transforms. Now, what if the convergence factor becomes e−σte−σte^{-\sigma t}? It will NOT make the function x(t) absolutely converging, since it has high values on the negative side. If you notice, the σσ\sigma of this e−σte−σte^{-\sigma t}, multiplied with e−jωte−jωte^{-j\omega t}, is what makes e−ste−ste^{-st}, the kernel for Laplace. Using the convergence factor e−σte−σte^{-\sigma t} with the original function x(t) and then carrying out the Fourier transform operation gives you the Bilateral Laplace transform. It is NOT the Fourier transform anymore because we are not even looking to satisfy Dirichlet's first condition. This allows us to analyse the sort of unstable functions that have ∫∞−∞|x(t)|dt→∞∫−∞∞|x(t)|dt→∞\int_{-\infty}^{\infty} |x(t)| dt \rightarrow \infty, as well as those with damping. Bilateral laplace transforms of sinwt, cos wt and signum t DOES exist - the regions of convergence are just different for the parts t>0 and t<0. The issue with the bilateral Laplace is that it does not consider the initial conditions of a system, and has a smaller ROC. For that reason, we generally end up using the Unilateral Laplace Transform, which works only on the causal parts of functions and provides for the initial conditions as well. This is superbly useful in circuit analysis and solving differential equations in general. This is obviously equivalent to a bilateral laplace of x(t) * u(t). It cannot be compared to the Fourier transform directly because of different limits and conditions of existence. Check out this course if possible in OCW: http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-003-signals-and-systems-fall-2011/. This slide http://www.eng.uwaterloo.ca/~jzelek/teaching/syde252/Laplace-roc.pdf gives some idea of the ROC of Laplace transforms too.