2|c|Table 2: Fourier Transformations | |
e^{-| t |} | |
e^{-t}.H(t) | |
H(t) | |
t e^{-t}.H(t) |
It is very simple to determine the Fourier transforms of all the functions in Table 1 using these simple properties of the Fourier transformation:
We can see from the Table 2 that the Fourier transforms of functions with discontinuities of order 0 behave at as .Meanwhile, in case of a discontinuity of order 1, behaves asymptotically as . Let us take, for instance, the third function from Table 1: . In this case, where Then
and when More generally, it is simple to prove that, if (i.e., is continuous), and if (i.e., is summable), then .Consequently, when , ,this convergence being not slower than .We shall prove the following statement. Let f(t) be a right-sided function, and, let's suppose that, for all , (when ) in the same way as exp . Then:
(2) |
As f(t) is a right-sided function, we have:
(3) |
(4) | ||
(5) | ||
(6) |
Repeating this N times we obtain
(7) |
By definition of the order of discontinuity . From the condition concerning the behavior of f^{(k)}(t) at , the integral on the righthand side of equation (7) converges to at . This means that the last equation is equivalent to equation (2). So the asymptotic behavior of the Fourier-transforms for high frequencies reflects the sharpness of the discontinuities of the (original) functions. Both notions (asymptotic behavior and discontinuities) are equivalent ones.