Fourier transformations

Very simple examples of Fourier transformations for functions with discontinuities are given Table 2.

$${2\over 1+{\omega}^2}$$ e^{-| t |}

$$(1-i\omega)/(1+{\omega}^2)$$ e^-t.H(t)

$${1\over i\omega}$$ H(t)

$$1/(1+i\omega)^2$$ 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 $\omega \rightarrow \infty$ as ${A_{0}\over i\omega }$.Meanwhile, in case of a discontinuity of order 1, $\vert F(\omega )\vert$ behaves asymptotically as ${A_{1}\over \omega^{2}}$. Let us take, for instance, the third function from Table 1: $g(t)=\exp (-\alpha t)\sin {\omega}_{0}t H(t)$. In this case, $g(t) =f(\alpha t) \sin {\omega}_{0}t$ where $f(t)=\exp (-t) H(t).$ Then

$$G({\omega}) = {\bf F}_{t}g(t)={1 \over 2i\alpha} \left[ {1-i... ...over 1 + {({{\omega}+ {\omega}_{0} \over \alpha})}^{2}} \right]$$

and when $\omega \rightarrow \infty$

$$G({\omega}) \simeq {1 \over 2i \alpha} {{2i{\omega}_{0} \ove... ...} \over \alpha ^{2}}} \simeq {{\omega}_{0} \over {\omega}^{2}}.$$

More generally, it is simple to prove that, if $f^{(r-1)}(t) \in {\rm C}$ (i.e., is continuous), and if $f^{(r)}(t) \in {\rm L}_1$ (i.e., is summable), then $(i\omega)^rF(\omega) \in {\rm L}_1$.Consequently, when $\omega \rightarrow \infty$, $F(\omega)\rightarrow 0$,this convergence being not slower than $1\over (i\omega)^{r+1}$.

We shall prove the following statement. Let f(t) be a right-sided function, and, let's suppose that, for all $r=0,1,2,\cdots$, $f^{(r)}(t)\rightarrow 0$ (when $t\rightarrow \infty$) in the same way as exp $(-\alpha_k\vert t\vert)$. Then:  

$$F(\omega) = \sum_{n=r}^{N}{A_n \over{{(i \omega)}^{n+1}}} + ... ...omega^{N+1}}} \big) \mbox{\hspace{2.0cm}} (A_r=f^{(r)}(0^{+})).$$ (2)

As f(t) is a right-sided function, we have:  

$$F(\omega) = \int_{0}^{\infty} f(t) e^{-i \omega t} dt.$$ (3)

Let us integrate this integral by parts:

$$\begin{eqnarray} F(\omega) & = & -{1 \over{i \omega}} \int_{0}^{\infty} f(t)d e^... ...er{i \omega}} \int_{0}^{\infty} f^{\prime}(t) e^{-i \omega t} d t.\end{eqnarray}$$ (4) (5) (6)

Repeating this N times we obtain  

$$F({\omega}) = \sum^{N}_{n=0} {A_{n} \over {(i{\omega})}^{n+1... ...ga})}^{N+1}} \int^{\infty }_{0} f^{(N+1)} (t)e^{i{\omega}t} dt.$$ (7)

By definition of the order of discontinuity $A_{n}=0 , {\:}n=0, {\:}1 \ldots, {\:}r-1$. From the condition concerning the behavior of f^(k)(t) at $t\rightarrow \infty$, the integral on the righthand side of equation (7) converges to at $\omega \rightarrow \infty$. 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.