0.4 Examples

Examples

• The pure oscillation (containing only one frequency)
  $$sin(2\pi t/T)\mathcal{F}\left\{sin,(,2,\pi ,t,/,T,)\right\}\left(f\right)=\frac{j}{2}(\delta (-1)-\delta \left(1\right))$$
  This formula can be obtained without computing integrals by noting that $sin\left(x\right)=(e^{jx}-e^{-jx})/\left(2j\right)=(j/2)(e^{-jx}-e^{jx})$ . Its power is $P=1/2$ (see [link] ).
• The perfect low (frequency) pass function:
  $$\text{sinc}\left(t\right):=\frac{sin\left(\pi t\right)}{\pi t}\mathcal{F}\left\{\text{sinc},\left(,t,\right)\right\}\left(f\right)=\text{Rect}\left(f\right)=\left\{\begin{array}{cc}1\hfill & \text{if}-1/2<f<1/2\hfill \\ 0\hfill & \text{else.}\hfill \end{array}\right)$$
  and more general (to pass exactly the frequencies $f\in ]-f_c,f_c[$ )
  $$2f_c·\text{sinc}\left(2f_{c}t\right)\mathcal{F}\left\{2,f_c,\text{sinc},\left(2f_{c}t\right)\right\}\left(f\right)=\text{Rect}\left(\frac{f}{2f_c}\right)=\left\{\begin{array}{cc}1\hfill & \text{if}-f_c<f<f_c\hfill \\ 0\hfill & \text{else.}\hfill \end{array}\right)$$
  Both, $x$ and $X$ are symmetric and real. Plancherel's formula allows to compute the energy of the $\text{sinc}$ :
  $${\left|\right|\text{sinc}\left(t\right)\left|\right|}^2={\int }_{-\infty }^{\infty }{X}^2\left(f\right)df={\int }_{-1/2}^{1/2}1df=1.$$
  More generally, the energy of $2f_c·\text{sinc}\left(2f_{c}t\right)$ amounts to $2f_c$ . Note that the sinc is not time-limited; it can't be by the Heisenberg principlesince it is bandlimited.
• The Dirac function $\delta \left(t\right)$ is often symbolically written as
  $$\delta \left(t\right)=\left\{\begin{array}{cc}``{\infty }^{\text{'}\text{'}}\hfill & \text{if}t=0\hfill \\ 0\hfill & \text{else.}\hfill \end{array}\right)\mathcal{F}\left\{\delta ,\left(,t,\right)\right\}\left(f\right)\equiv 1$$
  Clearly, the Dirac function is not really a function, and it has infinite energy. However,most manipulations work fine also for $\delta \left(t\right)$ . This is again an illustration of the Heisenberg principle. The Dirac function isthe extreme case which is sharply located in time, but has no characteristic frequency (all frequencies are present with equal strength).The properties of the Dirac function are best understood in terms of integrals:
  $${\int }_u^{v}g\left(t\right)\delta \left(t\right)dt=\left\{\begin{array}{cc}g\left(0\right)\hfill & \text{if}u<0<v\hfill \\ 0\hfill & \text{else}\hfill \end{array}\right).$$
  As a special case, the convolution with a function $g$ is again $g$ :
  $$\{\delta *g\}\left(a\right)={\int }_{-\infty }^{\infty }\delta (a-t)g\left(t\right)dt={\int }_{-\infty }^{\infty }\delta \left(t\right)g(a-t)dt=g\left(a\right)$$
  short: $\delta *g\left(t\right)=g\left(t\right)$ . For the shifted Dirac function ${\delta }_b\left(t\right)=\delta (t-b)$ we have
  $$\{{\delta }_b*g\}\left(a\right)={\int }_{-\infty }^{\infty }{\delta }_b\left(t\right)g(a-t)dt={\int }_{-\infty }^{\infty }\delta (t-b)g(a-t)dt={\int }_{-\infty }^{\infty }\delta \left(s\right)g(a-(s+b))ds=g(a-b)$$
  short, convolution with ${\delta }_b$ produces a shift by $b$ : ${\delta }_b*g\left(t\right)=g(t-b)$ .
• Double exponential:
  $$x\left(t\right)=e^{-\left|t\right|}X\left(f\right)=\frac{2}{1+4{\pi }^2{f}^2}$$
  Note that $x$ and its Fourier transform $X$ are real and symmetric. The power spectrum is ${\left|X\left(f\right)\right|}^2=4/{(1+4{\pi }^2{f}^2)}^2$ . Since $x$ is not differentiable at 0, the Fourier transform $X$ decays somewhat slowly: high frequencies are quite strong in this signal in order to make the sharp peak at 0.With this example, we may compute the energy directly:
  $$\left|\right|e^{-\left|t\right|}{\left|\right|}^2={\int }_{-\infty }^{\infty }{\left(e^{-\left|t\right|}\right)}^{2}dt=2{\int }_0^{\infty }{e}^{-2t}dt=-e^{-2t}{|}_0^{\infty }=1.$$
• One-sided Exponential:
  $$x\left(t\right)=\left\{\begin{array}{cc}{e}^{-t}\hfill & \text{if}t>0\hfill \\ 0\hfill & \text{else.}\hfill \end{array}\right)X\left(f\right)=\frac{1}{1+2\pi jf}$$
  The Fourier transform $X$ is complex with power spectrum ${\left|X\left(f\right)\right|}^2=1/(1+4{\pi }^2{f}^2)$ . Since $x$ is not even continuous at 0, the Fourier transform $X$ decays even slower than for the double exponential: high frequenciesare even stronger in this signal in order to make the jump at 0. With this example, we may verify Plancherel's theorem:
  $${\left|\right|x\left|\right|}^2={\int }_0^{\infty }{\left(e^{-t}\right)}^{2}dt={\int }_0^{\infty }{e}^{-2t}dt=-e^{-2t}/2{|}_0^{\infty }=1/2.$$
  $$\left|\right|\frac{1}{1+2\pi jf}{\left|\right|}^2={\int }_{-\infty }^{\infty }\frac{1}{1+4{\pi }^2{f}^2}df=\frac{1}{2\pi }arctan\left(2\pi t\right){|}_{-\infty }^{\infty }=\frac{1}{2\pi }(\pi /2-(-\pi /2))=1/2.$$
• The Gaussian Kernel is practically invariant under the Fourier transform (see Comment 5 )
  $$x\left(t\right)=e^{-t^2/2}X\left(f\right)=\sqrt{2\pi }{e}^{-{\left(2\pi f\right)}^2/2}$$
  Here, it is easy to verify $\left|\right|x\left|\right|=\left|\right|X\left|\right|$ via a substitution $t=2\pi f$ . The computation is somewhat harder and yields ${\left|\right|x\left|\right|}^2=\sqrt{\pi }\simeq 1.7725$ .

  Comment 5 From Probability theory, we know that (see “characteristic function of a Gaussian distribution”)

  $$\int e^{itf}\frac{1}{\sqrt{2\pi }}{e}^{-t^2/2}dt=e^{-f^2/2}$$
  Now replace $f$ by $2\pi f$ and multiply with $\sqrt{2\pi }$ to find the Fourier transform. For the energy:
  $${\left|\right|x\left|\right|}^2={\int }_{-\infty }^{\infty }{\left(e^{-t^2/2}\right)}^{2}dt=\sqrt{\pi }\frac{1}{\sqrt{\pi }}{\int }_{-\infty }^{\infty }{e}^{-t^2}dt=\sqrt{\pi }$$
  where we use in the last step, that $\frac{1}{\sqrt{\pi }}{e}^{-t^2}$ constitutes the probability density of a Gaussian variable with variance $1/2$ , and thus integrates to 1.

• The Mexican Hat (also called Ricker Wavelet in Geophysics) is the negative second derivative of the Gaussian:
  $$x\left(t\right)=(1-t^2)e^{-t^2/2}X\left(f\right)=4{\pi }^2\sqrt{2\pi }{f}^2{e}^{-{\left(2\pi f\right)}^2/2}$$
  Both, the Gaussian kernel and the Mexican hat are very useful since they are well located both in space and in frequency (see [link] ), meaning that the main portion of their energy stems from a narrow range of locations as well as a narrow range of frequencies.Thus, the may be used as low-pass, respectively band-pass filters. The Mexican hat is a wavelet; wavelets are used to determine which frequenciescontribute the main portion of the energy at a specific time. To this end, a wavelet needs to be well localized in time as well as in frequency.
• The Dirac Comp (peigne de Dirac) of step $\tau$
  $$\tau ·{\Delta }_{\tau }\left(t\right):=\tau \sum _n\delta (t-n\tau )\mathcal{F}\left\{\tau ,·,{\Delta }_{\tau }\right\}\left(f\right)=\sum _{k=-\infty }^{\infty }\delta (f-k/\tau )$$
  To verify this formula, we choose the integration interval $[-\tau /2,\tau /2]$ and write
  $$X_k=\frac{1}{\tau }{\int }_{-\tau /2}^{\tau /2}\tau ·{\Delta }_{\tau }\left(t\right)e^{-j2\pi kt/\tau }dt=\sum _n{\int }_{-\tau /2}^{\tau /2}\delta (t-n\tau )e^{-j2\pi kt/\tau }dt={\int }_{-\tau /2}^{\tau /2}\delta \left(t\right)e^{-j2\pi kt/\tau }dt=1$$
  The Fourier-representation becomes
  $$\tau ·{\Delta }_{\tau }\left(t\right)=\sum _{k=-\infty }^{\infty }{e}^{j2\pi kt/\tau }$$

  

  

  

  This formula will be crucial in the reconstruction formula in the Nyquist-Shannon Theorem. Note: One should not try to evaluate $\tau ·{\Delta }_{\tau }\left(t\right)$ ; indeed, it is not really a function, since it is formed by Dirac terms. Even its power is infinite.