1

Suppose we consider $\Pi((x-3)/2)$ where $$\Pi(x) = \begin{cases} 1 \ \ \text{if} \ |x| < 1/2 \\ 0 \ \ \text{if} \ |x| \geq 1/2 \end{cases}$$

So $$\Pi((x-3)/2) = \begin{cases} 1 \ \ \text{if} \ \left|\frac{x-3}{2}\right| < 1/2 \\ 0 \ \ \text{if} \ \left|\frac{x-3}{2}\right| \geq 1/2 \end{cases}$$

We can rewrite this as $$\Pi((x-3)/2) = \begin{cases} 1 \ \ \text{if} \ 2 < x < 4 \\ 0 \ \ \text{if} \ \left|\frac{x-3}{2}\right| \geq 1/2 \end{cases}$$

The Fourier Transform is $$F(s) = \int_{2}^{4} e^{-2 \pi i s x} \ dx$$

$$= -\frac{1}{2 \pi i s}(e^{-8 \pi i s}-e^{-4 \pi i s})$$ $$ = 2e^{-6 \pi i s} \text{sinc} \ 2s$$

What information does the graph of $|2e^{-6 \pi i s} \text{sinc} \ 2s|^{2} = 4 \text{sinc}^{2} \ 2s$ tell us about $\Pi((x-3)/2)$? I know that in image processing, the change in frequencies signifies changes in color.

NebulousReveal
  • 13,935

1 Answers1

1

Define:$$G(f)=\int_{-\infty}^{\infty}g(t)e^{-i2\pi ft}{dt}$$ then: $$|G(f)|^2$$ is known as the energy spectral density. It tells you how much energy is in frequency $f$. If you think of Fourier series, you build up the function by adding an infinite sum of $\text{sin}$ and $\text{cos}$ waves and the coefficient of each tells you how much of that wave is needed to build the function, well this is analogous to that; this says how much of frequency $f$ is needed to build the function $f(t)$. Also you have Plancherel's theorem which states that:$$\int_{-\infty}^{\infty}|g(t)|^2{dt}=\int_{-\infty}^{\infty}|G(f)|^2{df}$$ which essentially tells you that integrating the power over all time gives you the energy in $g(t)$, as does integrating the energy spectral density over the entire spectrum.

If we return to the analogy of Fourier series, we know to represent a function with sharp edges or a function that changes rapidly over its period we often have to weight the higher $\text{sin}$ and $\text{cos}$ harmonics more than we do in a slowly varying smooth function, effectively this means the Fourier series does not converge as quickly. The Fourier transform analogy of this is the concept of confined and spread out functions, if a function in the time domain is more confined, e.g. a narrow width Gaussian function, it can be thought of as changing rapidly in a local sense as the function quickly goes from zero to one and back to zero. Spectrally this rapid change requires more high frequency components to represent it and the spectrum becomes more spread out. In your case the $\text{rect}$ function transitions sharply from zero to one and back to zero again and this means it it spectrally spread out, the energy distribution falls off according to the envelope $\frac{1}{f}$ which is pretty slow compared to say the exponential decay of Gaussian functions.

This can cause problems in practical scenarios because we often need to represent a function using a limited region of the spectral domain (for example when we try to cram many data channels in to the spectrum of a fiber optic cable). If we truncate the spectral representation using a filter so that channels do not overlap, this may not give a good representation of the function in the time domain due to the highly weighted absent spectral components. For this reason $\text{rect}$ functions are not ideal for signalling when spectral space is limited.

Graham Hesketh
  • 5,409