The Fourier spectrum is in the Figure, how to find the essential bandwidth? The Fourier spectrum

  • $\begingroup$ How do you define 'essential bandwidth'? Anyway, I think that in this case it will be hard to find a useful definition which does NOT yield $\infty$ as a result. As a sidenote, you can't define the magnitude of this function because the term $\delta^2(\omega)$ has no meaning. $\endgroup$ – Matt L. Oct 13 '14 at 7:44
  • $\begingroup$ The unit step function is time-scale invariant, which means there is no way to define a unique time scale, and as such also no unique bandwidth of any kind. Your question is therefore absolutely meaningless. $\endgroup$ – Jazzmaniac Oct 13 '14 at 10:45
  • $\begingroup$ Also apart from the point Matt correctly states, your plot is quite wrong too. $\endgroup$ – Jazzmaniac Oct 13 '14 at 10:47
  • $\begingroup$ @Jazzmaniac, i like how the function turns up a little as "$|w|$" increases. $\endgroup$ – robert bristow-johnson Oct 13 '14 at 14:35

there can be no essential bandwidth of the unit step function. defining such would lead you to self-contradiction.

suppose you have a function $x(t)$ which has Fourier Transform

$$ X(f) = \int\limits_{-\infty}^{+\infty} x(t) e^{-j 2 \pi f t} \ dt $$

and from $X(f)$, you defined some consistent measure of bandwidth $B_x$ in such a way that is independent of the amplitude of $X$ and is dimensionally consistent with $f$.

now suppose you scale the time argument of $x(t)$ to speed it up or slow it down:

$$ y(t) = x(a t) \quad \text{for } a > 0 $$

we know that the the Fourier Transform of $y(t)$ would be

$$ Y(f) = \frac{1}{a} X\left(\frac{f}{a} \right) $$

and the bandwidth of $Y(f)$ defined in the same manner as $B_x$ would be

$$ B_y = a B_x $$

so, suppose $a>1$, then you speed up $x(t)$ by the factor $a$ and the bandwidth is increased by the same factor.

but if you "speed up" (or slow down) the unit step function, you still have the very same unit step function:

$$ u(t) = u(a t) \quad \text{for } a > 0 $$


$$ u(t) \triangleq \begin{cases} 1, & \text{if } t > 0 \\ 0, & \text{if } t < 0 \\ \end{cases} $$

so scaling time for the unit step does not change it, yet if it had a bandwidth, scaling the time must change that bandwidth. a contradiction results.


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