0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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:

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

where

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.