Here is a bandlimited function f(t) with bandwidth Ω:

enter image description here

The function f(t) is bounded in [-A,A].

Then the bound of the derivative of f(t) is bounded as: |f'(t)|≤2πΩA. So, what is the bound of its n-th order derivatives?

Is it as follows?

enter image description here

where n is the order of derivative.

If yes, then are the bounds of higher order derivatives much larger than the bound of f(t)?

I tried MATLAB simulations but the results were contrary.

  • $\begingroup$ See my comment on the answer below. $\endgroup$
    – Jazzmaniac
    Mar 15 at 11:02

What simulations did you try? I think what you mention is right.

If $\mathcal{F}[f(t)] = F(\omega)$, according to the inverse Fourier transform, we have

$$ f(t) = \frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{j\omega t}d\omega $$ $$ f'(t) = \frac{d}{dt}\Big(\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{j\omega t}d\omega\Big)= \frac{1}{2\pi}\int_{-\infty}^\infty j\omega F(\omega)e^{j\omega t}d\omega $$

So the Fourier transform of derivative should be

$$\mathcal{F}[f'(t)] = j\omega F(\omega)$$

and the Fourier transform of $n$-th order derivative is

$$ \mathcal{F}[f^{(n)}(t)] = (j\omega)^nF(\omega) $$

  • 1
    $\begingroup$ A frequency-domain bound does not imply a time-domain bound. The time-domain derivative of a bandlimited signal can be expressed as the convolution of the derivative of the sinc kernel with the signal. At the sample positions, the relevant term of the kernel of the first derivative is $\cos(k \pi)/k$, whose sum is not absolute convergent. That means even a bounded sequence of sample values can generate an unbounded first derivative. $\endgroup$
    – Jazzmaniac
    Mar 15 at 11:01
  • $\begingroup$ Thank you for your answer. I firstly generated a random bandlimited signal using a random array a low-pass filter. But when I calculate its derivative, I discovered that the level of the derivative is always smaller than the level of f(t) if the filter's pass-over frequency is not large. $\endgroup$
    – Y. H.
    Mar 15 at 11:18
  • 1
    $\begingroup$ @Y.H. Try the sequence (...,-1,1,-1,1,-1,1,0,-1,1,-1,1,-1,1,...). The derivative at the zero-sample is unbounded and grows with the length of the sequence. $\endgroup$
    – Jazzmaniac
    Mar 15 at 11:19
  • $\begingroup$ @Jazzmaniac yes you are right! $\endgroup$
    – ZR Han
    Mar 16 at 1:11

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