For a continuous signal (function), we have Bernstein inequality : $$ |{df(t)}/dt| \le 2AB\pi $$ where $A=\sup|f(t)|$ and $B$ is the bandwidth of $f(t)$. The question is: is there a relationship for a discrete function $x[n]$ like this? $$ |x[n] -x[n-1] | \le\ \mu\ W $$ where $$ X[k] = \sum\limits_{n = 0}^{N - 1} {x[n]{e^{ - j\frac{{2\pi }}{N}nk}}} $$ is the DFT for $x[n]$, $X[k]=0$ for $k> W$.

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    $\begingroup$ How do you define the bandwidth of $x[n]$? $\endgroup$ – AlexTP Oct 29 '18 at 18:17
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    $\begingroup$ I'm sitting – tired – in front of a paper; my understanding of the Bernstein ineq is: $$P(\omega)=\sum_{n=0}^N\left(a_k\cos(n \omega)+b_k\sin(n\omega)\right)$$ $$\max_{-\pi\le\omega < \pi} \left\lvert P'(\omega)\right\rvert\le \max_{-\pi\le\omega < \pi} \left\lvert P(\omega)\right\rvert $$ From that follows that continuous bandwidth would be $2N\omega$; assuming Nyquist-limiting $W < 2N\omega$. $\endgroup$ – Marcus Müller Oct 29 '18 at 18:35
  • $\begingroup$ @OlliNiemitalo edited $\endgroup$ – alwaystudent Oct 30 '18 at 8:06
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    $\begingroup$ If $A=\max |x[n]|$, then we simply have $|x[n]-x[n-1]|\le 2A$. So any formula involving the bandwidth should be tighter than that, if it's any good. $\endgroup$ – Matt L. Oct 30 '18 at 13:41
  • $\begingroup$ An alternative and not equivalent way to define $A$ is that it is the supremum of the absolute value of the band-limited continuous-time function the samples of which are $x[n].$ $\endgroup$ – Olli Niemitalo Oct 30 '18 at 14:02

As shown in the answers to this question, for continuous-time signals, the bound predicted by Bernstein's inequality is achieved with equality by a sinusoidal signal with a frequency equal to the upper frequency limit.

In analogy to this, in this answer I'll show a bound on $\big|x[n]-x[n-1]\big|$ for a discrete-time sinusoidal signal $x[n]$ at angular frequency $W$:


For the signal $x[n]$ given by $(1)$, the largest value of $\big|x[n]-x[n-1]\big|$ for all $n\in\mathbb{Z}$ occurs for two values $x[k]$ and $x[k-1]$ symmetrical to the point of the largest derivative of a (continuous) sinusoid, i.e., for $x[k]=A\sin(W/2)$ and $x[k-1]=A\sin(-W/2)$. Consequently, the bound for a sinusoid with frequency $W$ is given by

$$\big|x[n]-x[n-1]\big|\le 2A\sin\left(\frac{W}{2}\right)\tag{2}$$

This bound is achieved with equality for the signals

$$x[n]=A\sin\left(nW+\frac{(2l+1)W}{2}\right),\qquad l\in\mathbb{Z}\tag{3}$$

Note that the constant $A$ does not generally equal the maximum amplitude $B=\max |x[n]|$ of $x[n]$. Depending on the sampling phase, the maximum amplitude $B$ lies in the following interval:

$$A\cos\left(\frac{W}{2}\right)\le B\le A\tag{4}$$

Consequently, we have

$$A\le\frac{B}{\cos\left(\frac{W}{2}\right)},\qquad 0<W<\pi\tag{5}$$

Combining $(2)$ and $(5)$ we get

$$\big|x[n]-x[n-1]\big|\le 2B\tan\left(\frac{W}{2}\right),\qquad B=\max_n\big|x[n]\big|\tag{6}$$

Note, however, that $(6)$ is only useful for $\tan(W/2)<1$, i.e., for $W<\pi/2$ because for any $x[n]$ with $\max|x[n]|=B$ we must have

$$\big|x[n]-x[n-1]\big|\le 2B\tag{7}$$

I believe that the bound $(6)$ holds for all discrete-time band-limited signals with a maximum frequency $W$ (i.e., with $X(e^{j\omega})=0$ for $|\omega|\in (W,\pi]$), but I don't know how to show it.

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  • $\begingroup$ Hi! What do you mean by discrete-time band-limited signals ? Are there non band-limited discrete-time signals as well ? $\endgroup$ – Fat32 Oct 31 '18 at 10:21
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    $\begingroup$ Consider $x[n] = \frac{2}{\sqrt{2}}\sin\left(\frac{\pi n}{2}+\frac{\pi}{4}\right) = [1, 1, -1, -1, 1, 1, -1, -1, \ldots].$ Eq. 2 gives $\left|x[n] - x[n-1]\right| \le \sqrt{2}A, $ which is correct if $A = \frac{2}{\sqrt{2}}$ (see my comment to the question) but incorrect if $A = \max|x[n]| = 1.$ $\endgroup$ – Olli Niemitalo Oct 31 '18 at 11:33
  • $\begingroup$ @OlliNiemitalo: Yes, I need to clarify the definition of $A$. $\endgroup$ – Matt L. Oct 31 '18 at 11:46
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    $\begingroup$ @OlliNiemitalo: Fixed, kind of ... $\endgroup$ – Matt L. Oct 31 '18 at 15:51
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    $\begingroup$ That can be a useful result. It may well be a tight bound in cases that band limit $W = \frac{\pi}{N},$ where $N$ is an integer, because then each extrema of a sinusoid of frequency $W$ can be located just between the samples. With other band limit values, things may get complicated and the "worst case" sinusoid may have a frequency lower than the band limit, or perhaps the worst case is not a sinusoid at all. $\endgroup$ – Olli Niemitalo Oct 31 '18 at 22:18

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