Let $f(t)$ be a function with properties:

$$\begin{array}{ll} t\in\mathbf{R}&t\text{ is in reals}\\ f(t)\in\mathbf{R}\text{ for all } t&f(t)\text{ is in reals}\\ |f(t)|<A\text{ for all }t&\text{absolute value of }f(t)\text{ is bounded above by }A\\ \int_{-\infty}^{\infty} f(t) \ e^{- i \omega t} \ {\rm d}t = 0\text{ for all }|\omega|\ge B&f(t)\text{ is band-limited by frequency B in radians} \end{array}$$

Given $A$ and $B,$ what is the tight upper bound for $|f'(t)|,$ the absolute value of the derivative of the function?

Nothing else shall be assumed about $f(t)$ than what has been stated above. The bound should accommodate for this uncertainty.

For a sinusoid of amplitude $A$ and frequency $B,$ the maximum absolute value of the derivative is $AB.$ I wonder if this is an upper bound, and in that case also the tight upper bound. Or maybe a non-sinusoidal function has a steeper slope.

  • $\begingroup$ Have you checked this? $\endgroup$ – Tendero Aug 30 '18 at 13:24
  • $\begingroup$ @Tendero thanks. There, the signal energy is known, rather than the peak absolute value as in my question. $\endgroup$ – Olli Niemitalo Aug 30 '18 at 13:39
  • 1
    $\begingroup$ See my answer for the bound you seek. It says More generally, a result due to Bernstein says that if the maximum frequency in a generic $x(t)$ bounded within $[-1,1]$ is $f_0$, that is, $X(f) = 0$ for $|f| > f_0$, then $$\max \left| \frac{\mathrm dx}{\mathrm dt}\right| \leq 2\pi f_0.$$ $\endgroup$ – Dilip Sarwate Aug 30 '18 at 14:50
  • 1
    $\begingroup$ Based on the sharp version of Bernstein's inequality, from Dilip's linked answers, MBaz's edited answer and the literature cited, $AB$ is indeed the sharp (I called it tight meaning the same) upper bound for the maximum absolute value of the derivative, a full-scale sinusoid at exactly the band limit (not strictly allowed by the constraints I give) making the inequality an equality. $\endgroup$ – Olli Niemitalo Aug 31 '18 at 7:02

You'll be interested in Bernstein's inequality, which I first learned about in Lapidoth, A Foundation in Digital Communication (page 92).

With a well-behaved signal $f(t)$ as you defined it above (in particular, $f(t)$ is integrable and bandlimited to $B\,\text{Hz}$, and $\text{sup}\,|f(t)| = A$), then $$\left|\frac{\text{d}f(t)}{\text{d}t}\right| \leq 2AB\pi. $$

Note that the original result by Bernstein established a bound of $4AB\pi$; later, that bound was tightened to $2AB\pi$.

I have spent some time reading Zygmund's "Trigonometric Series"; all I'll say is that it is the perfect remedy for those under the impression that they know trigonometry. A full understanding of the proof is beyond my mathematical skill, but I think I can highlight the main points.

First, what Zygmund calls Bernstein's inequality is a more limited result. Given the trigonometric polynomial $$T(x) = \sum_{-\infty}^\infty c_k e^{jkx}$$ (with real $x$), then $$\max_x |T'(x)| \leq n \max_x |T(x)|$$ with strict inequality unless $T$ is a monomial $A \cos(nx+\alpha)$.

To generalize this we need a preliminay result. Consider a function $F$ that is in $\text{E}^\pi$ and in $\text{L}^2$. ($\text{E}^\sigma$ is the class of integral functions of type at most $\sigma$ -- this is one of the places where my math starts to fray at the edges. My understanding is that this is a mathematically rigorous way of stating that $f=\text{IFT}\lbrace F \rbrace$ has bandwidth $\sigma$.)

For any such $F$ we have the interpolation formula $$F(z) = \frac{\sin(\pi z)}{\pi}F_1(z),$$ where $z$ is complex and $$F_1(z) = F'(0) + \frac{F(0)}{\pi} + \sum_{n=-\infty}^\infty {^\prime} (-1)^nF(n) \left( \frac{1}{z-n}+\frac{1}{n} \right).$$ (This is theorem 7.19.)

Now we can state the main theorem. If:

  • $F$ is in $\text{E}^\sigma$ with $\sigma>0$
  • $F$ is bounded on the real axis
  • $M=\sup |F(x)|$ for real $x$

then $$|F'(x)| \leq \sigma M$$ with equality possible iff $F(z) = a e^{j\sigma z} + b e{-j\sigma x}$ for arbitrary $a,b$. We suppose that $\sigma=\pi$ (otherwise we take $F(z\pi/\sigma)$ instead of $F(z)$.)

To prove this, we write the derivative of $F$ using the interpolation formula above: $$F'(x) = F_1(x)\cos(\pi x)+\frac{\sin(\pi x)}{\pi} \sum_{n=-\infty}^\infty \frac{(-1)^nF(n)}{(x-n)^2}.$$ Setting $x=1/2$ we get $$F'(1/2) = \frac{4}{\pi} \sum_{n=-\infty}^\infty \frac{(-1)^nF(n)}{(2n-1)^2}$$ which implies $$|F'(1/2)| \leq \frac{4}{\pi} \sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2} = \frac{4M\pi^2}{4\pi} = M\pi.$$

Now we need a nice little trick: Take an arbitrary $x_0$ and define $G(z) = F(x_0+z-1/2)$. Then, $$|F'(x_0)| = |G'(1/2)| \leq M\pi.$$

(TODO: Show the proof for the case of equality. Define $\sum \prime$.)

| improve this answer | |
  • 1
    $\begingroup$ @OlliNiemitalo As pointed out in MattL's answer, the sinusoid $\sin(2\pi Bt)$ has maximum derivative $2\pi B$. This meets Bernstein's bound, as stated in my answer here on dsp.SE (cited in a comment on your question) and in my answer on math.SE that you found, with equality. $\endgroup$ – Dilip Sarwate Aug 30 '18 at 15:11
  • 1
    $\begingroup$ @OlliNiemitalo I found the proof given by Pinksy here (I hope that link works!). He definitely uses $4AB\pi$ as the bound, not $2AB\pi$. $\endgroup$ – MBaz Aug 30 '18 at 15:46
  • 2
    $\begingroup$ @MBaz Your link works indeed! At the end of the section 2.3.8 they say that the best known version of Bernstein's inequality has the factor 2 instead of 4, which is sharp, and that for details consult Zygmund (1959) Vol. 2, p. 276. I think that's Zygmund, A. Trigonometric series. 2nd ed. Vol. II. Cambridge University Press, New York 1959. $\endgroup$ – Olli Niemitalo Aug 30 '18 at 19:14
  • 2
    $\begingroup$ RP Boas, Some theorems on Fourier transforms and conjugate trigonometric integrals, Transactions of the American Mathematical Society 40 (2), 287-308, 1936 cites the relevant articles by Bernstein, Szegö, and Zygmund, already with the sharp bound, as far as I can tell. $\endgroup$ – Olli Niemitalo Aug 30 '18 at 20:26
  • 2
    $\begingroup$ @OlliNiemitalo Excellent! I had missed that note at the end of section 2.3.8. I'll update my answer. Also: that book by Zygmund is in my university's library, but it's not online. I'll take it out tomorrow and see what it says. $\endgroup$ – MBaz Aug 30 '18 at 22:15

In general you would get something like this, but it might not be tight:

$$\begin{align}|f'(t)|&=\left|\frac{1}{2\pi}\int_{-\infty}^{\infty}j\omega F(j\omega)e^{-j\omega t}d\omega\right|\\&\le \frac{1}{2\pi}\int_{-\infty}^{\infty}|\omega||F(j\omega)|d\omega\\&=\frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}|\omega||F(j\omega)|d\omega\\&\le \frac{|\omega_c|}{2\pi}\int_{-\omega_c}^{\omega_c}|F(j\omega)|d\omega\\&=\frac{\omega_c}{\pi}\int_{0}^{\omega_c}|F(j\omega)|d\omega\tag{1}\end{align}$$

The upper bound on $|f(t)|$ is of course implicit in $|F(j\omega)|$.

For a sinusoid $A\sin(\omega_ct)$, $(1)$ gives $A\omega_c$ as an upper bound, as expected.

| improve this answer | |
  • $\begingroup$ @Olli Niemitalo , I had derived the sinusoid case I think this is the general case we were looking at. Thanks Matt L. $\endgroup$ – MimSaad Aug 30 '18 at 18:22

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.