In class, we have defined bandwidth of a signal as the difference between the maximal and minimal frequency. Also, our tutor explained that the bandwidth indicates how fast the signal is changing.

We have been given an example of a single sinusoid which has zero bandwidth (since it's both the maximum and minimum frequency)

What isn't clear to me is why a rectangular shaped signal has infinite bandwidth.

How can it be explained rigorously and intuitively?

  • $\begingroup$ Do you know the Fourier transform of a rectangular signal? That knowledge will answer your question. $\endgroup$
    – Matt L.
    Mar 7, 2018 at 12:53

2 Answers 2


"tutor explained that the bandwidth indicates how fast the signal is changing"

This is not correct even in a hand-waving fashion. As you were told, a pure sinusoid $x(t) = \sin(2\pi f_0t + \theta)$ has zero bandwidth (according to the definition of bandwidth that your tutor is using) but that does not mean that the sinusoidal is not changing at all! It is changing and its maximum slope (a.k.a. "how fast the signal changing") is $$\max \left| \frac{\mathrm dx}{\mathrm dt}\right| = \max \left|2\pi f_0 \cos(2\pi f_0t+\theta)\right| = 2\pi f_0.$$ 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 $$\max \left| \frac{\mathrm dx}{\mathrm dt}\right| \leq 2\pi f_0.$$ In short, the minimum frequency (in a bandpass signal) is irrelevant; it is the maximum frequency that matters, and the "how fast the signal is changing" depends on the maximum frequency and not on the bandwidth.

Turning to the rectangular pulse that puzzles you, note that the signal doesn't have a derivative (as far as mathematicians, including Bernstein, are concerned) when the signal makes the abrupt transition from high to low or vice versa but here on dsp.SE we have no qualms about saying that the derivative is $\pm \infty$ at these points. As MattL says in a comment, if you know the Fourier transform of a rectangular pulse (hint: $\text{rect}(t) \leftrightarrow \text{sinc}(f)$), then you know that the rectangular pulse has no maximum frequency (there is no finite $f_0$ such that $\text{sinc}(f) = 0$ for all $|f|>f_0$) and that the infinitely fast changing of the rectangular pulse is well-supported by the infinite bandwidth of the rectangular pulse.

  • 1
    $\begingroup$ I really liked that answer. $\endgroup$
    – Royi
    Aug 6, 2018 at 7:49

I wanted to put it as a comment but couldnt do that and I am by no means an expert.

As a matter of fact "bandwidth" is ambiguous. There are many parameters that define bandwidth. 3 db, 6db, to name a few.

You see that an infinite sinusoid has zero bandwidth while a time limited sinusoid has finite bandwidth.

It feels as if you havent been through fourier analysis yet so I will try to explain it in different way.

Let us write Heisenberg's uncertainty principle in terms of frequency and time.

Δw*Δt ≥ 1/2

Now suppose that we have an infinite sinusoid running on entire time axis. The frequency of this sinusoid can be determined very closely to the exact value (remember, the principle states that you can measure one of the conjugate variable accurately if other one is not accurate).

Now if we go on to decrease the length of this sinusoid we will be going away from the exact value of frequency. In extreme case let us assume we have a "sinusoid" such that it exists only for a quarter cycle and even less than that, (its not exactly an sinusoid, BTW) we see that we can never guess the exact frequency of sinusoid. Instead, a range of frequencies (that range too will increase if you just keep on cutting your sinusoid away). Thus a time limited signal has associated with it a range of frequencies.

I am sure this explanation is far from being perfect. I am open to criticisms.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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