11
$\begingroup$

I would like to know how to go about calculating the bandwidth of:

  1. A constant (real) sinusoidal tone

  2. A (real) sinusoidal pulse.

The question is as simple as that, but I am having a hard time with the concept of what exactly the bandwidth of a constant tone should be to begin with, and from there, what the bandwidth of a pulse should be.

  • In the frequency domain, a constant real tone of frequency $f$ exists as two delta functions, located at $f$ and $-f$, but how does one go about calculating its bandwidth?
  • Furthermore, in regards to the pulse, this is rectangular function in time, and thus a sinc in frequency domain, so would not its bandwidth simply be $\frac{1}{T}$, where $T$ is the duration of the pulse?
$\endgroup$
  • 1
    $\begingroup$ The term "bandwidth" by itself is ambiguous. It's unfortunate, but when you see the term used, it's not typically described more specifically; there are often application-specific definitions that are commonly assumed. However, on a question like this, you need to choose a definition: 3-dB bandwidth? 6-dB? 99% bandwidth? Absolute occupied bandwidth (only finite for infinite-length signals)? Gabor bandwidth? There are a lot of choices. $\endgroup$ – Jason R Jul 1 '12 at 1:16
  • $\begingroup$ @JasonR Thanks, yes that makes sense. The question had come up as part of how to calculate the SNR of a signal, where the signal has some bandwidth, and the noise has some other bandwidth. Naturally the 0 bandwidth of a tone threw me off in this regard. In light of this, I think I will have to make a new question. $\endgroup$ – Spacey Jul 1 '12 at 17:47
15
$\begingroup$

The spectrum of a continuous tone is, as you said, of the form $\delta(f-f_0) + \delta(f+f_0)$: 2 impulses at frequencies $f_0$ and $-f_0$.

As a lowpass signal, this is said to have bandwidth $f_0$ (the one-sided spectrum has components up to $f_0$).

As a bandpass signal, it has zero bandwidth (there's nothing around the carrier frequency $f_0$).

If you multiply the sine wave by a pulse, this makes it time-limited, and therefore frequency-unlimited. Infinite bandwidth in theory.

In practice, you must define some criteria for estimating your bandwidth. Examples are:

  • 3 dB drop (of the sinc function around $f_0$)
  • 10 dB drop
  • drop below the noise level
$\endgroup$
8
$\begingroup$

The bandwidth of a theoretical infinite length sinusoid of a perfectly constant frequency is zero.

The bandwidth of a time-limited sinusoidal pulse is the transform of the pulse envelope. For a rectangular time window, that transform is a Sinc function. The main lobe of that Sinc is about 2/t in bandwidth, but that contains only a portion of that Sinc's total energy. Since a Sinc has infinite extent, so does the total bandwidth. In a more realistic situation, the Sinc will fall below some noise floor at some width from the main lobe. Pick your noise floor.

For CW modulation, usually one shapes the pulse window less sharply (less clicky) so that less of the energy is spread out far from the main lobe in the frequency domain.

$\endgroup$
3
$\begingroup$

By definition, the bandwidth in a spectrogram is a measure of how many components you will need to describe your signal. Let's look on the positive side of the frequency range: you are using a real signal and the other half is just a reflexion of what you see on the positive frequencies scale (and certainly more intuitive).

In a discrete setting (as usual on computers) an infinite sinusoid is described by one component, all other components up the Nyquist frequency are null. As you move to a continuous formulation -and as you mentioned- the spectogram is a pulse and the bandwidth gets to zero.

What is interesting is that if your sinusoid is included in a pulse (that is modulated for instance by a gaussian bump), then the bandwidth gets broader, proportionally to the inverse of the length of the temporal bump. Note that at the extreme, a very narrow pulse (a click) will cover the whole spectrum.

$\endgroup$

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.