From the video lecture on noise analysis by Razavi here. Can anyone explain in more detail what Razavi said here? Why is the energy at a single frequency (zero bandwidth) zero?

Lets look at the energy carried by a woman's voice at 20 kHz. Can I say that this dot represents the energy carried at 20 kHz? Answer: well not really because at a single frequency the energy that we carry is zero because the bandwidth is zero.

enter image description here

  • 1
    $\begingroup$ Side note: that diagram looks wrong (or at least, not a log scale on the frequency axis). The peak at ≈200Hz for men's voice is a bit high, and the peak at ≈2kHz for women's voice is wild. $\endgroup$
    – wizzwizz4
    Jan 10 at 16:57

2 Answers 2


The curves represent energy (or power) densities; thus, energy is what you get when you integrate them over a range. For example, you could get the power in the range 1000 Hz to 2000 Hz by integrating over the curve:

$$P_{1\,\text{kHz}\to2\,\text{kHz},\text{ woman}} = \int_{1000}^{2000} S_{\text{woman}}(f)\,\mathrm{d}f,$$

i.e. the area under the curve.

Now, since this curve is bounded at every point, the integral for a single point (so, the integral from 20 kHz to 20 kHz) is zero. That's a basic property of integrals.

  • $\begingroup$ Note that having a signal occur within a band, and to have a power density within that band, is characteristic of any signal. Even a real-world AM radio station's carrier wave isn't perfectly steady, and so occupies a non-zero spectral width. It may almost always be far more sensible to pretend that it is at one frequency (i.e., if you're modeling a communications system), but that doesn't mean that it actually is. $\endgroup$
    – TimWescott
    Jan 10 at 4:35
  • $\begingroup$ Thanks, that's one way to look at it. I'm looking for an intuitive explanation. Can you explain how power spectral density for a pure sinusoidal signal becomes zero in a different way? I read that power spectral density is related to autocorrelation, so I think it can probably be explained that way. $\endgroup$
    – hana
    Jan 14 at 12:22
  • $\begingroup$ what's more intuitive than "the area of a zero-width slice of anything is zero"? $\endgroup$ Jan 14 at 16:41
  • $\begingroup$ I'm sorry, talking about distributions / functionals might simply not be intuitive. Can't force math to be intuitive, if the relations aren't. "I've read that PSD and autocorrelation are related": yes! Please read the full thing where you read that, because that is not a complicated relationship, and you can read that just as well there instead of me repeating it. $\endgroup$ Jan 14 at 16:43

Power density graph does not work for signals that have infinitely narrow peaks.

If you had a signal with a continuous static frequency and plotted the power density of it in infinite resolution, the graph would have a spike of infinite height.

Practical power density graphs, like the one shown, have a limited horizontal resolution. The frequency domain is sampled at some interval, for example 1 Hz. Even if the signal had a peak narrower than this, it would get averaged out in the graph. When the graph has limited horizontal resolution, it does not provide useful information of bandwidths narrower than this.

It is easy to define a mathematical signal that has non-zero power at 0 Hz bandwidth: sin(x). We can even get quite close to pure sine wave in real-world signal generators. But any spectrum measurement for finite length of time will have a limit on the frequency resolution, causing the peak to get averaged out to more than 0 Hz. Mathematically the power spectrum of this signal would be Dirac delta: δ(f).


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.