I am trying to generate an audio signal that, like white noise, has "equal intensity at different frequencies, giving it a constant power spectral density", but unlike white noise, can be expressed without any random character and in a representative way even when just maximally brief burst.

What is the energy distribution of a an immediate discontinuity (typically experienced in the real world as a pop/click) where the signal goes for example from 0 to 1 over a one sample, then either stays there or gradually returns to 0 at an inaudible rate (eg. with a high pass filter for example at low frequency of say 1-5 Hz to return back to 0)?

I believe this must be like a square wave, since it is 1 half of a square wave:

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Where the amplitude of any given excited frequency relative to the base frequency $nf_0$ is $1/n$.

So does the discontinuity have an exponentially higher energy distribution towards 0 Hz, and near 0 energy at the 1/2 sample rate?

A white noise signal has equal energy at all frequencies but, in my experience, only as it is averaged over time. If the signal is too brief, it cannot express this due to the random evolving nature.

Is there any type of signal that can be expressed more immediately than white noise (like a discontinuity) but that has equal energy at all frequencies? If not, why is this not possible?

Is it a matter of needing more samples of random data to express the high frequency excitation?

For example, could one pass a discontinuity through some type of filter to raise the higher frequency energy and decrease the lower frequency energy to "even it out"?

Or is white noise only capable of doing this?

If white noise is the only way, at a given sample rate (say 44100 Hz) what is the minimum sample duration of white noise typically required to get an even excitation at all audio frequencies (to 20 kHz)?

Is it possible to write a sequence of the minimum effective duration of sample data that will do this reliably? Ie. Shortest pseudo-"white noise" burst effective at all audio frequencies of interest? Or effective up to 1/2 sample rate?

Thanks for any help.

  • 1
    $\begingroup$ Check the chirp signal. It is basically a harmonic signal with changing frequency. And by this property it is quazi-flat in a band, like a white noise $\endgroup$ Feb 28 at 8:46

2 Answers 2


A step has a $f^{-2}$ power spectrum, not white. A delta function (impulse) has a white noise spectrum. You may approximate it with a narrow pulse.

A pseudorandom binary sequence or pseudonoise sequence is white up to the cutoff determined by its pulse width, but composed of discrete lines at multiples of its repeat frequency (which may be below your resolution).

There are infinite possibilities here, since any signal created by arbitrary phase adjustment of the components of a white noise signal has a white spectrom.

  • $\begingroup$ That is quite helpful. To clarify further, you mean the width determines the lowest frequency excited, correct? So if you want white noise to excite all frequencies equally down to 20 Hz, at 44100 Hz sample rate, you would need a minimum of 2205 samples (0.05 seconds). Is this correct? Anything less than that will not excite 20 Hz equally to the high frequencies which are more easily excited? $\endgroup$
    – mike
    Jan 23 at 22:22
  • $\begingroup$ That is also helpful to know a delta function (impulse) has a white noise spectrum. I would have expected it to behave like a step or square wave, given it is similar to a square wave. I guess the asymmetry (only going positive from 0 to 1 then 0) is what differs versus a square wave which alternates + and - symmetrically? Thanks again. $\endgroup$
    – mike
    Jan 23 at 22:24
  • $\begingroup$ @mike The width determines the highest frequency. For a periodic, repeating approximation of white noise, the period determines the lowest frequency. $\endgroup$
    – John Doty
    Jan 23 at 22:28
  • $\begingroup$ I am not sure I understand John. (1) If we are talking about a simple burst of truly white noise, and want to excite evenly 20 Hz to 20 kHz (or up to Nyquist frequency) at a sample rate of 44100 kHz, what is the minimum burst width in samples or time that will do this? I would have thought high frequencies would be easier to excite with short bursts as they have shorter periods. (2) If a Dirac can do this with one single sample, why does a noise signal need many samples? Is it the periodic (symmetric +/-) nature of the noise? $\endgroup$
    – mike
    Jan 23 at 22:49
  • $\begingroup$ I am not sure how much energy you are willing to commit to this, but I made a related post here as you answer raised more questions for me I am curious about and need to figure out. I'd appreciate any thoughts on that too if you have the time/interest. physics.stackexchange.com/questions/798626/… $\endgroup$
    – mike
    Jan 23 at 22:51

An impulse function in the time domain, which is basically a very narrow pulse of energy, will excite all frequencies within the bandwidth given by the reciprocal of the pulse width of the impulse and any other bandwidth limiting components in the system.


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