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Sin harmonic components are used in sound synthesis ?

  1. Apr 27, 2005 #1
    pyschoacoustically is it important wether cos or sin harmonic components are used in sound synthesis ?
  2. jcsd
  3. Apr 27, 2005 #2


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    I can't imagine any way that it would make a difference. The 2 only differ by phase.
    Last edited: Apr 27, 2005
  4. Apr 27, 2005 #3


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    if the phases of the harmonic components are static and if the harmonic waveform does not pass through some non-linear processing element (which might be difficult, considering practically any loudspeaker system), then the phase or phase differences is nearly always not detectable. that is, in general:

    [tex] x(t) = \sum{r_{n} \cos \left(n \omega_0 t + \phi_{n} \right)} [/tex]

    will sound about the same, no matter what [itex] \phi_n [/itex] is.

    BUT the waveform will not look the same for all possible values of [itex] \phi_n [/itex], and if [itex] x(t) [/itex] is passed through some non-linearity (that might flatten out the peaks a little), a variant of [itex] x(t) [/itex] that is has more prominant peaks will sound different (after the non-linearity) than the variant of [itex] x(t) [/itex] that is less peaky.

    also, if you are doing musical synthesis, then [itex] r_n [/itex] and [itex] \phi_n [/itex] are (slowly changing) functions of time. that is

    [tex] x(t) = \sum_{n=1}^{N}{r_{n}(t) \cos \left(n \omega_0 t + \phi_{n}(t) \right)} [/tex]

    then your phase of each harmonic changes in time, and the derivative w.r.t. time of [itex] \phi_{n}(t) [/itex] is a frequency offset or "detuning" of the [itex] n^{th} [/itex] harmonic. that is perceptually salient. how each harmonic is slightly detuned from the exact harmonic frequency (that is an integer multiplying the fundamental frequency) is something that affects the sound of a tone. if they all are perfectly harmonic, the tone sounds a little "dead" compared to a tone with detuned "harmonics". in a real piano, the upper harmonics get sharper and sharper as you get to the 20th harmonic and higher.

    so absolute phase, might not matter, but the change of phase does matter. i would not throw phase information away.

    try Googling a paper i wrote: "Wavetable Synthesis 101". it lives at the harmony-central.com website somewhere, to get my spin on this issue.

    r b-j
  5. Apr 27, 2005 #4


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    Except in rather odd cases, the ear is not sensitive to phase.

    - Warren
  6. Apr 27, 2005 #5


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    it's more an issue of what the brain decodes. the ear is a sorta really "only" a sophisticated transducer. of course a phase change that is equivalent to a pure delay is inaudible (unless you notice the delay, but that is not at issue here).

    also some waveshapes are audibly indistinguishable from others where the difference is phases of different harmonics. such as

    [tex] x_1 (t) = \sum_{n=0}^N { \frac{(-1)^n}{2n+1} \cos \left( (2n+1) \omega_0 t \right)} [/tex]

    is usually audibly indistiguishable from

    [tex] x_2 (t) = \sum_{n=0}^N { \frac{1}{2n+1} \cos \left( (2n+1) \omega_0 t \right)} [/tex]

    yet the waveforms clearly look different. x1(t) approaches a square wave as N -> infinity but x2(t) is much more spiky.

    however, what would happen if x1(t) and x2(t) are passed through an identical gentle non-linearity, such as

    [tex] y_n (t) = \frac{1}{\alpha} \arctan \left( \alpha x_n (t) \right) [/tex] ?

    as alpha gets larger, this nonlinearity will start to kick in and you will hear a clear difference between what happens to x1(t) and x2(t).

    well, here's another odd case for you, Warren (and this one is purely linear):

    All-pass Filter (APF):

    [tex] H(z) = \frac{z^{-N} - p}{1 - p z^{-N}} [/tex]

    the frequency response of a digital filter (more precisely called a "discrete-time filter") is evalutated as

    [tex] H(z) \mid_{z=e^{i \omega}} = H \left( e^{i \omega} \right) [/tex]

    where [tex] \omega = \frac{2 \pi f}{F_s} [/tex] , [itex] f [/itex] is the frequency, and [itex] F_s [/itex] is the sampling frequency. [itex] F_s / 2[/itex] is the so-called "Nyquist" frequency and all frequencies must be less than Nyquist in magnitude (or you get aliasing).

    [itex] z^{-N} [/itex] represents a delay element of N samples and [itex] 0 \leq p < 1 [/itex] is a number that represents a "pole" if N were 1. (there are N poles, in reality.)

    it turns out that

    [tex] | H \left( e^{i \omega} \right) | = 1 [/tex]

    for all [itex] |f| \leq F_s / 2 [/itex] so this all-pass filter changes nothing (other than possibly phase).

    for [itex] F_s / 2 [/itex] equal to, say, 44100 Hz, and if p = 0.95, then if N = 1, then i would agree that, for the most part, the inclusion or removal of this APF would normally be inaudible. however, if N = 22055 (the delay element was 1/2 second) and p = 0.95 , i must steadfastly disagree with any notion that the inclusion or removal of this APF would be inaudible.

    so here is an example of where changing nothing other than phase, creates a clearly audible difference.

    r b-j
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