Sine animation

Discussion in 'Math' started by Art, Oct 17, 2015.

  1. Art

    Thread Starter Distinguished Member

    Sep 10, 2007
    Hi Guys,
    I started a sine animation to eventually become a Fourier animation.
    Could anyone tell me what relation to the circle the horizontal length of the sine is supposed to be?
    This is just arbitrary:

    The below pic is a similar animation. I can’t see the relation of the horizontal lengths.

  2. nsaspook

    AAC Fanatic!

    Aug 27, 2009
  3. Motanache


    Mar 2, 2015
    View topic in which I answered:

    <<I can’t see the relation of the horizontal lengths.>>
    Well, how do you not find?

    There you have the fundamental frequency, 'repetition frequency' of the signal analyzed.

    <<horizontal lengts>>"of the red line is the same for black line"

    Said rigorous: frequency of the rectangular signal, it is the same as for fundamental sin().
    Sin(x) - red line...from 0 deg to 360 deg in 2L for x axis

    this means:
    360 deg...2Pi 2L => u(x)=V Sin(2Pi*x/(2L)) ; x=t=time

    thereafter the following components:
    Sin (2X)
  4. sailorjoe


    Jun 4, 2013
    There is no more relationship other than this: one rotation around the circle equals one full cycle of the sine wave, no matter how large or small the circle. Therefore, one easy solution is to make the horizontal distance of the sine wave equal to the circumference of the circle. Then you have to scale it to fit on the screen. If you superimpose another circle rotating at a higher frequency, the wavelength of the new circle's sine wave would scale accordingly per the frequency/wavelength equation.
    You can see this here:
  5. Art

    Thread Starter Distinguished Member

    Sep 10, 2007
    Ok I think I understand the English part :D
    Coming from electronics/radio, but weak understanding of math,
    I’d be best off copying an oscilloscope display with variable time scale.
    The ratio of different circles will matter though.

    This is one size circle, but double frequency, then summed amplitude for mixed frequency.
    But to calculate ratios for the Fourier transform for square wave,
    can someone explain how to calculate with a scientific calculator, the radius of outer circles?

    This would be easy for mixed audio tones.. just the same ratio as the two frequencies.

    Say the first circle is 10.0 units of measurement to make it simplest.

    ps... ok can that.. I have the answer :D
    Last edited: Oct 20, 2015
  6. Art

    Thread Starter Distinguished Member

    Sep 10, 2007
    Ok, now it would be nice to sample a cycle of each wave
    and scale each of them to send to an audio unit as they appear on the screen.
    This one should sound bad as it gets near a square wave.

    Last edited: Oct 28, 2015
    nsaspook likes this.
  7. sailorjoe


    Jun 4, 2013
    Art, very nicely done. Good job.
  8. BR-549

    Distinguished Member

    Sep 22, 2013
    I don’t mean to be a mean old fart, but I disagree with Art’s animations, especially the mechanical rotational part.

    The second harmonic is just another helix with twice the pitch. (When you look at the side of the helix like need to change the time to see each helix.) It can be greater or less in amplitude than the fundamental helix, but not the same. It slides inside or outside the fundamental helix. AND phase locks.

    The same with all the harmonics. A higher pitched helix can slide in and fill the gaps of a lower pitch helix.

    When I voice modulate(AM) a sine wave, my voice helixes(frequencies) are locked and loaded the carrier helix.

    These harmonic “slots” can be controlled with electrical resonance or lack of. They also can be controlled with physical distance and angle.

    Frequency is a helix, not a sine.

    We need 3D oscilloscopes.

    This is what happens when you believe in math.

    The equations will give you the right values, but not understanding.

    The process of harmonics works because it is referenced to the same origin and time. It’s a spectrum of fixed rates based on the fundamental rate.
    Last edited: Dec 9, 2015
  9. BR-549

    Distinguished Member

    Sep 22, 2013
    I believe this is an important point. Let’s say you have a signal that is rich in harmonics. And let’s ay that you want to eliminate only one and leave the rest alone. We’ll say the 2nd one.

    Do you recall when I was explaining earlier that the helixes couldn’t be the same size? That’s because they are rotating in the same direction. We can insert a reverse rotating helix, the exact same size, because a reverse helix can rotate at the same size with it. They fit perfectly, rotating in opposite directions. That will cancel the 2nd harmonic and leave the rest alone.

    The hard part is understanding the BOTH helixes are moving in the same direction, but one is rotating backwards while doing this.

    That takes a little study.
  10. Wendy


    Mar 24, 2008
    A simple high Q filter band stop would do it. Models are just that, they don't always match reality.