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Euler's relations & 'Epicycloids'

  1. May 10, 2005 #1
    The idea
    An epicycloid is a superimposed circle on another circle. http://www.math.dartmouth.edu/~dlittle/java/SpiroGraph/ [Broken] can you find a java applet to show you. These epicycloids are tied to each other at their circumferences. But, what does change when using a slightly easier method, and superimpose the second circle (the referent) using it's centre!?

    A point on the first circle moves along its circumference, and because it is the centre of the second circle, the whole second circle does move with along it. Now comes the clue: do the same with the second circle. Take a point at the circumference of the second circle and move in the opposite direction (with a negative frequency). Like you can see will this point move along the horizontal axis. Not much have to be imagined to realize that this traject will be equal to [tex]2 \cos{\omega}[/tex]. Of course is this equals the sum of [tex]\exp{j \omega t}[/tex] and [tex]\exp{-j \omega t}[/tex], but it's cool that with rotating in the other direction a meaning is assigned to the concept "negative frequency".

    http://www.annevanrossum.nl/pictures/science/Epicycloid.gif [Broken]

    Does anyone know of a clarification of Euler's relations using these drawing techniques?

    Edit: Changed math to Latex.

    Observation: Odd, that the image can't be displayed. That's for paying members?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 10, 2005 #2
    Amplitude modulation

    Amplitude modulation
    Now I'm thinking about it. The frequency [tex]f_1[/tex] of the point on the first circle can be the carrier frequency in AM modulation. The frequency [tex]f_2[/tex] of the point of the second circle can be a frequency that modulates the carrier. Projecting the second point on the horizontal x gives the resulting AM signal. If I had time I'd like to make a java applet to show that. The complex envelope is approximated a circle (radius equals sum of radius of circle 1 and 2)! [tex]g(t)=A_c \exp{j \theta(t)}[/tex]
    Last edited: May 11, 2005
  4. May 11, 2005 #3
    Complex Fourier series

    Another thought. Even the complex Fourier series can be visualized as superimposed circles with different frequencies isn't it?
    [tex]f(t)=\sum_{n=-\infty}^{n=\infty}{A_n \exp{jn\omega_0t}}[/tex] with [tex]A_n[/tex] the radius of the circle n.

    Does nobody know of this manner of visualization?
  5. May 13, 2005 #4
    Can a moderator move this thread if it's not appropriate here? It's not homework...
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