Euler's relations & 'Epicycloids'

  • Context: Graduate 
  • Thread starter Thread starter saviourmachine
  • Start date Start date
  • Tags Tags
    Relations
Click For Summary
SUMMARY

This discussion centers on the mathematical concept of epicycloids and their relation to Euler's formulas, particularly in the context of amplitude modulation (AM). The user explores how a point on a circle can represent a carrier frequency in AM, while another point on a superimposed circle can modulate this frequency. The discussion also touches on visualizing complex Fourier series as superimposed circles with varying frequencies. The user seeks clarification on Euler's relations using these geometric drawing techniques.

PREREQUISITES
  • Understanding of epicycloids and their geometric properties
  • Familiarity with Euler's formulas and complex numbers
  • Knowledge of amplitude modulation (AM) principles
  • Basic grasp of Fourier series and their applications
NEXT STEPS
  • Research the visualization of Euler's relations using geometric techniques
  • Explore Java applets for simulating epicycloids and AM signals
  • Study the relationship between complex Fourier series and circular motion
  • Investigate advanced concepts in signal modulation and its mathematical foundations
USEFUL FOR

Mathematicians, electrical engineers, and students interested in signal processing and geometric representations of mathematical concepts.

saviourmachine
Messages
96
Reaction score
0
The idea
An epicycloid is a superimposed circle on another circle. http://www.math.dartmouth.edu/~dlittle/java/SpiroGraph/ 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 2 \cos{\omega}. Of course is this equals the sum of \exp{j \omega t} and \exp{-j \omega t}, 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

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:
Physics news on Phys.org
Amplitude modulation

Amplitude modulation
Now I'm thinking about it. The frequency f_1 of the point on the first circle can be the carrier frequency in AM modulation. The frequency f_2 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)! g(t)=A_c \exp{j \theta(t)}
 
Last edited:
Complex Fourier series

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

Does nobody know of this manner of visualization?
 
Can a moderator move this thread if it's not appropriate here? It's not homework...
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K