# AC Generator

1. Nov 25, 2008

### f77126

Consider a simple alternating current generator such as Hippolyte Pixii's dynamo. The current generated by such dynamo can be described as simple oscillating cosine signal. Assuming for simplicity that the max current generated each time a pole of Pixii's magnet passed the coil is (2/pi)0.5 amperes and that the magnet rotate with a characteristic angular frequency ω0.

Write the mathmatical expression for the time-domain signal (current vs. time) from such AC generator. For simplicity, assume that the magnet was already spinning at the beginning of the observation and the magnet's poled happened to be just below the coils at time zero.

I am not really sure how to start, I just need know that I need something relate current with time and equation should have cosine.
Thanks for any help

2. Nov 26, 2008

### gabbagabbahey

You are told that it is a "simple oscillating cosine" with amplitude (2/pi)0.5Amp and angular frequency $\omega_0$....Surely you can think of an expression for that.

3. Dec 1, 2008

### striday1

so i think i fiqured out the first part, i'm not to sure. the mathematical expression should be y(t)= sqrt(2/pi) cos (w_o x t). Usually theta would be in the equation but because the magnet is already spinning, there is no initial phase, therefore you leave the theta our of the equation.

then for the second part, when you draw the graph from 0 to 10 seconds, there should be 4 full wavelengths. you start your graph at (0,-1) and at 1 second (180 degrees, n) you should have passed (1,1), and at 360 degrees, which is 2 seconds at 2n, you should have gotten (2,-1). not sure about this tho

i'm not about how to start the three part. Could the Fourier transform of the equation be\sqrt{2 /pi}\cdot\frac{\delta(\omega-a)+\delta(\omega+a)}{2}\,. I have no clue!

4. Dec 1, 2008

### striday1

so i think i fiqured out the first part, i'm not to sure. the mathematical expression should be y(t)= sqrt(2/pi) cos (w_o x t). Usually theta would be in the equation but because the magnet is already spinning, there is no initial phase, therefore you leave the theta our of the equation.

then for the second part, when you draw the graph from 0 to 10 seconds, there should be 4 full wavelengths. you start your graph at (0,-1) and at 1 second (180 degrees, n) you should have passed (1,1), and at 360 degrees, which is 2 seconds at 2n, you should have gotten (2,-1). not sure about this tho

i'm not about how to start the three part. Could the Fourier transform of the equation be\sqrt{2 /pi}\cdot\frac{\delta(\omega-a)+\delta(\omega+a)}{2}\,. I have no clue!