FFT

x = 0 : 0.1pi : 6pi
y(x) = sin(x) + sin(2x) + sin(3x)

The mininum sampling spacing dx = 0.1pi.
Niquist frequency = 1 / (2*dx) = 5 / pi.
fy = | fft(y - ybar) |, where ybar is mean of y.

N is the number of samples = 61.
The dominant terms in the frequency domain are at N = 4, 7, 10.

N	cycles	period	which signal?
1	0	infinite	ybar
4	3	6pi/3=2pi	sin(x)
7	6	6pi/6=pi	sin(2x)
10	9	6pi/9=2pi/3	sin(3x)
31	30	6pi/30=pi/5	1/(Niquist freq)