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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)

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