Fast Fourier Transform

[Eddie W. Shore]

I hope you can help me. I am trying to understand how the fast Fourier transform is calculated (FFT).

Sample problem:
n_0 = 0.54
n_1 = 0.66
n_2 = 0.52
where N = 3

The formula to the FFT (I think) is:
X_k = Σ( x_n * e^(-i * 2 * π * k * n / N) ) for n = 0 to N-1

Using the formula above I get:
X_0 = 1.62
X_1 = 0
X_2 = 0

But the fft function on the HP Prime returns:
1.72
-0.05 - 0.12124355653i
-0.05 + 0.12124355653i

However Wolfram Alpha returns:
0.993042
-0.0288675 + 0.07i
-0.0288675 - 0.07i

I am confused. Are there different fast Fourier transforms or am I missing something obvious? I want to understand the basic calculation before I attempt to understand the Tukey and Cooley algorithm. Any help and insight is appreciated. Thanks!

[toshk]

since we don't time interval you are sampling i am assuming 1sec;
hence the frequency is 1; N=3; n=0..N-3; w*n=2*pi*f*i*n
e^(2*π*i*n/N)
your summing formula edited (Σ( x_n * e^(i * 2 * π * k * n / N) ) for n = 0 to N-1) in matrix form on Prime
X_K=vandermonde([1,e^(2*π*i/3),e^(4*π*i/3)])*[[0.54],[0.66],[0.52]]; and the inverse is true for X_n=inv(vandermonde([1,e^(2*π*i/3),e^(4*π*i/3)]))*[[1.72],[−0.05-0.12124355653*i],[−0.05+0.12124355653*i]]
or
simply fft([0.54,0.66,0.52])
Yes there are formulae for fft all depending how samples data are handle;(math, physics, statistics, signals...etc )

However Wolfram Alpha returns: normalize of Prime ans;
fft([0.54,0.66,0.52])/1.7211976093

[Thomas Klemm]

(12-09-2018 02:54 AM)Eddie W. Shore Wrote:  Any help and insight is appreciated.

Using your sample problem where N = 3:

\(x_0 = 0.54\)
\(x_1 = 0.66\)
\(x_2 = 0.52\)

This formula is the definition given for the discrete Fourier transform:

\(X_{k}=\sum _{n=0}^{N-1}x_{n}\cdot e^{-{\frac {2\pi i}{N}}kn}\)

We can use the following abbreviations for the 3rd roots of 1:
\(\Phi = e^{-{\frac {2\pi i}{3}}} = 1 ∠-120°\)
\(\Psi = \Phi^2 = \Phi^{-1} = 1 ∠120°\)

\(X_{0}=0.54 \cdot e^{-{\frac {2\pi i}{3}} 0\cdot 0}+0.66 \cdot e^{-{\frac {2\pi i}{3}} 0\cdot 1}+ 0.52 \cdot e^{-{\frac {2\pi i}{3}} 0\cdot 2} = 0.54 + 0.66 + 0.52 = 1.72\)
\(X_{1}=0.54 \cdot e^{-{\frac {2\pi i}{3}} 1\cdot 0}+0.66 \cdot e^{-{\frac {2\pi i}{3}} 1\cdot 1}+ 0.52 \cdot e^{-{\frac {2\pi i}{3}} 1\cdot 2} = 0.54 + (0.66 + 0.52\cdot\Phi)\cdot\Phi = -0.05000 -i0.12124\)
\(X_{2}=0.54 \cdot e^{-{\frac {2\pi i}{3}} 2\cdot 0}+0.66 \cdot e^{-{\frac {2\pi i}{3}} 2\cdot 1}+ 0.52 \cdot e^{-{\frac {2\pi i}{3}} 2\cdot 2} = 0.54 + (0.66 + 0.52\cdot\Psi)\cdot\Psi = -0.05000 +i0.12124\)

This is in accordance with the result that the HP Prime returns.

HTH
Thomas

[Eddie W. Shore]

Thank you so much! You are amazing!

Eddie