[SOLVED] inverse of FFT not the same as original function

inverse of FFT not the same as original function

I don't understand why the ifft(fft(myFunction)) is not the same as my function. It seems to be the same shape but a factor of 2 out (ignoring the constant y-offset). All the documentation I can see says there is some normalisation that fft doesn't do, but that ifft should take care of that. Here's some example code below - you can see where I've bodged the factor of 2 to give me the right answer. Thanks for any help - its driving me nuts.

import numpy as np
import scipy.fftpack as fftp
import matplotlib.pyplot as plt
import matplotlib.pyplot as plt

def fourier_series(x, y, wn, n=None):
    # get FFT
    myfft = fftp.fft(y, n)
    # kill higher freqs above wavenumber wn
    myfft[wn:] = 0
    # make new series
    y2 = fftp.ifft(myfft).real
    # find constant y offset
    myfft[1:]=0
    c = fftp.ifft(myfft)[0]
    # remove c, apply factor of 2 and re apply c
    y2 = (y2-c)*2 + c

    plt.figure(num=None)
    plt.plot(x, y, x, y2)
    plt.show()

if __name__=='__main__':
    x = np.array([float(i) for i in range(0,360)])
    y = np.sin(2*np.pi/360*x) + np.sin(2*2*np.pi/360*x) + 5

    fourier_series(x, y, 3, 360)

Solution

You are killing the negative frequencies between 0 and -wn.

I think what you mean to do is to set myfft to 0 for all frequencies outside [-wn, wn].

Change the following line:

myfft[wn:] = 0

to:

myfft[wn:-wn] = 0