0
$\begingroup$

I consider an array:

import numpy as np
from scipy.fft import fft
from scipy.signal import hilbert

a=np.random.rand(5)

First I manually compute the fourier spectrum of the analytic signal,i.e. by zeroing the negative frequency terms and doubling the positive frequency terms:

a_spec_manual = fft(a)
a_spec_manual[1:len(a)//2+1] = 2*a_spec_manual[1:len(a)//2+1]
a_spec_manual[len(a)//2+1:] = 0

Then I auto-compute the fourier transform of the analytic signal derived from a:

a_fft_analytic = fft(hilbert(a))

I was expecting a_fft_analytic to be equal to a_spec_manual. However, it is not the case. a_fft_analytic does have negative frequency components and I wonder why?

If indeed the first case is correct, is it possible to obtain the fourier spectrum of the analytic signal from real using single fft operation?

$\endgroup$
2
  • $\begingroup$ They are equal. Did you mean to use an even len(a)? See assert np.allclose(a_fft_analytic, a_spec_manual) $\endgroup$ Apr 5, 2022 at 22:35
  • 1
    $\begingroup$ I am sorry, I got confused with the negligible terms in the negative frequencies of the order of 10-17 for a_fft_analytic case. Probably these are rounding errors. However, I am actually trying to compute cross-correlations. I will edit the question and post again. Thanks for your reply. $\endgroup$
    – Arnautovic
    Apr 5, 2022 at 22:47

1 Answer 1

0
$\begingroup$

They aren't negative.

For the odd-length case, your code is already correct: assert np.allclose(a_spec_manual, a_fft_analytic).

For the even-length case, the Nyquist bin should not be doubled (briefly, it is already doubled in some sense, see link).

Following your comment: yes, 1e-16 is "float zero" (and for float32 that's 1e-7). FFT takes no shortcuts (always does compute), so even "perfect" examples won't have exact zeros.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.