There is probably something trivial behind this, but I am missing something. I need to create a stationary random time series data v(t) which is the the sum of another time series u(t) and u(t) with a time shift $\tau$, i.e

$v(t) = u(t) + u(t - \tau)$

The series u(t) is itself a stationary colored Gaussian time-series with some PSD $S_u(f)$. No problem! I wrote a small function (using Python) to make u(t) for a given PSD. I calculate the PSD of the the generated time series using Welch's method and it checks out.

enter image description here

If we take the Fourier transform of u(t - \tau) it should just gain an extra phase compared to the Fourier transform of u(t)

$F[u(t - \tau)] = \int dt \, u(t -\tau) \, e^{2 \pi j f t } = \int dt \, u(t) \, e^{2 \pi j f t } e^{- 2 \pi j f \tau } = \tilde{u}(f) e^{- 2 \pi j f \tau }$

Since $v(t) = u(t) + u(t - \tau)$,

$\tilde{v}(f) = \tilde{u} (f) \, (1 + e^{- 2 \pi j f \tau }) $

And the PSD of V should then be (using Euler identity)

$S_v (f) = 4 S_u (f) \cos^2 (\pi f \tau )$

In particular $S_v$ should vanish when $f = \frac{2n + 1}{2 \tau}$ with the smallest zero being at $\frac{1}{2 \tau}$. However when I test my this with the u(t) series I generated above the PSD looks different. There are zeros low frequencies which shouldn't exist. For example here is the PSD of the v(t) series and the spectra I would expect for $\tau = 2$. (The first zero should be at 0.25 Hz)

enter image description here

Moreover the zeros in the data seems to change with the sampling frequency I am using. The v(t) series for the above plot has a sampling frequency of 0.25 Hz. Here is one with 0.5 Hz, keeping everything else the same.

enter image description here

I am kind of baffled by this. Any help would be greatly appreciated. Sorry about the long post, and thanks in advance!!

EDIT: After using a higher sample rate as suggested by Hilmar below, and fixing the error I described in the comment everything looks good. Spectrum for the fixed case below.

enter image description here

  • $\begingroup$ plot your data on a linear frequency scale. the nulls are periodic in frequency $\endgroup$
    – user28715
    Jun 22, 2019 at 12:10
  • $\begingroup$ At a sample rate of 0.25Hz and a delay of 2 seconds, there shouldn't be any null at all $\endgroup$
    – Hilmar
    Jun 22, 2019 at 12:11

1 Answer 1


Your math is correct, for a delay of 2 seconds, the first null in the PSD should be at 0.25 Hz.

I'm guessing the problem is in the code you use to generate the signals and/or the PSD. The sample rates seem awfully low. At a sample rate of 0.25 Hz, your 2 second delay is just half a sample, that means you would have to implement a fractional delay to generate the signal. T

  • $\begingroup$ You are totally right!! The low sampling frequency should have had required a fractional delay. But I had further messed up the time delay part. In the code the time delay should be by something like $f_s \tau$ samples of the time series, where as I was doing $\frac{\tau}{f_s}$ samples. While this gave me an integer delay for the time series, it was of course the wrong time shift. I added a plot when I do the right time shift with a higher sampling frequency and things look okay. Thank you very much! $\endgroup$
    – elenasto
    Jun 22, 2019 at 13:21

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