\$x(t)\$ is a function of time. Physically, it can be voltage, displacement, magnetization and so on. It can be real, complex, vectors or more fancy numbers.

\$C(t)\$ is the auto correlation function of \$x(t)\$ , defined as:

$$ C(t)=\langle x(t) x^*(0) \rangle := \lim_{T\rightarrow \infty}\frac{1}{T}\int_0^T x(t'+t) x^*(t') dt' \tag{1} $$

\$I(\omega)\$ is the spectrum of \$x(t)\$, it is the Fourier transformation of \$C(t)\$, or it is the square of Fourier transformation of \$x(t)\$. These two definition are equivalent (apply convolution theorem): $$ I(\omega)= \int_{-\infty}^{+\infty} C(t) e^{-i\omega t} dt \tag{2} $$

$$ I(\omega)= \begin{vmatrix} \int_{-\infty}^{+\infty} x(t) e^{-i\omega t} dt \end{vmatrix}^2 \tag{3} $$

non-negativity of \$I(\omega)\$

We can show that, spectrum is non-negative: $$ I(\omega)= \begin{vmatrix} \int_{-\infty}^{+\infty} x(t) e^{-i\omega t} dt \end{vmatrix}^2 \geqslant 0 $$ And physically it makes sense, because \$I(\omega)\$ means the energy density within frequency range \$[\omega , \omega+d\omega) \$.

Energy density shouldn't be negative.

Question (1): what function \$C(t)\$ makes its Fourier transform \$I(\omega)\$ non-negative?

In practice, definition equation (2) is used rather than equation (3).

\$C(t)\$ is the inverse Fourier tranform of non-negative function \$I(\omega)\$.

I know some basic law like this: $$ \text{FT}[real,even]=real, even $$ $$ \text{FT}[real,odd]=imag, odd $$

I have no idea what should \$C(t)\$ be? $$ \text{FT}[???]=real, non \ negative $$ Understand the function class of \$C(t)\$ might help us solve Question (2), because later on, you will see, I'm doing a transformation to \$C(t)\$ (essentially it's a discretization), then perform Fourier transform. This discretization transformation might take \$C(t)\$ out of that class, therefore makes \$I(\omega)\$ no longer non-negative.

subtlety in numerics: \$I(\omega)\$ is no longer non-negative!

In numerical calculations, things become finite and discrete.

Equation (2) is modified to its numeric version:

$$ \int \xrightarrow{\text{numerics }} \sum $$ $$ \qquad \quad t \xrightarrow{\text{numerics }} t_i \quad i=0,1,2,\cdots,L $$

$$ \qquad \quad \omega \xrightarrow{\text{numerics }} \omega_n \quad n=0,1,2,\cdots,L $$ Or, an alternate way to understand numerics is to keep the continues integration, but sample the function \$C(t)\$ discretely and finitely: $$ C(t) \xrightarrow{numerics} C(t) \times \sum_{i=0}^{L-1} \delta(t-t_i) $$

Under this change, \$I(\omega)\$ is no longer non-negative.

I did some experiments, those negative value of \$I(\omega)\$ can be as large as 10% of the positive peak value of \$I(\omega)\$. Adding window function makes things better, but still have 0.1% negative spectrum density.

Only in some commensurate case, \$I(\omega)\$ is non-negative.

Question (2): is there any natural way to make \$I(\omega)\$ non-negative, in numerics?

I find adding absolute value might help $$I(\omega)\rightarrow |I(\omega)|$$ but don't see any reason, such as conservation of energy, so on

So, is there a natural way, to generate non-negative spectrum density, in numerics?

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    \$\begingroup\$ This sounds purely like a math question \$\endgroup\$ – PlasmaHH Jun 13 '18 at 6:48
  • \$\begingroup\$ Could I convince you to move this to math.se? \$\endgroup\$ – a concerned citizen Jun 13 '18 at 6:48
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    \$\begingroup\$ I think that dsp.stackexchange.com is a better place to ask the question \$\endgroup\$ – filo Jun 13 '18 at 7:02
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    \$\begingroup\$ Think about the use of the complex conjugate in the autocorrelation. That's putting the rabbit into the hat right in front of your eyes. \$\endgroup\$ – Neil_UK Jun 13 '18 at 7:10
  • \$\begingroup\$ Negative frequencies come about because of the direction of phase updating. \$\endgroup\$ – analogsystemsrf Jun 18 '18 at 5:07

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