# Can one generate a signal with an asymmetric power spectral density?

My question is as follows. I know that classically speaking, if you have some time varying signal $X(t)$ its power spectral density $S_X(f)$ will be symmetric around $f=0$ as the signal is real. This is why your spectrum analyzer only shows the positive part, the negative part is redundant.

But if one has a complex signal (such as often encountered in quantum mechanics) the PSD can be asymmetric around 0. This for example leads to the asymmetry between absorption and emission in atoms: atoms have spontaneous emission (emission in the absence of photons in the environment) but not spontaneous absorption.

What I am interested in is if there is a way to physically make such a complex signal with asymmetric PSD. Maybe this can be done by using two channels that output the signal with some phase difference? I've tried looking for some source material on this but I have not been successful.

@user129412, there are many ways to generate both continuous signals and discrete sequence having asymmetric PSDs. One way, for discrete sequences, is shown in the following diagram.

$x_r(n)$ is a narrow bandpass signal having spectral energy in the vicinity of $±f_c$ Hz. The complex-valued $x_c(n)$ output sequence is your desired asymmetrical sequence having spectral energy only in the vicinity of $+f_c$ Hz. (If your Hilbert transformer has M taps, where M is odd, then the upper path delay must be (M-1)/2 samples to achieve time synchronization between the two signal paths.) If you desire sequences whose asymmetrical spectra are centered at zero Hz, many ways to obtain such sequences can be found at:

https://www.dsprelated.com/showarticle/153.php

You are right that you need two signal paths: a complex signal is in fact two (real-valued) signals, combined as the real and imaginary parts of a complex-valued signal:

$$z(t)=x(t)+j\cdot y(t)\tag{1}$$

where $x(t)$ and $y(t)$ are real-valued.

Assuming that $x(t)$ and $y(t)$ are jointly wide-sense stationary (WSS) random processes, it can be shown that the the PSD $S_{zz}(\omega)$ of $z(t)$ is given by

$$S_{zz}(\omega)=S_{xx}(\omega)+S_{yy}(\omega)-2\,\text{Im}\{S_{xy}(\omega)\}\tag{2}$$

where $S_{xx}(\omega)$ and $S_{yy}(\omega)$ are the PSDs of $x(t)$ and $y(t)$, respectively, and $S_{xy}(\omega)$ is the cross-power spectral density of $x(t)$ and $y(t)$. Note that in $(2)$, $S_{xx}(\omega)$ and $S_{yy}(\omega)$ are even functions, but $\text{Im}\{S_{xy}(\omega)\}$ is odd, so $S_{zz}(\omega)$ is generally asymmetric.