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No, there is no standard way to represent a complex passband signal in baseband. However, this is not often a problem, at least in communications, because most passband signals of interest are real (see counter-examples in the comments).
In general, the spectrum of a complex passband signal $s(t)$ has no symmetries. This is an example of the magnitude ...
3
Yes and no.
If you recall from Calculus, if $g(t) = \frac{d}{dt} f(t)$, then $\int g(t)\ dt = f(t) + C$. So, since the derivative of a sinusoid is a sinusoid of the same frequency, if $g(t)$ sinusoidal, its integral must be a sinusoid of the same frequency, plus a constant.
Whether that "plus a constant" bit is going to make you consider the ...
3
From what you've said, you have two sample sets:
$$x_n, n = 0 \ldots N-1$$ and $$y_m, m = 0 \ldots M-1$$
where $M \ne N$ and you want to compare the distributions of the underlying processes.
Do you know anything about the expected distribution of the measurements?
If they're Gaussian, then you can just calculate the sample mean and sample standard deviation ...
2
Synchrosqueezing. Diminishes Wigner-Ville interaction disadvantages as described here.
2
The problem with the first point is that if we cannot restrict the set of possible answers, then we will not be able to identify the ground truth components. So we will have a solution that is as useful as the others, but we cannot guarantee that what we found are the components that generated the data.
E.g. if you have the mix of two people talking, you ...
1
It's easy to tell if you can find passages in the track that are "hard panned". If there is mostly time difference, it's A/B, if it's mostly level difference it's X/Y.
You can also look at the average power spectrum of the sum and difference channel. I'm suspecting that A/B will show some signs of comb filtering correlated to the microphone spacing....
1
What I need to do is rectifying the signal becomes no negative values but still has same mean and standard deviation as the original signal as seen here in Fig 4
You can't. If you rectify it the mean MUST change (unless the signal is all positive to start with). The paper does just a simple rectification and does NOT preserve mean or standard deviation.
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There is no free lunch here.
Can the time between signal pulses be used to encode additional bits of information over and above that encoded in the standard way?
I believe, this assumes that there are only very short signal pulses and the signal is zero otherwise. That's typically not the case in a "normal" communication system. The communication ...
1
From the DSP point of view it is possible, and if you produce and detect pulses this fast you could also transmit $10^{16}$, what is the minimum/maximum interval between two pulse?
Let's assume you have a distribution for the pulse length $p_k = p(k\, T)$. The maximum information can be transmitted per pulse is given by the entropy of this distribution.
$$ -\...
1
If the Fourier transform
$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$
of a function $x(t)$ exists, then the function $x(t)$ can be expressed in terms of its Fourier transform:
$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\tag{2}$$
Eq. $(2)$ is also called inversion formula. Note that $\omega$ is a real ...
1
There’s yet another possible answer if the positive and negative frequencies in a complex signal have non overlapping spectrum. Just take the real part.
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Yes. This is commonly done in IQ SDRs which use a (often Tayloe IQ mixer) sampling frequency offset in an earlier wider bandwidth intermediate frequency or IF stage IQ data stream from the signal frequency of interest (to help reduce the effects of first stage IQ mixer IQ imbalance or offset on the signal of interest).
To convert the spectrum at some offset ...
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All time-frequency representations that can be formulated as part of Cohen's class (bilinear ones) will have cross-terms. There are post-processing / filtering approaches (as mentioned on the Wikipedia page) that allow them to be filtered out.
OverLordGoldDragon's approach seems more sensible than anything in Cohen's class --- including the Wigner-Ville, ...
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Here is a demonstration that confirms what @Hilmar said:
We recall the initial relationship:
\begin{equation}
s_{hp}[n]=(−1)^ns_{lp}[n]
\end{equation}
Let us calculate its discrete Fourier transform:
\begin{equation}
DTFT\{s_{hp}[n]\}=S_{hp}[k]=\sum_{n=0}^{N-1}\{(−1)^ns_{lp}[n]\}e^{-j2\pi \frac{nk}{N}}
\end{equation}
But $(-1)^n$ is also equal to $e^{j\pi n}$...
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