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Let $x[n]$ be aliased sampling of real-valued $x(t)$ over $t_0 \leq t \leq t_1$. Can $|x(t)|^2$ be recovered more accurately than $x(t)$, over $t_0 \leq t \leq t_1$? If so, how?

For $|x[n]|^2$, superficially it's "yes", as aliasing can circularly shift certain $X[k]$, which $|x[n]|$ is invariant to. For $|x(t)|^2$, I see potential as it's lower in information.

Feel free to assume the amount of aliasing to be "limited", by whatever amount necessary (but do specify it).

Clarifying

"Standard theory", meaning sinc interpolation, says "no", and that much is clear. All modulus is aliased, since it's many-to-one. Yet this same flaw is a strength in the general framework, as $|x|$ is lower-information, hence inherently takes less to recover. For example, synchrosqueezed STFT can beat sinc interpolation for non-uniform sampling.

For complex $x$ that's a bandpass signal, $|x|$ can often be recovered more accurately, as modulus heavily shifts energy toward lower frequencies, even if new higher ones are introduced. Different with real $x$, but case in point, I'd not conclude it can't be done just because sinc can't do it.

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Is square of signal more recoverable than signal itself?

Generally no. The spectrum of $x^2(t)$ is the convolution of $X(\omega)$ with itself. This also means that if $x(t)$ has a bandwidth of B, $x^2(t)$ will have a bandwidth of 2B. It has twice the bandwidth so it's much more likely to create more aliasing.

Practical example: in audio processing, the signal is often up-sampled before doing non-linear processing (like squaring) to reduce the aliasing caused by the extended bandwidth.

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  • $\begingroup$ Consider modulated BPSK (0 180 degree modulation) with no filtering. It has infinite BW with a Sinc shape but on a practical level the BW of the main lobe is twice the modulation rate. Interestingly when we square this, all modulation is stripped and we get a single tone at twice the carrier frequency and DC (in the original modulated signal the carrier was suppressed). For typical filtered BPSK, the modulation is also suppressed, and 2x carrier recovered, but the doubled modulation will exist at a much lower level at twice the BW as you suggest. Doubling QPSK produces a same BW BPSK sig. $\endgroup$
    – Dan Boschen
    Jan 9 at 13:16
  • $\begingroup$ ah, but that assumes symbol-synchronous sampling, @DanBoschen; that's not the same as being able to reconstruct an arbitrary signal from a sampled version :) $\endgroup$
    – Marcus Müller
    Jan 9 at 13:23
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    $\begingroup$ @MarcusMüller No I don't think so? We can arbitrarily sample the signal as long as we obey Nyquist (and oversampled enough to also include the 2x frequency, otherewise get an aliased version of that) and this effect will occur. Such carrier recovery using squaring can be done in the continuous time domain or at any sample rate prior to timing recovery. So basically bottom line for any modern modulation, squaring destroys any chance of recovering the signal. $\endgroup$
    – Dan Boschen
    Jan 9 at 13:44
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Can |x(t)|² be recovered more accurately than x(t), over t0≤t≤t1? If so, how?

The situation for a sampling of $|x|^2$ can, in the best case only be as most as good as for $x$, unless frequency components in $x$ cancel upon squaring, which is a huge exception, and needs spectrum to be rationally related, and phases to be definite - that's such a significant restriction that I wouldn't call it a "continuous-time reconstruction from a sampled signal".

You can spectrally motivate that: squaring introduces intermodulation products of all the oscillations in $x$, thus adding new and higher frequencies to your aliasing.

Yet this same flaw is a strength in the general framework, as |x| is lower-information,

It's exactly 1 bit lower in information – a sign information. But being continuous-valued, it has infinite information anyway, so this argument doesn't work.

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  • $\begingroup$ Don't think it's worth staining an otherwise valid answer with the last paragraph's oversimplification. Sign is arguably half the information. $\endgroup$
    – OverLordGoldDragon
    Jan 9 at 14:32
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    $\begingroup$ no, it's not half the information, it's literally 1 bit, assuming the signal is as likely to be positive as negative at any given time; $$\begin{align}I(\text{sign}) &= -P(\text{sign} = +) \cdot \log_2 [P(\text{sign} = +) ] -P(\text{sign} = -) \cdot \log_2 [P(\text{sign} = -)]\\ &= - \frac 12 (-1)+ \frac12 (-1) \\&= \frac12 +\frac12 \\ &= 1.\end{align}$$ (any other distribution of signs has even less entropy in the sign.) $\endgroup$
    – Marcus Müller
    Jan 9 at 15:00
  • $\begingroup$ What if we encode ALL the information in the sign (such as BPSK)? Is that valid example to say there are cases where we could lose ALL the information? $\endgroup$
    – Dan Boschen
    Jan 10 at 3:51
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    $\begingroup$ @DanBoschen sure, but then we're not talking about the reconstruction of an arbitrary analog signal, but the reconstruction of the digital signal that last to that $\endgroup$
    – Marcus Müller
    Jan 10 at 9:15
  • $\begingroup$ @MarcusMüller Excellent point, which brings us back to just being 1 bit! And when you square QPSK which is 2 bits, you get BPSK, again losing 1 bit. Cool. Very insightful. $\endgroup$
    – Dan Boschen
    Jan 10 at 11:20

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