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If x1 = x0[::2] is unaliased subsampling, then $E(x_1) = E(x_0)/2$, which can be proven via Parseval's theorem.

For same $x_0, x_1$, however, E(x1[:a]) = E(x0[:2*a])/2 doesn't hold: that is, unpadding identical portions of the two waveforms that are identical in information contents:

Unpadding aliases (proof omitted) - but even so, above is strange: full signal's energies are equal (within 1/2), but not half.

Is there a relationship, similar to subsampling, but for unpadding, when $x_1$ is an unaliased subsampling of $x_0$? If not, can this be proven? -- Code -- $E(x) = \sum |x|^2$

Note: answer accepted with disagreement on last paragraph, stated in a comment.

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    $\begingroup$ It would help if you define the term "unpadding" with a precise equation and not just a code snippet. The way it's written it's hard to understand what exactly you find "strange". Unpadding works fine if you maintain the length of the sequence. If you truncate, the signals are not bandlimited anymore hence you get aliasing: The spectra of $x_1$ and $x_0$ are different and $x_1$ is NOT an unaliased subsampling anymore. That feels obvious to me (and I could easily prove it) but maybe I misunderstand what you are asking? $\endgroup$
    – Hilmar
    May 18, 2022 at 12:22
  • $\begingroup$ @Hilmar I failed to realize that trimming and subsampling commute... that's the answer, feel free to post -- thanks. $\endgroup$
    – OverLordGoldDragon
    May 18, 2022 at 21:33

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From the comments:

Unpadding works fine if you maintain the length of the sequence. If you truncate, the signals are not bandlimited anymore hence you get aliasing: The spectra of x1 and x0 are different and x1 is NOT an unaliased subsampling anymore.

Expanding a bit:

For $x_1[n] = x_0[2n]$ to be an "unaliased subsampling", both $x_1$ and $x_0$ must be infinitely long. In order to not alias, $x_0$ must be band-limited which means it must be infinitely long in time (and so would be $x_1$). And truncation will result in infinite bandwidth.

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  • $\begingroup$ This answer could do without the last paragraph, we don't need to imply there's some sort of distortion despite DFT being zeros N//4 to N//2. If we insist on pedagogical correctness, $x_1$ and $x_2$ are "implicitly periodic". $\endgroup$
    – OverLordGoldDragon
    May 19, 2022 at 20:04

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