Here I answer the question, "Can time aliasing in convolution cause peaks?". It showed up in a recent question, Random Peak at the end Impulse Response, where users suggested time aliasing as a potential cause. The answer is - meaningfully, "no" - strictly, "no" for 'full' as the reference convolution, and "yes" for 'same' but the "caused" peak should be there. That's per a major definition of "peak" - a general discussion is in a later section ("Mini-peaks?"), but long story short: if in doubt, try non-circular convolution, but if the peak is the absolute maximum, it probably should be there.
Convolution is a locally stable operator: outputs close in time are correlated, and most filters will prohibit "wild variations". It's an intuition that can be formalized - instead, I prove the exact claim.
My implied and equivalent claim is, "circular convolution cannot cause peaks that don't already exist with a linear (zero-padded) convolution". What this means is, if we take sequences $a$ and $b$ that are separated in time, with more zeros between them than the filter's support; $a$ is left, $b$ is right. The "yes" case means that joining these sequences can create a peak, without peaks already being created by convolving with the separated $a$ and $b$.
The 'same' vs 'full' distinction is due to the "peaks already being created" part; circular cross-correlation cannot make peaks that are already present in 'full' convolution, but it can make peaks that are not present in 'same' convolution; put differently, if the existing peaks are located inside of bounds defined by 'same' convolution, then circular cross-correlation cannot make new peaks.
For those who want a quick read, the proof actually ends in equivalently three sections, as will be stated.
The setup, and Case 0
On left, we begin with a certain summation - $\sum_n a[n] \cdot h[n]= K_1$. If we slide the filter left, we get another summation, that's potentially larger - the greatest potential is with it fully inside, producing an identical output vs. joined case. Now imagine sliding $b$ closer and closer left. As it begins to overlap $h$, the sum becomes $K_1 + \sum_n b[n] \cdot h[n] = K_3$. Let $K_2$ be the result with the filter over $b$'s edge (still unjoined). Clearly, $K_3 = K_1 + K_2$.
We define "peak" as a value that's more than double the next highest value - put differently, "a peak is a point with more than double the max value of a sequence without the peak". Now, clearly, $|K_3| \leq |K_1| + |K_2|$. To simplify, let all be positive. The highest value that $K_3$ can be, is if $K_1$ and $K_2$ are both maxima of the respective outputs. Following our definition of "peak", since $K_3 = K_1 + K_2$ and $K_1$ and $K_2$ are the maximum and next-maximum (or vice versa) of the convolution before joining, the best case scenario is if $K_3 / \texttt{max}\{K_1, K_2\}$ is maximized, which is achieved with $K_1 = K_2$. This makes $K_3$ exactly double the next maximum.
Now it's possible to skip to "Finalizing" and/or "Alternative proof", then "Out-of-bounds case" to end the proof, and "Peaks odds" and "Mini-peaks?" for extra info. I dug in arithmetic details before I took a bird's-eye view, and it took long enough. But below maybe has more insight.
Other shift cases
To see if better can be done, we exhaust our options. For now, assume $a,b,h\geq 0$.
- $K_1$ and/or $K_2$ aren't maxima. Then the ratio of $K_3$ to the existing maximum is even lower.
Let $J_1, J_2$ be $K_1, K_2$, except the filter is to left of $a$'s edge and to right of $b$'s edge by in respective cases, by arbitrary amounts but never fully inside of either so that joining makes a difference. In the joined case, let $s_1$ be the contribution from $a$, and $s_2$ from $b$, in the multiply-sum with $h$, for any arbitrary case, and $s_3 = s_1 + s_2$ (by definition). Note, now $J_3$ doesn't necessarily equal $J_1 + J_2$; they by definition correspond to arbitrary and independent shifts of $h$.
- Let $s_1=J_1$ and $s_2=J_2$. This is same as "Case 0".
- Let $J_1, J_2$ be maxima of convolutions with $a, b$, and let $s_1=J_1$ and $s_2 < J_2$. Since the best case is $J_1 = J_2$, hence $s_1 = s_2$, this cannot be double or more of $\texttt{max}\{J_1, J_2\}$.
- Case 3, except $s_2 > J_2$. This is impossible by definition.
- Let neither of $J_1, J_2$ be maxima of convolutions with $a, b$, but let them be maxima within the filter's support with respect to each edge. Since by definition $s_1 \leq J_1$ and $s_2 \leq J_2$ and the maximum $s_1 + s_2$ is $J_1 + J_2$, yet there's points greater than $J_1, J_2$, $J_1 + J_2$ can never reach double the maximum.
- Case 5, except $J_1, J_2$ aren't maxima within the filter's support with respect to each edge. This is a contradiction, such maxima exist per how $J_1, J_2$ are defined.
- Redefine $s_1$ to permit shifting $h$ to left of $a$ by more than the support of $h$, but $s_2$ stays the same. Then $s_2=0$, and $s_1$ can never exceed the existing maxima of convolutions with the unjoined $a, b$.
- Case 7, except swap $s_1, s_2$, same conclusion.
This exhausts every possible location of $h$ before or after $a$ and $b$ are joined, and all pre-join proximities of $a$ to $b$. That's all possibilities. Except, we've assumed $a, b, h \geq 0$.
Arbitrary $a, b, h$
The case for $a, b, h \leq 0$ is completely analogous, with the definition of "peak" amended by replacing "maximum" with "absolute maximum". Now let $a,b,h$ be anything, except complex-valued for now. If all positives (or sign-aligned) case is the best possible case, then all that remains is to prove that.
The proof is simple: if signs for any $a[n_1], h[n_1]$ or $b[n_2], h[n_2]$ aren't aligned, then they're opposite, meaning the sums are less than they would be with alignment: $\left|a[n_1]h[n_1] + b[n_2]h[n_2]\right| < \left|a[n_1]h[n_1]\right| + \left|b[n_2]h[n_2]\right|$. For the general case where we don't know of alignment status, $<$ becomes $\leq$. Hence, $|s_1 + s_2| \leq |s_1| + |s_2|$ in all cases above, yet above is $|s_1 + s_2| = |s_1| + |s_2|$. Since all of the above always strives to maximize $|s_1 + s_2|$, and the general case can do no better, this proves that the sign-aligned case is the best case, and that case itself has been proven, which exhausts all real-valued possibilities.
Now for complex values. Again, it suffices to prove that it's upper-bounded by real-valued sign-alignment. For this, we realize that positive real values are sign-aligned and phase-aligned complex values, whose absolute product doesn't care, but whose absolute sum is strictly greater than with sign-unaligned or phase-unaligned complex values, and hence likewise sum of products.
Finalizing
Thus, or rather it's all a long way of saying,
$$
|s_1 + s_2| \triangleq
\left|\sum_{n=0}^{N_a - 1} h[n] a[n] + \sum_{n=N_a}^{N - 1} h[n] b[n]
\right|
\leq
\sum_{n=0}^{N_a - 1} |h[n] a[n]| + \sum_{n=N_a}^{N - 1} |h[n] b[n]|
$$
and
$$
\sum_{n=0}^{N_a - 1} |h_{nn}[n] a_{nn}[n]| + \sum_{n=N_a}^{N - 1} |h_{nn}[n] b_{nn}[n]|
=
\sum_{n=0}^{N_a - 1} h_{nn}[n] a_{nn}[n] + \sum_{n=N_a}^{N - 1} h_{nn}[n] b_{nn}[n]
$$
where $a_{nn}, b_{nn}, h_{nn} \geq 0$, and,
$$
\frac{|s_1 + s_2|}{\texttt{max}\{|a * h|, |b * h|\}} \leq 2
$$
which finally, per Cases 7 & 8, is same as
$$
\boxed{
\frac{\texttt{max}\{|[a, b]*h|\}}
{\texttt{max}\{|a * h|, |b * h|\}}
\leq 2
}
$$
In words, convolution of a filter with the sequences joined can't have a value that's more than double the max value of convolution of a filter with the sequences separately.
Of course, by "peak", we usually mean something that's much more than just double the next-highest value - which is, of course, much more than what's already impossible. And, by "peak" we also mean more than just a single point - which just means amending the definition of "next maximum" to exclude all such points, which only lowers the reference "maximum" in above cases. This of course allows exceeding $2$, but not by a factor greater than the ratio of the actual next-maximum to new next-maximum, which simply exposes an oversimplification in the technicality of the definition: we still cannot create peaks that weren't already there.
The proof in continuous time is completely analogous, barring functions that warp reality and break space-time continuum. This concludes the proof.
Interestingly, while time aliasing in convolution cannon create peaks, it can very much destroy them: $K_1 = -K_2$.
Alternative proof
Output of convolution of the joined sequence equals the overlap-adding of outputs of the separate convolutions, which can at best double the maximum of either separate output. I didn't realize this immediately because I'm not very smart.
Out-of-bound case
The exhaustive proof forgot to amend the definition of "peak" to handle an important practical case: the peak being centered to right of $a$ and to left of $b$, but the "reference convolutions" with $a$ and $b$ excluding such points.
Consider the convolution of $e^t$, that's zero past some $t > t_0$, with itself; another remark is that our math so far was based on cross-correlation, and it didn't matter, but it does now. Let $a = h = e^t$, and forget $b$. At the right edge, we're looking at $e^t$ vs $e^{-t}$; now, the difference between the final output point of 'same' and 'full' is colossal. Circular convolution is same as joining $b = a$ onto $a$; the 'full' tails of both will fully overlap. In this example, the joining isn't what causes the peak, but simply the equivalent output domain extension; circ-conv always operates per 'full' in the sense that it guarantees a complete sweep over every input point.
In fact, this is always the case; joining cannot cause peaks - per "Alternative proof", we're just overlap-adding the bounds' tails. So only completed sweeps can cause peaks - and since the full-information convolution is meant to complete the sweep, this cannot be considered a "distortion". Put differently, the peaks cannot be caused by the filter "wrongly interacting" with the joined segment - if the peaks are there, they should be there.
The peak potential is unbounded; let $a=h=$ unit impulse on right edge; 'same' output is zero, 'full' and circular is non-zero, ratio $= \infty$.
Peaks odds
There's little guesswork - the answer to "did time aliasing cause peaks" is same as to "are there peaks near edges of left-zero-padded 'full'" (because, 'full' only fully sweeps to the right of input, not left) - or more precisely, outside of bounds of 'same'. A more important question is, should the "caused" peak be there - per the general definition of convolution, there's no such thing, but practically, it depends on whether the "completed sweep" isn't something that should be happening.
The more precise criterion is, "within $h$'s half-support of 'same''s bounds", since that's how much 'same' discards relative to the left-zero-padded 'full'.
Mini-peaks?
We get it, no new dramatic absolute maxima. What of simply sharp rise-falls, or pulses? All hell breaks loose if we don't define these precisely, but let's just say, if you're worried about detrimental peaks of any sort, try non-circular convolution. But if the peaks aren't within $h$'s full support of the boundaries, it can't possibly be the cause, in any sense at all!
Below, any convolution discussion doesn't concern itself with 'same', only 'full'.
Can time aliasing (general) cause peaks?
Absolutely: Subsampling in time <=> Folding in Fourier. "Folding" is fancy for "overlap-add", and addition is unbounded - so just subsample more, or sample more sparsely, in other domain, to explode the other domain.
"But by convolution theorem, $x*h \Leftrightarrow X\cdot H$, and $Z=X\cdot H$ can be anything?" $X\cdot H$ can't be anything. The continuous-time equivalent is for $x$ to be a joining of $a, b$, and "joining" means not overlapping, which constrains $X$ in terms of $A$ and $B$, and the objective is defined in terms of $X, A, B$. In discrete-time (infinite discrete), it means the amount of permitted time aliasing is limited - note,
$s_1 + s_2$ is identically the overlapping of the outputs of the separate convolutions with $a, b$.
Fourier arguments (failed)
I initially tried to prove this by working purely in Fourier domain. It either can't be done, or I've not managed - the Fourier domain is too general and there appears to be no clear way to encode the given restriction (except, I suppose, in the way I did in the previous section, but that doesn't handle CT or D). Just for sake of informativeness, I describe some of it.
The setup is $O_s(\omega) = H(\omega) \cdot \left(A(\omega) + e^{-j\omega s}B(\omega)\right)$, where $s$ is the amount of shift, and $a(t), b(t)$ are originally separated by more than the support of $h(t)$. The idea is to run metrics on $s=0$ (unshifted) vs joined, $s_j = -\texttt{supp}\{h\}$. We don't know what $s_j$ is, and it doesn't matter (for what I tried), but we know $|s_j| > 0$. Failed:
- Find $\texttt{max}\{|o(t)|\}$ as described here - unbounded ratio in absence of constraints
- Parseval-Plancherel's theorem: forgot, something about bounding and uninformativeness
- (1) but using $|O_s(\omega)|$ - this went far, but failed, and I forgot why for both
