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I was reading a relatively old paper from the 1970s on smoothing by FT methods (chemistry applications), where the authors show that if we do rotation translation operation on the signal (y- values) before filtering the data by cutting off higher frequencies, the output is free from various artifacts and oscillations. What that rotational-translation does is that it makes the start point and the end-point equal to zero (see equations in the original text).

I have not seen this type of operation mentioned in modern texts. Here is the explanation by the author of Digital smoothing of electroanalytical data based on the Fourier transformation, 1973.

I did not see any major advantage in MATLAB if we use an example of a noisy Gaussian peak. Was this suggestion an artifact of how FT algorithms worked in the 1970s or there is a theoretical advantage? I assume the author is trying to periodize the signal?

In the figure, please follow the arrows from the bottom to see the rotation-translation operation. Has anyone seen this equation elsewhere and what could be the fundamental advantage of this operation? Thanks.

enter image description here

enter image description here

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  • $\begingroup$ Um, I'm going to be that person: this doesn't look like they understand what they're doing, and you shouldn't try to understand where there's no sense. $\endgroup$ – Marcus Müller Oct 17 at 15:07
  • $\begingroup$ That group was a very respectable group of spectroscopists in the country and I see this operation used even in chemical applications of wavelets today as a data pre-treatment/denoising to make the start and ends of the data set equal to zero before doing FT to remove artifacts using equations 1,2, and 3. $\endgroup$ – M. Farooq Oct 17 at 15:22
  • $\begingroup$ yeah, but the problem is this: we can describe the operation "rotate the signal" (rotation in a very geometrical sense) very well in terms of matrix descriptions that involve trigonometric functions – and these can be very well understood in Fourier domain. Especially, this rotation means you frequency-shift the whole signal by a time-dependent beat-frequency. But that changes the frequency content of the signal, which makes the filtering step more than questionable, as that rotation is a non-linear operation and can't be undone like they do it – I honestly fail to see the strict mathematical $\endgroup$ – Marcus Müller Oct 17 at 17:41
  • $\begingroup$ relation between in- and output, especially in the presence of noise on the last, least-amplitude samples. Also, "a simple, satisfactory approach was found" literally says that this works for them, but that they don't want to give a mathematically clean analysis of what they're doing. And: a (specific) wavelet might be far less sensitive to the effects of the rotation than the plain Fourier transform. $\endgroup$ – Marcus Müller Oct 17 at 17:45
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Upon closer inspection, this doesn't even seem to be a rotation:

$R_n$ being the output sample corresponding to the input sample $A_n$, we can understand their equation

$$R_n = A_n - \Delta_n \tag1$$

as the statement that we add sequence $(-\Delta_n)_{n=1,\ldots,k}$ to the input signal $a$ and get the output signal $r$. I'll call that sequence $s$, $S_n := -\Delta_n$.

Let's look at this not on the individual sample level, but from a signal level:

$$r = a + s\tag{*}\label{additive}$$

From $\eqref{additive}$ we can see that this is just adding another signal to the input signal.

Let's have a closer look at $S$, as they claim it compensates spectral components present in $R$:

\begin{align} s_n &= \underbrace{-A_1}_{\text{const.}} - \frac{\overbrace{(A_k - A_1)}^{\text{const.}}(n-1)}{\underbrace{k-1}_{\text{const.}}}& \text{eq. (2) from paper}\\ &= -A_1 + \frac{A_1-A_k}{k-1}(n-1)\\ &=\underbrace{\frac{A_1-A_k}{k-1}}_{c_1}n -\underbrace{\left(\frac{A_1-A_k}{k-1}+A_1 \right)}_{-c_2}\\ &=c_1n+c_2 \end{align}

That is... underwhelming. The magical rotation is just an addition with a ramp of slope $\frac{A_1-A_k}{k-1}$. So let's look at the FFT of that:

\begin{align}\DeclareMathOperator{\dft}{\mathrm{DFT}_k} \dft\{s\}[l] &= \dft\{c_1n\}[l] + \dft\{-c_2\}[l]\\ &= \begin{cases} -kc_2+c_1\displaystyle \frac{k(k-1)}{2},&\quad l=0\\ -c_1\displaystyle\frac{k}{1-e^{-j2\pi l/k}},&\quad l\in[1,k-1] \end{cases} \label{matt}\tag{;}\\ \end{align}

Eq. $\eqref{matt}$ follows according to Matt.

I see absolute no sense in these operation – it actively introduces an error term, that will skew an estimate the original exponential decay signal model, to the signal – without that actually being there.

So, that "pseudoration" actually fakes an exponential decay component that depends less on the actual exponential decay of the signal than on the observation length $k$. That transform is bad and you shouldn't use it.


Generally, applying a rectangular window in frequency domain – which seems to be what they're aiming for – is simply a bad idea for circular convolution reasons. We've got a pretty good answer explaining why that's the case. The fact that this isn't bad here is that there's little signal at the end of the observation anyway – but I don't see how folding that over your beginning actually improves the data! It just makes it look smoother.

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  • $\begingroup$ Thanks for analyzing that. I was indeed wondering about this "rotation" because I was treating a noisy Gaussian peak and it "operation" was doing no good! Yes, it did make the first and the last point zero. $\endgroup$ – M. Farooq Oct 17 at 21:06

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