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

  • $\begingroup$ 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 '19 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 '19 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 '19 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 '19 at 17:45

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.

| improve this answer | |
  • $\begingroup$ 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 '19 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.