Talk:Analytic signal

Error

The definition is not correct. The expression

${\begin{aligned}x_{\mathrm {a} }(t)&={\mathcal {F}}^{-1}\{X(f)\}\ *\ {\mathcal {F}}^{-1}\{2U(f)\}\end{aligned}}$

looks wrong: The left hand side depends on $t$ while the right hand side does not. In addition, the right hand side depends on $f$ while the left hand side does not. Please, specify set of values of $t$ and $f$, at which the equality takes place. I suspect, such a set has measure zero. dima (talk) 08:43, 16 June 2009 (UTC)

As the ${\mathcal {F}}^{-1}$ is the inverse Fourier transform operator, it's "output" is implied as function of t. I guess we could be more explicit and write the relationship as

$x_{\mathrm {a} }(t)={\mathcal {F}}^{-1}\{X(f)\}(t)\ *\ {\mathcal {F}}^{-1}\{2U(f)\}(t)$

but that looks more cluttered. Oli Filth^{(talk|contribs)} 10:23, 16 June 2009 (UTC)

And if we replace some of the context that your observation omitted:

${\begin{aligned}x_{\mathrm {a} }(t)&={\mathcal {F}}^{-1}\{X(f)\}\ *\ {\mathcal {F}}^{-1}\{2U(f)\}\quad \quad {\mbox{convolution}}\\&=x(t)*\left[\delta (t)+j\cdot {1 \over \pi t}\right],\,\end{aligned}}$

Oli's answer should be apparent to anyone who has enough background to be seriously reading this article.

--Bob K (talk) 13:36, 16 June 2009 (UTC)

Wiki is supposed to be understandable for the people without background. If one goes to wiki, it is usually because one needs to get this background. The appearence of argument $f$ in the right hand side looks just as a contradiction to the very basic concepts of mathematical function, although I believe it is a tradition to write such senseless letter among the specialists who work in this area. The correct expression for the first line might have form

${\begin{aligned}x_{\mathrm {a} }(t)&={\mbox{convolution}}{\Big (}{\mathcal {F}}^{-1}\{X\},{\mathcal {F}}^{-1}\{2U\}{\Big )}(t)\end{aligned}}$

under assumption that $2U$ is function such that $2U(z)=2\cdot U(z)~\forall z\in \mathbb {C}$. dima (talk) 06:24, 19 June 2009 (UTC)

Article limited to real valued signals?

The process of eliminating negative frequencies from a signal is more general, and also applies to complex valued signals. I believe the article could be more general, since currently it explicitly says the pre-analytic signal should be real. -Roger (talk) 19:12, 28 May 2010 (UTC)

What source are you relying on for such a generalized usage? Everything I find (like this) suggests that "the analytic signal" comes from a real signal. You're right that the process of removing negative frequencies is more general, and if you do that, what you're left with is the analytical signal of its real part, but that's already implied, and nobody does that, right? Dicklyon (talk) 06:28, 29 May 2010 (UTC)

I've only come across one reference[1](2.3). If you look at the Harley receiver, it seems to be doing just this: computing the "analytic signal" (and frequency shifting) to remove negative frequencies, then taking the real part. This only agrees with part of the definition given in this article though, since the process isn't reversible when the input is complex. -Roger (talk) 16:26, 29 May 2010 (UTC)

Interesting. It says "A signal with positive-frequency components only is called an Analytic Signal." But that seems to me to be an idiosyncratic definition (that is, one unique to that author). He admits that "Most texts only consider the special case of real input signals," (in reference to the filter, not the definition) and doesn't provide any examples of sources that use his interpretation. So probably it's not a good idea to propagate that novel interpretation here. If you want to describe the Hartley reciever in an article like single sideband, that would be good. Dicklyon (talk) 13:30, 30 May 2010 (UTC)

If you want to remove negative frequencies from complex-valued functions, then this transform will not be reversible (injective), so it will lose its main property. Incnis Mrsi (talk) 14:55, 29 May 2010 (UTC)

True, but regardless I think the complex input case is worth a mention. -Roger (talk) 16:31, 29 May 2010 (UTC)

Small edit: To me, "former" and "latter" is confusing in the first paragraph. It would be clearer to rewrite the sentence in a way that states directly what former and latter reference. Since I came to this article to understand what an analytic signal is, I can't suggest how it should be rewritten.

Archive created

A small number of messy posts prior to 2009.

• Talk:Analytic signal/Archive 1 — MaxEnt 09:28, 7 April 2014 (UTC)

Analytic representation

I was wondering why this page highlights the wrong title phrase at the outset of the lead.

It does seem on a Google search that the present title better discriminates. But then it's odd the text focuses instead on the title not used. — MaxEnt 09:30, 7 April 2014 (UTC)

My answer is that it's one way to make the point that both are acceptable / customary / synonymous, just as Fourier Series, Fourier Series Expansion, and Fourier Series Representation are all synonymous. If you prefer an explicit statement, I have no objection.

--Bob K (talk) 19:48, 8 April 2014 (UTC)

A possible addition to the paragraph discussing passband signals

There is an additional possibility for illustrating a similar operation for envelope determination and that is in the application where a passband is not encountered. For example instead of the 'center of the desired passband' which may be confusing to some students, especially those not in EE, one could introduce the fairly simple concept of spectrum centroid of an analytic signal, without reference to filtering or another class of signals (passband signals). This is especially easy with analytic signals, as the same concept in real signals would require the second moment visualization, which is a disadvantage of real signals. In an operation of baseband determination along these lines, inputs to a multiplicator would include the analytic signal and a sinusoid equivalent to its spectrum centroid. In a paper I'm writing I don't have the freedom to employ analytic signals and as such have to refer to the second moment. Groovamos (talk) 00:18, 14 April 2020 (UTC)

In my experience, EEs prefer their mathematics with application examples, such as passband signals and filtering. I don't think it's expecting too much for "non-EEs" to grasp those concepts and generalize them. If you are still concerned, I would prefer that you change 'center of the desired passband' to 'center of the desired passband' (note the WikiLink) or that you import one of the images from there to here.

--Bob K (talk) 13:16, 14 April 2020 (UTC)

recent insertion by user 188.193.103.199

The insertion in question is this:

[The analytic representation] is used to represent a causal signal, that starts at a certain time and propagates as the time evolves, i.e. not backwards in time. This leads to the absence of negative frequencies, which is a side effect, reflecting causality.[2]

The problem, as I see it, is that the analytic representation of any function, causal or not, has an absence of negative frequencies. The association with causality is misleading.

--Bob K (talk) 14:09, 12 May 2021 (UTC)

It is exactly the point: The whole reason is, that it has starting point, which is reflected by the convolution with the step function. This causes the negative part A(e^(-jwt)) of the regular baseband description (that includes negative frequencys and goes from -inf to +inf) to disappear. The cause is causality. This is the major difference between one sideded Laplace Tranform (evaluated at 0 real coordinates) and the regular Fourier Transform.

So the point is, and this is the reason for my edit: An analytic signal (which reflects the physics) is always causal. This is essential to understand it, but is not in the text. It is unbelievable, how many people make this wrong, often leading to very "funny" publications (which again is one of the reason for these edits). These people tend to inform themselves via wikipedia (no matter how often they told not to do this, for exactly this reason). You know in real life: Somebody has to switch it on.

188.193.103.199 (talk) 11:00, 13 May 2021 (UTC)

There is no "convolution with the step function". There is multiplication (frequency domain) whose time-domain equivalent is convolution with an infinite, non-causal impulse response. And now you say "An analytic signal is always causal."  Nonsense. For instance  $e^{i\omega t},\omega >0$ is analytic and not causal.
"It is unbelievable, how many people make this wrong." -- Count me as a non-believer. You are the one who's wrong.
--Bob K (talk) 11:55, 13 May 2021 (UTC)

In the time domain, it is convolution with the "Hilbert Transform of dirac delta", that produces the imaginary part, which is redundant. This is exactly the point. There is nothing to believe and no religion. It is simply about cause and effect. If you restrict your example to be zero for negative times, then it is analytic and causal. Without this restriction it is not analytic, since you loose the coupling between real and imaginary part for negative times. Dont waste your time, you will not find a counterexample. But dont worry, usually this takes years to understand this. It is a very deep philosophical issue, that also has implications to probability theory. This is a relatively cool effect of the Hilbert Tranform. These are some of the strange effects of these distributions and tranformations. I have already seen people implementing the Hilbert Tranform...— Preceding unsigned comment added by 188.193.103.199 (talk) 12:18, 13 May 2021 (UTC)

"Dont waste your time," -- Exactly... you are not worth it.
--Bob K (talk) 14:12, 13 May 2021 (UTC)