User talk:Bob K

Welcome!

Hello, Bob K, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:

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I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out Wikipedia:Where to ask a question, ask me on my talk page, or place {{helpme}} on your talk page and someone will show up shortly to answer your questions. Again, welcome!  DV8 2XL 11:38, 19 November 2005 (UTC)[reply]

Is there any way to list all the subpages of a specific page?

To find all subpages of Manifold, for instance, go to Special:Allpages and type "manifold/" in the box labelled "Display all pages starting with:" (I learned this trick from R. Koot) -- Jitse Niesen

[2]

I also went there and changed PD to PD-self, but I didn't realize that I was not logged in, because I wasn't a member of commons. It seemed to work anyway. Is that true? And did I need to do that? If so, shouldn't someone have notified me?... like maybe the person who moved the image? Is is going to stay in one place now? Should I have uploaded it differently than I did? -- Bob K

Since the image contains English text not suitable to http://simple.wikipedia.org/ I think User:Tiaguito's move to commons was wrong, and your original upload was fine. It's not a problem since it still shows up fine in the articles. -- Jeandré, 2006-01-11t11:00z

Hi, your edit summary said that some characters do not appear correctly in Internet Explorer. They seem fine on my copy, which has no special fonts installed. But if this is generally a problem for people running internet explorer, then something should be done to make changes across several articles. Is it possible to get a screen shot to see what it looks like on your copy? Thenub314 (talk) 11:04, 19 January 2009 (UTC)[reply]

When User:Bdmy changed ƒ*g_T to ƒ∗g_T, the asterisk became an empty square (aka "box") on my IE screen. Upon further reading, such as Talk:Set_(mathematics)#Box-fixing and Template_talk:SpecialChars#Internet_Explorer, I learned that the problem is associated with IE. I have Firefox, and it works, but I use IE for everything else, as do many other readers.

--Bob K (talk) 14:28, 19 January 2009 (UTC)[reply]

This is good to know. I when I tried IE on my vista machine it worked fine, when I tried on my XP machine it displayed a box. I wonder why they only fixed the bug on Vista? Oddly, as an a side, on my XP system the PNG graphics also looked very pixelated. I am surprised the issue about boxes has never come up on the WT:WPM. I will try to keep this in mind when editing articles, and make sure to use * instead of ∗. Thenub314 (talk) 15:05, 19 January 2009 (UTC)[reply]

Ah Ha! I discovered an easy fix. I changed Times New Roman to Lucida Sans Unicode in a pull down menu at Tools -> Internet Options -> Fonts -> Webpage Font:.  That fixes the problem for me. But how do we disseminate that information to the readership?

--Bob K (talk) 14:55, 19 January 2009 (UTC)[reply]

Well, one could include a link on the special characters template that discusses possible fixes. Thenub314 (talk) 15:05, 19 January 2009 (UTC)[reply]

Yes, I was just looking into that. But I haven't figured out where the "fix" page should exist. Any suggestions?

--Bob K (talk) 15:20, 19 January 2009 (UTC)[reply]

More progress... I just created section Changing Internet Explorer's (IE) default font

--Bob K (talk) 17:03, 19 January 2009 (UTC)[reply]

Nice work! It is looking much better. Thenub314 (talk) 19:07, 19 January 2009 (UTC)[reply]

From Talk:Convolution : The "reflection" of g(t) is h(t) = g(-t). E.g., h(0.2) = g(-0.2), and h(-0.2) = g(0.2). Sleep on it. --Bob K (talk) 13:56, 10 August 2018 (UTC)

The fabricated reflection of g(t) about the vertical axis is h(t). It is only feasible to say that g(-t) = g(t) after that fabrication. Otherwise, it's wrong (and vague) to simply and generally say that the 'reflection of g(t) is g(-t)'. Sleep on that one. KorgBoy (talk) 12:52, 15 August 2018 (UTC)[reply]

Hi. Thank you for your recent edits. Wikipedia appreciates your help. We noticed though that when you edited Fourier analysis, you added a link pointing to the disambiguation page DFT (check to confirm | fix with Dab solver). Such links are almost always unintended, since a disambiguation page is merely a list of "Did you mean..." article titles. Read the FAQ • Join us at the DPL WikiProject.

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Спасибо! Привет тебе из России. Не хворай ;-) http://translate.google.com/#ru/en/ Xakepp35 (talk) 20:03, 7 February 2013 (UTC)[reply]

You're very welcome. I'm glad to know the answer actually reached you. And thanks for the well wishes. --Bob K (talk) 21:21, 8 February 2013 (UTC)[reply]

I saw the wiki page, but I couldn't find any examples using actual numbers evaluating the formula. Could you give some examples of convolution, please? Mathijs Krijzer (talk) 22:14, 9 March 2013 (UTC)[reply]

The convolution of f and g is written f∗g, using an asterisk or star. It is defined as the integral of the product of the two functions after one is reversed and shifted. As such, it is a particular kind of integral transform:

${\displaystyle (f*g)(t)\ \ \,}$	${\displaystyle {\stackrel {\mathrm {def} }{=}}\ \int _{-\infty }^{\infty }f(\tau )\,g(t-\tau )\,d\tau }$
	${\displaystyle =\int _{-\infty }^{\infty }f(t-\tau )\,g(\tau )\,d\tau .}$       (commutativity)

The convolution of two complex-valued functions on R^d

${\displaystyle (f*g)(x)=\int _{\mathbf {R} ^{d}}f(y)g(x-y)\,dy}$

is well-defined only if f and g decay sufficiently rapidly at infinity in order for the integral to exist. Conditions for the existence of the convolution may be tricky, since a blow-up in g at infinity can be easily offset by sufficiently rapid decay in f. The question of existence thus may involve different conditions on f and g.

When a function g_N is periodic, with period N, then for functions, f, such that f∗g_N exists, the convolution is also periodic and identical to:

${\displaystyle (f*g_{N})[n]\equiv \sum _{m=0}^{N-1}\left(\sum _{k=-\infty }^{\infty }{f}[m+kN]\right)g_{N}[n-m].\,}$

When a function g_T is periodic, with period T, then for functions, f, such that f∗g_T exists, the convolution is also periodic and identical to:

${\displaystyle (f*g_{T})(t)\equiv \int _{t_{0}}^{t_{0}+T}\left[\sum _{k=-\infty }^{\infty }f(\tau +kT)\right]g_{T}(t-\tau )\,d\tau ,}$

where t_o is an arbitrary choice. The summation is called a periodic summation of the function f.

For complex-valued functions f, g defined on the set Z of integers, the discrete convolution of f and g is given by:

${\displaystyle (f*g)[n]\ {\stackrel {\mathrm {def} }{=}}\ \sum _{m=-\infty }^{\infty }f[m]\,g[n-m]}$

${\displaystyle =\sum _{m=-\infty }^{\infty }f[n-m]\,g[m].}$       (commutativity)

When multiplying two polynomials, the coefficients of the product are given by the convolution of the original coefficient sequences, extended with zeros where necessary to avoid undefined terms; this is known as the Cauchy product of the coefficients of the two polynomials.

Hi. Thank you for your recent edits. Wikipedia appreciates your help. We noticed though that when you edited Nyquist rate, you added a link pointing to the disambiguation page Bandwidth (check to confirm | fix with Dab solver). Such links are almost always unintended, since a disambiguation page is merely a list of "Did you mean..." article titles. Read the FAQ • Join us at the DPL WikiProject.

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I thought your comment on the technical sophistication of a signls processing article I was reading was very sensible. David Lloyd-Jones (talk) 22:26, 4 February 2014 (UTC)[reply]

Please explain the criteria for your assertion that (quote): ""starred transform" receives the higher percentage of relevant hits on the internet", which was your argument for renaming the article "star transform". Consider the following results:

• "star transform"
  • Google Scholar: 125 results
  • IEEE Xplore: 4 results
• "starred transform"
  • Google Scholar: 36 results
  • IEEE Xplore: 1 result

Based on these results, I expect you to revert your changes. --Sagie (talk) 13:04, 6 February 2014 (UTC)[reply]

I saw the Google Scholar results, but my informal inspection of the hits revealed that the "star transform" hits were mostly about text compression algorithms, which I assume are irrelevant here. There were many other irrelevant hits as well.

--Bob K (talk) 14:32, 6 February 2014 (UTC)[reply]

While your contribution to said articles is very appreciated, you have also been systematically deleting information from them. Specifically, you have deleted (twice) information about relationship between the Z-transform and the Laplace transform, by means of the star transform:

1. in your first edits that relationship was removed altogether.
2. in your last edits (following my retrieval of said information) you've removed information regarding the conversion from a continuous-time Laplace representation to a discrete time Z/star representation, which is what the star transform is all about, since from before it was even known as star/starred transform (see E. Jury, "Analysis and Synthesis of Sampled-Data Control Systems", Sept. 1954)

While the information about star transform could probably have been presented more succinctly in the Z-transform article, I find removing it altogether is completely unacceptable. I'd appreciate your reinstating of the removed information.

--Sagie (talk) 13:09, 6 February 2014 (UTC)[reply]

Regarding the second point I made, note that you've removed this information from both articles (Z-transform and Star transform)

--Sagie (talk) 13:11, 6 February 2014 (UTC)[reply]

Comparing the current version of Starred transform to the 23Mar'13 version, it appears that what you're concerned about is merely this:

${\displaystyle x(t)\rightarrow X^{*}(s)\rightarrow x^{*}(t)}$

Really?

--Bob K (talk) 15:41, 6 February 2014 (UTC)[reply]

I don't know why you're playing dumb. Your changes to said article are far more than the above mentioned deletion. What I'm mainly concerned about is you deleting the following paragraph, which the main way to convert between the transforms:

The star transform can be related to the Laplace transform, by taking the residues of the Laplace transform of a function, as such:

${\displaystyle X^{*}(s)=\sum {\bigg [}{\text{residues of }}X(\lambda ){\frac {1}{1-e^{-T(s-\lambda )}}}{\bigg ]}_{{\text{at poles of }}X(\lambda )},}$

--Sagie (talk) 12:30, 19 February 2014 (UTC)[reply]

First of all, your point no. 1 is irrelevant. It's quibbling about the route I took, rather than the result. Regarding your point no. 2, indeed I seem to have made a mistake when I tried to compare the Feb 6 version (https://en.wikipedia.org/w/index.php?title=Starred_transform&oldid=594159008) version with the Mar 2013 version (https://en.wikipedia.org/w/index.php?title=Starred_transform&oldid=546556726). Had I done it correctly, my suggestion would have been to better-support the deleted, isolated fragment, as you have now apparently done (Starred_transform#Relation_to_Laplace_transform) on your own. Better. Not sure if your text changes are an improvement though. Not sure that I really care anymore. Perhaps others are watching who will weigh in.
--Bob K (talk) 17:09, 19 February 2014 (UTC)[reply]

Question for you on [Instantaneous phase]]. Does it make sense to you that in the figure, the y-axis has different scale in the top vs the bottom graphs? Thanks. 22:06, 16 June 2014 (UTC)

Since they are being compared side-by-side, one can certainly make a good case for using one scale for both. And by-the-way, phase unwrapping algorithms are not fool-proof when looking at angle-modulated signals with possible smeared 180 degree phase shifts. The approach I've used in systems since the 1970's, is to simply stack three "copies" of the wrapped phase plot vertically, and label the y-axis [-3π,3π]... works like a charm. Another good trick is to not try to "connect the dots" of the graph. Instead, vertically elongate each dot (for instance 1 pixel wide and 8 pixels high), which tends to make them look connected without having to make error-prone algorithmic decisions.

--Bob K (talk) 11:35, 17 June 2014 (UTC)[reply]

Thanks for that, and for the extra advice too! Jytdog (talk) 11:59, 17 June 2014 (UTC)[reply]

The article A derivation of the discrete Fourier transform has been proposed for deletion because of the following concern:

Unsourced alternative-pov content fork of discrete Fourier transform, written in an essay-like and unencyclopedic way, not suitable for merging.

While all constructive contributions to Wikipedia are appreciated, content or articles may be deleted for any of several reasons.

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Hello Bob K,

I've read your posts in the section: Talk:Window function. I posted a question yesterday in that section under the title identical to this post. I would appreciate if you could take a look at it. Thanks much. --AboutFace 22 (talk) 14:50, 17 June 2015 (UTC)[reply]

Hello Bob K, please comment your deleting of my text regarding:

"In the easer case can be use the algorithm:^[1]

${\displaystyle y[n]=\sum _{k=0}^{M-1}x[k]\ e^{-i2\pi fkT}}$,

where T is interval between samples of signal."

This Algorithm is very importent for decimation of ADC samples before digital beamforming in digital antenna arrays in radars and MIMO systems in communications. I'm sorry, but I should cancelled your deleting of this text. It's very imporetent for Wiki-readers. Thank you for your understanding. Best regards,Swadim (talk) 08:32, 24 August 2017 (UTC)[reply]

The formula you inserted is not even a function of n. It is just one sample of a discrete Fourier transform, X(f). In the future, please discuss your ideas on the talk page, before changing the article. --Bob K (talk) 12:11, 24 August 2017 (UTC)[reply]

Thank you! I has some mistake on this formula. The true formula is

${\displaystyle y[n]=\sum _{k=0}^{M-1}x[nM+k]\ e^{-i2\pi fkT},n=0,1,..,N}$. For example by ${\displaystyle 2\pi fkT=k\pi /2}$ we have

${\displaystyle e^{-i2\pi fkT}=1,e^{-i\pi /2},e^{-i\pi },...}$ and^[1]

${\displaystyle Re(y[n])=\sum _{k=0}^{M-1}(Re(x[nM+k])cos{k\pi /2}+Im(x[nM+k])sin{k\pi /2})\,}$,

${\displaystyle Im(y[n])=\sum _{k=0}^{M-1}(Im(x[nM+k])cos{k\pi /2}-Re(x[nM+k])sin{k\pi /2})\,}$.

For ${\displaystyle Im(x[nM+k])=0}$ we have

${\displaystyle Re(y[n])=x[nM]-x[nM+2]+x[nM+4]-...,}$,

${\displaystyle Im(y[n])=-x[nM+1]+x[nM+3]-x[nM+5]-...}$.

For video signals (${\displaystyle Im(x[nM+k])=0}$ and ${\displaystyle Re(x[nM+k])>=0}$)

${\displaystyle y[n]=\sum _{k=0}^{M-1}x[nM+k]\ }$.

It's enough for your understanding? Swadim (talk) 10:00, 25 August 2017 (UTC)[reply]

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	The Editor's Barnstar
	Thanks, Bob K, for resolving once and for all (I hope!) the four-year-old controversy about the convolution and cross-correlation graphs in File:comparison_convolution_correlation.svg. Cheers, cmɢʟee⎆τaʟκ 00:59, 27 November 2016 (UTC)[reply]

Bob,

"Introduction to Signal Processing", by Sophocles J. Orfanidis shows the equation for the Hamming Window like the one shown in Equation \ref{eqn:orfanidis_hamming}. However, MATLAB, and "Digital Signal Processing Using MATLAB", by Ingle and Proakis show the Hamming Window as shown in Equation \ref{eqn:ingle_hamming}. \begin{equation}

 \label{eqn:orfanidis_hamming}  w(m)=0.54+0.46cos(\frac{\pi m}{M}) 

\end{equation} \begin{equation}

 \label{eqn:ingle_hamming}  w(m)=0.54-0.46cos(\frac{\pi m}{M}) 

\end{equation}

If I plot both of these, only Equation \ref{eqn:ingle_hamming} appears and behaves like a Hamming Window. Have you tried plotting Equation \ref{eqn:orfanidis_hamming}?

The Orfanidis formula is zero-phase, w₀(n). The Matlab formula is w[n], as clearly defined in the article. The zero-phase formula will also look and behave like a Hamming window, when you are windowing data between -N/2 and N/2.

--Bob K (talk) 14:39, 26 September 2017 (UTC)[reply]

Thank you sincerely for the edification. I'm definitely a newly at DSP and did not recognize the condition, just that I've tried entering that equation several times in MATLAB and other tools and don't get what I need.

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Bob_K, please see https://en.wikipedia.org/wiki/Talk:Single-sideband_modulation#Bandwidth_of_an_AM_signal

Bob, expansion is a good concept to add, but in practice upsampling is not implemented in two steps like that. I'd rewrite the lead to describe the net process instead of the steps. Dicklyon (talk) 16:51, 17 November 2018 (UTC)[reply]

Hi Dick. I think we're covered, because the very next paragraph begins with "Rate increase by an integer factor L can be explained as a 2-step process, with an equivalent implementation that is more efficient:"

I recall at the time we were writing the lede, it seemed very likely that people would come to Upsampling expecting to read about Expansion. So it seemed important to be clear about the terminology right up front. Anyhow, feel free to take your own suggestion. I'm sure you'll do a great job.

--Bob K (talk) 22:00, 17 November 2018 (UTC)[reply]

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Hi Bob,

I've noticed you undid my contribution of the animation on the wikipeida page witht the comment "I am familiar with the subject, but can't make sense out of the gif. Sorry. Looks like you put a lot of effort into it!"

I have to say that someone else may find the animaion helpful, I suggest adding a discussion on the talk page about it and see if a general concensous can be reached in regards to it.

For now i've added it back in and started the topic on the talk page

Davidjessop (talk) 09:10, 24 February 2019 (UTC)[reply]

RMaung (WMF) 16:01, 6 September 2019 (UTC)[reply]

RMaung (WMF) 15:38, 20 September 2019 (UTC)[reply]

RMaung (WMF) 20:39, 3 October 2019 (UTC)[reply]

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--User:Martin Urbanec (talk) 21:59, 23 November 2019 (UTC)[reply]

Hi, Bob. Congratulations on the Yankee Dutch Crossing article! You should tell Gaye. I'll bet she'll be delighted. I assume you are Bob Kettig, and that you contra dance at Glen Echo. We haven't really met, but we've seen each other on the dance floor. BTW, I recently moved all the external links in the contra article. They need to be cleaned up and, especially, pruned. Maybe we could collaborate on that. Paulmlieberman (talk) 14:42, 6 May 2020 (UTC)[reply]

Thanks Paul, and yes I am Kettig, and I know you too. My suggestion for what to do next is move the band links to the External links section.

--Bob K (talk) 22:02, 6 May 2020 (UTC)[reply]

Congrats, Bob K! @Paulmlieberman: it looks like everything is good here, but do please be careful not to accidentally out anyone (WP:OUTING) — people prefer varying degrees of anonymity on Wikipedia, and for privacy it's best to assume they may not want to disclose more than they have on their user page. Off-wiki communications are fine, though; we need all the social connection we can get during this time . Cheers, {{u|Sdkb}} ^talk 00:51, 9 May 2020 (UTC)[reply]

Actually, you are the one who has mentioned both my surname and a heretofore unmentioned first name (Gaye). Have I mentioned any names that give rise to your concern? --Bob K (talk) 01:24, 9 May 2020 (UTC)[reply]

@Bob K: I was addressing Paulmlieberman there (I'm a different user than him, although also a dancer!). I wish talk pages were better formatted to make it easier to tell everyone apart haha. Cheers, {{u|Sdkb}} ^talk 01:30, 9 May 2020 (UTC)[reply]

Doh! Thanks for clearing that up. Now of course, it's obvious. Your sig just did not "compute", and I just disregarded it. --Bob K (talk) 01:43, 9 May 2020 (UTC)[reply]

BTW, I have also authored a Salty Dog Rag article... just trying to stay relevant, during the pandemic.

--Bob K (talk) 22:46, 8 May 2020 (UTC)[reply]

Thanks guys! My apologies for naming names here. If I ever feel the urge again, I'll do it on FB. God, I miss contra! Paulmlieberman (talk) 02:01, 9 May 2020 (UTC)[reply]

No worries, Paul. When I started on Wikipedia, 15 years ago, I was reluctant to use my full name, and that just became my "normal". With 20/20 hindsight, if I had it to do over again, I'd use my full name, not because I care about recognition (I don't), but just because Wikipedia has gained my trust.
--Bob K (talk) 02:12, 9 May 2020 (UTC)[reply]

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An automated process has detected that when you recently edited Discrete-time Fourier transform, you added a link pointing to the disambiguation page Normalized frequency.

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You might be interested to know that User:Gkl329Hf293 has also been reverting edits to Dirac delta function with no explanation and I have reported them for edit warring at [3]. Elwoz (talk) 20:44, 25 September 2021 (UTC)[reply]

Thank you. I was hoping someone else would step in. I didn't (and still don't) know any other way to handle that situation. --Bob K (talk) 12:24, 26 September 2021 (UTC)[reply]

It's a little tedious, but I just followed the instructions in the "create a new report" form at https://en.wikipedia.org/wiki/Wikipedia:Administrators%27_noticeboard/Edit_warring . Elwoz (talk) 12:36, 28 September 2021 (UTC)[reply]

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If you wish to participate in the 2023 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 00:24, 28 November 2023 (UTC)[reply]

Please do not introduce links in actual articles to user pages or sandboxes, as you did at Fourier transform. Since these pages have not been accepted as articles, user pages, sandboxes and drafts are not suitable for linking in articles. and such links are contrary to the Manual of Style. These links have been deleted, please do not re-add any such links, thank you - Arjayay (talk) 14:14, 31 December 2023 (UTC)[reply]

In this edit you removed non-breaking spaces between numbers and units. These are generally desirable per MOS:NBSP. The edit includes other formatting changes that I assume are good so I did not not revert. Would you mind cleaning this up? ~Kvng (talk) 15:50, 17 August 2024 (UTC)[reply]

Yes sir, and I thank you. I will try. I was using the edit, not edit source, and I did not understand the grey colored things that just seemed annoying at the time. Please feel free to help me where I inevitably fall short. --Bob K (talk) 22:11, 17 August 2024 (UTC)[reply]