Vector 0 to 2pi with 1000 point equal spacing

Hi,

it might be trivial but I struggle to create a vector that contrains 1000 points equally spaced between 0 and 2pi.

any help will be appreciated!

Answer 1

[Azzi Abdelmalek]

linspace(0,2*pi,1000)

[Walter Roberson]

Be careful, though: the spacing between the items above will not be 2*pi/1000, it will be 2*pi/999 .

When you talk about a particular number of points "between" a range, you have to think carefully about whether one endpoint or both endpoints or neither endpoints is included in that particular number.

linspace(0,2*pi,1000) would not be appropriate for 1000 Hz: 1000 Hz would be linspace(0,2*pi,1001), because for a sine or cosine wave, 0 and 2*pi overlap, like counting the same point twice.

If you want one full cycle exactly then,

t = linspace(0, 2*pi, 1001);  %generate 1000 points _before_ the endpoint
t(end) = []; %throw away the endpoint

[Vaibhav]

thank you sir for helping us ..........

Answer 2

[Jan]

When I ask my favorite internet search engine for the terms "matlab vector equally spaced", I get a link to the documentation of linspace as first hit. Of course, this forum is smart also, Google took 0.35 seconds only for this answer. I'm afraid we cannot beat this speed. :-)

[John D'Errico]

This is what amazes me. Rather than look for the solution using a simple, easy search, some would rather ask a question here only to then wait for possibly hours to get an answer. +1 anyway, for pointing out that a search would have solved the problem so quickly.

[Jan]

@John: Human brains are designed for a person to person communication. Therefore it seems to be more reasonable to ask experienced Matlab users than a general but dull search engine. In my opinion this is a pleasant and congenial feature, although it is not efficient.

I hesitate to mention, that Google is better than the active members of the Matlab community, but it is true here. You know that I still believe in human power for many questions about face recognition. But here moving the focus to searching the internet might help Marcel to solve many other problems by his own also.

[Alison Haire]

So being dicks about it because he may not have hit the exact right search term for the feature he was looking for is ok Why not just answer the question if it was soooo simple for you? -1 for arrogance

[Stephen23]

"because he may not have hit the exact right search term "

One of the nice things about internet search engines is that one does not need an "exact term" to search for something. In fact, when I tried searching using the complete question title and several combinations of its words the MATLAB documentation for linspace was always in the first two results. Not only that but the more people practice searching for information the better they get at it.

The MATLAB documentation is the most important source of information on how to use MATLAB. While it is nice to be able to ask other people for help, this is certainly no replacement for actually reading the documentation, nor a replacement for learning how to search for information in the documentation.

[Walter Roberson]

Alison Haire: Jan linked to linspace(), which is the appropriate routine for this purpose.

[Jan]

@Alison: "Arrogance"? I'm less emotional when posting answers than you assume. I've posted the link to the solution linspace and explained, how such problems can be solved much more efficiently than waiting for the forum to answer. The idea behind it is not to solve only this problem, but allowing for more efficient solutions in the future also.

When I take the relevant words from the title: "Matlab Vector equal spacing" or from the body: "Matlab points equally spaced", Google finds the link to linspace as first hit. It is hard here, not to find matching terms for a search.

Alison, note that we find many questions every day in this forum, which could be solved in few seconds by searching in the net or in the forum at first. Perhaps I've posted about 2000 or 3000 answers, which could have been found by the OP by investing a tiny bit of time with searching at first. And this does not concern only me, but all of the frequent contributors. Many users do not show any own effort, but try to forward the required work to others in the forum. With this background, I ask you to read my answer again: I've simply explained, how I would find the solution of the problem, included a link and admitted, that Google is unfortunately more efficient than the forum (include my person). Finally I added a smiley. Now you confuse this with "Google it by your own, you idiot". Sigh. I'm not here to make the other members happy, but only to solve Matlab problems.

Answer 3

[maryam Haider]

t=linspace(0,2*pi,1000)