why cannot this simple graph be plotted correctly?

[rlyang]

Please find why I cannot graph this correctly in period

In the graph, why is the period T not equat 2 pi / 6283.2 = 0.001second ?

How to make this graph correctly?

Thanks

[ValeryOchkov · Accepted solution]

Do not use the quick plot - see the attach!

[rlyang]

Thank you very much.

I am beginer. I got another plot, I do not know why I cannot plot all red line ?

There suppose to be a continuous red line. Please give me advice.

Thank you

[ValeryOchkov]

Attach please the file!

[rlyang]

Thank you Valery for the help

[MikeArmstrong]

It's because you have only defined t up to 0.01.

Have a look at the attached.

Mike

[VladimirN]

I think, it is better to use such record for function:

[rlyang]

Please look, why my graph is different from yours. Which one is correct?

[MikeArmstrong]

Ruilin Yang wrote:

Please look, why my graph is different from yours. Which one is correct?

Have a look at my example.

Mike

[rlyang]

Where is your example? Please showm me why different.

[MikeArmstrong]

Below.

Mike

[MikeArmstrong]

Sorry, worksheet attached.

Mike

[rlyang]

My question is why we got different results ? Why?

[MikeArmstrong]

Your scale on the graph. Have a look at the attached.

Mike

Message was edited by: Mike Armstrong - Oh, I also removed the definition of t.

[rlyang]

I still do not understand this. What wrong with t definition ?

My t scale is the same as yours.

Thanks

[rlyang]

why my t definition:

could change the graph ? I found when I plot sin ro cos wave graphs, I need that definition.

[RichardJ]

Your problem is that your function has singularities. You have defined the range variable such that some of the points coincide with those singularities, and so all you see are huge negative spikes. Mike didn't use a range variable, he just used a Quickplot, and when he chose those axis limits it happens that no points coincide exactly with a singularity (or got too close to them). In your range variable definition change 1000 to 1100.

[RichardJ]

Here's another solution that gets around the problem of having to choose an axis where no points coincide with the singularities.

[rlyang]

Thank you. I removed the t range definition.

But please look this graph. When I change the range of the t axis, I got different graph

Please check if this is right, see the attached please.

Thank you again

[RichardJ]

It doesn't matter whether you use a range variable or the quickplot feature. The graph is plotted at certain points, and exactly where those points are depends on how you set it up. Your function has singularities, so the points might or might not be close to them. If they are close you get a very large positive or negative number. It doesn't help that you are looking at such a tiny graph. If you expand the graph so it's the size of the screen you will see what is happening.

[rlyang]

Does those spikes represent ? 75 or infinity?

Please see the attached, top graph means 75? The bottom means 2.5 ?

Or all represent infinity ?

Please help to verify for me.

Thanks

[RichardJ]

If it plots as 75, it's 75.

[rlyang]

I do not know whhat is wrong with this graph ? please see the attached file.

[AlanStevens]

You are using a frequency of 1000 cycles per unit time with a range of 20 time units. That means you are asking the plot to display 20000 sinusoids. This is asking too much! Change the plot range to, say, 0 to 10^-2 and you will see the sinusoids.

Alan

[RichardJ]

In addition to what Alan has pointed out, because of the way you have defined your function it happens that the values of t for which Qickplot plots the graph are values for which omega*t is an integer multiple of PI. Therefore r(t) is always zero (or very close to it).