Solved: Re: There are too many ranges variables

TN_9676358 · ‎Oct 19, 2020

Hello everyone,

In scenario 4, I got everything correctly. However, when I tried to graph the plot. There were an issue with "too many range variables".

I read some posts that recommend to change the number of points. However, it was invisible to click on. Would you guys please help me with this?

Thank you so much

Werner_E · ‎Oct 20, 2020

You still have the wrong definition of sigma! You compare 0.5*1 with 0.5K and the result is zero (because they are not equal) and this zero is assigned sigma. Your sigma is zero! Give it a try. Furthermore in the definition of y.n you used sigma as if it were a function.

And the way you defined and used x.n your "noisy" signal would have used the very same random number for every time - not very noisy 😉

Here is a way to do it. Do you really mean T.n to be 0 (zero what - K, °F, Delta°F,..?) for times smaller than 2 minutes? You may replace the 0 by NaN (which suppresses plotting) or by Tsf(2 min).

View solution in original post

ValeryOchkov · ‎Oct 20, 2020

TN_9676358 · ‎Oct 20, 2020

Do you have any recommendation for this? 😞 The Tsf was calculated in scenario 3. I'm not sure why the Tn(ti) doesn't work

MartinHanak · ‎Oct 20, 2020

Martin Hanák

TN_9676358 · ‎Oct 20, 2020

Hello Martin,

Sigma unit is K so yn is also K. I have added the unit to K and Tn(i) still doesnt work. It probably because of my the function or something.

thank you so much for the feedback

MartinHanak · ‎Oct 20, 2020

Hi,

you must define yn as function !!!

Martin Hanák

Werner_E · ‎Oct 20, 2020

Looking for something like this?

TN_9676358 · ‎Oct 20, 2020

Hello,

it looks really good. But the Tn is T noise so it would start at 2 and the time interval is ti = 2,3...99. I have some new updated here.

You are life saver for sure 🙂

Thank you so much.

Werner_E · ‎Oct 20, 2020

You still have the wrong definition of sigma! You compare 0.5*1 with 0.5K and the result is zero (because they are not equal) and this zero is assigned sigma. Your sigma is zero! Give it a try. Furthermore in the definition of y.n you used sigma as if it were a function.

And the way you defined and used x.n your "noisy" signal would have used the very same random number for every time - not very noisy 😉

Here is a way to do it. Do you really mean T.n to be 0 (zero what - K, °F, Delta°F,..?) for times smaller than 2 minutes? You may replace the 0 by NaN (which suppresses plotting) or by Tsf(2 min).

TN_9676358 · ‎Oct 20, 2020

Hello,

Got it. I'm working on it again now. I mean at Tn won't generate the first signal at 2mins. so from 0-2 mins, Tsf is the actual sensor temperature (Tsf).

I think the issue here is t(time) range, because the Tn will send the mixed signal at t=2min while Tsf start at 1min. I also attached the picture that I supposed ti generate

Thank you so much

TN_9676358 · ‎Oct 20, 2020

i GOT ITTTTTT. We just need to graph this with ti not t. yayayayayayayya