cancel
Showing results for 
Search instead for 
Did you mean: 
cancel
Showing results for 
Search instead for 
Did you mean: 

Community Tip - Want the oppurtunity to discuss enhancements to PTC products? Join a working group! X

Plotting Points ?

lvl107
20-Turquoise

Plotting Points ?

   Hello, Everyone.

   From :

( B ).PNG

  How to plot 8 upper-corner-points ?
     Thanks in advance.
         Regards.

1 ACCEPTED SOLUTION

Accepted Solutions
LucMeekes
23-Emerald III
(To:lvl107)

Like this:

LM_20181019_points.png

 

Success!

Luc

View solution in original post

13 REPLIES 13

point-1.png

LucMeekes
23-Emerald III
(To:lvl107)

Like this:

LM_20181019_points.png

 

Success!

Luc

lvl107
20-Turquoise
(To:LucMeekes)

   Your method to solve it, this is a new to me ! Smiley HappySmiley HappySmiley Happy Many thanks for your time and help, Luc. I got it. Smiley HappySmiley HappySmiley Happy

( B' ).PNG

        Best Regards.

lvl107
20-Turquoise
(To:LucMeekes)

Luc, I don't know the reason the plot doesn't show up in this case.

( B''' ).PNG( B'''' ).PNG

    Regards.
         Loi.

Werner_E
24-Ruby V
(To:lvl107)

You have disabled the plot region!

Right click the plot region and chose "Enable Evaluation"

B.PNG

As you can see the poles at +/-sqrt 2 and 0 are no valid solutions to your problem this time. This is because they are poles of second order (the square in d(x)) and so there is no change of signs at those positions. A sign change only occurs at poles of odd order.

lvl107
20-Turquoise
(To:Werner_E)

   Many thanks, Werner. And how to keep the same "procedure" and eliminate 3 points are not 6 lower-intersection-points ? ( maybe just edit ? ). ( And how to plot local minimum-points of the f(x) ? )

   Best Regards.

lvl107
20-Turquoise
(To:lvl107)

Correction : plot local minimum-points of the f.0(x) 

Werner_E
24-Ruby V
(To:lvl107)


@lvl107 wrote:

   Many thanks, Werner. And how to keep the same "procedure" and eliminate 3 points are not 6 lower-intersection-points ? ( maybe just edit ? ).


You may think beforehand and don't ask for the zeros of the denominator which are of even order. In your example obviously all are of even order and so you would not ask for d(x)=0 at all.

As an alternative you may consider writing a function which eliminates all points of zeros (nominator and denominator separately) which occur an even number of times. This routine also must consider removable discontinuities like when nominator and denominator have a common zero. Good luck!

lvl107
20-Turquoise
(To:LucMeekes)

Luc, after I have got all of your hints, I try to plot all lower-intersection-points of logical-plot, but I don't know it's true for all of cases :

( C ).PNG

       Best Regards.

           Loi.

lvl107
20-Turquoise
(To:lvl107)

  Oh, my trial is fail.

( D ).PNG

      Missing point.

-MFra-
21-Topaz II
(To:lvl107)

Hi,

You could also use the "numer" and  "denom" functions as shown:

liv107 19102018.jpg

Werner_E
24-Ruby V
(To:-MFra-)

or more compact:

B.PNG

 

lvl107
20-Turquoise
(To:Werner_E)

  F.M. and Werner, with your "procedures" above, It seems that I just need "input" a new function ! Smiley HappySmiley HappySmiley Happy :

( B'' ).PNG

  Many thanks for all your time and help.
      Regards.

Top Tags