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-MFra-21-Topaz II
21-Topaz II
October 13, 2022
Solved

Cut-set based Network analysis

  • October 13, 2022
  • 4 replies
  • 2272 views

Good evening everyone!

I have already presented the photos of the attached worksheet (which I give to this community) in a previous post. However, there is a small problem: why does transformation 28), at the bottom of the worksheet, go wrong?

I thank those who will answer and also those who will find something wrong in all the work that has requested  "Stunde und Stunden der Arbeit"...

I hope that, with the download of the worksheet, the bitmap images do not move from their original position, otherwise there will be a bit of chaos.

 

Best answer by LucMeekes

Operation 28) fails most probably because the invlaplace of V.b6(s) contains the delta function, which the numeric processor doesn't (want to) know.

 

Success!
Luc

4 replies

L
LucMeekes23-Emerald IVAnswer
23-Emerald IV
October 13, 2022

Operation 28) fails most probably because the invlaplace of V.b6(s) contains the delta function, which the numeric processor doesn't (want to) know.

 

Success!
Luc

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    -MFra-21-Topaz IIAuthor
    21-Topaz II
    October 13, 2022

    I really don't think the cause of the malfunction is as you say. So much so that the Laplace anti-transformations are previously calculated individually.

    W
    Werner_E25-Diamond I
    25-Diamond I
    October 13, 2022

    Nonetheless Luc is perfectly right in his assumption. If you define the Dirac Delta function (I boldly set it constant 0) so it can be used in numerics, too, your functions works for numeric plotting as well. It just turns red because Mathcad decides that the symbolic result is too large and should not be displayed.

    Werner_E_0-1665686234061.png

    EDIT: Here is a more sophisticated Delta function

    BTW, you had a typo. Your plot tries to plot vb.67 but you had assigned the result of invlaplace to a function v.b67

    Werner_E_1-1665688499199.png

    Furthermore it looks to me that you were aware of the Delta(t) problem when you plotted vb6 in Fig.14. So why were you puzzled by expression 28?

    The problem with expression 28 is that you can't see the result and so you can't manually remove the Delta function as you did in Fig.14, but defining a new Delta-Function should solve that problem. You may use Delta:=Delta to keep the original symbolic Delta function for symbolic calculations.

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    -MFra-21-Topaz IIAuthor
    21-Topaz II
    October 13, 2022

    It would have been better to say "Stunden und Stunden lang Arbeit". Alas! I got really rusty with the German language ... I need a German friend (female to be clear) to practice a little ...

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    -MFra-21-Topaz IIAuthor
    21-Topaz II
    October 13, 2022

    How to use -Equation Format-.jpg

    L
    LucMeekes23-Emerald IV
    23-Emerald IV
    October 13, 2022

    Here's your work converted to Prime.

    The results in Prime 8 are the same. The file is saved for Prime 4 (and higher).

     

    Success!
    Luc

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