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I'am trying to plot the function:

BodePlot[1/(1 + 0.1 s), s]

The phase plot is wrong (I'm using Mathematica 11.2):

enter image description here

Why does this happens and how can I get the right result?

Thank you so much for your time.

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  • $\begingroup$ Works fine on my machine (Mma 11.3.0.0). $\endgroup$ – David G. Stork Aug 8 '18 at 17:56
  • $\begingroup$ Hello @DavidG.Stork it is very strange. Maybe the version 11.2 has some bugs. $\endgroup$ – Gennaro Arguzzi Aug 8 '18 at 17:58
  • $\begingroup$ Works fine on my machine (Mma 10.2.0.0). Windows 8.1 64 bit $\endgroup$ – Mariusz Iwaniuk Aug 8 '18 at 17:58
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    $\begingroup$ Try BodePlot[1/(1 + 0.1 q)]. $\endgroup$ – Hector Aug 8 '18 at 18:08
  • $\begingroup$ Hello @Hector, it works. It is very strange. $\endgroup$ – Gennaro Arguzzi Aug 8 '18 at 18:38
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BodePlot[
 TransferFunctionModel[
  {
   {1/(1 + 0.1 s)}
   }, s]
 , s
 ]

Mathematica graphics

| improve this answer | |
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  • $\begingroup$ Good solution @rhermans. In your opinion is it a bug? $\endgroup$ – Gennaro Arguzzi Aug 8 '18 at 18:39
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    $\begingroup$ @GennaroArguzzi No, in my opinion the documentation for BodePlot is very clear and says "The system lsys can be TransferFunctionModel or StateSpaceModel, including descriptor and delay systems." $\endgroup$ – rhermans Aug 8 '18 at 18:55
  • $\begingroup$ According to the documentation of BodePlot, the Op's syntax should work. Look at Generalizations & Extension, 1er example $\endgroup$ – andre314 Aug 8 '18 at 19:03
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    $\begingroup$ Though, it's not exactly the same in the doc : there's only one argument. So BodePlot[1/(1 + 0.1 s)] should work $\endgroup$ – andre314 Aug 8 '18 at 19:06
  • $\begingroup$ Hello @andre, yes your solution works. $\endgroup$ – Gennaro Arguzzi Aug 8 '18 at 19:08

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