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I was doing the TransferFunctionModel and BodePlot. But the result from Mathematica and MATLAB are different.

The function is as follows:

Bi = TransferFunctionModel[(1 + 3*I*ω)/(1 + 1.5*I*ω), ω];

BodePlot[Bi, GridLines -> Automatic, ImageSize -> 400, 
         FrameLabel -> {{{"Magnitude (db)", None}, {None, 
         "Frequency Transfer Function Inside"}}, {{"Phase(deg)", 
          None}, {"Frequency (rad/sec)", None}}}, 
         ScalingFunctions -> {{"Log10", "dB"}, {"Log10", "Degree"}}, 
         PlotRange -> {{{0, 100}, Automatic}, {{0, 100}, {0, 20}}}, 
         PlotStyle -> {Red, Thick}];

Mathematica result MATLAB result

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  • $\begingroup$ How is the phase treated in each case? Phase can be wrapped between 0 to 2Pi or unwrapped to go to infinity. $\endgroup$
    – Hugh
    Commented Sep 21, 2015 at 18:01
  • $\begingroup$ Actually in both cases I didn't specify the phase range. The phase range in MM has negative part but in Matlab the phase is positive. $\endgroup$
    – yangyang
    Commented Sep 21, 2015 at 19:11
  • $\begingroup$ Without all the options in BodePlot,you'll get the same. $\endgroup$
    – Phab
    Commented Oct 1, 2015 at 11:55

1 Answer 1

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I'm getting a minimum phase of 0 in Mathematica as well.

Manipulate[tf = TransferFunctionModel[eq, s];

 BodePlot[tf, GridLines -> Automatic, ImageSize -> 500, 
  FrameLabel -> {{{"Magnitude (db)", None}, {None, 
      "Bode Plot"}}, {{"Phase(deg)", None}, {"Frequency (rad/sec)", 
      None}}}, 
  ScalingFunctions -> {{"Log10", "dB"}, {"Log10", "Degree"}}, 
  PlotRange -> {{{0.1, 100}, Automatic}, {{0.1, 100}, 
     Automatic}}], {eq, (1 + 3 s)/(1 + 1.5` s)}]

enter image description here

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  • $\begingroup$ But eq should be (1+3*is)/(1+1.5*is). Did you miss the "i"? $\endgroup$
    – yangyang
    Commented Sep 22, 2015 at 6:33
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    $\begingroup$ @yangyang Your use of "i" is non standard. Usually there are no imaginary units in the Laplace domain and you only get them when you replace s with i w. The Mathematica Bode plot does this for you. It assumes you are putting in an s-plane transfer function and want to plot it on the frequency axis. You need to clarify what you are trying to do. $\endgroup$
    – Hugh
    Commented Sep 22, 2015 at 13:56
  • $\begingroup$ @yangyang As Hugh said, I substituted iw with s so that it would work with a document that I had written earlier that calculates from the Laplace space. As he also pointed out, your mistake was that TransferFunctionModel[] assumes Laplace domain and you didn't make the substitution of wi ->s. Now that I try it again using s*I, I'm getting the -300 you were mentioning earlier. $\endgroup$
    – Brian G
    Commented Sep 22, 2015 at 14:31
  • $\begingroup$ @BrianG I understand your point. Thank you for your help!!! $\endgroup$
    – yangyang
    Commented Sep 22, 2015 at 14:48
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    $\begingroup$ @BrianG I know what is the problem. If I use b=(1+3*s)/(1+1.5*s); then I got the same as MM. It seems the imaginary part "i" should not be there. $\endgroup$
    – yangyang
    Commented Oct 1, 2015 at 11:28

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