My transfer function is enter image description here

with Omega_n = 1000.8 and Lambda = .5004 and k = 3

But when I try to individually plot the phase I get something like thisenter image description here

I believe my equation for finding the Phase using ArcTan function is right. Does anyone know why my individual phase plot does not look like phase plot in BodePlot?

  • \$\begingroup\$ For starters the x axis in the upper plots is logarithmic. Try using a log axis on your plot and see if it becomes identical. \$\endgroup\$
    – John D
    Sep 20, 2016 at 14:37
  • \$\begingroup\$ You should use something like atan2 which is inverse tangent but not restricted to -90 to 90 degrees as the one you used. ;) \$\endgroup\$
    – carloc
    Sep 20, 2016 at 14:41
  • \$\begingroup\$ Don't know which language you are using, but in some there is a arctan2, which unwinds the angles, so that they are not constraint between -90 and 90 degree. \$\endgroup\$
    – Arsenal
    Sep 20, 2016 at 14:42
  • 2
    \$\begingroup\$ As the others have said, atan2() is the four quadrant version of atan(). What you're seeing is the wrapping, that is, if the phase will vary more than +/-90deg the output will "reset the counter". As can be seen in the phase graph, your min/max are +/-90deg (+/-pi/2). atan2() will wrap it around +/-180deg. For more, you'll need to unwrap the phase. For your simple experiment, I'd suggest using LTspice with a Laplace expression, in an .AC analysis. \$\endgroup\$ Sep 20, 2016 at 17:20
  • 3
    \$\begingroup\$ I'm voting to close this question as off-topic because it's about maths primarily and how to use the correct tan function in excel. \$\endgroup\$
    – Andy aka
    Jul 7, 2017 at 6:54

1 Answer 1


Take care in arctg(A/B) when B < 0 (add 180 degrees in result). In your case, when x > 1000.8. When x = 1000.8, occurs a division by 0.


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