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The classic Bode plot (2D Bode) represent value of the transfer function (a complex number) evaluated at \$j\omega\$ (imaginary axis). It provides information on system closed-loop stability margins and also some means to estimate pole/zero of the transfer function, if not available (for experimental data).

Still, given that the classic Bode plot (2D Bode) provides transfer function values only on the \$j\omega\$-axis (i.e. only the Sinusoidal pieces of system characteristics), what information of the system characteristics (as far as modes and stability) are missing by not looking at the 3D Bode plot (Sinusoidal (\$j\omega\$) + Exponential (\$\sigma\$) information)?

In other words, what useful/practical information does the 3D Bode add to the 2D Bode?

Also, found this question useful but not enough to address my question.: What does a Bode plot represent and what is a pole and zero of a Bode plot?

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    \$\begingroup\$ Bode plot plot is providing the full information - the amplitude and the phase versus the frequency. It is a 3D plot, just divided into two separate plots. \$\endgroup\$
    – Eugene Sh.
    Oct 20, 2020 at 14:17
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    \$\begingroup\$ Most signals of interest can be decomposed into sinusoids. So, the plot along \$j\omega\$ axis is enough to get info about the system. Moreover, one can reconstruct the pole zero locations if the plot along the \$j\omega\$ axis is available. So IMO, there is no information lost. \$\endgroup\$
    – AJN
    Oct 20, 2020 at 14:30
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    \$\begingroup\$ Well, you didn't explicitly specify a linear system. Often we take bode plots of non-linear systems with a small-signal disturbance so you get the needed information at a single bias point, but you don't know what the response will be at a different operating point. \$\endgroup\$
    – John D
    Oct 20, 2020 at 15:38
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    \$\begingroup\$ @EugeneSh. I think OP's point was that if you consider the Laplace domain, the Bode plot doesn't directly show the behavior for s off the imaginary axis. I think if you consider only systems defined by a finite number of poles and zeros, all the information is still there. But if you consider systems (which I don't know how you would build) with other functional dependency (\$H(s) = \exp{[-a(s-s_0)^2]}\$, for example) then only knowing about the imaginary axis might lose some information. \$\endgroup\$
    – The Photon
    Oct 20, 2020 at 15:54
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    \$\begingroup\$ @ThePhoton But Bode plot is not providing the data from imaginary axis only. It is providing the amplitude of the complex value. And it is providing its phase. These two values are fully describing this number. \$\endgroup\$
    – Eugene Sh.
    Oct 20, 2020 at 16:22

1 Answer 1

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This pole zero diagram: -

enter image description here

Tells you everything about the transfer function and it also tells you everything about what the bode plot will look like. It's just a plan view of this: -

enter image description here

You cannot always (or even accurately) reverse the bode plot i.e. this: -

enter image description here

Into the pole zero diagram (top picture in this answer).

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  • \$\begingroup\$ As I mentioned in comments to the question, this assumes you have a TF that can be defined by a finite set of poles and zeros. \$\endgroup\$
    – The Photon
    Oct 20, 2020 at 16:34
  • \$\begingroup\$ @ThePhoton explanations were the most useful for me. Appreciate it. \$\endgroup\$
    – Alborz
    Oct 20, 2020 at 16:38
  • \$\begingroup\$ @Alborz if you are done with this Q and A you should think about accepting my answer. If you don't know about this requirement read this. You should also consider upvoting useful answers (not just on this question but on others you have asked. You've been a member for 8 months so it's about time you were reminded of the etiquette on this site. \$\endgroup\$
    – Andy aka
    Jul 1, 2021 at 15:29
  • \$\begingroup\$ @Andyaka. Thanks for the reminder. I'm afraid your "answer" does not really answer my question and that clearly explains why I did not accept your answer. "ThePhoton" indeed gave correct answer but I don't seem to be able to accept his comment. And to your "upvote" comment, if you look through the comments you will see that I did indeed upvote "ThePhoton" comments as they were useful. \$\endgroup\$
    – Alborz
    Jul 1, 2021 at 15:41

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