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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?

Alborz
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    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. – Eugene Sh. Oct 20 '20 at 14:17
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    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. – AJN Oct 20 '20 at 14:30
  • @Alborz what was it on the 3D pole-zero/bode plot diagram that you were unable to find? I ask because you said it was *useful but not enough*. - so please be clear about what you are looking for. – Andy aka Oct 20 '20 at 15:38
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    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. – John D Oct 20 '20 at 15:38
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    @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. – The Photon Oct 20 '20 at 15:54
  • @Andyaka, I found your explanation on 3D Bode plot vey insightful but at the same time I meant to mention that the discussion on the question did not explicitly address my question here. Otherwise, my question here would have been redundant. I edited the question to better clarify. – Alborz Oct 20 '20 at 16:18
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    @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. – Eugene Sh. Oct 20 '20 at 16:22
  • @EugeneSh. It provides both the magnitude and phase of H(s), but only for values of s that are on the imaginary axis. Its range is the whole complex plane, but its domain is only the imaginary axis. – The Photon Oct 20 '20 at 16:23
  • @ThePhoton Maybe I misunderstand something... but magnitude is \$|z| = \sqrt{zz^*}\$ - calculated over both imaginary and real axis. The phase is given by \$atan(Im(z)/Re(z))\$ - again calculated over both axes. – Eugene Sh. Oct 20 '20 at 16:27
  • Exactly! you can't decipher the TF from the bode plot without some guesswork. – Andy aka Oct 20 '20 at 16:28
  • @EugeneSh., Suppose I give you the Bode plot and ask you what's \$H(2j)\$? Then you can read the result (both imaginary and real part) off the Bode plot. But what if I ask you what's \$H(3+2j)\$? – The Photon Oct 20 '20 at 16:29
  • @ThePhoton OK, I see what you mean. So it is fourth dimension not the third one – Eugene Sh. Oct 20 '20 at 16:31
  • @EugeneSh., I'd say its the dimensionality of the domain of the function rather than the dimensionality of the range of the function. – The Photon Oct 20 '20 at 16:33
  • @ThePhoton Not sure what would be the physical meaning of it? So the "imaginary" axis is the frequency, what would the real axis correspond to? – Eugene Sh. Oct 20 '20 at 16:34
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    @EugeneSh. To be honest I rarely work in the Laplace domain, but whatever reasons people have for working with Laplace rather than Fourier, those would be the reasons to consider the real axis. (I suspect it's something like consider both power signals and energy signals) – The Photon Oct 20 '20 at 16:39

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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).

Andy aka
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  • 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. – The Photon Oct 20 '20 at 16:34
  • @ThePhoton explanations were the most useful for me. Appreciate it. – Alborz Oct 20 '20 at 16:38
  • @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](https://electronics.stackexchange.com/tour). 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. – Andy aka Jul 01 '21 at 15:29
  • @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. – Alborz Jul 01 '21 at 15:41