Time Domain from Bode Plot

So I'm finishing a little write-up for the presentation that I'm giving in March, and I realize that I can look at a Bode plot and have a pretty
good idea of what the system step response is going to look like. But what I can't do is explain why, or justify any conclusions without mathematics that go way beyond the scope of my talk.
On the one hand I have the utter precision of taking the inverse Fourier transform of the frequency response, which doesn't yield much intuition but does give you an impulse response thats as exact as your frequency response. On the other hand I have some hand-waving observations about looking for sharp amplitude or phase changes, or that long, low bump in the amplitude response you get from a well-damped PID controller that results in a long, small tail in your time-domain response, but all that'll do is speed up someone else's acquisition of their own intuition.
Is there a middle ground? Can anybody recommend any good references for anything in between? Something that would take 2 - 4 slides in a presentation would be absolutely perfect.
Thanks much.
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Tim Wescott
Wescott Design Services
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Tim Wescott wrote:

With a little practice, a Bode plot can yield the locations of poles and zeros by inspection, provided they aren't too close. At first at least, laying the asymptotes with a straightedge helps. Parlaying that information to impulse response (I always found step response easier) requires the implicit assumption of minimum phase, which is usually pretty accurate for most circuits.
There's a little-known gotcha. A lead-lag network to improve phase margin is usually designed so that its pole cancels the servo's second zero -- the one that tips the rolloff from 6 dB/octave to 12 and therefore the phase asymptote from 90 degrees to 180. By the time the networks pole kicks in, the loop gain is below unity and it doesn't hurt. All well and good, but the cancellation is never complete, so there's a hitch in the gain and phase curves that designers sometimes go to great lengths to minimize.
Don't. The time-domain detail caused by the mismatch is also a hitch. The step response shows a rise that extents most of the way to the asymptotic value pretty quickly, then rises the rest of the way with a longer time constant. The better the match, the closer the initial rise gets to the final value, and the longer it takes to go the rest of the way. I ran into this first with an op-amp circuit that settled to within 98% in 200 ns, and took nearly a millisecond to settle the rest of the way to within the noise. After days of thinking and tweaking, I ended up with 95% in 200 ns and all the way in 250. It turns out that the time constant of the tail is nearly one over the difference of the two frequencies one hopes will cancel. There's a price for everything.
Jerry
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Bode Plots show phase, by definition. No assumptions are necessary.
Scott
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Scott Seidman wrote:

Technically, yes. Computed ones, sure. When the information comes from the device, one often has only frequency-magnitude data plotted on log-log paper. If one is lucky. :-)
Jerry
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Tim,
For me, the nyquist plot is more intuitive than bode. It encapsulates both the frequency and phase plots into one curve, and things like phase and gain margin are easily visualised. One problem is that it hides frequency, but with a bit of clever graphics using colour this should be soluble.
On a more general level, I recommend that you take care in attempting to put across technical concepts like this in quickie presentations. You can easily crash and burn. Run your presentation by a test audience first. I find my (non technical) wife is a good critic. Good luck!

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bruce varley wrote:

I'm thinking that I'll just do a little handwaving, say "of course if it's shifted up in frequency then it'll settle faster, all else being equal", and leave it at that.
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Tim Wescott
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Tim, a presentation where? A trade show? March is trade show season.
I know what you are trying to do. It will be interesting to see what you come up with. I find it hard to pick out the zeros and poles accurately from a Bode plot even if I know where the poles and zeros are. How do you get a computer a computer to pick out the poles and zeros from a Bode plot? This should be interesting.
If you remember I use the least squares system identification method using data in the time domain. I then compute the model in the z domain then the s-domain and then I can make the bode plots. I do it backwards. However, I am still interested in your approach, especially if you have real applications where your method can be applied. One should always be willing to add another arrow to his quiver.
Peter Nachtwey
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Peter Nachtwey wrote:

Yes, see my other post "Shameless Plug".

Well you don't, but if you do an IFFT on a Bode plot (particularly one that's taken in sampled-time) you should get an impulse response.

Well, I was looking for arrows for my quiver, actually -- mostly I wanted to know if there's a way that you can look at a Bode plot and, by inspection, get a good idea of what the step response is going to look like. I can sort of do this intuitively, but that's from 10 years of looking at Bode plots and fiddling with time-domain responses and I don't always get it right.
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Tim Wescott
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I took some ideas from my mind, I know that these are elemtary ones, but the purpose is only to think like a newbie: Damp factor is the most representative variable from the time response, OK? So, if we have it, we can pre-visualize, on mind, the time response shape, alright??? For example:0.7<Damp<1 =>exponential | 0<Damp<0.7 => Means that overshoot exist (damped sinusoidial)| Damp=0 => Oscilatory response, looks like a system with non-energy consumption. Damp factor has a relation with the Phase Margin (PM) (if PM < 60 degree, a linear relation, remember??)The phase margin can be obtained from open-loop bode diagram analysis. But how to explain that damp factor values like that cause what you've said? Speak a little about root locus, show a root locus diagram of open-loop transfer function (graphs, sometimes, speak more clear that math expression, isn't???) and explain the relation between the "THETA" angle and the damp factor (something like that: Damp=cos(THETA), while THETA = arctan(Pole(Im)/Pole(Re))), this close the thought. Well, I know that sounded elementary for you (I read your resume), but this is a vision of a less-then-one-year junior electrical engineer (almost nothing, hehehe)

Yes, I'm sure that is a hard math work. I like maths, but the practical way of the engineering is more atractive and involves "feeling" (this word sound romantic not technical, hehe). To put Fourier's thought in 4 slides is not enough. I agree with you.

I hope I've added some new ideas. Good luck and fell free to contact
Roberto Feliciano Dias Filho Recife-PE/Brazil
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Tim,
We teach here that for systems with dominant slow poles there is a strong relationship between the Bode magnitude response M(w) (NB without logs) and the unit step response y(t) such that
y(t) is approximately equal to M(w=1/t)
It can be shown that for a class of real systems this is exact at the limits t=0 or w=0 (steady state) , and also that the rate of change also match at these limits (roll off vs speed of response).
Unfortunately inbetween the error for even a first order lag may be significant (after one time constant T a first order lag gets to 63% of the final value, the bode magnitude drops to 50% at the break frequency 1/T). Also clearly doesn't work well if the damping factor is less than 0.7. But it does work well for systems with responses that include slow (and therefore dominant) non-oscillatory poles.
So you can plot an open loop Nyquist diagram from frequency response data. Design a frequency domain compensator. Predict closed loop magnitude response using M-circles Predict closed loop step time response
All without a mathematical model of the plant/ process.
I do not recall ever seeing this in a text book bu it does appear in our students lecture notes! Hope the above is clear/ sensible...
Michael Tombs
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Michael Tombs wrote:

Yes, clear and sensible. I'm not sure if it'll fit into my presentation, but that's my problem.
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Tim Wescott
Wescott Design Services
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