Bode plot Vs. Root Locus

• Now this distance of roots from the imaginary axis determines the stability of the closed

• loop system.

Almost right. It is the relative distance from the imaginary axis as compared to the distance from the real axis that determines the stability. As you increase the gain, this relative distance becomes smaller, and the closed loop system will become more oscillatory, although with a second order system, the oscillations will eventually die away.

Pieter Steenekamp

Pieter, thanks for your reply.

For the system I described (G(s)=K/(s+a)*(s+a)) The distance from imaginary axis remain CONSTANT(=a/2) irrespective of the value of gain. In other words, as per root locus analysis the system will never oscillate .... so the question remains unanswered.

-- Alex

Alex,

Maybe you are confusing "oscillatory" with "stability". A system with a closed loop pole on the left hand plane, away from both axis, but relatively much closer to the imaginary axis than the real axis, is very oscillatory, yet stable, because the oscillations will eventually die away. It is true to say that it is for all practical purposes unstable, but according to the purists, it is still stable. This is exactly the same as for a systen with a very small phase margin.

Pieter Steenekamp

Right. The distance remains constant. But that doesn't matter.

What _does_ matter, as Pieter said, is the _angle_ (or damping ratio, which is monotonically related). That gets lower and lower and lower as the gain goes up. Not only will the low damping ratio make the system "practically" oscillate as you term it, it will tend to make it _really_ oscillate when the inevitable extra poles that always lurk out there in the high frequencies come in to their own.

Tim/Pieter, Thanks for your replies.

Actually a simple plot on a paper helped me understand the point you two have made. The damping ratio (cos(theta) ) will decrease as the OL gain, hence the imaginary part, increases.

I have two more questions:

1. On a root locus plot how is the damping ratio calculated ? From what I know it is the cosine of the angle made with real axis by the vector connecting the origin with a particular gain point. But what happens when the root locus has multiple branches giving multiple vectors for a given value of gain ..
2. How is phase margin calculated from a root locus plot ?

Thanks again.

-- Alex.

Multiple vectors for a given value of gain means that there will be multiple resonant pole pairs. In that case you would talk about the damping ratio of each pair, or you would pretend that one pair is dominant and ignore the rest (at your own risk, of course).

With difficulty. Actually, I've never needed to after I got out of school. If I _do_ need to, I'll look in a book on basic control theory.

Dnia Tue, 19 Dec 2006 07:10:28 +0100, snipped-for-privacy@yahoo.com napisa³:

No, no, no, no. lim G(s->0)= 10000/(100)=100 that's 40dB.

No, no, no. At 100 rad/s.

How do you calculate this? Phase will be about 168 deg. That's quite a difference.

(...)

This system is always stable after closing negative unity feedback if finite gain applied. Use any known stability criteria. Hourwitz or something.

Phase margin you read from Bode or Nyquist plot of *open loop* transfer function. You do not have to close loop to find if it will be stable or not.

Root locus is a plot of pool positons versus gain in *closed loop* system. It's handy if you want to see your results.

I would recommend you to use Matlab or similar (Octave, SciLab) and try your G(s). There are nice tools for that: sisotool, ltiviewer.

I hope I have putted some light on Bode and rlocus. Have fun.

Mikolaj, Thanks for your comments. I made a mistake by in my G(s) by putting the poles in the (s+p) format, for the gain-phase calculation in my original post to be correct the G(s) will have to be in a (t*s+1) format with t=1/10.

The most important thing I understood from your reply is that, by comparing Bode plot with Root Locus I was comparing apples to oranges. To 'emulate' Bode plot within a root locus I will have to do the following:

1. Based on open loop gain mark the two CLOSED LOOP poles on the root locus
2. Take a point on imaginary axis, say p1. Compute the maginitude of the vectors from this point to both the poles (say, m1 and m2).
3. Move p1 from origin to infinite frequency and plot m1 and m2 alongwith.

The quantity K/(m1*m2) will be the CLOSED loop gain plot I would get from the Bode plot.

Talking of phase margin, like Tim pointed out, it will be difficult to estimate using root locus.

To summarize my understanding: Root Locus plots how the CLOSED loop poles move w.r.t to OL gain and as such is better suited for time domain analysis. Bode plot on the other hand will not give the closed loop pole location, however it will tell the margins for stability.

Thanks to all.

-- Alex

...

I couldn't find my Spirule last time I looked for it, but it simply automates what can be done with a protractor, scale, and calculator. It's all laid out in most books I've seen.

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Fred

Why not? I think wrinkles are elegant. I'm still waiting for mine. :-)

Jerry

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I always assume that if you're playing in the controls playpen, then you either have Matlab on your computer or at least you have access somewhere.

So, if you know all the roots, which you would have to to do a root locus, then using Matlab you can easily get to the transfer function and then to the phase margin.

Dnia Wed, 20 Dec 2006 10:50:51 +0100, snipped-for-privacy@yahoo.com napisa³:

That's the point :).

Just thought of sharing the good link on root locus.

-- Alex