# Analysing bode plots. Stability

1. Mar 19, 2014

### james1234

Hi all,

I would be greatful is someone could kindly enlighten me as to the correct interpretation of the appended bode plots.

My understanding when interpreting bode plots is that we desire 0 gain where the phase is equal to or exceeds 180 degrees (marginally stable / unstable).

In the attached figure 'bode simple' the relationship is clear. Good phase margins can be observed i.e. where the phase shift exceeds 180 deg the magnitude is less than one. where the magnitude exceeds 1 the phase is less than 180.

In the second figure ('bode_conf') the relationship is not at all clear to me.
Two things appear to be evident in this plot.
1) The phase is increasing below 5 rad/s and therefore
a. there is either a minimum phase zero in the system at aprox 0.1 rad/s; or,
b. there is an unstable pole in the system at this same frequency (positive phase)
2) For the frequency range 0.001-1.5 rad/s the magnitude of the response is greater than 0db (a gain ratio in exces of one) while the phase shift over this same frequency range exceeds 180 degrees.

I would therefore deduce from the bode plot that this system is unstable. Not so! All poles of the system remain in the left hand half plane of the s-domain.

I would be greatful of any insight into the correct interpretation of the latter bode plot.
The magnitude of the resposne clearly exceeds one at the initial phase crossover (w_180) and indeed for all frequencies 0-1.5 rad/s.
The significance of the second crossover?

Regards,

Jamie

#### Attached Files:

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• ###### bode_conf.JPG
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2. Mar 20, 2014

### milesyoung

It's unclear to me if you're talking about open-loop or closed-loop stability. Recall that we use the Bode plot (or Nyquist diagram in general) of the open-loop function to determine the stability of the closed-loop system.

Let's assume both of your Bode plots are of their respective open-loop functions. If you then want to determine if these systems are stable in closed-loop, you should apply the Nyquist criterion:

Z = N + P

where:
- Z is the number of unstable poles of the closed-loop system,
- N is the number of clockwise encirclements of the critical point by the Nyquist diagram (negative if counterclockwise),
- P is the number of unstable poles of the open-loop system.

Assuming your open-loop systems have no poles at the origin, their Nyquist diagrams are then just their Bode plots converted into polar form and superimposed on top of their complex conjugate.

For your first attachment, N = 0, so if it's stable in open loop it's stable in closed loop.

For your second attachment, N = -2, so it's only stable in closed-loop if the open-loop system has two poles in the RHP.

3. Mar 20, 2014

### james1234

Thanks Miles,

Sorry my explanation wasnt clear. I was refering to the closed-loop stability of the systems and wondering how this might be determined purely from the bode diagram of the broken loop. Is this possible for systems with multiples phase/gain crossovers?

The second system is indeed open-loop unstable. It was the excesive phase shift of the broken loop that threw me. I had always thought that large gains in the presence of greater than 180 deg phase was a bad thing (I'm refering again to the broken-loop response). Obviously not the case..

I gather that what we have in the second plot is a lower limit on the gain (GM at the first crossover)

Cheers!

4. Mar 20, 2014

### milesyoung

If your system is stable in open loop, the Bode plot is often adequate for determining closed-loop stability. In general, though, the Bode plot is a very poor substitute for a Nyquist diagram.

Try, for instance, to determine from its Bode plot if this open-loop stable system:
$$G(s) = \frac{1}{s^2 (s + 1)}$$
is stable in a unity-feedback configuration.

That depends on the number of poles in the RHP of the open-loop function. The open-loop gain can change N, but you still need to know P to determine if Z > 0.

The Nyquist criterion works in general. In terms of stability, gain/phase margin only makes sense for "simple" systems.

5. Mar 20, 2014

### james1234

I dont know why; I've always avoided nyquist plots like the plague. Looks like its time I convert.

Cheers for the input!

6. Mar 20, 2014

### milesyoung

I've heard that a lot before. Mainly I think it has to do with the following:

A lot of systems are very "well behaved", e.g. they're stable in open loop and have very little phase lag at low frequencies. This type of system has a very specific kind of Nyquist diagram that makes it easy to tell if the system is stable in closed loop. Since the Nyquist diagram and Bode plot are very intimately related, that translates into the stability criterion (gain at -180° crossover frequency) you're familiar with. It's really just a way of evaluating the Nyquist criterion that only works for this special class of systems.

If you're not regularly exposed to systems outside this class, Nyquist diagrams just seem like an unnecessary hassle. It would be easy to forget why the Nyquist diagram is not only useful, but necessary.

I can recommend a book if you're interested. I've used 'Feedback Control Systems' by Phillips and Harbor extensively, as I think it strikes a nice balance between mathematical detail and intuitive examples. In addition to the basics of control theory, it also includes some great chapters on digital control systems.

7. Mar 21, 2014

### james1234

Hi Miles,

Thanks for the reference. Much appreciated. Its always a pleasure tracking down a good text. I'm not familiar with Phillip's and Harbor's book. I will most certainly pick up a copy.

Best,

Jamie