Question about automatic control system compensation

about three frequency compensation in automatic control open loop system, we have these describe: at low frequency section, the slope should be greater than 40db/dec for precision, at middle frequency section, be 20db/dec and across x- axis, then at hight frequency section again greater 40db/dec, Now the question is at low frequency, it ampletude is greater than 1 and still we get a phase shift greater than 180 degree. Shall we get an oscillating state at present? if it is not, how shall we judge the stablility at the section? Thanks

Reply to
wcn
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I suggest that you do a little bit of studying on the representation of systems using the Laplace transform and frequency domain design techniques.

By "180 degrees" phase shift I assume you mean a feedback loop with 180 degrees phase shift _before_ the added 180 degrees shift of the subtraction, as in the figure:

_ .---------. + / \\ | | ------->| + |----->| G |-----o---->

\\_/ | | | - ^ '---------' | | | '------------------------' created by Andy´s ASCII-Circuit v1.24.140803 Beta

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There is no reason that the transfer function G can't have more than unity gain at 180 degrees phase shift and still have a stable system.

The classic way to determine the stability of a system from it's frequency response is to generate a Nyquist plot. This requires that you know how many unstable poles in G before you generate the plot, however. IMHO the _best_ way to determine stability of a system is to start with a Bode plot of the system in a stable configuration and make darn sure that any modifications to the system _keep_ the system stable.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

If the zero gain point occurs at a frequency that is within the "middle" section, then the system will be stable. The phase shift at this frequency wil be 90 Degrees.

Reply to
Jon

Sorry if this is a multiple reply (finger blight). If the frequency at which the open loop gain is within the "middle" section (20 db/dec) then the system will be stable. The phase shift at this frequency will be approximately 90 deg.

Reply to
Jon

Except for cases that have excess phase ;-)

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

The often quoted "gain greater than 1 with net positive feedback and it will oscillate" is bogus.

Stability is determined whether or not there is a pole in the right hand plane. If such a pole exists, it results in an ever increasing output until a non linearity clamps it.

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Kevin Aylward snipped-for-privacy@anasoft.co.uk

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SuperSpice, a very affordable Mixed-Mode Windows Simulator with Schematic Capture, Waveform Display, FFT's and Filter Design.

Reply to
Kevin Aylward

| / | / | / 40db/dec | / | \\ | \\ | \\ 20db/dec __|_______|____\\_____|_______ | a0 a1 \\ a2 | \\ | \\ | \\ | / | / 40db/dec | / |

reasonable compensation curve in open looop system. Most control theroy textbook says that.In the way compensation,we can get a excellent stable close loop system and precision. But as we say, at the section of

Reply to
wcn

Now we are taking accout of muxium phase system only and whole open loop sytem was compensated to get a socalled reasonable curve illustrated as above. Its very kind of you offer me your paper at above site. I read and havnt find the information reletivily. | / | / | / 40db/dec | / | \\ | \\ | \\ 20db/dec _ _|_______|____\\_____|_______ | a0 a1 \\ a2 | \\ | \\ | \\ | / | / 40db/dec | / |

reasonable compensation curve in open looop system. Most control theroy textbook says that.In the way compensation,we can get a excellent stable close loop system and precision. But as we say, at the section of

Reply to
wcn

No it is not going to oscillate there. It wasn't going to oscillate there when Winfield Hill gave you the same answer. It wasn't going to oscillate there when Kevin Aylward gave you the same answer. It wasn't going to oscillate there when 'Jon' gave you the same answer.

You've asked essentially the same question four or five times here, gotten essentially the same answer, and still you're not satisfied! If you can't accept technical answers to technical questions perhaps you should be posting on one of the religious newsgroups.

If you are stuck on the idea that a system will oscillate with a phase shift of 0 degrees around the whole loop (your 180 degrees -- understand that and maybe other things will become clear) with a gain over 1 YOU ARE MISTAKEN, and you need to get over it. If you just can't believe us here then YOU NEED TO STUDY STABILITY THEORY. The Barkhausen criterion says that a system with no phase shift around the _whole_ loop and a gain of _exactly_ 1 will be on the _verge_ of oscillation, but it does _not_ say whether bumping the gain up or down will be the direction that will cause the system to go unstable.

I suggest that if you want to go beyond this you accept all of the answers you've gotten at face value and instead of just repeating your question hoping that you'll luck out you ask "how can a system with a gain greater than one and a total loop phase shift of zero be stable? I don't get it".

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

google symmetric optimum. and maybe learn how to read german. W. Leonhard has a nice section in his book.

Cheers Terry

Reply to
Terry Given

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