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

 

[Tim Wescott]

wcn@public.zz.ha.cn wrote:

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

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 www.tech-chat.de

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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Jon]

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.

 

[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.

 

[Jim Thompson]

On 11 Aug 2005 08:05:04 -0700, "Jon" <jon.lark@l-3com.com> wrote:

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.

Except for cases that have excess phase ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Kevin Aylward]

wcn@public.zz.ha.cn wrote:

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

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.

http://www.anasoft.co.uk/EE/feedbackstability/feedbackstability.html

Kevin Aylward
informationEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Terry Given]

wcn@public.zz.ha.cn wrote:

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

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

Cheers
Terry

 

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

From a0 to a2 is of 20db/dec slope, a2=5a1=5a0, at the section of <a0,
is of 40db/dec.

and at the section of >a2,is of 40db/dec else. this figure is socalled
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 <a0, the phase shift will be 180 degrees. is it
oscillation?

 

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
| /
|

From a0 to a2 is of 20db/dec slope, a2=5a1=5a0, at the section of <a0,
is of 40db/dec.

and at the section of >a2,is of 40db/dec else. this figure is socalled
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 <a0, the phase shift will be 180 degrees. is it
oscillation?

 

[Tim Wescott]

wcn@public.zz.ha.cn wrote:

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

From a0 to a2 is of 20db/dec slope, a2=5a1=5a0, at the section of <a0,
is of 40db/dec.
and at the section of >a2,is of 40db/dec else. this figure is socalled
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 <a0, the phase shift will be 180 degrees. is it
oscillation?

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".

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com