Competition

Yeah

Nowadays we just use A Relativity Drive

Try generating 10 Hz sine waves with an actuator with a 10 Hz band width.

Peter Nachtwey

On 18 Jan 2007 17:56:20 -0800, "Peter Nachtwey" proclaimed to the world:

Stop quoting out of context and go listen to some 60 cycle humm. :-)

The solution of higher order continuous DE's is intriguing but ......what does it represent.....

A bit like humans!

Well, despite what many people might expect, in my experience for the process industries the profile of energy as a control improvement driver has stayed pretty much where it always was in the scheme of things. What has become hugely more important for CEs, and instrument guys as well, is 'licence to operate' issues. In the old days if your product was off spec you just snuck it out into the marketplace, if your plant was smelly then that was tough. Now either of these situations can put you out of business if you're caught out by a regulatory authority. If you're looking for a current-day hot area, then I'd look no further than LTO. However, the transfer functions and smart algos still tend to be pretty much 'second tier' behind the people and business issues, as they've always been.

You are striking on some very common issues. The first step in ANY control problem is to identify the true process objectives. This may be energy, quality, response to upset, or one of many other conditions.

Consider altitude control on an aircraft. At 30,000 feet, with a full load of passenger receiving dinner, the main concern is smooth control response. But when you are landing, it is precise control, and definitely no overshoot!!!

You can find a number of studies on control performance at:

In fact, a control conference reviewing practical applications is scheduled for April, 2007. I am currently writing a text for the bio-pharmaceutical industry, which will consume the next 6 montsh. Such a practical book for the general industry is needed. If you are seriously interested in publishing a book on practical applications of control, including REAl sensors, REAL controllers, and REAL valves, contact me via email.

-George

I think you need to add a quantizing function to the PV so the PV has no more resolution that the 'real' feedback device you would be using. Then run your simulations using PID^5 and tell us what happens.

Peter Nachtwey

"Peter Nachtwey" schrieb im Newsbeitrag news: snipped-for-privacy@m58g2000cwm.googlegroups.com...

See equation (25) and Fig. 9 c for simple implementation:

I stated before G1*G7 = 1

1 F(s) closed loop = --------------- 1 1 + ----------- K1*K7*G1*G7

Total open loop amplification = K1*K7 (not time-relevant)

P-Band of controller ~ 1/K1

That doesn't answer the question about the quantizing of the feed back and how you can take the 5th derivative of the PV using your PD^5.

I also find it odd that on your website that your process has only zeros and your controller has only poles. A PD has a gain and a zero which is in consistent with the your equations on your website.

Peter Nachtwey

"Peter Nachtwey" schrieb im Newsbeitrag news: snipped-for-privacy@m58g2000cwm.googlegroups.com...

That's all about.

Therefore I have written a program that can find the cofficients for differential equations up to PD^5:

page 2 for finding DE (second line from bottom) from data points page 1, and see example implementation on page 3 for PD^3 with 3 integrators and calculated coefficients.

Again:

for PD^n: K1 is P-amplification Gain for process^n: K7 PD^n control will have a remaining error (eps) depending on K1. K1 can be set. With high precision compensation eps can be very small. The integrational part (I) normally used in PID controllers will not be necessairy.

Sorry, I don't see problems so far. See for more explanation:

A_i coefficients belong to process^3 B_i coefficients belong to PD^3

A_i coeffcients (process^n) are not known, but 'can be calculated' by a programm I've written. Then I use them for compensating the process (B_i=A_i).

Result:

F1(p)*F2(p) is equal to G1(s)*G7(s) Next step: consider K1 and K7 K1*F1(p)*K2*F2(p) is equal to K1*G1(s)*K7*G7(s)