PID Controller - ID combination

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Thread Starter

Anonymous

I'm new to this forum and I'm fresher. I have a question on PID. generally all combinations are used like PI, PD, PID, etc. But NOT "ID". Why? Can anybody tell me this?
 
It is all to do with the maths of control.

Generally most systems can cope well just Proportional control (P) the I and D assist with getting to the setpoint quicker and them maintaing without over shoot, or minimal overshoot.

I may be wrong but i don't think you could control without an P term in your equation.

Never really though about it!!

Jim
 
M
Because P (proportional) is really what is needed as baseline - to minimize (counteract) the error - whereas I (Integral) and D (Differential or Derivative) are just to compensate for respective system dynamics. You may want to study first 2nd order differential equations and control systems (e.g., servo) in general. You'll meet names like Nyquist, Bode, Root-locus, Ruth-Hurwitz and others there.

Meir
 
Hi.

At the beginning we have P.
P is the proportional regulation parameter in the V = e*P equation.

The others, I (Integrative) and D (Derivative), are used only to increase the quality of the regulation loop but the P remains fundamental.
 
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The professor

If you had any idea how the algorithm works you wouldn't have asked this question. Since you seem to lack this basic information, I will enlighten you.

In all PID algorithms with which I am familiar the Proportional (P) is always multiplied directly against the rest of the calculation, regardless of whether you are including Integral (I) and/or Derivative (D). I'm sure you realize, then, that a proportional of 0 (zero) would result in a null value for the entire calculation. When the process matches up well with the controller, you will find that a gain of 1 is suitable and as such the proportional has no multiplicative effect. O.K.?

The Professor
 
C
Hi,

Your question tweaked my curiosity and I tried this in simulation and was able to control with integral control alone, though not with out some under damping (it oscillated before settling). Because of this it may take a long time to settle. I'll work through the math if you are curious.

FYI: It is mathematically impossible (this is Basic control theory) to reach your set point with proportional or derivative control alone UNLESS the plant you are controlling is itself an integrator (a water tank will a fill valve is mathematically an integrator a.k.a pure capacity system). For non-integrating or pure capacity plants you MUST have an integrator to reach SP.

If you analyze the closed loop between a summing stage and the integrator you will find that it actually constitutes a low pass filter: that is, it passes DC or steady state values but blocks high frequency components.

Chris Schene
 
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Chris Schene

"In all PID algorithms with which I am familiar the Proportional (P) is always multiplied directly against the rest of the calculation, regardless of whether you are including Integral (I)"

In most actual practice cases that is probably true, but from a purely mathematical sense the proportional control is simply the error (SP-PV) gained up by P.

If you take one form of the equation:

P(1 + Ti/s + Tds)*e(s)

The question of whether or not you can leave out a particular term (P in this case) has some practical implications. If you could control with a K/s term alone, you could save processing cycles in an MCU, or DSP or a components and connections in analog circuits.

It is sometimes useful to forget the practice, which often obscures theory, and go back to basic principles to see what is really happening under the hood.

You can control some plants with K/s alone. You CANNOT reach SP without an 1/s term unless the plant itself is an integrator (a.k.a pure capacity system). So, for most plants, P alone will never drive you to SP.
 
Most of the answers to this student are not very helpful.

A controller may be set without the P term by setting its gain to zero. The effect would be that the controller would not have any immediate response to a change in setpoint, but the Integral term would tend to eventually move the output so that the error between the PV (measurement) and Setpoint will eventually become zero - slowly. The Derivative term is usually computed from changes only in the PV so that Setpoint changes do not affect the output.

When the load on the process changes so that the PV drifts away from Setpoint, it is the P term that reacts quickly to restore control at the Setpoint. ID control would work eventually to reduce the error to zero, but ever so slowly.

I terms of evaluation of PID control tuning, the established criteria is to minimize IAE (Integral Absolute Error). Optimal tuning always results in some positive value of the Proportional Gain constant.
 
Although is not so usual, integral action can be used alone if the process is open loop stable and very fast if compared with the intended closed loop response. In a control loop (say, current control, for example), if the open loop has a settling time of 10ms, and the desired settling time is 100ms, the process can be aproximated by a gain K; if the controler transfer function is 1/(Ti.s), the closed loop transfer function will be 1/( (Ti/K).s + 1 ).

In most cases, it is expected that the closed loop response would be faster or at least the same as the open loop process. That's the real need for proporcional action.
 
It depends in which PID algorithm you choose.

According to wiki (yes.. Wikipedia).
The PID algorithm is Kp + Ki/s + Kds.

Which shows that Kp easily can be 0, and still have the other part functionable.
 
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Peter Nachtwey

This is easy. One gain is needed for every pole of an open loop system if you wish to place all the poles. If you want a critically damped 2nd order system then 2 gain are required. This is usually the P and D gain. If the system is a type 0 or non-integrating system the one must add the integrator but the integrator gain comes with its own pole so it can't be counted as one of the poles used to place the two poles of the 2nd order plant.

I answer this question a little more thoroughly on another recent thread.

<b>Moderator's Note:</b> Thread is http://www.control.com/thread/1400089214
 
H
a)Self-regulating process: Neither P nor D can be used alone or in combination as a PD since they never can bring the controlled variable to the setpoint.

The I-controller could be used alone for fast processes with little or no dead time, although the loop will – depending on the tuning – be either much slower than PI-control or produce overshoot and swings.

Adding D-action, i.e. using an ID-controller, gives no advantage: The overshoot is reduced and swings are dampened but the loop is even slower than with I-control only.

b)Integrating process. Using an I-controller alone gives perfect oscillations which cannot be stopped by adding D-action. Also here, ID gives no advantage.
 
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