Chapter 26.1 Introduction
Don’t be afraid of the chapters 26…30 big sizes! Majority of them are repetitive.
Controller P (other name-Proportional controller) is the simplest of the PID controllers family. It has one adjustment only-gain Kp.
We will test controller P with the objects:
Chapter 26.2 Closed loop transfer functon Gz(s) and steady state gain Kz
It’s a short reminder of the Chapter 20 How does feedback work?
Fig. 26-1a -open loop diagram
Fig. 26-1b -closed loop diagram
Theory confirms the practice. The steady state gain Kz in particular!
Note-there are different oscilloscope parameters–> The real close loop dynamic is much better than you see above!
The formula K*e(t)=y(t) is fulfilled in steady state. It’s very important to understand the feedback principle! All the Chapter 20 is a justification.
Chapter 26.3 Controller P with the one-inertial object
Chapter 26.3.1 Introduction
We said in the 20 Chapter that static objects are equipped with the steady gain K=1 mostly. It means that Kp is a steady state gain of all the open loop object (objest+controller). It’s easier to adjust Kp parameter now, more generally-all PID parameters.
Chapter 26.3.2 One-inertial unit as a open loop
K=1 and T=10 sec
The K and T of the one-inertial unit calculation.
Chapter 26.3.3 Controller P with Kp=10
Controller P executes very simple algorithm. It calculates the control error e(t)=x(t)-y(t) and calculates the control signal s(t) then:
s(t) = Kp*e(t) = Kp*[x(t) – y(t)]. Here Kp=10.
The RC circuit is in the steady state when s(t)=y(t). The steady state with the feedback is analogously. All static and stable units with the feedback are going to the state K*e(t)=s(t)=y(t). The example is Fig. 26-5. The reason of the signal y(t) growth or decline is a
difference s(t)-y(t)! It’s very important to understand the feedback. When you have problems, return to chapter 20!
The diagram confirms the theoretical Kz=0.91.
The oscilloscope cuts the y(t) on the 2 level. Let’s change the oscilloscope parameters to see the whole y(t) signal
The closed loop s(t) signal is much bigger than in open loop! See Fig. 26-6. It’s the reason that y(t) changes faster the in closed loop. The bigger is Kp the more dynamic is the system. The steady state is when y(t)=s(t).
Let’s compare open and closed loop systems in the common diagram.
Let’s compare open y1(t) and closed loop y2(t) diagrams.
The closed loop system y2(t) is 11 times faster than the opened loop system y1(t). But there is a steady state error e=1-0.91=0.09. I hope you don’t come to splendid conclusion “Open loop system is slowly but it enables steady state error e=0 because x=y”. Ok. You are right but the open loop system isn’t resistant to disturbances!
Chapter 26.3.4 Controller P with Kp=100
System is faster and the steady state error is very small. The red y(t) almost squares with with black x(t)! The theory confirms too–>Kz=0.99 i Tz=0.1 sec. The oscilloscope cuts the y(t) on the 2 level. Let’s change the oscilloscope parameters to see the whole y(t) signal.
You see the whole s(t) control signal. Very, very big! It may be a technical problem. An example. The furnace require 10 kW steady power, but 1MW at the start! It assures fast y(t) response. The crazy engineer will project this power only. The wise engineer gives 30 kW. The system will be not so fast then. We return to this subject in the Chapter 34 Nonlinearity influence for regulation.
Chapter 26.3.5 Summary P controller with the one-inertial object
It’s easy to control. Big Kp causes fast responses and small steady error. We can prove (Nyquist or Hurwitz) that one-inertial object with the feedback is always stable. By big Kp too!
Chapter 26.4 Controller P with the two-inertial object
Chapter 26.4.1 Introduction
We will repeat these same experiments but with the two-inertial object.
Chapter 26.4.2 Two-inertial unit as a open loop
K1=K2=1 and T1=3 sec, T2=5 sec.
There is a typical for multiinertial unit so-called point of inflection. It’s easy to obtain steady state gain Ku. But it isn’t so nice with the T1 and T2 parameters. There are the methods of course but let’s give it a rest.
Rozdz. 26.4.3 Controller P with Kp=10
The two-inertial object is a only difference of the Fig. 26-4
There are oscillations now. The steady state is in circa t=35 sec when:
2. there is “no movement” (all derivatives are null)
There are moments when s(t)=y(t) (circa t=5 sec for example) but there is “movement” here! So it isn’t steady state.
Let’s look the whole s(t) signal.
This is the same block diagram but there a different oscilloscope parameters.
The s(t) control signal is crazy now. There is big steady error e(t)=0.09 here and in Fig. 26-15. We can minimize this error (but not e(t)=0!) by Kp increasing.
Chapter 26.4.4 Control Kp=100
Call Desktop PID/12_regulacja_typu_P/10_2T_Kp100.zcos
There is a steady error e(t)=0.01 but the oscillations and long control time (t=45 sec. circa) are indecent!
Chapter 26.4.5 Summary P controller with the two-inertial object
It’s more difficult to control than a one-inertial object. There are oscillations now but we can prove (Nyquist or Hurwitz) that
two-inertial object with the feedback is always stable. By big Kp too!
Chapter 26.5 Controller P with the three-inertial object
Chapter 26.5.1 Three-inertial unit as a open loop
The one-inertial with a small time constant T3=0.5sec is added to the two-inertial (Fig. 26-12). We have a three-inertial unit now. The change is small. I expect that system with the feedback behaviour will be simillar as for two-inertial. Let’s check it.
The response is similar to Fig. 26-13, but a bit more lazy. The To parameter increased.
Chapter 26.5.2 Control Kp=100
What will be the influence of this small T3=0.5 sec?
Terrible oscillations and a long control time! The increased Kp will worse the situation probably.
Chapter 26.5.3 Control Kp=30
The system is instable! By the way. Has a calculated gain Kz=0.97 a sense? It has a bit of sense.
It’s a constant component of the oscillations! Let’s look not cutted s(t) control signal yet.
The block diagram is the same as Fig. 26-23 but the oscilloscope parameters are different now.
The increasing s(t) and y(t) amplitudes are seen. Note that s(t) amplitudes are bigger than y(t).
Chapter 26.5.4 Summary P controller with the three-inertial object
It’s more difficult object to control than the previously. The Amplitude Phase Characteristic of the three-inertial objects crosses 3 quadrants–> This object may be instable–>see Nyquist benchmark.
Chapter 26.6 How does P controller suppress the disturbances?
Chapter 26.6.1 Introduction
The disturbances suppression is the main job of all the controllers. The earlier experiments lasted 1 min. The next will be 2 min. The step x(t) is the first and the step type disturbance z(t) is the next in 70 sec. Imagine that you put the additional heater or cooler to the liquid tank. The “healthy” system should compensate this disturbance by the additional cooling or heating. “Healthy” means that the Kp is good adjusted.
This 2 min isn’t too much but some people may be nervous. Scilab and all simulation oriented language have a good option for these nervous people–>time shortening. The 1 sec is shortened up to 0.1 sec. I don’t recommend this method for automatic beginners. The main advantage of this course is a intuition and time shortening isn’t good for it.
How to shorten the time from 120 sec up to 12 sec? Make steps 1,2 and 3 under.
We will test controller P with the objects :
The input signals are
– x(t) as usual
– z(t) positive or negative disturbance (heating or cooling)
Chapter 26.6.2 One-inertial object, Kp=10 and positive disturbance z(t)=+0.2
z(t)=+0.2 The additional heater is put in the liquid tank.
The P controller reaction for heating is right. The z(t)=+0.2 increasing is compensated by the control signal s(t) decreasing. This compensation isn’t ideal–>Δs(t)=-0.182 instead Δs(t)=-0.2. It’s typical for Proportional Controllers. The bigger is Kp the better is compensation. But the ideal compensation isn’t possible for P controllers!
You see the small disturbance z(t)=+0.2 influence for the y(t) signal. There is a addtional +0.2 heating but the otput signal increases only about +0.018! The absolute compensation is possible when integration algorithm is included. We will speak about it later.
Chapter 26.6.3 One-inertial object, Kp=10 and negative disturbance z(t)=-0.2
The negative dosturbance z(t)=-0.2 (cooling)
The P controller reaction for heating is right. The z(t)=-0.2 decreasing is compensated by the control signal s(t) increasing. The compensation isn’t ideal too. The situation will be better when the Kp parameter is increasing. For example Kp=100.
Chapter 26.6.4 One-inertial object, Kp=100 and positive disturbance z(t)=+0.2
Beautifully! The y(t) stands almost a lancers lanca! Regardless of z(t) disturbance! There is a small disturbance z(t) influence, but it’s not seen at these oscilloscope parameters.
Chapter 26.6.5 One-inertial object, Kp=100 and negative disturbance z(t)=-0.2
Chapter 26.6.6 Two-inertial object, Kp=10 and positive disturbance z(t)=+0.2
Two-inertial object, Kp=10 and positive z(t) = +0.2
The experiment is an example that response for x(t) is different than for disturbance z(t)! The response for z(t) is better than for x(t)! Many people are adjusting the P (more generally PID) controler when x(t) is an input signal. They are glad that is ok but they forget the above effect!
When x(t) changes often–>the input x(t) is most important, otherwise the disturbance z(t).
Chapter 26.6.7 Two-inertial object, Kp=10 and negative disturbance z(t)=-0.2
The P controller reaction is right. It tries to compensate the negative (cooling) z(t)=-0.2 disturbance by the control increase signal (heating) s(t).
Chapter 26.6.8 Three-inertial object, Kp=10 and positive disturbance z(t)=+0.2
Chapter 26.6.9 Three-inertial object, Kp=10 and negative disturbance z(t)=-0.2
We will not test the Kp=30 (and more) because the system is instable then (see Fig. 26-24 and 26-25)
Chapter 26.7 Why the closed loop system is better than the open loop?
For majority of readers the question is like “Who want you be? Young rich and healthy or old, poor and ill?” So the rest of this subchapter is for a doubtful minority.
Let’s compare the open and closed loop system!
These same signals x(t) and z(t) are acting for:
– open loop system (higher object.)
– closed loop system (lower object)
Come to your own conclusions.
The fault of the P control type is no null steady error signal and that they aren’t absolutely resistant to z(t) disturbances. The PI and PID controllers are better in this area. We will discuss it later.