# PI Controller – Proportional Integral – Control System

Steady-state error can be improved by placing an open-loop pole at the origin,
because this increases the system type by one
. For example, a Type 0 system
responding to a step input with a finite error, will responds with zero error if the system
type is increased by one. But, we want to do this without affecting the transient response.

However, if we add a pole at the origin to increase the system type, the angular contribution of the open-loop poles at hypothetical point A is no longer 180, and the root locus no longer goes through point A, as shown in Figure 1.a and 1.b:

Figure 1.

To solve the problem, we also add a zero close to the pole at the origin, as shown
in Figure 2:

Figure 2.

Now the angular contribution of the compensator zero and compensator pole cancel out, point A is still on the root locus, and the system type has been increased. That is how we can improve the steady-state error without affecting the transient response.

A compensator with a pole at the origin and a zero close to the pole is called an ideal integral compensator, or Proportional-plus-Integral PI compensator, which transfer function Gc(s)  is:

Next example allows to find how PI compensation works.

For control system of Figure 3, it is required to reduce steady-state error to zero, through a PI controller, keeping damping at ξ=0.173. The plant transfer function is G(s) and its original controller is represented by the gain k:

Figure 3.

The first step is to evaluate the system before the compensation, then to find the location of the two closed-loop second-order dominant poles  in order to get the damping requiered by the design specifications.

Figure 4 shows the Root-Locus of the system before compensation:

>> sgrid(z,0)
>> s=tf(‘s’);
>> G=1/((s+1)*(s+2)*(s+10));
>> rlocus(G);

Figure 4.

Using the damping line in Matlab, we can find the intersection point between the root-locus and the value ξ=0.173as we can see in Figure 5:

>> z=0.173;
>> sgrid(z,0)

Figure 5.

The intersection of Figure 5 shows us that adjusting the gain to k=165 of the original controller, we obtain the damping requiered: ξ=0.173. We also see in Figure 5 that the closed-loop second-order dominant poles s1 and s2, before compensation are:

Now we look for the third pole in the root locus. In Figure 6 we must set the same gain k=165 at the third pole line, in consequence s3 is located at:

Figure 6.

With k=165 we calculate the steady-state error e1(∞) for a step input, before compensation:

Where kp1 the position constant before compensation:

Where kG(s) is the system forward transfer function multiplied by the adjusted gain, before compensation, as in Figure 3. Therefore:

We add a PI controller in cascade into the system, as in Figure 7:

Figure 7.

Here, we have matched the gain constant of the compensator with the original gain constant, that is to say k=ki. The constant a is determined by the location of compensator zero, wich must be near the compensator pole. That is why we set the compensator zero at s=-0.1 , that is to say  a=0.1. The root locus of this compensated system is in Figure 8:

>> G=(s+0.1)/(s*(s+1)*(s+2)*(s+10));
>> rlocus(G);

Figure 8.

In view of the fact that we want to maintain the transient response as unchanged as possible, in Figure 9 we draw the damping line in the root locus and search for the point of intersection between the lines of the root locus and ξ=0.173:

>> z=0.173;
>> sgrid(z,0);

Figure 9.

Adjusting the gain to k=159 in Figure 9, we obtain the damping ξ=0.173. We see that closed-loop second-order dominant poles s1 and s2, after compensation, are:

Looking for the third pole in the root locus,  we must set the gain k=159 at the third pole line. After that, s3 is located at:

These results show that approximately the values ​​of the 3 poles before and after the PI compensation have been conserved, indicating a similar transient response after correcting the error in steady state from 0.108 to 0, as shwon later.

The forward transfer function G2(s)  of the system after compensation is:

One more time, we calculate steady-state error e2(∞) for a step input, after compensation:

In consequence:

Figure 10 compares the step response of the closed-loop system  before and after compensatio PI:

>> G1=165/((s+1)*(s+2)*(s+10));
>> sys_antes=feedback(G1,1);
>> G2=(159*(s+0.1))/(s*(s+1)*(s+2)*(s+10));
>> sys_despues=feedback(G2,1);
>> step(G1,G2)

Figure 10.

Figure 10 shows that through PI compensation we have managed to improve the steady-state error without considerably modifying the transient response of the original system.

`Compensación en Cascada - Lag Compensation`

In construction…

Source :

Written by Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer

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