Control chap8

CONTROL SYSTEMS THEORY

Design Using the Graphical Tool

CHAPTER 8STB 35103

Objectives To use the root locus to design cascade

compensators to improve the steady-state error.

To use the root locus to design cascade compensator to improve the transient response.

To use the root locus to design cascade compensator to improve the steady-state error and the transient response.

Introduction In our previous chapter we learnt how to

sketch the root locus. The root locus allows us to choose the proper loop gain (value of K) to meet a transient response specification (e.g. %OS).

Setting the gain at a particular value will produce the transient response dictated by the poles at that point on the root locus. Thus we are limited to those response that exist along the root locus.

Introduction When we design a control system the two

most important things that we will consider are: Transient response

Settling time, percent overshoot, peak time

Steady-state error The difference between the input and the output

Introduction – improving transient response Example

We have a control system that produces 20% overshoot with settling time, Ts = 20 seconds. Let’s say we want to speed up our process to make it produce a stable output within 10 seconds instead of 20 seconds (Ts = 10) but still maintain 20% overshoot. Figure below shows our root locus and the radial line.

0.45

Introduction – improving transient response

Based on the previous example, poles on the same radial line will have the same overshoot value. Pole at A has the same overshoot as pole at B because both poles are on the same radial line.

A point on the radial line that is further from origin will have smaller settling time compared to point on radial line that is close to origin. Point at A has larger settling time compared to point at B.

Introduction – improving transient response Let us assume point at A produce Ts = 20

and point at B produce Ts = 10. Our objective is to produce transient response with 20% overshoot and Ts = 10. So, the best point in the s-plane to get 20% OS and Ts =10 is at point B.

0.45

Introduction – improving transient response However, based on the figure below, the root

locus only crosses point A, even if we increase the gain the root locus will not reach point B.

One way to solve this problem is to replace the existing system with a system that can reach point B but this replacement is expensive.

Introduction – improving transient response A better method is compensate the

system with additional poles and zeros, so that the compensated system has a root locus that goes through the desired pole location for some value of gain.

Disadvantage for this method is the system order will increase and could effect our transient response.

Introduction – improving steady-state error Compensator can also be used to improve the

steady-state error. In our previous chapter we varied the gain of

our system to meet the transient response specification. When gain is varied our steady-state error will also change.

The higher the gain, the smaller the steady-state error but the larger the percent overshoot.

Reducing the gain to reduce overshoot will increase the steady-state error.

Introduction- Compensators In summary, transient response is improved

with the addition of differentiation, and steady state error is improved with the addition of integration in the forward path.

Our focus is only for cascade compensator for a unity feedback system.

Introduction – modes of control There are actually three “Modes of control”.

Proportional (P) Integral (I) Derivative (D)

Previously we learnt to change the value of gain, K, to get a certain transient response specification. The mode of control for this type is Proportional or P-controller

Proportional

Improving steady-state error via cascade compensation Proportional-plus-integral (PI) controller

Steady state error can be improved by placing an open-loop pole at origin, because this will increase the system type by one.

For example, a Type 0 system with step input produces finite error. The error becomes zero when we increase the type by one.

Preferably, we improve the steady-state-error without affecting the transient response.

Improving steady-state error via cascade compensation

Type 0 has finite error for step input

When we increase Type 0 by one into Type 1, the error becomes zero for step input

Improving steady-state error via cascade compensation Example for PI controller:

Figure below is an example of a unity feedback system with gain, K. Point A is on the root locus which means the [sum of zeros angle] – [sum of poles angle] = odd multiples of 180 ̊.


Assume the input, R(s) for this system is step input, the steady state error for this system is a finite value. We can reduce the steady-state error to zero by increasing the system type. This can be done by putting integration (poles at s=0) at gain.


If you notice, the root locus for our compensated system no longer goes through point A, meaning our transient response has changed.

In order to get the same transient response similar to the transient response before we put integration (poles at s=0) we must make some changes to the compensated system to make sure the root locus goes through point A again.


We know in root locus, the poles will always move to zeros. Previously we introduce a pole at zero thus changes our transient response. In order to undo the change and still maintain the increase in system type, we will need to put a zero very close to the compensated pole. Both zero angle and pole angle will cancel each other thus making the root locus resemble the original root locus but with an increase in system type (pole-zero cancellation)

Improving steady-state error via cascade compensation Example 9.1

Given the system below with a step input and 57.4% overshoot, reduce the steady-state error to zero without greatly affecting the transient response.


Analysis (solving the proportional part) We will first investigate the transient response and

steady-state error for this system.

First draw the poles and zeros using the open loop transfer function of the unity feedback system.

Next, we need to find the value of gain where our system produce 57.4% overshoot.


Next, we need to find the value of gain where our system produce 57.4% overshoot.

1 32

0.174

A CB

K A B C

The gain we found using calculation is K=164.6 (hint: r = 3.987)

r

X XX


We are trying to improve the steady-state error so we need to know the steady state error of our current system. Since the input is a step, we need to calculate the final error value for step.

The value of step input is 1 but since we have error the output that we get from the system is (1-0.108 = 0.892)

0

1( ) , lim ( )

1step ps

p

e K KG sK

0

164.6lim 8.23

( 1)( 2)( 10) (1)(2)(10)

1( ) 0.108

1 8.23

ps

step

KK

s s s

e


The system has three poles instead of two poles, we need to determine whether the system is a valid second-order approximation. We must first find the coordinate of the system poles when gain, K =164.6 (57.4% overshoot). We can find the poles using the closed loop transfer function of the unity feedback system.

The closed loop transfer function, T(s) is

3 2( )

13 32 20

KT s

s s s K


We will substitute the value of gain where the root locus intersect the radial line to produce 57.4 % overshoot and get the poles of the system during this gain. From our previous calculation we know gain, K = 164.6, so

We will then plot the poles by factorizing the denominator.

3 2

164.6( )

13 32 20 164.6T s

s s s

164.6

( )11.613 0.693 3.926 0.693 3.926

T ss s j s j


We will include the poles during gain, K = 164.6 in our root locus.

The distance of the dominant second order poles from the jω-axis is 0.694 while the distance of the higher order poles is 11.61. The distance of the higher order pole is more than five times the distance of the dominant second-order poles from the jω-axis. Hence it is a valid 2nd order approximation

11.61/ 0.694 16.729 times farther

Improving steady-state error via cascade compensation Summary of the original (uncompensated) system is, the

system produce 57.4% overshoot when gain, K = 164.6. The error for this system is 0.108. The system can be approximated as a second-order system.

Parameter Uncompensated

Percent overshoot 57.4%

Gain, K 164.6

Dominant second-order poles

-0.693 + j3.926-0.693 - j3.926

Higher order poles -11.613

estep(∞) 0.108

Input =1 Output = 0.892

Improving steady-state error via cascade compensation Solution (solving the integral part):

The question : to reduce error to zero without greatly affecting the transient response. We reduce the error to zero by increasing the system type. In order to increase the system type we will need to add integration (poles at origin). To make sure the transient response of our original system does not change we will also put a zero into our system very close to the compensated pole ( pole at origin). The angular contribution of both compensated pole and zero will cancel each other thus maintaining the original transient response =pole-zero cancellation


We will now plot the zeros and poles using the open loop transfer function of the compensated unity feedback system, GOL(s)

0.1( )

1 2 10OL

K sG s

s s s s

Poles: (0,0), (-1,0), (-2,0), (-10,0)Zero : (-0.1, 0)


The root locus with the compensated poles and zeros

We redo the process to find the value of gain where radial line intersect the root locus. (Hint: r = 3.897)

The gain found from calculation is K = 158.2.


The gain is found using the same method as the uncompensated system.

1 32

0.174

A CB

A B C EK

D


r

X XX O4

5X

D E


Next we find the position of poles when gain, K = 158.2. We will use the poles to check whether we can approximate as a second-order system. The closed-loop transfer function of the compensated unity feedback system, T(s), is

We will then need to factorized the denominator and plot the poles and zeros. Since the largest s value is 4, it is hard for us to do hand calculation, we need to use software to solve for the factors.

4 23

( 0.1)( )

13 32 (10 ) 0.1

K sT s

s s s K s K


The coordinate of the poles when gain, K = 158.2 are plotted in the root locus


From the root locus, the higher order poles is far from the jω-axis compared to the dominant second-order poles. The zero is close to a close-loop pole which means the angle contribution of both zero and pole will cancel each other. In conclusion, the compensated root locus can be approximated as a 2nd order system.

The additional poles (integration) to the system has increase the system type from Type 0 to Type 1. Error for step input Type 1 is zero which means the system produce no error but still maintain the same overshoot value.

Improving steady-state error via cascade compensation Summary result for uncompensated and

compensated system

Parameter Uncompensated Compensated

Percent overshoot 57.4% 57.4%

Gain, K 164.6 158.2

Dominant second-order poles

-0.693 + j3.926-0.693 - j3.926

-0.678 + j3.386-0.678 - j3.386

Higher order poles -11.613 -11.55

estep(∞) 0.108 0

Input =1 Output = 0.892 Output = 1

Improving steady-state error via cascade compensation Based on the previous table, we can see

that our compensated system produces zero error and at the same time the coordinate of the dominant second order poles and higher order poles for the compensated an uncompensated system is almost the same, which means the transient response for both compensated and uncompensated is almost the same. The only difference is the compensated system has zero error.

Summary - PI Pole at A is:

a. on the root locus without compensator;b. not on the root locus with compensator pole added;

Summary - PI c. Pole at A is:

approximately on the root locus with compensator pole and zero added

* Find angle of zero, θzc

* Find position of zero on the real axis

Summary - PI Closed-loop system for Example 9.1:

a. before compensationb. after ideal integral compensation

Root locus for uncompensated system

Summary - PI

Summary - PI Root locus for compensated system

Summary - PI Ideal integral compensated system response and the

uncompensated system response of Example 9.1

Improving Transient Response via cascade compensation Since we have solved the problem of

improving the steady-state error without affecting the transient response, let us now improve the transient response itself.

The objective of improving the transient response is to design a response that has a desirable percent overshoot and a shorter settling time than the uncompensated system.

Improving Transient Response via cascade compensation Proportional-plus-derivative (PD)

controller We can select a transient response of a system

by choosing an appropriate closed-loop pole location on the s-plane.

If the point is on the root locus we can just adjust the gain in order to meet the transient response specification.

Improving Transient Response via cascade compensation

If the closed-loop poles is not on the root locus, then we need to reshape it so that the compensated root locus goes through the selected closed-loop pole location.

We need to reshape the root locus to make it go through point B if we want to have the transient response specification at point B.

Improving Transient Response via cascade compensation We can reshape the root locus by adding

poles and zeros into the forward path. One way to speed up the original system is to add a single zero to the forward path.

Based on Laplace theorem, the Laplace form of derivative is s, so adding an s is adding a derivative to the system which is why this method is called “proportional-plus-derivative controller”.

Adding derivative will speed up our system, (reduce the settling time, Ts)

Improving Transient Response via cascade compensation Example for PD controller

Assume that a root locus for a system intersect 25.38% overshoot radial line at A. The settling time, Ts, at point A is 4.26 seconds. We want to reduce the settling time, Ts, to 1.64 seconds but still maintain the 25.38% overshoot.

A


Point B on the radial line has the same %OS but the settling time, Ts = 4.26s . Since we want Ts=1.64s but maintain the %OS, we need to move the root locus intersection from A to B.


We can move the root locus by using a zero as our cascade compensator. Adding a zero to our system will change the transient response which is why we must select the zero that will actually move our root locus intersection exactly at B.

Improving Transient Response via cascade compensation Example 9.3

Given the system of Figure below, design an ideal derivative (PD) compensator to yield a 16% overshoot, with a threefold (3 times) reduction in settling time.

Improving Transient Response via cascade compensation Analysis (solving the proportional part)

The question wants us to design the value of K to yield 16% overshoot.

Additionally the question also wants us to reduce the settling time threefold from the original (proportional part) transient response.

First step is to draw the root locus of the uncompensated (proportional part) system.

Use the open loop transfer function of the unity feedback system. Plot the poles and zeros.

Find the asymptotes. Sketch the root locus.


Next, we need to find the value of gain, K, where the root locus intersect with the radial line of 16% overshoot. 16% overshoot is equal to zeta = 0.504


We calculate the value of gain using the cosine, sine, tangent rules.

1 32

0.504

A CB

K A B C


r

X XX

Improving Transient Response via cascade compensation Since our system is actually a 3rd order

system, we must check whether we can approximate it as a 2nd order system or not. This is because the equation for settling time, Ts, is only for 2nd order system with no zeros.

We need to get the coordinate of the poles then we will check if the distance of higher order poles from jω-axis is five times farther or more than the distance of the dominant second-order poles from jω-axis. If the higher order poles if farther then we can approximate our system as 2nd order system.

Improving Transient Response via cascade compensation We can get the coordinate of the poles when

gain, K = 43.35 (16% overshoot) by substitute K value into closed loop transfer function of the unity feedback system, T(s)

3 2( )

10 24

KT s

s s s K

3 2

43.35( )

10 24 43.35T s

s s s

43.35

( )( 7.59)( 1.205 2.064)( 1.205 2.064)

T ss s j s j


We need to know the settling time, Ts, for our original system.

The question asks us to reduce the settling time, Ts, by threefold (divide by 3)

We need to modify our transient response so that our system will have Ts = 1.107 seconds.

4 43.320 seconds

1.205sT real

3.3201.107 seconds

3sT

Improving Transient Response via cascade compensation Solution ( solving the derivative part)

We know that the new settling time for our system is Ts = 1.107 seconds. From the new settling time we can get the new coordinate of our dominant second order system when Ts = 1.107 seconds. From the settling time equation,

41.107

43.613

1.107

sT real

real

X

real

0.504

-3.613


Now that we know the real part of the dominant second order poles for the new settling time, Ts = 1.107, we need to calculate the imaginary part of the coordinate.

The coordinate of our dominant second order poles when Ts = 1.107 are

imaginarytan

realimaginary real tan

3.613 tan 59.74

6.193

3.613 6.193j

6.1

93

3.613 6.193j

Improving Transient Response via cascade compensation Our system will have Ts = 1.107 and 16% OS if

the root locus goes through 3.613 6.193j


We need to move the root locus and make it go through the desired compensated dominant pole. In order to change the transient response we must add derivative (or zero) to our system.

We must make sure that the added zero to our system will produce a transient response that goes through our desired dominant second order pole. To get the best coordinate of the added zero we must first find the angle of our zero in reference to the desired dominant pole


The value of the compensated zero angle can be calculated using the fact that the [summation of zeros angle] minus [summation of poles angle]=odd multiple of 180 ̊. We must use the sine, cosine and tangent rules to calculate the angles.

1 2

0.504

A CBr

X X O3

4

X

D

3.613 6.193j

angle zeros angle poles 180

hint: r = 7.167)


We need to find the angle value of the compensated zero, θ3

We must first calculate the value of angle θ1, θ2 θ4

1 2

0.504

A CB

hint: r = 7.167)

r

X X O3

4X

D

3.613 6.193j


Value of r can be calculated using trigonometry.

Using the r value, we can calculate the angle of the compensated zero.

6.193

3.613

r

3.613

cosr


We need to find all angle values and calculate the angle for compensated zero. Substitute the values of the poles and find the angle of the compensated zero

From the calculation you will find that

Based on the root locus the maximum angle is only 180, so the value of θ3 is 455.625-360 = 95.6

3 1 2 4

3 1 2 4

( ) 180

180 ( )

3 180 (68.90 86.46 120.265)

455.625

3Answer: 95.6


We now know the angle of the compensated zero in reference to the desired dominant pole (Ts = 1.107). The next step is to know the coordinate of the compensated poles.

3.613 6.193j

3.613

95.6

3.613


The real part of the compensated zero coordinate can be calculated by

The imaginary part of the compensated zero coordinate is zero.

6.913tan 180 95.6

3.6136.913

3.613tan 180 95.6

6.9133.613

10.1993.006

3.613-σ

6.913

84.4º95.6º

3.613 σ


Gain K can be calculated using the equation below

1 2

0.174

A CB

The gain we found using calculation is K=47.75(hint: r = 7.167)

r

X X O3

4X

D

3.613 6.193j

A B DK

C


Redraw the root locus using the compensated zero.


The coordinate of the higher order poles and the dominant second-order poles can be calculated using the close loop transfer function of the compensated unity feedback system, T(s).

3 2

( 3.006)( )

10 (24 ) 3.006

( 3.006)

( 2.775)( 3.613 6.913)( 3.613 6.913)

K sT s

s s K s K

K s

s s j s j


The higher order closed loop pole is not very close with the compensated zero. Maybe the system cannot be approximate as a 2nd order system. We must use simulation to determine whether the %OS is the same between uncompensated system and compensated system.

Improving Transient Response via cascade compensation Once we decide the location of the

compensating zero, we can implement the PD controller based on figure below

The transfer function of the controller is1

2 1 22

( )c

KG s K s K K s

K

Improving Steady-State error and Transient response We can improve both steady-state error and

transient response for a system by combining the PI controller and PD controller.

We can either improve the steady-state error first then improve the transient response or improve the transient response then improve the steady state error.

PID controller that we are going to learn improves the transient response then improves the steady-state error.

Improving Steady-State error and Transient response A PID controller is shown in figure below.

Its transfer function is

22 2 1 21 3 1 3 3

3 3

( )c

K K K KG s K K s K s K s K s

s s K K

Improving Steady-State error and Transient response The design technique for PID controller

consists of the following steps:1. Evaluate the performance of the

uncompensated system (P-controller system).2. Design the PD controller to meet the transient

response specification.3. Design the PI controller to yield the required

steady-state error.4. Determine the gains, K1, K2, and K3.

12 1 2

2

( )c

KG s K s K K s

K

Improving Steady-State error and Transient response Example 9.5

Given the system below, design a PID controller so that the system can operate with a peak time that is two-thirds that of the uncompensated system at 20% overshoot and with zero steady-state error for a step input.

Improving Steady-State error and Transient response

Based on the PID design step, the first step is to evaluate the performance of our system.

Sketch the root locus. Draw the radial line for 20% overshoot. Calculate the gain, K, for the system to produce 20%

overshoot and find the coordinate of the poles and zeros during that gain.

Determine whether we can approximate our system as 2nd order system or not.

Calculate the settling time, Ts, and the peak time, Tp

4s

p

Treal

Timaginary


The root locus for the uncompensated system

(p-controller)


Characteristic of the uncompensated system.


Second step is to improve in term of transient response. We need to use PD controller. The question requires us to reduce the peak time to two-thirds of that of the uncompensated system.

The peak time, Tp, for the uncompensated system is 0.297. We need to reduce it by 2/3.

2new peak time, 0.297

30.198

pT


We need to find the intersection point of the new peak time Tp on the 20% overshoot radial line.

15.8670.198

p

p

Timaginary

imaginaryT

X

real

0.456

tan

15.878.131

tan tan(62.871)

imaginary

realimaginary

real

intersection coordinate = -8.13 15.87j


In order to change the transient response so that the root locus will goes through the intersection point of the new peak time, we must add a compensated zero into our system.

To get the coordinate of our compensated zero we must first get the angle of the zero in reference to the intersection point of the new peak time.


1 3

AC

B

hint: r = 17.831

r

X X5

X

8.13 15.87j

2 4 1 3 5

angle zeros angle poles (2 1)180

( ) ( ) (2 1)180

k

k

0.456

O 2 4O

D E


The angle of the compensated zero is 18.37 ̊. Now that we know the angle of the

compensated zero we will then calculate the coordinate of the compensated zero.

8.13 15.87j

cz8.13

18.37

15.87tan18.37

8.13

15.878.13

tan18.3715.87

8.13tan18.3755.92

c

c

c

z

z

z

the coordinate of our compensated zero is at

(-55.92,0)


Sketch the root locus with the compensated zero.


After we design the PD controller, we design the PI controller to reduce the steady-state error to zero for a step input. When put an integral (compensated pole) into our system we must also put a zero close to the origin.

Based on the previous root locus for PD controller, we can reduce the steady state error to zero by putting a compensated pole (at origin) and a zero (at 0.5).


Next step is to sketch the root locus for the PI controller and find the value of K for the intersection between the root locus and the radial line for 20% overshoot.

This root locus is also called root locus for PID controller since we combine PD then PI controller.

Improving Steady-State error and Transient response The summary of the results

PID summary

Control chap8

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