compensator design to improve transient performance using root locus

8/13/2019 Compensator Design to Improve Transient Performance Using Root Locus

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Compensator Design to Improve TransientPerformance Using Root Locus

Prof. Guy BealeElectrical and Computer Engineering Department

George Mason UniversityFairfax, Virginia

C ONTENTS

I INTRODUCTION 2

II DESIGN PROCEDURE 2II-A Compensator Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2II-B Outline of the Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3II-C System Type N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

II-D Selecting a Dominant Closed-Loop Pole Location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4II-E Determining the Compensators Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6II-E.1 Plant Phase Shift at s1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6II-E.2 Compensator Phase Shift at s1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7II-E.3 Placing the Compensator Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8II-E.4 Placing the Compensator Pole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9II-E.5 Determining the Compensator Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10II-E.6 Compensator Phase Shift Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11II-E.7 Simultaneous Placement of Compensator Pole and Zero . . . . . . . . . . . . . . . . . . . . 13II-E.8 Multi-Stage Compensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

III Design Example 16III-A Phase Lead Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

III-A.1 Given System and Speci cations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16III-A.2 Selection of the Dominant Closed-Loop Pole . . . . . . . . . . . . . . . . . . . . . . . . . . 16III-A.3 Designing the Compensator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

III-B Phase Lag Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18III-B.1 Given System and Speci cations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18III-B.2 Selection of the Dominant Closed-Loop Pole . . . . . . . . . . . . . . . . . . . . . . . . . . 19III-B.3 Designing the Compensator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

References 22

L IST OF F IGURES

1 Allowable region for s1 in order to satisfy speci cations on overshoot, settling time, and frequency of oscillation. 62 Calculating the phase shift of G(s) at the chosen point s = s1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 One possible solution for locating the compensator zero and pole. . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Comparison of closed-loop step responses for 5 compensator designs with s1 = 4 + j 5.4575. . . . . . . . . . . 115 Root locus plot for four compensator designs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 Comparison of lag and lead compensation for a particular system. . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Selecting the compensator zero and pole to maximize = zc /p c . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Step response of the uncompensated system for the phase lead design example. . . . . . . . . . . . . . . . . . . 179 Root locus of the uncompensated system and the desired closed-loop pole s1 = 0.25 + j 0.488. . . . . . . . . . 1810 Lead compensated root locus and step response for the design example. . . . . . . . . . . . . . . . . . . . . . . 1911 Comparison of root locus plots with two lag compensator designs. . . . . . . . . . . . . . . . . . . . . . . . . . 2112 Comparison of closed-loop step responses for two lag compensators. . . . . . . . . . . . . . . . . . . . . . . . . 22

These notes are lecture notes prepared by Prof. Guy Beale for presentation in ECE 421, Classical Systems and Control Theory , in the Electrical and Computer Engineering Department, George Mason University, Fairfax, VA. Additional notes can be found at: http://teal.gmu.edu/~gbeale/examples.html.


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I. INTRODUCTION

In my approach to root locus design, the purpose of compensator design using root locus methods generally is to establish aspeci ed point in the s-plane, s = s1 , as a closed-loop pole. The assumption is that time-domain transient speci cations, suchas settling time and overshoot, will be satis ed if s1 is a dominant closed-loop pole. In the simplest case, s1 is already onthe root locus of the uncompensated system. The compensator is then just a gain K c that is chosen to satisfy the magnitudecriterion at the point s1 .

More often, the point s1 is not on the uncompensated root locus, so the compensator must add enough phase shift at the

point s1 to satisfy the phase angle criterion so that the compensated root locus does pass through s1 . This is done by choiceof the compensators poles and zeros. The compensator gain is then chosen to satisfy the magnitude criterion at the point s1 .In many cases, the speed of response and/or the damping of the uncompensated system must be increased in order to satisfy

the speci cations. This requires moving the dominant branches of the root locus to the left. A phase lead compensator (providing positive phase shift at s1 ) is used for this purpose. If the branches need to be moved to the right, a phase lag compensator (providing negative phase shift at s1 ) is used. The design techniques are identical for the two types of compensator; the roles of the compensators poles and zeros are just reversed. Because of this similarity in the design methods, phase lead compensationwill be discussed in detail here. An example of each type of compensation will be given after the general design procedure isdescribed.

Conceptually, the design procedure presented here is graphical in nature. The process of locating the compensators polesand zeros to satisfy the phase requirements can be visualized from the trigonometric relationships that must be satis ed at thedesired dominant closed-loop pole. The computations can be easily done by calculator. If data arrays representing the numerator and denominator polynomials of the open-loop system are available, then the procedure can be done using a software package

such as MATLAB, and in many cases it can be automated. The examples and plots presented here are all done in MATLAB,and the various measurements that are presented in the examples are obtained from the arrays storing the appropriate variables.

The primary references for the procedures described in these notes are [1][3]. Other references that contain similar materialare [4][11].

II. DESIGN PROCEDURE A. Compensator Structure

The basic phase lead or phase lag compensator consists of a gain, one real pole, and one real zero. Based on the usualelectronic implementation of the compensator [3], the circuit for a lead or lag compensator is the series combination of twoinverting operational ampli ers. The rst ampli er has an input impedance that is the parallel combination of resistor R 1 and capacitor C 1 and a feedback impedance that is the parallel combination of resistor R2 and capacitor C 2 . The second ampli er has input and feedback resistors R 3 and R 4 , respectively.

Assuming that the op amps are ideal, the transfer function for this circuit is

Gc(s) = V out (s)

V in (s) =

K c (s zc)(s pc)

=K x szc + 1 s pc + 1

= K x (s + 1)

(s + 1) (1)

= R4R2

R3R1 (sR 1C 1 + 1)(sR 2C 2 + 1)

= R4R2

R3R1 R1C 1R2C 2

(s + 1 /R 1C 1)(s + 1 /R 2C 2)

= R4C 1

R3C 2 (s + 1 /R 1C 1 )(s + 1 /R 2C 2 )

The zero and pole of the compensator are located at s = zc at s = pc , respectively. Therefore, zc and pc are negative if theyare located in the left-half of the s-plane and positive if they are in the right-half plane. The following relationships 1 can beobtained by inspecting Eq. (1).

K x = R4R2

R3R1, = R1C 1 , = R2C 2 , =

zc pc

= R2C 2R1C 1

(2)

zc = 1/R 1C 1 , pc = 1/R 2C 2 , K c = K x

=

R4C 1R3C 2

The zero zc is to the right of the pole pc for a phase lead compensator, and it is to the left of the pole for a phase lagcompensator. This is true whether the pole and zero are in the left-half plane (the usual case) or in the right-half plane.

1 In my descriptions of the design of compensators using Bode plots ( Phase Lag Compensator Design Using Bode Plots, Phase Lead Compensator DesignUsing Bode Plots ), a slightly different de nition for the compensator transfer function is used, although it refers to the same electronic circuit shown in [3].


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Assuming that zc < 0 and pc < 0 (both in the left-half plane), then < 1 for a lead compensator, and > 1 for a lagcompensator.

At any point s = s1 , the compensator provides a magnitude

|Gc (s1)| = |K c | | s1 zc |

|s1 pc | =

|K c | q [Re(s1) zc]2 + Im 2 (s1)q [Re(s1) pc]2 + Im 2 (s1)

(3)

and a phase angle

] Gc (s1) = ] K c + ] (s1 zc) ] (s1 pc) = ] K c + tan 1 Im (s1)Re (s1) zc tan

1 Im (s1)Re (s1) pc (4)Only positive values of compensator gain will be discussed in these notes, so with K c > 0, the magnitude and phase of thegain are |K c | = K c and ] K c = 0 . In applications where K c < 0, the we have |K c | = K c and ] K c = 180

.

B. Outline of the Procedure

The following steps outline the procedure that will be used to design either a lead or a lag compensator using root locusmethods in order to satisfy transient performance speci cations, such as settling time and percent overshoot. Compensator design to satisfy steady-state error requirements is discussed in separate notes, Compensator Design to Improve Steady-State Error Using Root Locus .

1) Determine if the System Type N needs to be increased in order to satisfy the steady-state error speci cation, and if necessary, augment the plant with the required number of poles at s = 0 . This should be done at this point in the design,even though numerically satisfying the steady-state error speci cation is done later. If the System Type is not taken careof now, the remainder of the design procedure may be ineffective when the System Type is changed at the end of thedesign.

2) Choose a point in the s-plane to be the location for a dominant closed-loop pole. The selection of this point is based onthe transient performance speci cations and should produce a closed-loop system that will satisfy those speci cations.

3) Design the compensator:a) Compute the phase shift of the plant (including any additional poles at s = 0 needed to satisfy the steady-state error

speci cation) at the chosen point s = s1 . If the phase shift is an odd integer multiple of 180 (180 mod360 ), thenthe selected point is on the uncompensated systems root locus for positive gain. If the phase shift is an even integer multiple of 180 (0 mod360 ), then the selected point is on the uncompensated systems root locus for negativegain. In either case, the only compensation needed for the transient performance speci cations is the proper gain,and the procedure can jump to step 3(e).

b) Assuming that the selected point s1 is not already on the root locus (for either positive or negative gain), computethe amount of phase shift that the compensator must provide at s = s1 in order to make that point lie on the rootlocus. The compensators phase shift (usually) will be the shortest distance from the plants phase shift at s1 toan odd integer multiple of 180 for positive gain (the most usual case) or to an even integer multiple of 180 for negative gain. It should be noted that having the compensator provide this amount of phase shift only guaranteesthat the point s1 will lie on the compensated root locus; it does not guarantee that the closed-loop system will bestable when s1 is a closed-loop pole. Additional factors must be taken into account for that.

c) Select the location for either the compensator zero zc or pole pc . If the compensator phase shift at s1 computed in the previous step is positive, phase lead compensation is required. This places the compensator zero to the rightof the pole. If the compensators phase shift is negative, then phase lag compensation is needed, and the zero is tothe left of the pole. One of these factors (zero or pole) can be placed with some freedom; once it is placed, thelocation of the other factor is xed.

d) Compute the horizontal distance from the point s = s1 to the location of the compensator factor not already placed

(pole or zero), and place that factor at that location. This pole/zero pair in series with the plant makes the rootlocus pass through the point s1 .

e) Compute the compensator gain K c . The value of the gain is chosen to satisfy the magnitude criterion at s1(|Gc (s1) G p (s1)| = 1) in order for s1 to be a closed-loop pole.

4) If necessary, choose appropriate resistor and capacitor values to implement the compensator design.

To illustrate the design procedure, the following system model and speci cations will be used:

G p(s) = 8s + 4

(5)

steady-state error speci cation for a unit ramp input is ess _ specified = 0 .1; step response settling time speci cation is T s specified = 1 second ; step response overshoot speci cation P Ospecified 10%.


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C. System Type N The rst step in the design of the compensator will be to determine if the plant G p(s) has the correct System Type to

satisfy the steady-state error speci cation. De ning the number of open-loop poles of a system that are located at s = 0 to be the System Type N , and restricting the reference input signal to having Laplace transforms of the form R(s) = A/s q , thesteady-state error and error constant are (assuming that the closed-loop system is bounded-input, bounded-output stable)

ess = lims 0 As

N +1 q

sN + K x (6)where

K x = lims 0 sN G(s) (7)

For N = 0 , the steady-state error for a step input (q = 1) is ess = A/ (1 + K x ). For N = 0 and q > 1, the steady-stateerror is in nitely large. For N > 0, the steady-state error is ess = A/K x for the input type that has q = N + 1 . If q < N + 1 ,the steady-state error is 0, and if q > N + 1 , the steady-state error is in nite. Therefore, for a system to have a non-zero, nitesteady-state error for a speci ed reference input, the System Type must satisfy N = N req = q 1. This is the total number of open-loop poles at the origin needed for the compensated system to satisfy the steady-state error speci cation. If the SystemType of the plant G p(s) is N sys , then the compensator must have (N req N sys ) poles at the origin. These poles would beincluded with the plant model during the design of the rest of the compensator, and then they would be implemented as partof the compensator after the design is complete. The system that will be evaluated during the rest of the design process will be

G(s) = 1

s (N req N sys ) G p(s) (8) Example 1:The plant transfer function in (5) has no poles at the origin, so it has System Type N sys = 0 . The steady-state error

speci cation is for a unit ramp input which has a Laplace Transform R(s) = 1 /s 2 , so for this input q = 2 . Therefore, in order for the steady-state error speci cation to be satis ed, the total number of poles at the origin must be N req = q 1 = 2 1 = 1 .Thus, the compensator must provide N req N sys = 1 0 = 1 pole at the origin. The system G(s) corresponding to Eq. (8)for this example is G(s) = 8 / [s (s + 4)] .

Note that satisfying the numerical value of the steady-state error is not done at this point in the design. That will be doneas a separate task.

D. Selecting a Dominant Closed-Loop Pole Location

The real engineering design takes place in this step. This is the mapping of the performance speci

cations that must besatis ed into the design parameter that in uences the rest of the steps in the design. For other than very simple systems, sound engineering judgement is needed in order for this mapping to yield a compensator design that actually allows the speci cationsto be satis ed.

In many cases, the dominant closed-loop poles are chosen to be a complex conjugate pair, so the design point s = s1 is acomplex number with negative real part and positive imaginary part. The starting point in choosing s1 generally is to make useof the relationships between time-domain characteristics and closed-loop pole locations that exist for the standard second-order system, shown in (9).

G(s) = 2n

s (s + 2 n ), T CL (s) =

2ns2 + 2 n s + 2n

(9)

where is the dimensionless damping ratio, and n is the undamped natural frequency (rad/ sec). For 0 < < 1, the closed-loop system is underdamped, the closed-loop poles are complex conjugates, and the step response exhibits overshoot and damped sinusoidal transient behavior. The closed-loop poles are at

p1 , p2 = n j nq 1 2 (10)In this case, speci cations on the amount of overshoot and the settling time for the step response are natural speci cations.These types of speci cations will be used in the design procedure presented in these notes. A steady-state error speci cationon the ramp response is also appropriate; that will be discussed in separate notes.

Percent overshoot (P O ) in the step response of the standard second-order system is only a function of the damping ratio.The value of overshoot (%) and the value of are related by

P O = he / 1 2

i 100%, = lnP O100 q 2 + ln 2 P O100

(11)


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Points in the s-plane that correspond to a constant value of lie on a radial line from the origin making an angle with respectto the negative real axis, with = cos 1 ( ). Therefore, a speci cation of percent overshoot establishes a xed relationship between the imaginary part of the dominant closed-loop pole and the real part of the pole. This relationship is given by tan (),which is the slope of the radial line corresponding to the value of , so that Im [s1] = Re [s1] tan cos

1 ( ).De ning settling time T s as the time required for the step response to get within and stay within a xed percentage of the nal value, the settling time for the standard second-order system is a function of the product of damping ratio and undamped natural frequency. In these notes, the band about the nal value used to de ne T s is 2%, and the settling time is

T s = 4n(12)

Comparing (12) and (10) shows that settling time is related to the real part of the closed-loop poles, so a settling timespeci cation corresponds to the dominant closed-loop poles lying on a vertical line in the s-plane located at

Re [s1] = 4T s (13)If the settling time speci cation is an upper bound, then the closed-loop poles must lie on or to the left of the vertical linegiven in (13).

The damped natural frequency is de ned to be d = np 1 2 , which is seen in (10) to be the imaginary part of theclosed-loop poles. This parameter is also the frequency of oscillation in the transient part of the step response. Therefore, aspeci cation on the frequency of oscillation imposes a restriction on the imaginary part of the dominant closed-loop poles,which corresponds to a horizontal line in the s -plane.

If two or more speci cations are imposed on the system, then there are multiple constraints on the location of the dominantclosed-loop pole s1 . An acceptable location for that pole must satisfy each of the constraints. Therefore, s1 must lie in theintersection of the regions de ned by the various speci cations.

Example 2:The following speci cations are to be imposed on a systems closed-loop step response: (1) 5% P O 25%, (2) 0.5 s

T s 2 s, (3) d 12.6 r/s. Assuming that the equations for the standard second-order system will hold for the actual system,then the following constraints are imposed on the location of s1 . From (11), the lower bound of 5% overshoot corresponds toa damping ratio = 0 .6901 and an angle = 46 .4 . The point s = s1 must lie on or above this radial line. The upper bound of 25% overshoot corresponds to a damping ratio = 0 .4037 and an angle = 66 .2 . The point s = s1 must lie on or belowthis radial line. From (13), the lower bound of T s = 0 .5 s corresponds to real part of Re [s1 ] = 8. The point s1 must lie onor to the right of this vertical line. The upper bound of T s = 2 s corresponds to real part of Re [s1] = 2. The point s1 mustlie on or to the left of this line. The upper bound of d = 12 .6 r/s corresponds to a horizontal line at j 12.6. The closed-loop pole must lie on or below this line. The dominant closed-loop pole s1 must lie in the intersection of these 5 constraint curves.The dotted region in Figure 1 is the set of acceptable locations for s1 for this example.

Example 3:In Example 1 the original G p(s) augmented with one pole at the origin to satisfy the steady-state error requirement is

G(s) = 8 / [s(s + 4)] . The transient speci cations are T s specified = 1 second and P O 10%. The damping ratio associated with 10% overshoot is = 0 .5912. The corresponding angle and slope of the radial line from the origin through s1 are = 53 .76 and 1.3644, respectively. From (13), the real part of s1 is s = 4. Combining these two requirements places thedominant closed-loop pole at s1 = 4 + j 5.4575.

The relationships between dominant pole locations and step response characteristics presented above are only for the standard second-order system of (9). These relationships may provide good starting points for the selection of s1 to satisfy speci cations, but they can only be used as general guidelines for more complex system models. Even if the given system G p(s) is modeled by (9), any compensation in series with G p(s), other than a pure gain, will result in a more complex model. The overshootequation (11) generally is not a good prediction of what the actual overshoot will be in a higher-order system or in a systemwith zeros. The actual overshoot may be less than predicted by (11), but it will more likely be larger. In choosing a value for s1to satisfy a percent overshoot speci cation, the recommended approach is to be very conservativethat is, use a signi cantlysmaller value of overshoot than the speci ed value when computing the effective damping ratio from (11).

The prediction of the real part of s1 computed from (13) to satisfy a settling time speci cation is generally fairly accurate.This is due to the fact that the decay of the transient response is controlled by the real part of the dominant closed-loop poles.This is how the expression in (13) was derived and accounts for its accuracy in predicting the settling time of higher-order systems.

In many actual design problems with speci cations on overshoot, settling time, frequency of oscillation, etc., the proper choice of location for the dominant closed-loop pole (or complex conjugate pair) will be achieved only through iteration.Engineering design is an iterative process, and the selection of s1 is a major step in the design of the compensator that willallow the speci cations to be satis ed.


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20 15 10 5 0 55

0

5

10

15

20

Real Axis

I m a g

A x

i s

Locating ClosedLoop Poles in the s plane to Satisfy Specifications

POmax

= 25%

min

= 0.4037

max

= 66.2

POmin

= 5%

max

= 0.6901

min

= 46.4

d

max

= 12.6 r/s

n = 2

n = 8

Ts

max

= 2 sTs

min

= 0.5 s

Fig. 1. Allowable region for s 1 in order to satisfy speci cations on overshoot, settling time, and frequency of oscillation.

E. Determining the Compensators Parameters

Once the point s = s1 is chosen to be a dominant closed-loop pole, the design of the compensator is an exercise intrigonometry. The pole(s) and zero(s) of the compensator are chosen to satisfy the phase angle criterion at s1 so that the total phase shift of the series combination of the plant and compensator at s1 is an odd integer multiple of 180 if K > 0 or aneven integer multiple of 180 if K < 0. This makes the root locus pass through the point s1 . Once the pole(s) and zero(s) are placed, the gain of the compensator is selected to satisfy the magnitude criterion. This makes the point s = s1 be a closed-loop pole. However, as previously mentioned, making s = s1 be a closed-loop pole does not guarantee that the closed-loop systemis stable. There may still be closed-loop poles in the right-half plane. Thus, there is more to choosing the locations of the

compensators pole(s) and zero(s) than just satisfying the phase angle criterion.1) Plant Phase Shift at s1 : The starting point in the compensator design is to determine the phase shift of the augmented

plant of Eq. (8) at the point s1 . The phase shift of G(s) at s1 is

] G (s1) = ] K + ] N (s1) ] D (s1) (14)= ] K +

m

Xi =1 ] (s1 zi ) n

Xi =1 ] (s1 pi )= ] K +

m

Xi =1 tan 1 Im ( s1) Im ( zi )Re (s1) Re (zi )

n

Xi =1 tan 1 Im ( s1) Im ( pi )Re (s1) Re ( pi )

where the zi and pi are the open-loop zeros and poles, respectively. We will assume that K > 0, so ] K = 0 .


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10 8 6 4 2 0 26

4

2

0

2

4

6

Real Axis

I m a g

A x

i s

G(s) = 8/[s(s+4)], s1 = 4 + j5.4575

s1

p2

p1

Fig. 2. Calculating the phase shift of G(s) at the chosen point s = s 1 .

Care must be used when evaluating the tan 1 function when using the normal atan function in MATLAB or on calculators.If the denominator of the tan 1 function is negative, the atan function will return the incorrect angle. To obtain the correctangle in this case, 180 ( rad ) must be added to the angle returned by the atan function. The function atan2 returns thecorrect value since it takes two input arguments; its syntax is = atan2 (Im, Re) .

Example 4:From Example 1, the augmented plant model is G(s) = 8 / [s (s + 4)] , so there are no zeros zi and the open-loop poles are

p1 = 0 and p2 = 4. We will assume that the desired dominant closed-loop pole is at s = s1 = 4 + j 5.4575, based onExample 3. The phase angle of G(s) at s1 is: ] G (s1) = ] 8

] s1

] (s1 + 4) = 0

tan 1 [(5.4575

0) / (

4

0)]

tan 1 [(5.4575 0) / (4 + 4)] . ] G (s1) = tan 1 [5.4575/ (4)] tan

1 (5.4575/ 0) = (53.76 + 180 ) 90

=

216.24 = 143 .76 . Both forms for the phase shift, ] G (s1) = 216.24

or ] G (s1) = 143 .76 , are acceptable waysto represent the angle. Figure 2 shows the relationships between the open-loop poles and the point s1 . The angles that arecomputed are the angles of the vectors drawn from the poles to the point s1 measured counter-clockwise from the positivereal axis.

Note the 180 added to the phase shift returned by the atan function in the above example for the pole p1 . This will be thecase when the point s1 is to the left of the pole or zero under consideration.

2) Compensator Phase Shift at s1 : If the point s1 is on the root locus of the uncompensated system G(s) (which is G p(s)augmented with any poles at the origin that were needed to produce the correct System Type), then ] G (s1) is an odd integer multiple of 180 (for K > 0). In this case ] Gc (s1) = 0 , and the only (additional) compensation that is needed in order tomake s = s1 a closed-loop pole is a gain K c . That would be chosen to satisfy the magnitude criterion |K cG (s1)| = 1 .


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Most often, s1 will not be on the original root locus, so the compensator must also provide a phase shift at s1 so that thetotal phase shift of the plantcompensator combination is an odd integer multiple of 180 at that point. This is done by choiceof the compensators pole and zero. Satisfying the phase angle criterion must be done rst. After that, the value of K c iscomputed to satisfy the magnitude criterion.

Given the value ] G (s1) from the previous step, the required compensator angle ] Gc (s1) can be easily computed. Generally,the compensator phase shift at s1 is chosen to be the angle with the smallest absolute value such that ] G (s1) + ] Gc (s1)is an odd integer multiple of 180 . Thus, if ] G (s1) = 150 or ] G (s1 ) = 210

, then ] Gc (s1) = +30 will place s1 onthe root locus, and the compensator is phase lead. Likewise if ] G (s1) =

150 or ] G (s1) = 210 , then ] Gc (s1) =

30

will place s1 on the root locus, and the compensator is phase lag. However, in some cases it might be necessary (or desirable)to use a particular type of compensator, such as phase lead, in order to satisfy other objectives. In a situation like this, if ] G (s1) = 60

for example, then the compensator phase shift at s1 would be chosen to be 180 (60 ) = 240 (phase

lead) rather than 180

(60 ) = 120

(phase lag). A more detailed example illustrating this situation is given in SectionII-E.6.

The necessary compensator phase shift to make s1 lie on the root locus is given by

] Gc (s1) = 180 (2l + 1) ] G (s1) (15)

= ] K c + ] N c (s1) ] D c (s1 )= ] K c +

m c

Xi =1 ] (s1 zc i ) n c

Xi =1 ] (s1 pc i )Assuming that K c > 0, then ] Gc (s1) = ] N c (s1) ] D c (s1).

Example 5:From Example 4, the phase shift of G(s) at s1 = 4 + j 5.4575 is ] G (s1) = 216.24

= 143 .76 . The phase angle of the compensator needed to place s = s1 on the root locus (using the smallest absolute value for the compensator phase angle)is ] Gc (s1) = 180

(216.24 ) = 180 143.76

= 36 .24 . Since this angle is positive, the compensator is phase lead.The pole and zero of the compensator will be placed on the real axis such that ] (s1 zc) ] (s1 pc) = 36 .24

. Methodsof choosing appropriate locations are presented in the next section. One possible solution is to place the compensator zero ats = 5, which provides 79.62

of phase shift at s1 , and to place the compensator pole at s = 9.78, which provides 43.38

at s1 . The difference between those angles is the required value of 36.24 . Figure 3 illustrates this solution.

3) Placing the Compensator Zero: There is some exibility in choosing the location of the compensator zero, but notcomplete freedom. Assuming that a single stage of compensation is desired, the left-most location of the zero is governed by

the compensators phase shift requirement at s1 . Because ] Gc (s1) = ] (s1 zc) ] (s1 pc ), the compensator zero must provide more phase shift at s1 than the total compensator. Therefore, the left-most allowed position of the compensator zerois constrained by ] Gc (s1).

The right-most location of the zero is generally constrained by transient performance speci cations, and ultimately by therequirement for closed-loop stability. Moving the compensator zero to the right also moves a closed-loop pole to the right.For some choices for the location of the zero, there may be a closed-loop pole closer to the j axis than the selected points1 . When this happens, s1 may no longer be the dominant pole in terms of settling time. If the compensator zero is in theright-half plane, then the closed-loop system may be unstable.

Unfortunately, there is not a rule for placing the compensator zero. The amount of overshoot can vary widely depending onits location. For systems with one open-loop pole at the origin and with one or more additional poles on the negative real axis,a rule of thumb is to place the compensator zero at or to the left of the second real-axis open-loop pole. This rule of thumbis intended to ensure that s = s1 is the dominant closed-loop pole, but it does not guarantee that overshoot speci cations will be satis ed.

Example 6:For the system G(s) de ned in the previous examples and the point s1 = 4 + j 5.4575, the required phase shift of thecompensator was found to be ] Gc (s1) = 36 .24 . Thus, the compensator zero must provide more than 36.24 at s1 . In order

to use a single stage of compensation, the left-most point for the zero is s = Re[ s1] [Im[s1] / tan(36 .4 )] = 11.45.The right-most point for the compensator zero is just to the left of the origin. The zero cannot be placed at s = 0 since

that would cancel the open-loop pole there, and internal stability would be lost. Placing the compensator zero in the right-half plane would also result in an unstable closed-loop system. Some possibilities for the location of the zero are zc = 9[] (s1 zc) = 47 .51

], zc = 7 [] (s1 zc) = 61 .20 ], zc = 5 [] (s1 zc) = 79 .62

], zc = 3 [] (s1 zc) = 100 .4 ],

zc = 1 [] (s1 zc) = 118 .8 ]. The effects of these different choices for zc will be shown in the following examples.

The rst three locations for the zero in the above example (zc = 9, zc = 7, zc = 5) satisfy the rule of thumb mentioned earlier. For these choices, the closed-loop poles at s = s1 and its complex conjugate will be closer to the j axis than the


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4

2

0

2

4

6

Real Axis

I m a g

A x

i s

Providing 36.24 degrees at s1 = 4 + j5.4575

s1

zc

pc

Fig. 3. One possible solution for locating the compensator zero and pole.

third pole. Therefore, the settling time of the step response will be governed by the choice of s1 , as desired. The remainingtwo choices for the compensator zero (zc = 3, zc = 1) do not satisfy the rule of thumb. The poles at s1 will no longer beclosest to the j axis, and so will not be controlling the settling time. This does not mean that the settling time speci cationwill be violated if the rule of thumb is not satis ed, but as the zero is moved farther and farther to the right, at some point,the settling time speci cation will be violated.

4) Placing the Compensator Pole: Once the location for the compensator zero has been selected, there is no more freedom inthe design; the compensator pole and gain are now xed. The location of the pole is constrained by the phase angle requirementon the compensator at the point s1 , and once the pole location is determined, the gain is constrained by the magnitude criterion

at s1 .Since the total phase shift of the compensator at s = s1 is the angle produced by the zero minus the angle produced by the

pole, the phase shift of the compensator pole at s1 is

] (s1 pc) = ] (s1 zc) ] Gc (s1) (16)The value of this angle can be computed once the location of the zero is chosen. There is only one location for the compensator pole that will produce the angle ] (s1 pc) at s = s1 . The distance d pc from the real-axis projection of s1 to the compensator pole can be computed easily from

d pc = Im [s1]

tan[ ] (s1 pc)] (17)


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The compensator pole is located to the left of s1 if d pc > 0 and to the right of s1 if d pc < 0. The location of the compensator pole at s = pc is given by

pc = Re [ s1] d pc (18)

Example 7:From Example 5, the compensator must provide 36.24 of phase shift at s1 . In Example 6, ve possible choices for the

compensator zero were presented, namely, zc = {9, 7, 5, 3, 1}. For each of these choices, the compensator pole can be located by using Eqs. (16) (18). The angles, distances, and locations of the compensator pole are shown in the followingtable for the various choices of compensator zero.

zc ] (s1 pc) d pc pc9 11.27

27.4 31.47 24.96

11.7 15.75 43.38

5.78 9.783 64.14

2.64 6.641 82.56

0.713 4.713The table shows that as the compensator zero moves to the right, the required phase shift of the pole increases, and the polealso moves to the right.

5) Determining the Compensator Gain: Now that the pole and zero of the compensator have been selected, the root locus plot will pass through the point s = s1 . Therefore, s1 is a potential closed-loop pole. In order for that point to actually bea closed-loop pole, the magnitude criterion must be satis ed at s1 by the series combination of plant and compensator. Thecompensator gain is used for this purpose.

The magnitude criterion states that

|K c | | s1 zc ||s1 pc |

|G (s1)| = 1 (19)

if s = s1 is a closed-loop pole, with G(s) de ned in (8). The compensator gain is the only parameter that is undetermined atthis point. Therefore, to make s = s1 be a closed-loop pole, the compensator gain must be

|K c | = |s1 pc |

|s1 zc | |G (s1)| (20)

Note that (20) only provides the magnitude (absolute value) of the compensator gain. If the system requires that the gain benegative, then K c = |K c | .

Example 8:For each of the combinations of compensator zero and pole from Examples 6 and 7, there is a unique value for the gain K c .

The value of gain that places one of the closed-loop poles at s1 is computed from (20). The magnitude of the augmented plantat s1 is |G (s1)| = 8 / (|s1 | | s1 + 4 |) = 0 .2166. Using the values for zc and pc from the table in Example 7, the correspondinggain values are given in the following table.

zc |s1 pc | | s1 zc | K c9 27.94 7.402 17.427 12.93 6.228 9.585

5 7.946 5.548 6.611

3 6.065 5.548 5.0451 5.504 6.228 4.080

For each of these combinations, the compensator transfer function, including the pole at s = 0 needed for the steady-stateerror speci cation, is Gc(s) = K c (s zc) / [s (s pc)]. With the values used here, the closed-loop system is stable, and twoof the closed-loop poles are at the speci ed location of s1 = 4 + j 5.4575 and its complex conjugate. The location of thethird closed-loop pole varies with the choice of compensator parameters. For the values used in these examples, the third poleis at s = {27.4, 11.7, 5.78, 2.64, 0.713}. Note that for the particular structure of Gc(s)G p(s) in these examples, theabsolute value of the third closed-loop pole is equal to the distance between the real-axis projection of s1 and the compensatorsopen-loop pole. The closed-loop step responses for these ve compensator designs are shown in Fig. 4. The settling times for the rst four compensator designs are all approximately the same, slightly less than 1 second. The settling time for the fthcompensator design is signi cantly longer and does not satisfy the speci cations. Of the compensator designs that do satisfy


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0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

O u t p u

t A m p l

i t u d e

ClosedLoop Step Responses

1.1

1.02

0.98

zc = 9

zc = 7

zc = 5

zc = 3

zc = 1

Fig. 4. Comparison of closed-loop step responses for 5 compensator designs with s 1 = 4 + j 5.4575 .

the settling time requirement, only the fourth design (zc = 3) also satis es the overshoot speci cation. The dashed lines inthe gure show the 2% interval for the settling time speci cation and the 10% overshoot speci cation. Figure 5 shows theroot locus plots for four of the compensator designs. The triangles shown in the plots represent the closed-loop poles. The point s1 is a closed-loop pole in each case. It is easily seen in the gure that the third closed-loop pole moves to the right aszc does, and when zc = 1 the third pole is far to the right of s1 . This causes the much longer settling time evident in Fig.4.

These examples have illustrated the procedure for mapping a set of transient response speci cations into a desired locations = s1 for the dominant closed-loop pole and for designing a compensator to make that desired location actually be a closed-loop pole. The examples have shown that not every compensator design that places a closed-loop pole at s1 will satisfy all of the speci cations. Closed-loop stability is not even guaranteed if the compensated system is third-order or higher. Therefore,

the design of the compensator will generally be an iterative process. Although there are general guidelines that can be followed,a design needs to be validated through simulation to determine whether or not it is an acceptable design.6) Compensator Phase Shift Revisited: It was mentioned in Section II-E.2 that the compensator phase shift at s1 is generally

taken to be the smallest absolute value that will yield an odd integer multiple of 180 for the plantcompensator combination.It was also mentioned that there are instances when such a choice would be inappropriate. The following example illustrateswhen closed-loop stability requirements force a change to this general procedure.

Example 9:Consider the system G(s) = 5( s + 2) / [s2(s + 0 .4)(s + 6)] and the desired closed-loop pole s1 = 1 + j 0.8. The phase

shift of G(s) at s1 is ] G (s1) = 19.98 . A compensator phase ] Gc1 (s1) = 160.02

would place s1 on the root locus for K c > 0. This would require a phase lag compensator. On the other hand, a compensator phase ] Gc2 (s1) = 199 .98 would also place s1 on the root locus for K c > 0. This would require a phase lead compensator. Which one should be used? Inanswering this question, it should be remembered that open-loop poles tend to repel branches of the root locus, and open-loop


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20 15 10 5 0 520

15

10

5

0

5

10

15

20

Real Axis

I m a g

A x i s

zc = 7

10 5 0 515

10

5

0

5

10

15

Real Axis

I m a g

A x i s

zc = 5

8 6 4 2 0 210

5

0

5

10

Real Axis

I m a g

A x i s

zc = 3

5 4 3 2 1 0 18

6

4

2

0

2

4

6

8

Real Axis

I m a g

A x i s

zc = 1

Fig. 5. Root locus plot for four compensator designs.


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13

zeros tend to attract them. The system G(s) has two open-loop poles at the origin, the boundary of stability. Using a phaselag compensator, with the pole to the right of the zero (closer to the j axis assuming they are in the left-half plane) would tend to move the branches of the root locus that begin at s = 0 into the right-half plane, resulting in an unstable closed-loopsystem. A lead compensator, on the other hand, would tend to draw those branches into the left-half plane. Figure 6 illustratesthis. The top two plots are the root loci for the system with phase lag and phase lead compensation. The compensator transfer functions are Gc lag (s) = 0 .39 (s + 5 .54)

2 / (s + 1) 2 and Gc lead (s) = 71 .9 (s + 0 .75)2 / (s + 7 .19)2 . It is clearly seen that

the lag compensator produces an unstable system for all K > 0. The lead compensator is able to stabilize the system. It should be noted that s = s1 is a closed-loop pole in both designs. The bottom two plots are the closed-loop step responses. Theinstability of the design with the lag compensator is clearly indicated.

7) Simultaneous Placement of Compensator Pole and Zero: As we discussed in the previous sections, there is some freedomin placing the compensator zero and pole, as long as the total phase angle of the compensator has the correct value. Thatfreedom can be utilized to vary the amount of overshoot or the value of the settling time. Unfortunately, there is generally nota clear decision procedure for selecting the location of the pole or zero.

There is a procedure that places both the pole and zero at the same time. There is no freedom of choice with this procedure, but it does have the advantage of maximizing the value of = zc /p c for a lead compensator or minimizing for a lagcompensator. This is advantageous because it minimizes the range of resistor and capacitor values used to implement thecompensator.

We will de ne the phase shift of the desired closed-loop pole at s1 to be the angle of the radial line drawn from the originto s1 , measured in the counter-clockwise direction from the positive real axis, and denote it by ] s1 . With that de nition, the phase shifts of the compensator zero and pole at s1 are

] (s1 zc) = ] s1 + ] Gc (s1)

2 , ] (s1 pc) =

] s1 ] Gc (s1)2

(21)

Therefore, the distances from the real-axis projection of s1 to the zero and pole are

dz c = Im [s1]

tan[ ] (s1 zc)], d pc =

Im [s1 ]tan[ ] (s1 pc)]

(22)

and the compensator zero and pole are located at

zc = Re [ s1]dz c , pc = Re[ s1] d pc (23)Examination of (21) shows that ] (s1 zc) ] (s1 pc) = ] Gc (s1) as it must. Since ] Gc (s1) is computed to providethe proper phase shift at s1 to make the root locus pass through that point, the compensator parameters in (23) provide a valid

solution to the design problem. This solution is not necessarily any better than any other solution other than the fact that itoptimizes the ratio zc/p c .

Example 10:From Example 5, the compensator must provide 36.24 at the point s1 = 4 + j 5.4575. The phase shift of s1 itself is] s1 = 126 .24 . From (21), the angles of the compensator zero and pole are 81.24 and 45 , respectively. The horizontal

distances from s1 to the zero and pole are dzc = 0 .8411 and d pc = 5 .4575. This places the zero zc at s = 4.8411 and the pole pc at s = 9.4575. The value of for this polezero combination is = zc/p c = 0 .5119. Figure 7 shows the geometryof this solution. 8) Multi-Stage Compensation: In the examples and discussion thus far, it has been assumed the compensator had one

pole and one zero. In some cases it may be desirable or necessary to use multiple stages of compensation. If the required compensator phase shift ] Gc (s1) is very large, then the left-most allowed location for the compensator zero may be closeto the j axis or even in the right-half plane. Moving the zero to the left of this point means that one zero will not provideenough positive phase shift, so that two or more zeros will be required.

Even with moderate values of ] Gc (s1) it may be desirable to use multiple stages of compensation. This would allow thezero and pole to be moved farther to the left, away from the dominant pole location. In this way, the equations that apply tosecond-order systems might be more applicable to the higher-order system.

The easiest way to design a multi-stage compensator is to assume that each stage provides the same amount of phase shiftat s1 . Since the phase shift of a product of complex numbers is the sum of the individual phase shifts, using this approachmeans that the phase shift of each stage of the compensator is the total compensator phase shift divided by the number of stages. The most usual case would probably be two stages; therefore, each stage would provide one-half of the total required phase shift.

Once the total phase shift is divided by the number of stages, the design of each stage follows the procedure discussed inthe previous sections. The location of the compensator zero is selected to provide a phase shift larger than the phase shiftrequired by an individual stage. The compensator pole is computed to provide the correct phase shift for an individual stage.


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8 6 4 2 0 2 4

4

3

2

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4

Real Axis

I m a g

A x

i ss

1

Lag Compensated

10 8 6 4 2 0 2

10

5

0

5

10

Real Axis

I m a g

A x

i s s1

Lead Compensated

0 2 4 6 8 1012000

10000

8000

6000

4000

2000

0

2000

Time (s)

S t e p

R e s p o n s e

Lag Compensated

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

S t e p

R e s p o n s e

Lead Compensated

Fig. 6. Comparison of lag and lead compensation for a particular system.


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2

0

2

4

6

8

Real Axis

I m a g

A x

i s

Alternative Placement of Compensator Pole and Zero

s1

zc

pc

Fig. 7. Selecting the compensator zero and pole to maximize = zc /p c .

The total compensator numerator then has the form Qnum _ stagesi 1 (s zc i ), and similarly for the compensator denominator.The gain K c is then computed to satisfy the magnitude criterion. In implementation, the gain can be assigned equally to each

stage by letting K c stage = num _ stages K c . Example 11:The compensator phase shift computed in the previous examples is ] Gc (s1) = 36 .24 . Although this compensator can be

designed with a single stage of compensation, it was shown in Example 8 that there was a restricted range of locations for the compensator zero that allowed both the overshoot and settling time speci cations to be satis ed. To use a single stage of compensation, the zero must provide more than 36.24 at s1 .

In order to move the additional closed-loop poles far to the left of s1 , a two-stage compensator will be designed. This meansthat each stage of compensation must provide 36.24 / 2 = 18 .12 at s1 . The phase angle for the zero of each stage will bechosen as ] (s1 zc) = 19

. Therefore, the angle of each pole must be ] (s1 pc) = 0 .88 . The horizontal distances from

s1 to the compensator zero and pole are 15.85 and 355.1, respectively. Therefore, the zero and pole are zc = 19.85 and pc = 359.1. Using two poles and zeros at these locations, the gain is K c = 2071 .9, so the compensator is

Gc = 2071.9 (s + 19 .85)2

(s + 359 .1)2 = 45.52 (s + 19 .85)(s + 359 .1)

2

(24)

Using this compensator with the system G(s) = 8 / [s (s + 4)] produces an overshoot of 11.6%, which is less than three of the single-stage compensators. The settling time is approximately the same as the rst four designs presented in Example 8.With the two-stage compensator, there are now four closed-loop poles, rather than three as before. Two of them are at s1 and


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16

its complex conjugate. The remaining two closed-loop poles are at s = 357.1 j 123. Although the overshoot is still a bithigher than speci ed, it is smaller than the other designs that placed the compensator zero to the left of s1 , and the additionalclosed-loop poles are much farther to the left of s1 than in any of the single-stage designs.

III . D ESIGN EXAMPLE

Both phase lead and phase lag examples will be presented in this section. The lead example will be presented rst since thatis the compensator type most often used because it increases the system damping. After that, the lag compensator examplewill be given. It will be seen that the design procedures for the two types of compensators are identical; just the role of the pole and zero are reversed.

A. Phase Lead Example

1) Given System and Speci cations: The open-loop transfer function for the system to be controlled is

G p(s) = 0.375(s + 0 .8)

s(s + 0 .2)(s + 1)( s + 1 .5) (25)

The system is Type 1, so it will have zero steady-state error for a step input and a non-zero, nite steady-state error for aramp input. If unity feedback is placed around G p(s), the closed-loop poles are located at s = {0.054 j 0.4405, 0.9004,1.6916}. Therefore, the original system is closed-loop stable with unity feedback and Gc(s) = 1 . However, the dampingratio of the complex conjugate pair of closed-loop poles is = 0 .122, which corresponds to 68% overshoot for the standard second-order system. The settling time is dominated by those complex conjugate poles since they are the closest to the jaxis, and the predicted settling time is 74 seconds. The uncompensated (plant with unity feedback) step response is shown inFig. 8, and it is clear that the predictions on overshoot and settling time are accurate for this system.

The speci cations that are imposed on the system are: overshoot in the response to a step input: P O 20%; settling time for the response to a step input: T s 16 seconds.

Thus, the compensator needs to reduce the overshoot and the settling time signi cantly. This requires that the effective dampingof the system be increased, which requires a phase lead compensator.

2) Selection of the Dominant Closed-Loop Pole: Using the equations for the standard second-order system, the dampingratio that corresponds to an overshoot of 20% can be computed from (11), and is = 0 .4559. The angle and slope of the radialline associated with this value of are = 62 .87 and 1.952, respectively. The settling time speci cation requires the dominantclosed-loop poles to have a real part computed from (13) of Re [s1] = 4/ 16 = 0.25. Combining these requirements placesthe desired closed-loop pole at s = s1 = 0.25 + j 0.488.The uncompensated root locus is shown in Fig. 9, along with the point s1 . It is clear from the plot that the dominant branchesof the root locus need to be moved to the left in order to pass through s1 , again indicating the need for a lead compensator.

3) Designing the Compensator: The rst step in designing the compensator is to determine the phase shift of G p(s) ats = s1 . This angle is ] G p (s1) = tan 1 [0.488/ 0.55]tan

1 [0.488/ 0.25]tan 1 [0.488/ 0.05]tan

1 [0.488/ 0.75]tan 1 [0.488/ 1.25], so ] G p (s1) = 41 .58 (62.87 + 180 )(84.15

+ 180 )33.05

21.33 = 225.77

= 134 .23 .Therefore, in order for s1 to be on the root locus, the compensator must provide a phase shift of ] Gc (s1) = 45 .77 at s = s1 .This can be done with a single polezero pair.

In order to use a single stage of compensation, the compensator zero must provide more than ] Gc (s1) = 45 .77 at s1 , so theleft-most location for the zero is s = 0.725. Following the rule of thumb that the zero should be at or to the left of the second real-axis open-loop pole, the compensator zero will be placed at the second pole, namely at s = 0.2. This location producesan angle of 95.85 at s1 , so the compensator pole must have an angle at s1 of ] (s1 pc) = 95 .85

45.77 = 50 .08 .

The distance from the real-axis projection of s1 to the compensator pole needed to provide this angle is equal to d pc =

Im [s1] / tan[50 .08

] = 0.4083. Therefore, the compensator pole is located at s = 0.6583. At this stage in the design, thecompensator is Gc(s) = K c (s + 0 .2) / (s + 0 .6583).The compensator gain is determined from the magnitude criterion |Gc (s1) G p (s1 )| = 1 . The gain calculation is

K c = |s1 | | s1 + 1 | | s1 + 1 .5| | s1 + 0 .6583|

0.375 |s1 + 0 .8| = 1 .5192 (26)

Note that the plant pole and compensator zero at s = 0.2 have been omitted from the calculation since they would cancelout exactly. The nal lead compensator for this example is

Gc(s) = 1.5192(s + 0 .2)

(s + 0 .6583) (27)


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0 10 20 30 40 50 60 70 80 90 1000

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Time (s)

O u t p u

t A m p

l i t u d e

Uncompensated Step Response

PO = 67.6%

Ts = 72.6 s

Fig. 8. Step response of the uncompensated system for the phase lead design example.

The compensated root locus and step response are shown in Fig. ?? . The overshoot is 19.4%, and the settling time is 15.7seconds, so the speci cations have been satis ed. For the particular G p(s) in this example, the dominant closed-loop polelocation s1 is suf ciently farther to the right than the other closed-loop poles to make the second-order equation for overshoothold for this higher-order system.

The following table summarizes the performances of the original system and the nal compensated system. It is seen thatthe compensator de ned in (27) allows both of the transient performance speci cations to be satis ed.

Uncompensated Compensated Speci cationP O 67.6% 19.4% 20%T s 72.6 sec 15.7 sec 16 secT r 2.60 sec 2.98 sec Nonee

ss

ramp 1 2.17 None

closed-loop poles 0.054 j 0.4405 0.25 j 0.488 s1 = 0.25 + j 0.4880.9004 0.2 None1.6916 0.8285 None

1.8298 None Note that one of the compensated closed-loop poles is at s = 0.2, the location of one of the plants open-loop poles. Thereason for this is the fact that the compensator zero was placed at that same location. A compensator zero and a plant pole

(or vice versa) at the same location always results in a closed-loop pole at that location also. That pole does not affect thesettling time since there is a closed-loop zero at that point also. Two of the closed-loop pole for the compensated system areat s = s1 and its complex conjugate as desired. The lead-compensated system has larger steady-state error for a ramp inputthan the uncompensated system, but since that characteristic was not speci ed, this does not cause a problem. If there was aspeci cation on steady-state error, it would be considered at this point in the design process, using the technique described in


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1.5

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0.5

0

0.5

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1.5

2

Real Axis

I m a g

A x

i s

s1

Uncompensated Root Locus

Fig. 9. Root locus of the uncompensated system and the desired closed-loop pole s 1 = 0.25 + j 0.488.

my notes Compensator Design for Steady-State Error Using Root Locus .

B. Phase Lag Example

1) Given System and Speci cations: The system to be controlled has the open-loop transfer function

G p(s) = 2(s + 1)

(s + 2) (28)

The system is Type 0 so it will have a non-zero, nite steady-state error for a step input. Speci cally, ess = 0 .5 from Eqs.

(6) and (7). If unity feedback is placed around G p(s), the closed-loop pole is at s = 1.33. Therefore, with Gc(s) = 1 theclosed-loop system is stable.The speci cations that are imposed on the system are: overshoot in the response to a step input: P O < 5%; settling time for the response to a step input: T s 2 seconds; steady-state error for a step input: ess = 0 .

The steady-state error speci cation requires a Type 1 system (or higher). Therefore, the compensator will have to have a poleat the origin as its minimum con guration. For purposes of designing the complete compensator, this pole may be included with the plant model to form the following augmented model corresponding to Eq. (8).

G(s) = 2(s + 1)s(s + 2)

(29)


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19

3 2 1 0 12

1.5

1

0.5

0

0.5

1

1.5

2

Real Axis

I m a g

A x i s

s1

Lead Compensated Root Locus

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

O u t p u

t A m p l

i t u d e

Lead Compensated Step Response

PO = 19.4%

Ts = 15.7 s

Fig. 10. Lead compensated root locus and step response for the design example.

The transfer function in (29) is the one that will be evaluated relative to the transient performance speci cations for the designof the compensator.

2) Selection of the Dominant Closed-Loop Pole: Once again, the equations for second-order systems may be used to choosethe location of the dominant closed-loop pole. The overshoot speci cation imposes the constraint on the damping ratio of > 0.6901, and the settling time speci cation imposes the constraint Re [s1] 2. For convenience, the damping ratio will be = 0 .707 since that corresponds to an angle of 45 and an overshoot of 4.3%. The dominant closed-loop pole will be placed at s1 = 2+ j 2. For the standard second-order system, this choice for s1 would allow both of the transient performancespeci cations to be satis ed. Although the system in (29) is not the model of (9), and will be further modi ed by any additionalcompensation other than a gain, we will use this choice as at least a starting point.

Since the root locus for G(s) lies entirely on the real axis (for both positive and negative K ), the point s1 is not on theroot locus. Therefore, the compensator will need to provide phase shift at s1 in order to make a branch of the root locus pass

through that point.3) Designing the Compensator: The rst step will to compute the phase shift of the augmented system at s1 . This value is

] G (s1 ) = ] (s1 + 1) ] s1 ] (s1 + 2) (30)= 116 .57 135

90

= 108.43 = 251 .57

In order to place s = s1 on the root locus, the compensator must provide a phase shift at s1 of


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20

] Gc (s1) = 180

(108.43 ) (31)

= 180 251.57

= 71.43

Since ] Gc (s1 ) < 0, phase lag compensation is needed, so the compensator pole will be to the right of the compensator zero. The method described in Section II-E.7 will be used to choosing the pole and zero so that the value of = zc /p c will be as small as possible. The phase angle of the point s = s1 is ] s1 = 135

, so the necessary angles of the compensator poleand zero are

] (s1 zc) = 135 + ( 71.43

)2

= 31 .72 (32)

] (s1 pc) = 135 (71.43

)2

= 103.29

The distances from the real-axis projection of s1 to the zero and pole and the locations of the compensator zero and pole are

dzc = 2

tan[31 .72 ] = 3 .2361 (33)

d pc

= 2

tan[103 .29 ] =

0.4721

and

zc = 5.2361, pc = 1.5279 (34)so the total compensator at this point in the design (including the pole needed for the steady-state error speci cation) is

Gc1 = K c1 (s + 5 .2361)

s (s + 1 .5279) (35)

The gain K c1 is computed from the magnitude criterion for the series combination of the original plant model in (28) and the compensator in (35). The value of the gain is K c1 = 0 .6833, so the nal compensator is

Gc1

= 0.6833(s + 5 .2361)

s (s + 1 .5279) (36)

A second compensator will be designed before the results of this rst compensator are evaluated. Then a comparison of thetwo designs will be made. Since ] Gc (s1) < 0, and the total phase shift of the compensator is the angle of the numerator minus the angle of denominator, this compensator can be implemented without a zero, using only a pole (in addition to theone at s = 0 ) and a gain. If this is done, the phase angle of the numerator is 0, so the phase angle of denominator is the total phase shift of the compensator.

With this approach, ] (s1 pc2) = 0 ] Gc (s1) = 71 .43 . The distance from the real-axis projection of s1 to the

compensator pole is d pc = 0 .6667, so the pole is located at s = 2.6667. From the magnitude criterion, the gain is computed to be K c2 = 2 .6667, so the nal form of this compensator isGc2 =

2.6667s (s + 2 .6667)

(37)

Both compensators produce third-order systems when placed in series with G p(s), and s = s1 is a closed-loop pole witheach of the designs. Figure 11 shows the compensated root locus plots for each of these designs. The small triangles on the plots are the actual closed-loop poles. It is clear that the design parameter of having s1 be a closed-loop pole is satis ed. Two points should be noted when comparing these root locus plots. The rst point is the differences in the numbers of asymptotesin the plots. The closed-loop system with Gc1(s) has only 1 more pole than zero (relative degree = 1), so the only asymptote is180 . The system with Gc2(s) has 2 more poles than zeros, so the asymptotes have angles of 90 . This may be an importantdifference if something causes the plant or compensator gain to increase in value. This would reduce the effective dampingratio, leading to more overshoot. Although the closed-loop system would remain stable with this G p(s) for all K > 0, thiswould not be true for other systems. If the second compensator design increased the relative degree of a system from 2 to 3(rather than from 1 to 2 as in this example), increases in gain would lead to an unstable closed-loop system if those increaseswere suf ciently large. Therefore, the presence of the zero in Gc1(s) provides additional stability robustness for the closed-loopsystem.


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22/22

22

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

O u t p u

t A m p l

i t u

d e

Lag Compensated Step Responses

Gc1

(s); Ts = 1.08 s

Gc2

(s); Ts = 4.71 s

Gc(s) = 1/s; T

s = 5.5 s

Fig. 12. Comparison of closed-loop step responses for two lag compensators.

be considered at this point in the design process, using the special lag design described in my notes Compensator Design for Steady-State Error Using Root Locus .

R EFERENCES[1] J. DAzzo and C. Houpis, Linear Control System Analysis and Design . New York: McGraw-Hill, 4th ed., 1995.[2] R. C. Dorf and R. H. Bishop, Modern Control Systems . Reading, MA: Addison-Wesley, 7th ed., 1995.[3] K. Ogata, Modern Control Engineering . Upper Saddle River, NJ: Prentice Hall, 4th ed., 2002.[4] G. Franklin, J. Powell, and A. Emami-Naeini, Feedback Control of Dynamic Systems . Reading, MA: Addison-Wesley, 3rd ed., 1994.[5] G. Thaler, Automatic Control Systems . St. Paul, MN: West, 1989.[6] W. A. Wolovich, Automatic Control Systems . Fort Worth, TX: Holt, Rinehart, and Winston, 3rd ed., 1994.[7] J. V. de Vegte, Feedback Control Systems . Englewood Cliffs, NJ: Prentice Hall, 3rd ed., 1994.[8] B. C. Kuo, Automatic Controls Systems . Englewood Cliffs, NJ: Prentice Hall, 7th ed., 1995.[9] N. S. Nise, Control Systems Engineering . New York: John Wiley & Sons, 3rd ed., 2000.

[10] C. Phillips and R. Harbor, Feedback Control Systems . Upper Saddle River, NJ: Prentice Hall, 4th ed., 2000.[11] G. C. Goodwin, S. F. Graebe, and M. E. Salgado, Control System Design . Upper Saddle River, NJ: Prentice Hall, 2001.

compensator design to improve transient performance using root locus

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