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Compensator Design to Improve TransientPerformance Using Root Locus

Prof. Guy BealeElectrical and Computer Engineering Department

George Mason UniversityFairfax, Virginia

CONTENTS

I INTRODUCTION 2

II DESIGN PROCEDURE 2II-A Compensator Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2II-B Outline of the Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3II-C System Type N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4II-D Selecting a Dominant Closed-Loop Pole Location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4II-E Determining the Compensator’s Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

II-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 Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18III-B.2 Selection of the Dominant Closed-Loop Pole . . . . . . . . . . . . . . . . . . . . . . . . . . 19III-B.3 Designing the Compensator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

References 22

LIST OF FIGURES

1 Allowable region for s1 in order to satisfy specifications 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 + j5.4575. . . . . . . . . . . 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 α = zc/pc. . . . . . . . . . . . . . . . . . . . . . . . . . . 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 + j0.488. . . . . . . . . . 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 andComputer 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. INTRODUCTIONIn my approach to root locus design, the purpose of compensator design using root locus methods generally is to establish a

specified point in the s-plane, s = s1, as a closed-loop pole. The assumption is that time-domain transient specifications, suchas settling time and overshoot, will be satisfied 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 Kc 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 thepoint s1 to satisfy the phase angle criterion so that the compensated root locus does pass through s1. This is done by choiceof the compensator’s 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 satisfythe specifications. This requires moving the dominant branches of the root locus to the left. A phase lead compensator (providingpositive 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 ofthe compensator’s 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 compensator’s polesand zeros to satisfy the phase requirements can be visualized from the trigonometric relationships that must be satisfied at thedesired dominant closed-loop pole. The computations can be easily done by calculator. If data arrays representing the numeratorand denominator polynomials of the open-loop system are available, then the procedure can be done using a software packagesuch 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 PROCEDUREA. 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 amplifiers. The first amplifier has an input impedance that is the parallel combination of resistor R1 andcapacitor C1 and a feedback impedance that is the parallel combination of resistor R2 and capacitor C2. The second amplifierhas input and feedback resistors R3 and R4, respectively.

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

Gc(s) =Vout(s)

Vin(s)=

Kc (s− zc)

(s− pc)=

Kx

µs

−zc+ 1

¶µ

s

−pc+ 1

¶ =Kx (τs+ 1)

(ατs+ 1)(1)

=R4R2R3R1

· (sR1C1 + 1)(sR2C2 + 1)

=R4R2R3R1

· R1C1R2C2

· (s+ 1/R1C1)(s+ 1/R2C2)

=R4C1R3C2

· (s+ 1/R1C1)(s+ 1/R2C2)

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 relationships1 can beobtained by inspecting Eq. (1).

Kx =R4R2R3R1

, τ = R1C1, ατ = R2C2, α =zcpc=

R2C2R1C1

(2)

zc = −1/R1C1, pc = −1/R2C2, Kc =Kx

α=

R4C1R3C2

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.

1In 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 definition 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 α < 1 for a lead compensator, and α > 1 for a lagcompensator.

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

|Gc (s1)| =|Kc| · |s1 − zc||s1 − pc|

=|Kc| ·

q[Re (s1)− zc]

2+ Im2 (s1)q

[Re (s1)− pc]2+ Im2 (s1)

(3)

and a phase angle

]Gc (s1) = ]Kc +] (s1 − zc)− ] (s1 − pc) = ]Kc + 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 Kc > 0, the magnitude and phase of thegain are |Kc| = Kc and ]Kc = 0

◦. In applications where Kc < 0, the we have |Kc| = −Kc and ]Kc = 180◦.

B. Outline of the ProcedureThe following steps outline the procedure that will be used to design either a lead or a lag compensator using root locus

methods in order to satisfy transient performance specifications, such as settling time and percent overshoot. Compensatordesign to satisfy steady-state error requirements is discussed in separate notes, Compensator Design to Improve Steady-StateError Using Root Locus.

1) Determine if the System Type N needs to be increased in order to satisfy the steady-state error specification, and ifnecessary, 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 specification 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 specifications and should produce a closed-loop system that will satisfy those specifications.

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

specification) 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 system’s root locus for positive gain. If the phase shift is an even integermultiple of 180◦ (0◦mod360◦), then the selected point is on the uncompensated system’s root locus for negativegain. In either case, the only compensation needed for the transient performance specifications 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 compensator’s phase shift (usually) will be the shortest distance from the plant’s 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◦ fornegative 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 computedin the previous step is positive, phase lead compensation is required. This places the compensator zero to the rightof the pole. If the compensator’s 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 fixed.

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 Kc. The value of the gain is chosen to satisfy the magnitude criterion at s1(|Gc (s1)Gp (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 specifications will be used:

Gp(s) =8

s+ 4(5)

• steady-state error specification for a unit ramp input is ess_specified = 0.1;• step response settling time specification is Ts−specified = 1 second;• step response overshoot specification POspecified ≈ 10%.

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C. System Type N

The first step in the design of the compensator will be to determine if the plant Gp(s) has the correct System Type tosatisfy the steady-state error specification. Defining the number of open-loop poles of a system that are located at s = 0 tobe the System Type N , and restricting the reference input signal to having Laplace transforms of the form R(s) = A/sq, thesteady-state error and error constant are (assuming that the closed-loop system is bounded-input, bounded-output stable)

ess = lims→0

∙AsN+1−q

sN +Kx

¸(6)

whereKx = lim

s→0

£sNG(s)

¤(7)

For N = 0, the steady-state error for a step input (q = 1) is ess = A/ (1 +Kx). For N = 0 and q > 1, the steady-stateerror is infinitely large. For N > 0, the steady-state error is ess = A/Kx 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 infinite. Therefore, for a system to have a non-zero, finitesteady-state error for a specified reference input, the System Type must satisfy N = Nreq = q− 1. This is the total number ofopen-loop poles at the origin needed for the compensated system to satisfy the steady-state error specification. If the SystemType of the plant Gp(s) is Nsys, then the compensator must have (Nreq −Nsys) 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 willbe

G(s) =1

s(Nreq−Nsys)·Gp(s) (8)

Example 1:The plant transfer function in (5) has no poles at the origin, so it has System Type Nsys = 0. The steady-state error

specification is for a unit ramp input which has a Laplace Transform R(s) = 1/s2, so for this input q = 2. Therefore, in orderfor the steady-state error specification to be satisfied, the total number of poles at the origin must be Nreq = q−1 = 2−1 = 1.Thus, the compensator must provide Nreq −Nsys = 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 LocationThe real engineering design takes place in this step. This is the mapping of the performance specifications that must be

satisfied into the design parameter that influences the rest of the steps in the design. For other than very simple systems, soundengineering judgement is needed in order for this mapping to yield a compensator design that actually allows the specificationsto be satisfied.

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-ordersystem, shown in (9).

G(s) =ω2n

s (s+ 2ζωn), TCL(s) =

ω2ns2 + 2ζωns+ ω2n

(9)

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

p1, p2 = −ζωn ± jωn

q1− ζ2 (10)

In this case, specifications on the amount of overshoot and the settling time for the step response are natural specifications.These types of specifications will be used in the design procedure presented in these notes. A steady-state error specificationon the ramp response is also appropriate; that will be discussed in separate notes.

Percent overshoot (PO) 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 ζ are related by

PO =he−πζ/

√1−ζ2

i· 100%, ζ =

¯̄ln¡PO100

¢¯̄qπ2 + ln2

¡PO100

¢ (11)

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

£cos−1 (ζ)

¤.

Defining settling time Ts as the time required for the step response to get within and stay within a fixed percentage of thefinal value, the settling time for the standard second-order system is a function of the product of damping ratio and undampednatural frequency. In these notes, the band about the final value used to define Ts is ±2%, and the settling time is

Ts =4

ζωn(12)

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

Re [s1] =−4Ts

(13)

If the settling time specification 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 defined to be ωd = ωnp1− ζ2, which is seen in (10) to be the imaginary part of the

closed-loop poles. This parameter is also the frequency of oscillation in the transient part of the step response. Therefore, aspecification 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 specifications 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 defined by the various specifications.

Example 2:The following specifications are to be imposed on a system’s closed-loop step response: (1) 5% ≤ PO ≤ 25%, (2) 0.5 s≤ Ts ≤ 2 s, (3) ωd ≤ 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 ζ = 0.6901 and an angle θ = 46.4◦. The point s = s1 must lie on or above this radial line. The upper boundof 25% overshoot corresponds to a damping ratio ζ = 0.4037 and an angle θ = 66.2◦. The point s = s1 must lie on or belowthis radial line. From (13), the lower bound of Ts = 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 Ts = 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 ωd = 12.6 r/s corresponds to a horizontal line at j12.6. The closed-looppole 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 Gp(s) augmented with one pole at the origin to satisfy the steady-state error requirement is

G(s) = 8/[s(s+ 4)]. The transient specifications are Ts−specified = 1 second and PO ≈ 10%. The damping ratio associatedwith 10% overshoot is ζ = 0.5912. The corresponding angle and slope of the radial line from the origin through s1 areθ = 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 + j5.4575. ¨

The relationships between dominant pole locations and step response characteristics presented above are only for the standardsecond-order system of (9). These relationships may provide good starting points for the selection of s1 to satisfy specifications,but they can only be used as general guidelines for more complex system models. Even if the given system Gp(s) is modeledby (9), any compensation in series with Gp(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 specification, the recommended approach is to be very conservative—that is, use a significantlysmaller value of overshoot than the specified 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 specification 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-ordersystems.

In many actual design problems with specifications on overshoot, settling time, frequency of oscillation, etc., the properchoice 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 specifications to be satisfied.

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−20 −15 −10 −5 0 5−5

0

5

10

15

20

Real Axis

Imag

Axi

s

Locating Closed−Loop Poles in the s−plane to Satisfy Specifications

POmax

= 25%

ζmin

= 0.4037

θmax

= 66.2ο

POmin

= 5%

ζmax

= 0.6901

θmin

= 46.4ο

ωd

max

= 12.6 r/s

ζωn = 2ζω

n = 8

Ts

max

= 2 sTs

min

= 0.5 s

Fig. 1. Allowable region for s1 in order to satisfy specifications on overshoot, settling time, and frequency of oscillation.

E. Determining the Compensator’s ParametersOnce the point s = s1 is chosen to be a dominant closed-loop pole, the design of the compensator is an exercise in

trigonometry. The pole(s) and zero(s) of the compensator are chosen to satisfy the phase angle criterion at s1 so that the totalphase 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) areplaced, the gain of the compensator is selected to satisfy the magnitude criterion. This makes the point s = s1 be a closed-looppole. 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 thecompensator’s 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 augmentedplant 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 +mXi=1

] (s1 − zi)−nXi=1

] (s1 − pi)

= ]K +mXi=1

tan−1∙Im (s1)− Im (zi)Re (s1)−Re (zi)

¸−

nXi=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 2−6

−4

−2

0

2

4

6

Real Axis

Imag

Axi

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 = s1.

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◦ (π 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 θ = 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 + j5.4575, 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 Gp(s)augmented with any poles at the origin that were needed to produce the correct System Type), then ]G (s1) is an odd integermultiple 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 Kc. That would be chosen to satisfy the magnitude criterion |KcG (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 plant–compensator combination is an odd integer multiple of 180◦ at that point. This is done by choiceof the compensator’s pole and zero. Satisfying the phase angle criterion must be done first. After that, the value of Kc 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◦ (phaselead) 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)= ]Kc +]Nc (s1)− ]Dc (s1)

= ]Kc +

mcXi=1

] (s1 − zci)−ncXi=1

] (s1 − pci)

Assuming that Kc > 0, then ]Gc (s1) = ]Nc (s1)− ]Dc (s1).

Example 5:From Example 4, the phase shift of G(s) at s1 = −4 + j5.4575 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 flexibility 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 bythe compensator’s phase shift requirement at s1. Because ]Gc (s1) = ] (s1 − zc) − ] (s1 − pc), the compensator zero mustprovide 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 specifications, 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 jω 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 specifications willbe satisfied.

Example 6:For the system G(s) defined in the previous examples and the point s1 = −4 + j5.4575, the required phase shift of the

compensator 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 sincethat 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 first three locations for the zero in the above example (zc = −9, zc = −7, zc = −5) satisfy the rule of thumb mentionedearlier. For these choices, the closed-loop poles at s = s1 and its complex conjugate will be closer to the jω axis than the

9

−10 −8 −6 −4 −2 0 2−6

−4

−2

0

2

4

6

Real Axis

Imag

Axi

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 jω axis, and so will not be controlling the settling time. This does not mean that the settling time specificationwill be violated if the rule of thumb is not satisfied, but as the zero is moved farther and farther to the right, at some point,the settling time specification 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 fixed. 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 criterionat s1.

Since the total phase shift of the compensator at s = s1 is the angle produced by the zero minus the angle produced by thepole, 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 compensatorpole that will produce the angle ] (s1 − pc) at s = s1. The distance dpc from the real-axis projection of s1 to the compensatorpole can be computed easily from

dpc =Im [s1]

tan [] (s1 − pc)](17)

10

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

pc = Re [s1]− dpc (18)

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

compensator zero were presented, namely, zc = {−9,−7,−5,−3,−1}. For each of these choices, the compensator pole canbe 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) dpc pc−9 11.27◦ 27.4 −31.4−7 24.96◦ 11.7 −15.7−5 43.38◦ 5.78 −9.78−3 64.14◦ 2.64 −6.64−1 82.56◦ 0.713 −4.713

The 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 locusplot 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 satisfied at s1 by the series combination of plant and compensator. Thecompensator gain is used for this purpose.

The magnitude criterion states that

|Kc| · |s1 − zc||s1 − pc|

· |G (s1)| = 1 (19)

if s = s1 is a closed-loop pole, with G(s) defined 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

|Kc| =|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 Kc = − |Kc|.

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 Kc.

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| Kc

−9 27.94 7.402 17.42−7 12.93 6.228 9.585−5 7.946 5.548 6.611−3 6.065 5.548 5.045−1 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 specification, is Gc(s) = Kc (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 specified location of s1 = −4 + j5.4575 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)Gp(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 compensator’sopen-loop pole. The closed-loop step responses for these five compensator designs are shown in Fig. 4. The settling times forthe first four compensator designs are all approximately the same, slightly less than 1 second. The settling time for the fifthcompensator design is significantly longer and does not satisfy the specifications. Of the compensator designs that do satisfy

11

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

Out

put A

mpl

itude

Closed−Loop 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 s1 = −4 + j5.4575.

the settling time requirement, only the fourth design (zc = −3) also satisfies the overshoot specification. The dashed lines inthe figure show the ±2% interval for the settling time specification and the 10% overshoot specification. Figure 5 shows theroot locus plots for four of the compensator designs. The triangles shown in the plots represent the closed-loop poles. Thepoint s1 is a closed-loop pole in each case. It is easily seen in the figure 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 specifications 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 ofthe specifications. 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 generallytaken to be the smallest absolute value that will yield an odd integer multiple of 180◦ for the plant–compensator 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 + j0.8. 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 forKc > 0. This would require a phase lag compensator. On the other hand, a compensator phase ]Gc2 (s1) = 199.98

◦ wouldalso place s1 on the root locus for Kc > 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

12

−20 −15 −10 −5 0 5−20

−15

−10

−5

0

5

10

15

20

Real Axis

Imag

Axi

s

zc = −7

−10 −5 0 5−15

−10

−5

0

5

10

15

Real AxisIm

ag A

xis

zc = −5

−8 −6 −4 −2 0 2−10

−5

0

5

10

Real Axis

Imag

Axi

s

zc = −3

−5 −4 −3 −2 −1 0 1−8

−6

−4

−2

0

2

4

6

8

Real Axis

Imag

Axi

s

zc = −1

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

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 jω axis assuming they are in the left-half plane) wouldtend 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 transferfunctions 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 shouldbe 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 α = zc/pc for a lead compensator or minimizing α for a lagcompensator. This is advantageous because it minimizes the range of resistor and capacitor values used to implement thecompensator.

We will define 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 definition, thephase 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

dzc =Im [s1]

tan [] (s1 − zc)], dpc =

Im [s1]

tan [] (s1 − pc)](22)

and the compensator zero and pole are located at

zc = Re [s1]− dzc , pc = Re [s1]− dpc (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 validsolution to the design problem. This solution is not necessarily any better than any other solution other than the fact that itoptimizes the ratio zc/pc.

Example 10:From Example 5, the compensator must provide 36.24◦ at the point s1 = −4 + j5.4575. 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 horizontaldistances from s1 to the zero and pole are dzc = 0.8411 and dpc = 5.4575. This places the zero zc at s = −4.8411 and thepole pc at s = −9.4575. The value of α for this pole–zero combination is α = zc/pc = 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 onepole and one zero. In some cases it may be desirable or necessary to use multiple stages of compensation. If the requiredcompensator phase shift ]Gc (s1) is very large, then the left-most allowed location for the compensator zero may be closeto the jω 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 ofstages. The most usual case would probably be two stages; therefore, each stage would provide one-half of the total requiredphase 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.

14

−8 −6 −4 −2 0 2 4−4

−3

−2

−1

0

1

2

3

4

Real Axis

Imag

Axi

s

s1

Lag Compensated

−10 −8 −6 −4 −2 0 2−10

−5

0

5

10

Real Axis

Imag

Axi

s s1

Lead Compensated

0 2 4 6 8 10−12000

−10000

−8000

−6000

−4000

−2000

0

2000

Time (s)

Step

Res

pons

e

Lag Compensated

0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

Step

Res

pons

e

Lead Compensated

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

15

−10 −8 −6 −4 −2 0 2−4

−2

0

2

4

6

8

Real Axis

Imag

Axi

s

Alternative Placement of Compensator Pole and Zero

s1

zc

pc

Fig. 7. Selecting the compensator zero and pole to maximize α = zc/pc.

The total compensator numerator then has the formQnum_stages

i−1 (s− zci), and similarly for the compensator denominator.The gain Kc is then computed to satisfy the magnitude criterion. In implementation, the gain can be assigned equally to eachstage by letting Kc−stage =

num_stages√Kc.

Example 11:The compensator phase shift computed in the previous examples is ]Gc (s1) = 36.24

◦. Although this compensator can bedesigned with a single stage of compensation, it was shown in Example 8 that there was a restricted range of locations forthe compensator zero that allowed both the overshoot and settling time specifications to be satisfied. To use a single stage ofcompensation, 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 andpc = −359.1. Using two poles and zeros at these locations, the gain is Kc = 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 ofthe single-stage compensators. The settling time is approximately the same as the first 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

16

its complex conjugate. The remaining two closed-loop poles are at s = −357.1 ± j123. Although the overshoot is still a bithigher than specified, 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. DESIGN EXAMPLE

Both phase lead and phase lag examples will be presented in this section. The lead example will be presented first 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 thepole and zero are reversed.

A. Phase Lead Example1) Given System and Specifications: The open-loop transfer function for the system to be controlled is

Gp(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, finite steady-state error for aramp input. If unity feedback is placed around Gp(s), the closed-loop poles are located at s = {−0.054± j0.4405, −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 ζ = 0.122, which corresponds to 68% overshoot for the standardsecond-order system. The settling time is dominated by those complex conjugate poles since they are the closest to the jωaxis, 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 specifications that are imposed on the system are:• overshoot in the response to a step input: PO ≈ 20%;• settling time for the response to a step input: Ts ≈ 16 seconds.

Thus, the compensator needs to reduce the overshoot and the settling time significantly. 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 ζ = 0.4559. The angle and slope of the radialline associated with this value of ζ are θ = 62.87◦ and 1.952, respectively. The settling time specification 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 + j0.488.

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 first step in designing the compensator is to determine the phase shift of Gp(s) ats = s1. This angle is ]Gp (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 ]Gp (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 pole–zero pair.

In order to use a single stage of compensation, the compensator zero must provide more than ]Gc (s1) = 45.77◦ at s1, so the

left-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 secondreal-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 dpc =

Im [s1] / tan [50.08◦] = 0.4083. Therefore, the compensator pole is located at s = −0.6583. At this stage in the design, the

compensator is Gc(s) = Kc (s+ 0.2) / (s+ 0.6583).The compensator gain is determined from the magnitude criterion |Gc (s1)Gp (s1)| = 1. The gain calculation is

Kc =|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 final lead compensator for this example is

Gc(s) =1.5192 (s+ 0.2)

(s+ 0.6583)(27)

17

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)

Out

put A

mpl

itude

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 specifications have been satisfied. For the particular Gp(s) in this example, the dominant closed-loop polelocation s1 is sufficiently 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 final compensated system. It is seen thatthe compensator defined in (27) allows both of the transient performance specifications to be satisfied.

Uncompensated Compensated SpecificationPO 67.6% 19.4% ≈ 20%Ts 72.6 sec 15.7 sec ≈ 16 secTr 2.60 sec 2.98 sec Noneess−ramp 1 2.17 Noneclosed-loop poles −0.054± j0.4405 −0.25± j0.488 s1 = −0.25 + j0.488

−0.9004 −0.2 None−1.6916 −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 plant’s 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 specified, this does not cause a problem. If there was aspecification on steady-state error, it would be considered at this point in the design process, using the technique described in

18

−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Real Axis

Imag

Axi

s

s1

Uncompensated Root Locus

Fig. 9. Root locus of the uncompensated system and the desired closed-loop pole s1 = −0.25 + j0.488.

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

B. Phase Lag Example1) Given System and Specifications: The system to be controlled has the open-loop transfer function

Gp(s) =2(s+ 1)

(s+ 2)(28)

The system is Type 0 so it will have a non-zero, finite steady-state error for a step input. Specifically, ess = 0.5 from Eqs.(6) and (7). If unity feedback is placed around Gp(s), the closed-loop pole is at s = −1.33. Therefore, with Gc(s) = 1 theclosed-loop system is stable.

The specifications that are imposed on the system are:• overshoot in the response to a step input: PO < 5%;• settling time for the response to a step input: Ts ≤ 2 seconds;• steady-state error for a step input: ess = 0.

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

G(s) =2(s+ 1)

s(s+ 2)(29)

19

−3 −2 −1 0 1−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Real Axis

Imag

Axi

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)

Out

put A

mpl

itude

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 specifications 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 specification imposes the constraint on the damping ratio ofζ > 0.6901, and the settling time specification imposes the constraint Re [s1] ≤ −2. For convenience, the damping ratio willbe ζ = 0.707 since that corresponds to an angle of 45◦ and an overshoot of 4.3%. The dominant closed-loop pole will beplaced at s1 = −2+j2. For the standard second-order system, this choice for s1 would allow both of the transient performancespecifications to be satisfied. Although the system in (29) is not the model of (9), and will be further modified 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 passthrough that point.

3) Designing the Compensator: The first 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

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 compensatorzero. The method described in Section II-E.7 will be used to choosing the pole and zero so that the value of α = zc/pc willbe 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)

dpc =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 specification) is

Gc1 =Kc1 (s+ 5.2361)

s (s+ 1.5279)(35)

The gain Kc1 is computed from the magnitude criterion for the series combination of the original plant model in (28) andthe compensator in (35). The value of the gain is Kc1 = 0.6833, so the final compensator is

Gc1 =0.6833 (s+ 5.2361)

s (s+ 1.5279)(36)

A second compensator will be designed before the results of this first 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 numeratorminus 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 totalphase shift of the compensator.

With this approach, ] (s1 − pc2) = 0 − ]Gc (s1) = 71.43◦. The distance from the real-axis projection of s1 to thecompensator pole is dpc = 0.6667, so the pole is located at s = −2.6667. From the magnitude criterion, the gain is computedto be Kc2 = 2.6667, so the final form of this compensator is

Gc2 =2.6667

s (s+ 2.6667)(37)

Both compensators produce third-order systems when placed in series with Gp(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 theplots are the actual closed-loop poles. It is clear that the design parameter of having s1 be a closed-loop pole is satisfied. Twopoints should be noted when comparing these root locus plots. The first 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 Gp(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 sufficiently large. Therefore, the presence of the zero in Gc1(s) provides additional stability robustness for the closed-loopsystem.

21

−12 −10 −8 −6 −4 −2 0 2−4

−3

−2

−1

0

1

2

3

4

Real Axis

Imag

Axi

s

Using Lag Compensator Gc1

(s)

s1

−3 −2 −1 0 1 2−5

0

5

Real Axis

Imag

Axi

s

Using Lag Compensator Gc2

(s)

s1

Fig. 11. Comparison of root locus plots with two lag compensator designs.

The second point to note is the location of the third-closed loop pole for the two designs. With Gc1(s), that pole is veryclose to the open-loop zero at s = −1. Therefore, its effect on the transient response will be small, even though the pole iscloser to the jω axis than s1. By contrast, the third closed-loop pole with Gc2(s) is farther away from the open-loop zero andalso even closer to the jω axis. The effect of this pole on the transient response might be significant.

Figure 12 shows the step responses for these two compensator designs, as well as for the closed-loop response of theaugmented system in (29). The only system that satisfies the settling time specification is the one with compensator Gc1(s).That system also has a much shorter rise time than the other two designs.

The following table summarizes the results from this example.

Gp(s) G(s) Gc1(s)Gp(s) Gc2(s)Gp(s) SpecificationPO 16.67% 0% 0.65% 0% < 5%Ts 2.12 sec 5.50 sec 1.075 sec 4.66 sec ≤ 2 secTr undefined 2.67 sec 0.749 sec 1.88 sec Noneess−step 0.5 0 0 0 0closed-loop poles −1.33 −0.5858 −2± j2 −2± j2 s1 = −2 + j2

−3.4142 −0.8944 −0.6667 None

The overshoot and settling time for the uncompensated system are based on the final value for the output of 0.5. Since theuncompensated system has the same number of poles and zeros, there is an impulse in the step response, and the maximumvalue (0.6667) occurs at t = 0. This value was used in the computation of overshoot. Since the output never takes on a valueequal to 10% of its final value, the rise time is undefined.

Augmenting the system with the pole at the origin satisfies the steady-state error and overshoot specifications, but not thesettling time specification. As previously mentioned, the system using Gc1(s) satisfies all of the specifications. The secondcompensator design fails on the settling time specification due to the closed-loop pole at s = −0.6667.

For the augmented system of (29) and the transient response specifications in this example, gain compensation can be used.Both the specifications are satisfied for all Kc ≥ 4.55. With this approach, the total compensator would be Gc(s) = Kc/s. Thepoint s1 would not be a closed-loop pole; both poles would be real. Very short rise and settling times can be achieved by usinglarge gain values. However, a potential drawback with this approach is large acceleration in the system output. The controlsignal would change values rapidly near the beginning of the response causing the output to change rapidly also, with theoutput signal approaching a discontinuity for large gains. If the output signal represented mechanical position of some object,a large acceleration might be objectionable. The rapid change in the value of the control signal might also exceed physicallimitations of the actuator providing motion to the system. The control signals generated by Gc1(s) and Gc2(s) change moresmoothly and limit the acceleration.

This example illustrates the design technique of satisfying transient performance specifications using root locus. Even thoughthe compensated system was third-order and contained one zero, the selection of the point s1 yielded a compensator designGc1(s) that more than satisfied the specifications. If there was a specification on steady-state error for a ramp input, it would

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)

Out

put A

mpl

itude

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 Designfor Steady-State Error Using Root Locus.

REFERENCES

[1] J. D’Azzo 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.