1

Design via Root Locus

2

1 Introduction

• The root locus graphically displayed both transient response and stability information.

• The locus can be sketched quickly to get a general idea of the changes in transient response generated by changes in gain k.

• Specific points on the locus also can be found accurately to give quantitative design information.

• The root locus typically allows us to choose the proper loop gain to meet a transient response specification.

3

1.1 Root Locus Techniques

• The features of a step response (Tr ,overshoot, Ts , are related to pole locations in the s-plane. • A graphic representation of the closed-loop poles as a system parameter is varied.• Gives information about the stability and transient response of feedback control systems.

Fig.1 pole locations and corresponding natural responses

Complex polestransfer function Correspondence between the parameters

Standard second-order system impulse response

Fig.2 s-plane plot for a pair of complex poles

Fig.3 Second-order system response with an exponential envelope

Definition of rise time tr, settling time ts, and overshoot Mp

Rise Time (Tr) Peak Time (Tp) Overshoot (Mp) Settling Time (Ts)

4

1.2 Defining the Root Locus

• The root locus technique can be used to analyze and design the effect of loop gain upon the system's transient response and stability. Assume the block diagram representation of a tracking system as shown in Figure 1 and 2

Fig.1 block diagram Fig.2 closed-loop transfer function

• the closed-loop poles of the system change location as the gain, K, is varied.

The representation of the paths of the closed-loop poles as the gain K (K≥0) is varied is called the root locus.

• The root locus shows the changes in the transient response as the gain, K, varies.

- the poles are real for gains less than 25. Thus, the system is overdamped

- At a gain of 25, the poles are real and multiple and hence critically damped.

- For gains above 25, the system is underdamped.

5

1.2. Root Locus Sketching

• The closed-loop transfer of system (fig.1) function is:

Fig.1 Basic closed-loop block diagram• The graph of all possible roots of this equation (K is the variable parameter) is called the root locus, and the set of rules to construct this graph is called the root-locus method of Evans.

root-locus form or Evans form of a characteristic equation

Characteristic equation

G(s)H(s)D(s)

Evans’s methodCompensation by selecting the best value of K or dynamic

Compensation (add zeros and poles)

1. Number of branches: The number of branches of the root locus equals the number of closed-loop poles (n).2. Real-axis segments: On the real axis, the root locus exists to the left of an odd number of real-axis open-loop poles (n) and zeros (m).3. Starting and ending points: The root locus begins at the poles of G(s)H(s) and ends at the zeros of G(s)H(s).4. Asymptote: The root locus approaches straight lines as asymptotes as the locus approaches infinity. the asymptotes (real-axis intercept at and angle Өa )

where k = 0, ±1, ±2, ±3 … (n-m-1) and the angle is given in

radians with respect to the positive extension of the real axis.

6

Example 2

Solution

Sketch the root locus for the system shown in Figure

the asymptotes: the real-axis intercept

The angles of the lines that intersect at -4/3,

Zeros: -3 and poles: 0, -1, -2, -4

• the locus begins at the open-loop poles and ends at the open-loop zeros.

• there are 3 more open-loop poles than open-loop zeros. Thus, there must be 3 zeros at infinity.

7

Example 31

1. Find the open-loop TF, the closed-loop TF and the open-loop poles and zeros.2. Determine root locus on the real axis3. Approximate the asymptotes on the root locus4. Approximate break-away and break-in points

Solution

Open loop: Open at the summing junction and stretch

Plot how the closed-loop poles change with Kp

Kp 0 CLD s2+2s+3 Open-Loop CEKp =1 CLD s2+2s+5

Kp CLD s2+4s+7 ∞ CLD Kp (s+2)Kp =2 Reach the zero

𝐶𝐿𝑇𝐹 :𝐺𝐶𝐿 (𝑠)=

𝐾 𝑃(𝑠+2)𝑠2+2𝑠+3

1+𝐾 𝑃(𝑠+2)

𝑠2+2𝑠+3

=𝐾 𝑃(𝑠+2)

𝑠2+ (2+𝐾 𝑃 )𝑠+3+2 𝐾 𝑃

Zeros = -2, 𝑂𝐿𝑇𝐹 :𝐺𝑂𝐿(𝑠)=

𝐾 𝑃 (𝑠+2)𝑠2+2 𝑠+3 𝑃𝑜𝑙𝑒𝑠=−1± 𝑗√2

CLD=Closed- loop denominator

𝐺𝐶𝐿(𝑠)=𝐾 𝑃(𝑠+2)

𝑠2+(2+𝐾 𝑃 ) 𝑠+3+2 𝐾 𝑃

1. open-loop TF

closed-loop TF

2. Determine root locus on the real axis

8

In general, breakaway (between poles on real axis) and break-in (between zeros on the real axis) points occur when branches come together.Break point between two poles on the real axis, between 0 and -1.

Example 32

3. Approximate the asymptotes on the root locus

Number of zero at infinity = 3 asymptotes.

= 60o, 180o and 300o = - 4/3= -1,33

4. Approximate break-away and break-in points

9

1.4 Selecting the Parameter Value

• The purpose of design is to select a particular value of K that will meet the specifications for static and dynamic response.• The gain selection is almost never done by hand because the determination can be accomplished easily by MATLAB.• For values of s on the root locus, the phase of L(s) is 180°, so we can write the magnitude condition as:

• The value of K is given by 1 over the magnitude of L(s0), where s0 is the coordinate of the dot.

• To use MATLAB: [K,p] = rlocfind(sys, s0)

Example for : has one integrator and, in a unity feedback configuration, will be a Type 1 control system with poles at 0, .

• Finally, once the gain selected, it is possible to compute the error constant ( related to ess)

of the control system (type 1).

• If no value of K satisfies all of the constraints, as is typically the case, then additional modifications are necessary to meet the system specifications.

Graphical calculation of the desired gain

Gain at s0 on the root locusp0=-1.63+3.69i

10

1.5 Configurations

• Two configurations of compensation are covered here: cascade compensation and feedback compensation.

• With cascade compensation, the compensating network, G1(s), is placed at the low-power end of the forward path in cascade with the plant.

• If feedback compensation is used, the compensator, H1(s), is placed in the feedback path.

• Both methods change the open-loop poles and zeros, thereby creating a new root locus that goes through the desired closed-loop pole location.

Compensation techniques: a. cascade; b. feedback

11

2 Design Using Dynamic Compensation

• If the process dynamics are of such a nature that a satisfactory design (point B) cannot be obtained by adjustment of the proportional gain k alone (current root locus at A), then some modification or compensation of the dynamics is indicated.

1. Improving Steady-State Error:

One objective of this design is to improve the steady-state error without appreciably affecting the transient response. (PI ideal compensator implemented with active networks, lag compensator implemented with passive networks )

2. Improving Transient Response:The objective is to design a response that has a desirable percent overshoot and a shorter settling time than the uncompensated system. (PD ideal compensator, Lead Compensation)

Desired transient response(overshoot and settling time)

Sample root locus, showing possible design point via gain adjustment (A) and desired design point that cannot be met via simple gain adjustment (B);

Responses from poles at A and B

• One way to solve this problem is to compensate the system with additional poles and zeros, so that the compensated system has a root locus that goes through the desired pole location for some value of gain.

12

2.1 Ideal Integral Compensation (PI)• Steady-state error can be improved by placing an open-loop pole at the origin, because this increases the system type by one

and eliminates the steady-state error for the unit step input.

Pole at A is:a. on the root locus without compensator; b. not on the root locus with compensator pole added; c. approximately on the root locus with compensator pole and zero added

system operating with a desirable transient response generated by the closed-loop poles at A

If we add a pole at the origin to increase the system type, the angular contribution of the open-loop poles at point A is no longer 180°, and the root locus no longer goes through point A,

To solve the problem, we also add a zero close to the pole at the origin, the angular contribution of the compensator zero and compensator pole cancel out, point A is still on the root locus, and the system type has been increased.

we have improved the steady-state error without appreciably affecting the transient response

13

Example11Given the system of Figure (a), operating with a damping ratio of 0.174, show that the addition of the ideal integral compensator shown in Figure (b) reduces the steady-state error to zero for a step input without appreciably affecting transient response.

SOLUTION

• We first analyze the uncompensated system to determine the locationof the dominant, second-order poles.

• The root locus for the uncompensated system is shown in Figure. A damping ratio of 0.174 is represented by a radial line drawn on the s-plane at 100.02°.

• Searching along this line we find that the dominant poles are 0.694 ± j3.926 for a gain, K, of 164.6. for the third pole on the root locus beyond —10 on the real axis.

• for the same gain as that of the dominant pair, K = 164.6, we find that the third pole is approximately at -11.61.

Closed-loop systema. before compensation;b. after ideal integralcompensation

This gain yields . Hence, the steady-state error is

14

Example12

• Adding an ideal integral compensator with a zero at - 0.1, we obtain the root locus shown in Figure. • Another section of the compensated root locus is between the origin and - 0.1.• Searching this region for the same gain at the dominant pair, K = 158.2 (the gain at the intersection with the line ) , the fourth

closed-loop pole is found at -0.0902.

Damping Ratio unchanged (with K = 158.2).Steady State Error is ZERO!.

Same transient responsePoles and gain are approximately the same

Desired response

15PI controller implementation

• The value of the zero can be adjusted by varying

• The error and the integral of the error are fed forward to the plant G(s).

• Made steady-state error zero.

• However, it is expensive to implement as the integrator requires active elements.

• We may want to use the solution presented next: Lag Compensation.

Example13

A method of implementing an ideal integral compensator is shown in the figure.

16

2.3 Lag Compensation• Similar to the Ideal Integrator, however it has a pole not on origin but close to the origin (fig c) due to the passive networks.• In order to keep the transient response unchanged, the compensator pole and zero must be close to each other near the origin.

Before compensation: The static error constant (old), , for the system is:

After compensation (New):

• Steady State Improvement:

• The effect on the transient response is negligible:

Root locus: a. before lag compensation; b. after lag compensation

If the lag compensator pole and zero are close together, the angular contribution of the compensator to point P is approximately zero degrees. point P is still at approximately the same location on the compensated root locus.

𝐾 𝑃 𝑜=lim𝑠→ 0

𝐺 (𝑠)=𝐾𝑧 1 𝑧 2…

𝑝1𝑝2 …

𝐾 𝑃 𝑁=𝑧 𝑐

𝑝𝑐

𝐾𝑧 1 𝑧2 …

𝑝1𝑝2 …=

𝑧𝑐

𝑝𝑐

𝐾 𝑃𝑜

The ratio must be large to improve the steady-state error

17

Example21

Compensate the system of Figure (a), whose root locus is shown in Figure (b), to improve the steady-state error by a factor of 10 if the system is operating with a damping ratio of 0.174.

SOLUTION

• The uncompensated system error from Example 1 was 0.108 with = 8.23. A tenfold improvement means a steady-state error of:

𝑒 (∞ )=0.10810

=0.0108 ,𝑠𝑖𝑛𝑐𝑒𝑒 (∞ )= 11+𝐾 𝑝

⇒𝐾 𝑝=91.59

• The improvement in from the uncompensated system to the compensatedsystem is the required ratio of the compensator zero to the compensator pole:

𝑧𝑐

𝑝𝑐

=𝐾 𝑝𝑁

𝐾 𝑃𝑜

=91.598.23

=11.13 Arbitrarily selecting 𝑝𝑐=0.01 𝑧𝑐=11.13𝑝𝑐 ≈ 0.111

• The compensated system

18

Example22

• Comparison of the Lag-Compensated and the Uncompensated Systems

Root locus for compensated system

On the ζ= 0.174 line the second-order dominant poles are at —0.678 ±j3.836 with gain K=158.1

The third and fourth closed-loop poles are at -11.55 and - 0.101 (on real axis)

The fourth pole of the compensated system cancels its zero.

19

• The transient response of both systems is approximately the same with reduced steady state error

Example23

Less Steady State Error 0.0108

the transient responses of the uncompensated and lag-compensated systems are the same

20

3.1 Ideal Derivative Compensation (PD)1

• The transient response of a system can be selected by choosing an appropriate closed-loop pole location on the s-plane.• If this point is on the root locus, then a simple gain adjustment is all that is required in order to meet the transient response specification.• If the closed loop root locus doesn’t go through the desired point, it needs to be reshaped (add poles and zeros in the forward

path).• One way to speed up the original system is to add a single zero to the forward path. 𝐺𝑐 (𝑠)=𝑠+𝑧𝑐

see how it affects by an example of a system operating with a damping ratio of 0.4:

For each compensated case, the dominant, second-order poles are farther out ( larger) along the 0.4 damping ratio line than the uncompensated system (same overshoot).closed-loop poles have more negative real parts than the uncompensated, dominant closed-loop poles, settling times for the compensated cases will be shorter than uncompensated case.

uncompensated compensated, zero at -2 compensated, zero at -3 compensated, zero at -4

21

3.1 Ideal Derivative Compensation (PD)2

Observations and facts:- In each case gain K is chosen such that percent overshoot is same.- Compensated poles have more negative real part (smaller settling time) and larger imaginary part (smaller peak time).- Zero placed farther from the dominant poles, compensated dominant poles move closer to the origin.

22

Example31

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

SOLUTION

the performance of the uncompensated system operating with 16% overshoot.• Root-Locus for the uncompensated system fig (b):

Fig (b) Root locus for uncompensated system

Fig (a) Feedback control system

16% Overshoot ζ = 0.504,along damping ratio line

Dominant second-order pair of poles

-1,205 ±j2.064.

• The gain of the dominant poles 𝑘=43.35

• Third pole at -7.59.

Settling time

𝑇 𝑠=4

ζ𝜔𝑛

=4

1.205=3.320 𝑠𝑒𝑐

Real part of the dominant poles

23

Example32

• We proceed to compensate the system1. location of the compensated system's dominant poles.

threefold reduction in the settling time𝑇 𝑠 (𝑛𝑒𝑤 )=𝑇 𝑠

3=1.107

real part of the compensated system's dominant, second-order pole

𝜎=4𝑇 𝑠

=4

1.107=3.613

2. design the location of the compensator zero

- The angle contribution of poles for the desired pole location(sum of angles of the desired pole to the open-loop poles):

- In order to achieve -180 the angle contribution of the placed zero should be 95.6.

- From the figure:

tan (180− 95.6 )=𝜔𝑑

𝜎 −𝜎 𝑧𝑐

= 6.1933.613−𝜎 𝑧𝑐

=¿ 𝜎 𝑧𝑐=3.006

Fig (b) Evaluating the location of the compensating zero

Fig (a) Compensated dominant pole superimposed over the uncompensated root locus

𝜔𝑑=3.613 𝑡𝑎𝑛 (180𝑜− 120.26  0 )=6.193

Imaginary part of the compensated system's dominant, second-order pole

on the line tan (𝛽 )=𝜔𝑑

𝜎

24

Example33

Root-Locus After CompensationImprovement in the transient response

Fig (b) Uncompensated and compensated system step responsesFig (a) Root locus for the compensated system

25

Example34

Implementation

Once the compensating zero located, the ideal derivative compensator used to improve the transient response is implemented with a proportional-plus-derivative (PD) controller.

The transfer function of the controller is:

equal the negative of the compensator zero,

loop-gain value.

Implementation of ideal differentiator is expensive. So we may use the next technique: Lead Compensation

Two drawbacks: - requires an active circuit to perform the differentiation.- differentiation is a noisy process.

Figure 1: PD controller

26

3.2 Lead Compensation1

• Passive element approximation of PD.• it has an additional pole far away on the real axis (angular contribution still positive to approximate the single an equivalent

single zero).• Advantage 1: Cheaper.• Advantage 2: Less noise amplification• Disadvantage: Doesn’t reduce the number of branches that cross the imaginary axis into the right-half-plane.

• The difference between 180° and the sum of the angles must be the angular contribution required of the compensator.

Basic Idea:

• For example, looking at the Figure, we see that:

Figure 1: Geometry of lead compensation

27

3.2 Lead Compensation2

• There are infinitely many choices of and providing same • The choice from infinite possibilities affects:

- Static Error Constants.- Required gain to reach the design point on the root locus.- Difficulty in justification a second order approximation.

28

Example41

Design three lead compensators for the system in Figure to reduce the settling time by a factor of 2 while maintaining 30% overshoot.

SOLUTION

• Characteristics of the uncompensated system operating at 30% overshoot

30% Overshoot ζ = 0.358,

along damping ratio line

Dominant second-order pair of poles

-1,007 ±j2.627.damping ratio

From pole's real part

𝑇 𝑠=4

1.007=3.972 s

settling time

• Design point

twofold reduction in settling time

𝑇 𝑠=3.972

2=1.986 s

Imaginary part of the desired pole location

−ζ 𝜔𝑛=− 4𝑇 𝑠

=− 2.014

real part of the desired pole location

𝜔𝑑=− 2.014 tan ( 110.980 )=5.252

29

• Lead compensator Design.Place the zero on real axis at -5 (arbitrarily as possible solution). sum the angles (this zero and uncompensated system's poles and zeros),

Example42

resulting angle

𝜃𝑝 𝑐=−1800+172.690=−7.310

the angular contribution required from the compensator pole

𝜃=− 172.690

Fig (a) 5-plane picture used to calculate the locationof the compensator pole

location of thecompensator pole

From geometry in fig(a)

5.252𝑝𝑐−2.014

=tan (7.310)compensator pole

𝑝𝑐=42.96

Fig (b) Compensated system root locusFig (c) Uncompensated system and lead compensation responses (zeros at a:-5, b:-4 c: -2)

approximation is not valid for case C

30

4 Improving Steady-State Error and Transient Response

• Combine the design techniques to obtain improvement in steady-state error and transient response independently.

- First improve the transient response.(PD or lead compensation).

- Then improve the steady-state response. (PI or lag compensation).

• Two Alternatives

- PID (Proportional-plus-Integral-plus-Derivative) (with Active Elements).

- Lag-Lead Compensator. (with Passive Elements).

31

4.1 PID Controller Design

• Transfer Function of the compensator (two zeros and one pole):

Fig (a) PID controller implementation

𝐺𝑐 (𝑠)=𝑘1+𝑘2

𝑠+𝑘3 𝑠=

𝑘1 𝑠+𝑘2+𝑘3 𝑠2

𝑠=

𝑘3(𝑠2+

𝑘1

𝑘3

𝑠+𝑘2

𝑘3

)

𝑠

• Design Procedure (Fig (a) )

1. From the requirements figure out the desired pole location to meet transient response specifications.

2. Design the PD controller to meet transient response specifications.

3. Check validity (all requirements have been met) of the design by simulation.

4. Design the PI controller to yield the required steady-state error.

5. Determine the gains, (Combine PD and PI).

6. Simulate the system to be sure all requirements have been met.

7. Redesign if simulation shows that requirements have not been met.

32

Example51

SOLUTION

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

Fig (a) Uncompensated feedback control system

• Evaluation of the uncompensated system

A third pole at -8.169

Third pole between - 8 and -10 for a gain equivalent to that at the dominant poles

• To reduce the peak time to two-thirds. (find the compensated system's dominant pole location)

The imaginary part𝜔𝑑=

𝜋𝑇𝑝

=𝜋

(23)(0 . 297)

=15 . 87

The real part 𝜎=𝜔𝑑

tan (117 .130)=−8 . 13

20% Overshoot 𝜁 =  0. 456  ( π − β=117 .130 ¿

Dominant second-order pair of poles

along damping ratio line

-5,415 ±j10.57 with gain of 121.5.

damping ratio

Peak time𝑇 𝑝=0 .297 𝑠𝑒𝑐𝑜𝑛𝑑𝑠=10.57

(

33

Example52

• Design of the compensator

Fig (a) Calculating the PD compensator zero

(sum of angles from the uncompensated system's poles and zeros to the desired compensated dominant pole is )

the required contributionfrom the compensator zero

198.370− 1800=18.370

From geometry in Fig(a)

compensating zero's location. 15.87

𝑧𝑐 −8.13=𝑡𝑎𝑛(18.37¿¿0)¿ 𝑧𝑐=55.92

the PD controller.

)

Fig (b) Root locus for PD-compensated system

gain at the design point

𝑘=5.34

• The PD-compensated system is simulated. Fig (b) (next slide) shows the reduction in peak time and the improvement in steady-state error over the uncompensated system. (step 3 and 4)

34Fig (b) Step responses for uncompensated, PD compensated, and PID compensated systems

Example53

• A PI controller is used to reduce the steady-state error to zero (for PI controller the zero is placed close to the origin)

PI controller is used as 𝐺𝑃𝐼 (𝑠)=

𝑠+0 .5𝑠

Fig (a) Root locus for PID-compensated system

Searching the 0.456 damping ratio line

−7 .516 ± 𝑗14 .67𝑤𝑖𝑡 h𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑𝑔𝑎𝑖𝑛𝑘=4 .6we find the dominant,

second-order poles

• Now we determine the gains (the PID parameters),

Matching: 𝑘1=256 .5 ,𝑘2=128 .6 ,𝑘3=4 . 6

35

4.2 Lag-Lead Compensator Design(Cheaper solution then PID)

• First design the lead compensator to improve the transient response. Next we design the lag compensator to meet the steady-state error requirement.

• Design procedure:

1. Determine the desired pole location based on specifications. (Evaluate the performance of the uncompensated system).

2. Design the lead compensator to meet the transient response specifications.(zero location, pole location, and the loop gain).

3. Evaluate the steady state performance of the lead compensated system to figure out required improvement.(simulation).

4. Design the lag compensator to satisfy the improvement in steady state performance.

5. Simulate the system to be sure all requirements have been met. (If not met redesign)

36

Example61

Design a lag-lead compensator for the system of Figure so that the system will operate with 20% overshoot and a twofold reduction in settling time. Further, the compensated system will exhibit a tenfold improvement in steady-state error for a ramp input.

Fig (a) Uncompensated systemSOLUTION

• Step 1: Evaluation of the uncompensated system

20% Overshoot ζ = 0.456 ,

along damping ratio line

Dominant second-order pair of poles

-1,794 ±j3.501 with gain of k=192.1.damping ratio

• Step 2 : Lead compensator design (selection of the location of the compensated system's dominant poles).

lead compensator design.

Twofold reduction of settling time

the real part of the dominant pole

−ζ 𝜔𝑛=− 2 (1 .794 )=−3 .588

the imaginary part of the dominant pole

𝜔𝑑=ζ 𝜔𝑛 tan (117 .130)=7 .003

Arbitrarily select a locationfor the lead compensator zero.

𝑧𝑐=−6

- compensator zero coincident with the open-loop pole to eliminate a zero and leave the lead-compensated system with three poles. (same number that the uncompensated system has)

37

Example62

Fig (a) Root locus for uncompensated system

Compensator pole.

- Finding the location of the compensator pole. - Sum the angles to the design point from the uncompensated system's poles

and zeros and the compensator zero and get -164.65°.- The difference between 180° and this quantity is the angular contribution

required from the compensator pole (—15.35°).- Using the geometry shown in Figure (b)

7.003𝑝𝑐− 3.588

=tan(15.35°) 𝑝𝑐=− 29.1

Fig (b) Evaluating the compensator pole

Fig (c) Root locus for lead-compensated system

- The complete root locus for the lead-compensated system is sketched in Figure (c)

38

Example63

• Steps 3 and 4: Check the design with a simulation. (The result for the lead compensated system is shown in Figure(b) next slide and is satisfactory.)

• Step 5: design the lag compensator to improve the steady-state error.uncompensated system's open-loop transfer function

𝐺 (𝑠)= 192.1𝑠 (𝑠+6 )(𝑠+10)

static error constant 𝑘𝑣=3.201

inversely proportional to the steady-state

error

𝐺𝐿𝐶 (𝑠 )= 1977𝑠 (𝑠+10 )(𝑠+29.1)

𝑘𝑣=6.794the addition of lead

compensation has improvedthe steady-state error by a

factor of 2.122

static error constant

Need of tenfold improvement

lag compensator factor for steady-state error

improvement

102.122

=4.713=𝑘0

Step 6: We arbitrarily choose the lag compensator pole at 0.01,lag compensator

Zero 𝑧𝑐=0.04713 𝐺𝐿𝑎𝑔 (𝑠)=(𝑠+0.04713)

(𝑠+0.01 )

lag compensator

lag-lead-compensatedOpen loop TF

𝐺𝐿𝐿𝐶 (𝑠)= 𝐾 (𝑠+0.04713)𝑠 ( 𝑠+10 )(𝑠+29.1) (𝑠+0.01 )

- The uncompensated system pole at - 6 canceled by the lead compensator zero at -6.- Drawing the complete root locus for the lag-lead-compensated system and by searching along the 0.456 damping ratio line

closed-loop dominant poles

𝑝𝐶𝐿=− 3.574± 𝑗6.976 with  a  gain   of   1971.

Lead Compensator: Eliminate the pole at -6 by and Add at -29.1

39

Example64

Fig (a) Root locus for lag-lead-compensated system Fig (b) Improvement in step response for lag-lead-compensated system

Step 7: The final proof of our designs is shown by the simulations of Figure (b)

- Fig (b) shows the complete draw of the lag-lead-compensated root locus. - The lag-lead compensation has indeed increased the speed of the system (settling time or the peak time).

40

6. Physical Realization of Compensation

Fig (a) Operational amplifier for transfer function realization

6.1 Active-Circuit Realization

• and are used as a building block to implement the compensators and controllers, such as PID controllers.

• The transfer function of an inverting operational amplifier

𝑉 0(𝑠 )𝑉 𝑖 (𝑠 )

=−𝑍2(𝑠)𝑍1(𝑠)

41Table 1 Active realization of controllers and compensators, using an operational amplifier

• Table1 summarizes the realization of PI, PD, and PID controllers as well as lag, lead, and lag-lead compensators using Operational amplifiers.

Fig (a) Lag-lead compensator implemented with operational amplifiers

• Fig (a) : lag-lead compensator can be formed by cascading the lag compensator with the lead compensator.

42

Example8Implement the PID controller of Example 5

SOLUTION

• The transfer function of the PID controller is𝐺𝑐 (𝑠)=4.6 (𝑠+55.92 ) (𝑠+0.5 )𝑠

• which can be put in the form 𝐺𝑐 (𝑠)=𝑠+56.42+27.96

𝑠

• Comparing the PID controller in Table 1 with this equation we obtain the following three relationships:

we arbitrarily select a practical value:

𝑅2

𝑅1

+𝐶1

𝐶2

=56.42 = 11

𝑅1 𝐶2

=1

𝐶2=0.1𝜇𝐹 𝑅1=357.65𝑘Ω ,𝑅2=178.891𝑘Ω𝑎𝑛𝑑𝐶1=5.59𝜇𝐹

• The complete circuit is shown in Figure (a) where the circuit element values have been rounded off.

Fig (a) PID controller

43

6.2 Passive-Circuit Realization

• Lag, lead, and lag-lead compensators can also be implemented with passive networks (Table 2) .

TABLE 2 Passive realization of compensators

44

Example9Realize the lead compensator designed in Example 4 (Compensator b zero at -4).

SOLUTION

• The transfer function of the lead compensator is 𝐺𝑐 (𝑠)= 𝑠+4𝑠+20.09

• Comparing the transfer function of a lead network shown in Table 2 with The equation, we obtain the following two relationships:

1𝑅1 𝐶

=4 𝑎𝑛𝑑1

𝑅1 𝐶+

1𝑅2 𝐶

=20.09

• Since there are three network elements and two equations, we may select one of the element values arbitrarily

𝑅1=250𝑘Ω𝑎𝑛𝑑𝑅2=62.2𝑘Ω𝐶=1𝜇 𝐹