FeedCon[Unit 6]

September 7, 2017 | Author: Dovie May | Category: Control Theory, Cybernetics, Applied Mathematics, Systems Theory, Physics & Mathematics

Short Description

FEEDBACK AND CONTROL SYSTEMS LECTURE 6...

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Feedback and Control Systems ANALYSIS AND DESIGN OF SYSTEMS VIA ROOT LOCUS

At the end of this chapter, the students shall be able to: 6.1. Define and state the properties of root locus of a closed-loop transfer function; 6.2. (a) Sketch the root locus of the closed loop poles of a system as the gain of the open-loop transfer function is varied, manually and with the aid of a computer; (b) interpret the root locus sketch and determine the value of the gain at the critical points of the root locus; 6.3. Design the gain of a negative unity feedback system to meet transient response specifications using root locus; 6.4. Sketch the root locus of a system as a closed-loop pole is being varied. 6.5. Design a cascade compensator to improve steady-state error, transient response and both with the aid of root locus; 6.6. Design feedback compensator to improve transient response; 6.7. Implement the physical realization of compensators.

6.1. Introduction to Root Locus Intended Learning Outcome: Define and state the properties of root locus of a closed-loop transfer function.

Root locus, a graphical presentation of the closed-loop poles as a system parameter is varied, is a powerful method of analysis and design for stability and transient response. The real power of the root locus lies in its ability to provide solutions for systems of order higher than 2. It can be used to describe qualitatively the performance of the system as various parameters are changed. Besides transient response, the root locus also gives graphical representation of a system’s stability. The ranges of stability, ranges of instability, and the conditions that cause a system to break into oscillation can be seen using the root locus graph.

The Control System Problem. A typical closed-loop feedback control system is shown below. Note that the poles of the open-loop transfer function can easily be evaluated but the poles of the closed-loop transfer function are not.

Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems

Figure 6. 1. Feedback control system and its equivalent closed-loop form.

The open-loop transfer function of 6.1a is KGsHs, which is simply a cascade of two transfer functions;

thus its poles can be easily determined. Furthermore, variations in K do not affect the location of the poles.  

On the other hand, one cannot determine the poles of Ts =    unless the denominator is factored out. Also, the poles of Ts varies with K. Since the location of the poles of Ts is a function of

K, the root-locus technique will give a vivid picture of how system parameters – transient response and stability – varies when the system gain is adjusted. Vector Representation of Complex Numbers. Any complex number σ + jω, described in Cartesian plane can be graphically represented by a vector as shown below. It can be described in polar form as M∠θ, where M is the magnitude while θ is the angle of the radial line from the positive real axis.

Figure 6. 2. Vector representation of complex numbers.

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Feedback and Control Systems If the complex number is substituted into a complex function, Fs, another complex number results. For example if Fs = s + a, then substituting the complex number s = σ + jω yields Fs = σ + a + jω, another complex number. This can be plotted on 6.2b. Translating the axis along the zero of Fs at

– a, the plot 6.2c terminates at s = σ + jω. Thus the alternate representation of 6.2b is 6.2c, where the vector originates from the zero of Fs and terminating at s = σ + jω. For a complicated function, the following examples explain the method.

Example 6.1 Evaluate Fs = Answer:

 

at the point s = −3 + j4.

Fs|

! "#

= 0.217∠ − 114.3°

Example 6.2   #

Evaluate the function Fs =  ! * at the point s = −7 + j9 Answer: 0.096∠ − 110.7° Defining the root locus. The root locus technique can be used to analyze and design the effect of loop gain upon system’s transient response and stability. For example, for a block diagram and its equivalent closed-loop form shown in figure 6.3a, the poles are computed for various values as shown in figure 6.3b and the pole plot and the root locus in figure 6.3c.

Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems

(a)

(b)

(c) Figure 6. 3. Deriving the root locus for a block diagram.

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Feedback and Control Systems From this, the root locus is defined as the representation of the paths of closed-loop poles as a parameter of the closed-loop transfer function (in this case, the gain) is varied. For most examples and problems, the discussion will be limited to positive gain K ≥ 0. For this system, many things can be said about the system’s transient response and stability as the gain is increased. The power of root locus analysis will be extended to higher-ordered system whose pole location cannot exactly establish a useful mathematical representation of the system transient response.

Properties of root locus. The properties of root locus can be derived from the closed-loop transfer function of figure 6.1. Referring to Ts =

KGs 1 + KGsHs

(6.1)

a pole, s exists when the characteristic polynomial in the denominator becomes zero, or KGsHs = −1 = 1∠2k + 1π; k = 0, ±1, ±2, …

Alternately, a value s is a pole if

(6.2)

|KGsHs| = 1

(6.3a)

∠KGsHs = 2k + 1π

(6.3b)

and Equations 6.3a and 6.3b implies that when a complex value s is substituted in the function KGsHs, a

complex number results. If the angle of the complex number is an odd multiple of π, that value of s is a system pole for some particular value of K. This value of K should satisfy 6.3a, or K=

1 |Gs||Hs|

(6.4)

The concepts are illustrated in the following examples. Example 6.3 For the unity feedback system shown below, find a value of s that is on the root locus and find the particular value of K at that value of s.

Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems

Example 6.4 Given a unity feedback system that has the forward transfer function Gs =

 

4 # !

, determine if the

point s = −3 + j0 is on the root-locus and if it is, find the value of K at that point. Answer: Yes, the point is on the root-locus with K = 10. 6.2. Sketching the Root Locus

Intended Learning Outcomes: (a) Sketch the root locus of the closed loop poles of a system as the gain of the open-loop transfer function is varied, manually and with the aid of a computer; (b) interpret the root locus sketch and determine the value of the gain at the critical points of the root locus.

The following five basic rules aid in sketching the root locus using minimal calculations. To refine the sketch, additional important points on the root locus can also be found. However, the use of computeraided methods will still be used in sketching the root locus. These rules are enumerated only for the purpose of understanding how the computer came up with the sketch. 1. Number of branches. The number of branches of the root locus equals the number of poles. 2. Symmetry.The root locus is symmetrical about the real axis. 3. Real axis segments.On the real axis, for K > 0, the root locus exists to the left of an odd number of real axis, finite open-loop poles and/or finite open-loop zeros. 4. Starting and ending points.The root locus begins at the finite and infinite poles of GsHs and ends at finite and infinite zeros of GsHs.

5. Behavior at infinity. The root locus approaches straight lines as asymptotes as the locus approaches infinity. The equation of asymptotes is given by the real-axis intercept σ6 and angle θ6 as follows Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems σ6 =

∑ 8inite poles − ∑ 8inite zeros #8inite poles − #8inite zeros

(6.5a)

θ6 =

2k + 1π #8inite poles − #8inite zeros

(6.5b)

where k = 0, ±1, ±2, ±3, … and the angle is given in radians with respect to the positive extension of the real-axis.

Example 6.5 Sketch the root locus for the system shown below.

Example 6.6 Sketch the root locus and its asymptotes for a unity feedback system that has the forward transfer function

Gs =

K s + 2s + 4s + 6

The basic rules in sketching root locus can be refined better. The following terms are considered in sketching the root locus.

Real-Axis Breakaway and Break-in Points. Root loci tend to break away from the real axis to the complex plane. At other times, the loci appear to return to the real axis as the pair of conjugate poles become real. Referring to figure 6.4, the root locus breaks away from the real axis towards the complex plane between -1 and -2 and breaks in between 3 and 5 to terminate at the zeros.

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Feedback and Control Systems

Figure 6. 4. An example of a root locus.

For breakaway and break-in points, the following statements hold: •

Breakaway points occur between poles while break-in points occur between zeros.

At the breakaway or break-in point, the branches of the root locus form an angle of π/n with the real axis, where n is the number of closed-loop poles arriving at or departing from the single breakaway or break-in point on the real axis.

At the breakaway point, gain is at relative maximum along the real axis. At the break-in point, gain is at relative minimum along the real axis.

Figure 6. 5. Variation of gain along the real axis.

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Feedback and Control Systems Since the breakaway and break-in points maximizes and minimizes the gain, respectively, these points can be found by finding an expression for K in terms of the real part of s , making the real part of s a part of the root locus, and then differentiating. Example 6.7 Find the breakaway and break-in points for the root locus of figure 6.4.

Answer: Breakaway point, σ = −1.45; break-in point, σ = 3.82 Another method, called the transition method, can also be used. Breakaway and break-in poles satisfy the relationship I

J

1 1 G =G σ + zH σ + pH

(6.6)

wherezH and pH are the negative of the zeros and pole values respectively of GsHs. Example 6.8 Repeat example 6.7 using the relationship in Equation 6.6 The KL-crossings. To find at what point the root locus will cross the jω-axis, a Routh table will be generated, where an entire rows will be forced to zero.

Example 6.9 For the system shown below, find the gain and the point at which the root locus crosses the imaginary axis.

Answer: K = 9.65; s = ±j1.59 Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems Angles of departure and arrival. In order to sketch the root locus more accurately, the angle of departure from the open-loop poles and the angle of arrival to the open-loop zeros can be calculated. Based on figure 6.6, these angles can be calculated as shown.

Example 6.10 Given a unity feedback system shown below, find the angle of departure from the complex poles and roughly sketch the root locus.

Plotting and calibrating the root locus. Once the root locus is sketched using the rules previously described, one may want to accurately locate points on the root locus, as well as find their associated gains. For example it is required to find the exact coordinates of the root locus as it crosses a radial line representing 20% overshoot. The following examples clarify this point.

Figure 6. 6. Calculation of angles of departure and arrival.

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Feedback and Control Systems Example 6.11 For a unity feedback system whose forward transfer function is Gs =

 !

   #

, find a point on

the radial line that corresponds to a damping ratio of 0.45 and find the gain K at this point. Example 6.12 Given a unity feedback system that has the forward transfer function Gs =

 

4 # !

, do the following:

(a) Sketch the root locus. (b) Find the imaginary-axis crossing. (c) Find the gain, K, at the jω-axis crossing. (d) Find the break-in point. (e) Find the angle of departure from the complex poles.

Answers: (a) Using MATLAB, the root locus is shown below Root Locus 5

4

3

2

Imaginary Axis

1

0

-1

-2

-3

-4

-5 -8

-6

-4

-2

0

2

4

Real Axis

(b) s = ±j√21

(c) K = 4

(d) Break-in point = −7

(e) Angle of departure = −233.1° Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems As stated before, computer-aided methods of plotting and analyzing root locus of various systems will be employed for more accuracy. The MATLAB command rlocus can be used for such purpose. Example 6.13 Sketch the root locus for the system shown below and find the following: (a) The exact point and gain where the locus crosses the 0.45 damping ratio line. (b) The exact point and gain where the locus crosses the jω-axis. (c) The breakaway point on the real axis. (d) The range of K for which the system is stable.

Answers: The root locus is sketched in MATLAB using the following commands:

The root locus is shown below: Root Locus 5 4

Imaginary Axis (seconds-1)

3 2 1 0 -1 -2 -3 -4 -5 -5

-4

-3

-2

-1

0

1

2

3

Real Axis (seconds -1)

(a) To find the point and the gain for which the root locus crosses the dO = 0.45 line, the following Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems MATLAB commands are used:

resulting in a line to be imposed in the root locus showing the line corresponding to dO = 0.45 as shown below Root Locus 5 0.45 4

Imaginary Axis (seconds-1)

3 2 1 0 -1 -2 -3 -4 0.45 -5 -5

-4

-3

-2

-1

0

1

2

3

Real Axis (seconds -1)

Zoom in first into the intersection of the line and the point for a more accurate reading, then after type in the following commands:

which prompts the user to select a point on the root locus. Point to the intersection as accurately as possible, like the one shown in the figure below.

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Feedback and Control Systems

Root Locus 0.45 3.6 3.4

Imaginary Axis (seconds-1)

3.2 3 2.8 2.6 2.4 2.2 2 1.8 -2

-1.8

-1.6

-1.4

-1.2

-1

-0.8

-0.6

Real Axis (seconds -1)

then read in the command window the gain and the pole location corresponding to the intersection of dO = 0.45 line and the root locus. Thus the gain of the system at dO = 0.45 is K = 0.417 and the location of the poles are at s , = 3.4∠116.7°.

(b) The above steps can be repeated to determine the location of the pole at the jω-axis. Thus, at the jωaxis crossing, the gain of the system is K = 1.5 and the intersection is at s , = ±j3.9.

(c) The above steps can be repeated to determine the breakaway point at the real axis. Thus, the breakaway point is found at σ = −2.88.

(d) From (b), the frequency of oscillation is ωJ = 3.9.

Example 6.14   # Given the unity feedback system that has the forward transfer function Gs = 4 * P , do the following: (a) Sketch the root locus. (b) Find the imaginary-axis crossing using MATLAB and Routh table. (c) Find the gain, K at the jω-axis crossing using MATLAB and Routh table. (d) Find the break-in point, using MATLAB and hand calculations. (e) Find the point where the locus crosses the 0.5 damping ratio line using MATLAB.

(f) Find the gain at the point where the locus crosses the 0.5 damping ratio line using MATLAB.

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Feedback and Control Systems (g) Find the range of gain K for which the system is stable. Answers: (a) Root Locus 5 4

Imaginary Axis (seconds-1)

3 2 1 0 -1 -2 -3 -4 -5 -4

-3

-2

-1

0

1

2

3

4

5

Real Axis (seconds -1)

(b) s , = ±j4.06 (c) K = 1

(d) σHJ = +2.89

(e) s = −2.42 + j4.18 (f) K = 0.108 (g) K < 1

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Feedback and Control Systems Drill Problems 6.1 1. For each of the root loci shown, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons.

2. Sketch the general shape of the root locus for each of the open-loop pole-zero plots shown below.

3. Sketch the root locus for the unity feedback system shown below where Gs is

(a) Gs =

  *

(b) Gs =

S 4 #T

4 R P 4

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Feedback and Control Systems (c) Gs =

S 4 T 4

(d) Gs =  U  # Check your sketch using MATLAB.

4. For the open-loop pole-zero plot shown below, sketch the root locus and find the break-in and breakaway points, as well as the angle of departure from the complex poles. (a)

(b) There are two possibilities for the sketch below. Sketch each of the two possibilities,

5. For each of the systems below, use MATLAB to sketch the root locus and find the following, checking the computer generated results using hand calculations:

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Feedback and Control Systems

(a) The breakaway and break-in points. (b) The range of K to keep the system stable.

(c) The value of K that yields a stable system with critically damped second-order poles.

(d) The value of K that yields a stable system with a pair of second-order poles that have a damping ratio of 0.707.

6. For the unity feedback system whose open-loop transfer function is Gs =  * V , roughly sketch the root locus and find all the critical points of the root locus.

7. Given the unity feedback system whose open-loop transfer functions are given in (a) and (b) below respectively, plot the root locus and find the breakaway points, the jω-axis crossing, and the range of gain for stability in each case. Find the angle of arrival for (a). (a) Gs =

S 4  T

(b) Gs =

  

  

  

 

8. Given the unity feedback system whose open-loop transfer function Gs =   ! #, do the following: (a) Sketch the root locus. (b) Find the asymptotes.

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Feedback and Control Systems (c) Find the value of gain that will make the system marginally stable. (d) Find the value of gain for which the closed-loop transfer function will have a pole on the real axis at −0.5.  W

9. For the unity feedback system whose open-loop transfer function is Gs =  ! * find the values of K and α that will yield a second-order closed-loop pair of poles at −1 ± j100.

10. For the unity feedback system whose open-loop transfer function is given as Gs =

    

,

sketch the root locus and find the following: (a) The breakaway and break-in points. (b) The jω-axis crossing. (c) The range of gain to keep the system stable. (d) The value of K to yield a stable system with second-order complex poles, with a damping ratio of 0.5

6.3 Transient Response Design via Gain Adjustment Intended Learning Outcome: Design the gain of a negative unity feedback system to meet transient response specifications using root locus.

After sketching the root locus, the transient response of the system can be designed with adjustments in gain. However, if the closed-loop transfer function contains third- or higher-ordered poles and contains zeros, a second-order approximation must be justified first. The conditions are: 1. Higher-ordered poles are much farther into the left half of the s-plane than the dominant secondorder pair of poles. 2. Closed-loop zeros near the closed-loop second-order pole pair are nearly canceled by the close proximity of higher-order closed-loop poles. 3. Closed-loop zeros not canceled by the close proximity of higher-ordered closed-loop poles are far removed from the closed-loop second-order pole pair.

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Feedback and Control Systems Thus, to design the transient response of higher-ordered systems via gain adjustment, the following should be done: 1. Sketch the root locus for the given system. 2. Assume the system is a second-order system without any zeros and then find the gain to meet the transient response specifications. 3. Justify the second-order approximation using the conditions stated above. 4. If the assumptions cannot be justified, the solution must be simulated. It is a good idea to simulate all solutions anyway.

Example 6.15 Design the value of the gain K for the system below to yield a second-order response with an overshoot of 1.52%.

Answer: The table below summarizes the information gathered from the sketch of the root locus using MATLAB

The first and second cases shows that, although the third pole is far from the dominant second-order closed-loop pole, the closed-loop zero cannot be cancelled; hence, a second-order approximation is not valid. The third case shows that the third closed-loop pole and the closed-loop zero are near each other, hence a pole-zero cancellation can be applied, and a second-order approximation is valid. To show these, the system with the provided gain was simulated in LabVIEW.

Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems

Case 1. K = 7.3021

Case 2. K = 12.834

Case 3. K = 39.6103

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Feedback and Control Systems Example 6.16

Given a unity feedback system that has the forward-path transfer function Gs =   # *, do the following: (a) Sketch the root locus. (b) Using a second-order approximation, design the value of K to yield 10% overshoot for a unit step input. (c) Estimate settling time, peak time, rise time, and steady-state error for the value of K designed in (b). (d) Determine the validity of the second-order approximation.

Answers: (a) Root Locus 10 8

-1

Imaginary Axis (seconds )

6 4 2 0 -2 -4 -6 -8 -10 -16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

(b) K = 45.55

(c) T = 1.97 sec, T[ = 1.13 sec, TO = 0.53 sec, and e \][ ∞ = 0.51 (d) Second-order approximation is not valid.

6.4 Generalized Root Locus Intended Learning Outcome: Sketch the root locus of a system as a closed-loop pole is being varied. In the previous discussions, the root locus has always been drawn as a function of gain K. In this section, the root locus is to be sketch as a function of one of the open-loop pole. To do this, the open-loop transfer function must be expressed as an equivalent system with the open-loop pole appearing as the forward-path gain, that is expressing the closed-loop transfer function has a denominator of the form 1 + p GsHs.

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Feedback and Control Systems Example 6.17 Sketch the root locus of the system shown below as a function of the open-loop pole p .

Answer: The system must be expressed as an equivalent closed-loop system having the open-loop pole appear as the forward path gain, as in Ts =

_

4  _ [`  

1 + 4  _

With this, the open-loop transfer function is KGsHs =

p s + 2 s + 2s + 10

and the root locus is sketched as follows:

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Feedback and Control Systems Example 6.18 Sketch the root locus for variations in the value of p , for a unity feedback system that has the following forward transfer function Gs =

100 ss + p 

Imaginary Axis (seconds-1)

10

5

0

-5

-10

-15 -18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

Drill Problems 6.2  _ _

1. For the unity feedback system shown below, where Gs =  !_ 4

_ __

, do the following:

(a) Sketch the root locus. (b) Find the range of gain K, that makes the system stable.

(c) Find the value of K that yields a damping ratio of 0.707 for the system’s closed-loop dominant poles. (d) Find the value of K that yields closed-loop critically damped dominant poles.

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Feedback and Control Systems 2. For the system shown below in (a), sketch the root locus and find the following:

(a) Asymptotes (b) Breakaway points (c) The range of K for stability

(d) The value of K to yield a 0.7 damping ratio for the dominant second-order pair.

To improve stability, it is desired that the root locus to cross the jω-axis at j5.5. To accomplish this, the open-loop function is cascaded with a zero, as shown in the above figure in (b). (e) Find the value of α and sketch the new root locus. (f) Repeat part (c) for the new locus. (g) Compare the results of part (c) and part (f). How did the transient response improved?

3. For the unity feedback system shown in number (1) of this drill problem exercise where Gs =

Ks − 2s + 2 s + 2s + 4s + 5s + 6

do the following: (a) Sketch the root locus. (b) Find the asymptotes. (c) Find the range of gain, K, that makes the system stable. (d) Find the breakaway points. (e) Find the value of K that yields a closed-loop step response with 25% overshoot.

(f) Find the location of higher-order closed-loop poles when the system is operating with 25% overshoot. (g) Discuss the validity of the second-order approximation.

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Feedback and Control Systems 4. The unity feedback system shown in number (1) of this exercise where Gs =

  !  

is to be

designed for minimum damping ratio. Find the following: (a) The value of K that will yield minimum damping ratio. (b) The estimated percent overshoot for that case. (c) The estimated settling time and peak time for that case. (d) Discuss the justification for a second-order approximation. (e) The expected steady-state error for a unit ramp input for the case of minimum damping ratio.

5. For the unity feedback system of number (1) of this exercise where Gs =

s 

K s + 6 + 10s + 26s + 1 s + α

design K and α so that the dominant complex poles of the closed-loop function have a damping ratio of V

0.45 and a natural frequency of rad/sec. R 6. For the following system shown in the figure below, do the following:

(a) Sketch the root locus. (b) Find the jω-axis crossing and the gain K at the crossing. (c) Find the real-axis breakaway point. (d) Find the angles of arrival to the complex zeros. (e) Find the closed-loop zeros. (f) Find the gain K for a closed-loop step response with 30% overshoot. (g) Discuss the validity of the second-order approximation.

7. Sketch the root locus for the system shown below and find the following: (a) The range of gain to yield stability. (b) The value of gain that will yield a damping ratio of 0.707 for the system’s dominant poles. Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems (c) The value of gain that will yield closed-loop poles that are critically damped.

8. Sketch the root locus as the open-loop pole α is valid for the system shown below.

6.5 Improving Steady-state Error and Transient Response via Cascade Compensation Intended Learning Outcome: Design a cascade compensator to improve steady-state error, transient response and both with the aid of root locus.

Improving transient response. To improve the transient response of the system, the designer can select the point that is outside of the root locus. For example, if the designer seeks to speed up the settling time without changing the overshoot, then, as shown in the root locus of figure 6.7, the closed-loop pole must be moved from point A to point B, which is not on the root locus. Notice that point B is not in the original root locus.

Figure 6. 7. Designing the settling time with the same overshoot.

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Feedback and Control Systems To do this, the designer can compensate the system with additional poles and zeros, so that the compensated system will have a root locus that goes through the desired pole location for some value of gain. The advantage of compensation is that the additional poles and zeros can be added at the low-power end of the system before the plant. The disadvantage is that it increases the system order and adds additional zeros which can affect the desired response. At the beginning, the designer can determine the position of the additional open-loop poles and zeros, but the location of the higher-ordered poles and zeros after placing the additional open-loop poles and zeros are not known after the design is complete. Thus, simulation is needed in order for the designer to determine if the desired response is indeed met.

One method of compensating for the transient response is to insert a differentiator in the forward path in parallel with the gain. The differentiator works for position control system, as an example, as follows: for a step input, the error in position becomes very large. Since differentiating the function means taking the slope of the function at a particular point, at the onset of the change in position, this slope becomes very large at the start of the change in position. Thus the output of the differentiator is also large, which drives the plant faster. As the plant reaches the desired position, the error becomes smaller making the slope approach zero. At this point, the output of the differentiator is now negligible compared to the pure gain.

Improving steady-state error. The steady-state error can also be independently improved using compensators. It can be shown that increasing the forward gain will decrease the steady-state error, but will increase the overshoot and vise versa. Dynamic compensators can be used to improve both simultaneously.

To improve steady-state error, the system type must be increased so that the associated error will be driven to zero. Thus, an open-loop pole at the origin may be added, and an integrator is used for its realization.

Configurations. There are two configurations of compensators to be discussed in this chapter: the cascade compensator in this section and the feedback compensator in the next section. The configurations are shown in figure 6.8. Both methods change the location of the open-loop poles and zeros, and thus this creates a new root locus that goes through the desired closed-loop pole location.

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Feedback and Control Systems

Figure 6. 8. Configurations of compensators: (a) cascade; (b) feedback.

Compensators. Compensators can be active or passive. Active compensators, or ideal compensators use active devices, such as op-amps, in its realization, as well as additional power sources. Passive compensators on the other hand use only passive components such as resistors or inductors when implemented. Active compensators offer the advantage of providing pure integration or differentiation to the open-loop transfer function of the uncompensated system, thus achieving a more superior performance than passive compensators. For example, an ideal integral compensator can reduce the steady-state error to zero, because it provides pure integration of errors. Passive compensators, on the other hand, are less expensive and these does not require additional power source, making the realization less complicated. Improving steady-state error via cascade compensation. In this section, the objective is to reduce steady-state error of a feedback system without appreciably changing the transient response. Two techniques will be discussed: •

Ideal integral compensation, which will be also called proportional-plus-integral (PI) control, uses active devices to provide pure integration and drive the steady-state error to zero.

Lag compensation, which uses passive components and does not employ pure integration.

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Feedback and Control Systems Ideal integral compensation (PI control). Steady-state error can be improved by placing an open-loop pole at the origin to increase the system type. To improve steady-state error without changing the transient response using PI control, a pole at the origin and a zero close to the origin can be added to the system. Figure 6.9 shows the implementation of a PI controller.

Figure 6. 9. Block diagram implementation of PI control.

The PI controller has the following transfer function 4

K  K bs + ` c Ga s = K + = s s

(6.7)

The location of the zero near the origin is controlled by the gain K  /K . The effect of the PI control on the steady-state error of the system is investigated in the next section. Example 6.19 Investigate the effect of the compensation to the system shown in (a). The compensator design is shown in (b). The uncompensated system operates with the dominant pole damping ratio of 0.174.

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Feedback and Control Systems Answer: Using MATLAB to sketch the root locus of the uncompensated in (a) and finding the closed-loop poles intersecting the dO = 0.174 line, it is found that the dominant complex poles are in −0.694 ± j3.926.

The gain at this point is K = 164.55. To justify second-order approximation, the third pole is found at −11.6129, which qualifies for the five-times rule of thumb.

Root Locus 10 0.174 8

4

-1

Im aginaryAxis(seconds )

6

2 0 -2 -4 -6 -8 0.174 -10 -16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

At this point, the steady-state error is evaluated as e∞ =

1 = 0.108 1 + K[

The step response of the system is shown below. Notice the presence of error, which is around 10.8% of the final value of the input which is unity. Step Response 1.4

1.2

Amplitude

1

0.8

0.6

0.4

0.2

0

0

1

2

3

4

5

6

7

8

9

Time (seconds)

With the compensated system in (b), the root locus is redrawn, and the above steps are repeated. In this Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems case, the dominant complex poles are found at −0.678 ± j3.836, the third order pole far at −11.554 and a fourth-order pole at −0.090, which is close to a zero at −0.1. Thus the second-order approximation is justified with the third-order pole at least five times apart from the dominant complex poles and the fourthorder pole canceled by a zero near it. The gain at this point is K = 158.11. Thus, the time response of the characteristics is approximately the same that of the uncompensated system. Root Locus 10 0.174 8

-1

Imaginary Axis (seconds )

6 4 2 0 -2 -4 -6 -8 0.174 -10 -16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

The time responses of the uncompensated and the compensated system are shown below for comparison. Notice that the compensated system will now respond with a zero steady-state error for a step input, since, because of compensation, the system type has increased (from type 0 to type 1). Step Response 1.4 compensated s ystem uncompensated sy stem 1.2

Am plitude

1

0.8

0.6

0.4

0.2

0

0

5

10

15

20

25

30

35

Time (sec onds )

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Feedback and Control Systems The performances of the uncompensated and compensated systems are summarized in the table below.

Lag compensation. With the use of passive circuits, an improvement in the steady-state error can be introduced. For the figure of 6.10, the transfer function of a lag compensator is shown in (b). With this, the improvement in the static error constant is za /pa where za and pa are the zeros and poles of the lag compensator.

Figure 6. 10. The uncompensated system in (a) is lag compensated in (b).

In order to improve the steady-state error of the system, a pole can be placed near unity and a zero near this pole. An improvement can be achieved using this pole-zero placement and at the same time the time response will not change appreciably much. Example 6.20 Design a lag compensator for the system shown below to achieve a ten-fold improvement in the steadystate error. The system has dominant complex poles whose dO = 0.174.

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Feedback and Control Systems Answer: This is the same as the previous example, with the steady-state error at 0.108 and K [ = 8.23. A ten-fold

improvement in the steady-state error means that the compensated system will have 0.0108 error and a K [ = 91.59 .The ratio of the compensated to the uncompensated K [ is the required ratio of the compensator zero to compensator pole za /pa , or

za 91.59 = = 11.129 8.23 pa

Arbitrarily selecting a pole close to unity in the left half plane at pa = −0.01, compensator zero is calculated at za = 11.129−0.01 = −0.111 Thus, the compensated system is shown in the figure below.

The root locus of the compensated system is shown below. Note that the second-order approximation is justified and the gain at this point is K = 158.051. Root Locus 10 0.174 8

Imaginary Axis (seconds-1)

6 4 2 0 -2 -4 -6 -8 0.174 -10 -16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

The step responses of the uncompensated and compensated system are shown in the plot below for comparison. Note the improvement in the steady-state error performance of the system and the relatively

Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems the same time response performance of the system. Step Response 1.4 uncompensated system compensated system 1.2

Amplitude

1

0.8

0.6

0.4

0.2

0

0

5

10

15

20

25

Time (seconds)

The compensated system now has a K [ = 87.718 and a steady-state error of 0.0113 , a 9.818 improvement in the error performance of the system. The table below summarizes the findings in this example.

Example 6.21

A unity feedback system with the forward transfer function Gs =  d is operating with a closed-loop step response that has 15% overshoot. Do the following: (a) Evaluate the steady-state error for a unit ramp input. (b) Design a lag compensator to improve the steady-state error by a factor of 20. (c) Evaluate the steady-state error for a unit ramp input to your compensated system. (d) Evaluate how much improvement in steady-state error was realized.

Improving transient response via cascade compensation. In designing the transient response of the system, the typical objective is to produce a response that has a desirable overshoot and a shorter settling time than the uncompensated system. Two ways will be discussed:

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Feedback and Control Systems •

Ideal derivative compensation, which will also be called proportional-plus-derivative (PD) control, adds a zero to the forward-path transfer function and is implemented using active networks. However, differentiation is a noisy process.

Lead compensation, which approximates differentiation with passive networks by adding a zero and a more distant pole to the forward transfer function.

Ideal derivative compensation (PD control). One way to speed up the original system is to add a single zero to the forward path. This zero can be represented by the controller transfer function

Ga s = K + K  s = K  es +

K f K

(6.8)

where the ratio K /K  is chosen to be the negative of the compensator zero. Figure 6.11 shows the implementation of a PD controller.

Figure 6. 11. PD controller implementation.

Example 6.22

A unity feedback system has the forward path transfer function Gs =    P. Compare the time response characteristics of this uncompensated system with the performance of the compensated system when the compensating zeros are added at −2, −3 and −4. Answer: The table below shows the performance of the uncompensated system and the compensated systems with different compensator zero locations.

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Feedback and Control Systems

The following shows the response of uncompensated and the compensated systems, showing in an improvement in transient response and in this case, steady-state error.

Example 6.23 Given the system shown below, design an ideal derivative compensator to yield a 16% overshoot, with a threefold reduction in settling time.

Answer: Evaluating first the performance of the uncompensated system, the root locus is sketched first, and the

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Feedback and Control Systems performance is evaluated at 16% overshoot, which corresponds to dO = 0.5039. The root locus is

shown below with the three closed-loop poles corresponding to the 16% overshoot marked. The gain at this point is K = 43.39.

Root Locus 10 0.504 8

Imaginary Axis (seconds-1)

6 4 2 0 -2 -4 -6 -8 0.504 -10 -16

-14

-12

-10

-8

-6

-4

-2

0

2

4

Real Axis (seconds -1)

Note that a second-order approximation is justified since the third-order pole at −7.592 is at least five

times apart from the dominant complex poles at −1.204 ± j2.065. The performance of the system, as simulated in LabVIEW, is summarized below.

Proceeding with the design of the compensator, since the settling time of the uncompensated system is T = 3.478 sec, the settling time of the compensated system with three-fold improvement will be T = 1.159. Therefore, the real part of the compensated closed-loop dominant complex poles would be σg =

4 4 = = 3.450 T 1.159

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Feedback and Control Systems and the imaginary part is computed as ωg = 3.450 tancos dr = 5.917 From here, the compensator zero is to be located. It can be located by determining the angular contribution of the compensator zero to the compensated closed-loop dominant complex poles. It will be located at σa = −3.283. The plot of the new root locus, showing the compensated operating points and the summary of the performance of the compensated system is shown. The performance of the uncompensated system is again shown for comparison. Root Locus 25 20 15 0.504

Im aginaryAxis

10 5 0 -5 -10 0.504 -15 -20 -25 -7

-6

-5

-4

-3

-2

-1

0

1

Real Axis

Notice that the settling time is reduced by a factor of 2.93, while the percent overshoot is slightly reduced

to 13.4%. Notice also that the steady-state error is reduced. However, this is not the general case. The step response of the uncompensated and the compensated systems are shown below for comparison.

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Feedback and Control Systems

Lead compensation. Active differentiation can be approximated by a passive network. This compensator, called the lead compensator, introduces a zero and a pole. However, the pole can be placed father away from the imaginary axis so that the angular contribution of the compensator to the point of interest is still positive and thus approximates differentiation by the addition of the zero.

Lead compensators offer the advantage of having less complex circuitry in its realization as well as a less noisy process compared to PD compensators. A design example is shown next. Example 6.24 Design three lead compensators for the system shown below. It is desired that the settling time improve by a factor of 2 while maintaining a 30% overshoot.

Answer: The root locus of the system and the uncompensated system characteristics are shown in the figure and table below, respectively. The performance of the uncompensated system was determined using LabVIEW. From here, the desired design point is to be determined.

Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems Root Locus 10 0.358 8

-1

Imaginary Axis (seconds )

6 4 2 0 -2 -4 -6 -8 0.358 -10 -16

-14

-12

-10

-8

-6

-4

-2

0

2

4

Real Axis (seconds -1)

A two-fold reduction for the settling time requires that the compensated system shall have T =

1.991 sec. From this, the desired operating point is set at −2.0090 ± j5.2415. Arbitrarily putting

compensating zeros at −5 , −4 and −2 and designing the required pole location, the root loci, the

performance of the compensated systems, and the step response of the compensated systems are shown below.

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Feedback and Control Systems Root Locus 6 0.358

-1

Imaginary Axis (seconds )

4

2

0

-2

-4

0.358 -6 -45

-40

-35

-30

-25

-20

-15

-10

-5

0

5

-1

Real Axis (seconds )

Root locus for a compensator zero at −5 and compensator pole at −42.3443 Root Locus

Imaginary Axis (seconds-1)

10

0.358

5

0

-5

0.358 -10 -25

-20

-15

-10

-5

0

5

Real Axis (seconds-1)

Root locus for a compensator zero at −4 and compensator pole at −19.9158 Root Locus 10 0.358 8

Imaginary Axis (seconds-1)

6 4 2 0 -2 -4 -6 -8 0.358 -10 -18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

Real A xis (seconds -1)

Root locus for a compensator zero at −2 and compensator pole at −19.9158

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Feedback and Control Systems

Step Response 1.4 Uncompensated Response Compensator 1 1.2

Compensator 2 Compensator 3

Amplitude

1

0.8

0.6

0.4

0.2

0

0

1

2

3

4

5

6

Time (seconds)

Plot of step responses of the uncompensated and the compensated systems.

Example 6.25

A unity feedback system with the forward transfer function Gs =  d is operating with a closed-loop step response that has 15% overshoot. Do the following: (a) Evaluate the settling time. (b) Design a lead compensator to decrease the settling time by three times. Choose the zero compensator at −10.

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Feedback and Control Systems Improving steady-state error and transient response. There can be two ways to do this. The system can be designed for transient response and then improve the steady-state error; or the steady-state error designed first before improving the transient response. One drawback of these methods is the possibility that the parameter designed first might deteriorate when the second parameter is improved. Overdesigning can address this problem, if overdesign does not result to other design problems.

In this section, the approach is to design the transient response first, followed by the design in the steadystate error. The design can use either passive or active components. These controllers are: •

Proportional-integral-derivative (PID) controller, which is a combination of an active PD and an active PI controller; and

Lag-lead compensator, which is a combination of a passive lead controller, followed by a lag compensator.

PID controller. The transfer function of a PID controller is given as

K ! bs + ` s + 4 c K U U Ga s = K + + K! s = s s

(6.9)

which has two zeros plus a pole at the origin. Its implementation is shown in figure 6.11.

Figure 6. 12. Block diagram implementation of a PID controlled unity feedback system.

One zero and the pole of this controller can be used to improve the steady-state error, while the other zero is used to improve the transient response.

The following are the steps in designing the PID controller. 1. Evaluate the performance of the uncompensated system to determine how much improvement in the transient response is required. Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems 2. Design the PD controller to meet the transient response specifications. The design includes the zero location and the loop gain. 3. Simulate the system to be sure all requirements have been met. 4. Redesign if the simulation shows that requirements have not been met. 5. Design the PI controller to yield the required steady-state error. 6. Determine the gains K , K  and K ! in figure 6.12. 7. Simulate the design to be sure all requirements are met. 8. Redesign if the simulation shows that requirements have not been met. Example 6.26 Given the system shown below, design a PID controller so that the system can operate with a peak time that is two-thirds that of the uncompensated system at 20% overshoot and with zero steady-state error for a step input.

Answer: Step 1: Evaluate the performance of the uncompensated system to determine how much improvement in the transient response is required. The root locus of the system is sketched and the performance of the system is evaluated. The performance of the uncompensated system is summarized in a table below. Root Locus 40

30 0.456

Im aginaryAxis

20

10

0

-10

-20 0.456 -30

-40 -12

-10

-8

-6

-4

-2

0

2

Real Axis

Root locus of the uncompensated system Analysis and Design of Systems Via Root Locus

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Feedback and Control Systems

Step 2: Design the PD controller to meet the transient response specifications. The design includes the zero location and the loop gain. The location of the new operating point that corresponds to the two-thirds of the peak time of the uncompensated system at 20% overshoot is first located. This location is at −7.9743 ± j15.5679. The compensator zero of the PD controller is then to be added. Determining its location using the angular contributions of the existing poles and zeros as well as the compensator zero angles to this new operating point, the compensator zero is found at za PD = −56.9663. The PD controller now has the transfer function Gjk s = s + 56.9663 Steps 3 and 4: Simulate the system to be sure all requirements have been met; Redesign if the simulation shows that requirements have not been met. Sketching the new root locus with the PD controller in place, the performance of the PD compensated system is compared to the performance of the uncompensated system.

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Feedback and Control Systems Root Locus 50

0.456

40 30

Imaginary Axis

20 10 0 -10 -20 -30 -40 -50

0.456 -90

-80

-70

-60

-50

-40

-30

-20

-10

0

10

Real Axis

Root locus of the uncompensated system

Note that there is a two-thirds improvement in the compensated peak time as well as an improvement in the error of the system.

Step 5: Design the PI controller to yield the required steady-state error. To design the PI controller, a compensator zero close to the origin can be chosen. Any location will do, as long as it is near the origin. Choosing the compensator zero to be at za PI = −0.5, the transfer function of the PI controller is Gjm s =

Analysis and Design of Systems Via Root Locus

s + 0.5 s

Page 47

Feedback and Control Systems Step 6: Determine the gains K , K  and K ! in figure 6.12. The transfer function of the designed PID controller is Gjmk s =

Ks + 56.9663s + 0.5 s

where K is the gain to be determined from the root locus of the PID compensated system. Sketching the

root locus and determining the gain, it is found to be K = 4.3749. The root locus is shown below. Root Locus 0.456 40 30

Imaginary Axis

20 10 0 -10 -20 -30 -40 0.456

-50 -30

-25

-20

-15

-10

-5

0

Real Axis

This gives Gjmk s =

4.3749s + 56.9663s + 0.5 4.3749s + 57.4663s + 28.4831 = s s

or K = 251.4093, K  = 124.6107 and K ! = 4.3749.

Steps 7 and 8: Simulate the design to be sure all requirements are met; Redesign if the simulation shows that requirements have not been met. The performance of the PID compensated system is shown for comparison with the previous systems. The step responses of each system are also shown.

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Feedback and Control Systems

Step Response 1.4 Uncompensated System PD Compensated System PID Compensated System

1.2

Amplitude

1

0.8

0.6

0.4

0.2

0

0

1

2

3

4

5

6

7

Time (sec)

Note that the PID compensated system now has zero error for a step input and has approximately twothirds peak time. There is however, a degradation on the settling time of the PID controlled system since it will take some time for the integrator to correct all the errors.

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Feedback and Control Systems Lag-lead compensation. If both the transient response and the steady-state error needs to be improved but with the use of lead and lag compensators instead of ideal PID control, a lag-lead compensator is to be designed. The design approach is basically the same as that of the PID controller; that is the transient response is to be improved first with a lead compensator then a lag compensator is used to improve the steady-state error. The following steps summarize the procedure.

1. Evaluate the performance of the uncompensated system to determine how much improvement in transient response is required. 2. Design the lead compensator to meet the transient response specifications. The design includes the zero location, pole location, and the loop gain. 3. Simulate the system to be sure that all requirements have been met. 4. Redesign if the simulation shows that requirements have not been met. 5. Evaluate the steady-state error performance for the lead-compensated system to determine how much more improvement in steady-state error is required. 6. Design the lag compensator to yield the required steady-state error. 7. Simulate the system to be sure all requirements have been met. 8. Redesign if the simulation shows that the requirements have not been met.

Example 6.27 Design a lag-lead compensator for the system shown below so that the system will operate with a 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.

Answer: Step 1: Evaluate the performance of the uncompensated system to determine how much improvement in transient response is required. The root locus of the uncompensated system is sketched and its performance is determined. The table below summarizes the performance of the system.

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Feedback and Control Systems

Root Locus 10 0.456 8

Imaginary Axis (seconds-1)

6 4 2 0 -2 -4 -6 -8 0.456 -10 -18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

Step 2: Design the lead compensator to meet the transient response specifications. The design includes the zero location, pole location, and the loop gain. There should be a twofold improvement in the settling time. Thus the real part of the new operating point is computed as σg =

4 = 3.6510 1/22.1912

and the imaginary part as ωg = 3.6510 tancos Analysis and Design of Systems Via Root Locus

0.4559 = 7.1277 Page 51

Feedback and Control Systems The lead compensator is now to be designed, which requires a compensator zero and a compensator pole farther away. Arbitrarily choosing the compensator zero at za LD = −6 to coincide with the open-loop

pole at −6, the location of the compensator pole is then computed. Using the angle property of root locus, the compensator pole is located at pa LD = −31.0729. The new root locus is sketch, with the gain determined as K = 2165.9.

Root Locus 10 0.456 8

-1

Imaginary Axis (seconds )

6 4 2 0 -2 -4 -6 -8 0.456 -10 -18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

Real Axis (seconds -1)

Steps 3 and 4: Simulate the system to be sure that all requirements have been met; Redesign if the simulation shows that requirements have not been met. The performance of the lead compensated system is summarized below.

Note that the settling time has improved by 2.1912/1.066 = 2.06 times.

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Feedback and Control Systems Step 5: Evaluate the steady-state error performance for the lead-compensated system to determine how much more improvement in steady-state error is required. The lead compensated system improved the error by 0.3123/0.1435 = 2.1763. If there should be a tenfold improvement in the error of the lag-lead compensated system, an additional 10/2.1763 = 4.5950 improvement must be introduced by the lag compensated system.

Step 6: Design the lag compensator to yield the required steady-state error. Arbitrarily choosing the pole at pa LG = −0.01, the compensator zero must be placed at za LG = 4.595−0.01 = −0.0459

Thus, the lag-lead compensated system will have an open-loop transfer function Gs =

Ks + 0.0459 ss + 10s + 31.0729s + 0.01

where the open-loop pole at −6 is canceled by the open-loop zero at the same location. The root locus of the lag-lead compensated system is sketched below. Root Locus 10 0.456 8 6

Imaginary Axis

4 2 0 -2 -4 -6 -8 0.456 -10 -35

-30

-25

-20

-15

-10

-5

0

5

Real Axis

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Feedback and Control Systems Step 7 and 8: Simulate the system to be sure all requirements have been met; Redesign if the simulation shows that the requirements have not been met. The table below summarizes the performance of the lag-lead compensated system.

Note that after the lead compensation, there is a tenfold improvement in the steady-state error of the system from the uncompensated system, but the time response of the system is almost the same as that of the lead compensated system.

The table below summarizes the different types of cascade compensators.

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Feedback and Control Systems

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