MAK 391 System Dynamics & Control. Presentation Topic. The Root Locus Method. Student Number: Group: I-B. Name & Surname: Göksel CANSEVEN

Size: px

Start display at page:

Download "MAK 391 System Dynamics & Control. Presentation Topic. The Root Locus Method. Student Number: Group: I-B. Name & Surname: Göksel CANSEVEN"

Bryan Horn
6 years ago
Views:

1 MAK 391 System Dynamics & Control Presentation Topic The Root Locus Method Student Number: Group: I-B Name & Surname: Göksel CANSEVEN Date: December 2001

2 The Root-Locus Method Göksel CANSEVEN University of Sakarya Mechanical Engineering Department Abstract: The analysis of a control system centers on answering the question: Is the system stable? A stable control system responds to any reasonable input with an error that diminishes with time. The error of a stable system may or may not oscillate, and it may or may not diminish to zero. If the error oscillates, the amplitude of the oscillation diminishes exponentially with time. If the error does not diminish to zero, it eventually reaches a small, steady value. In contrast, an unstable control system responds to at least some reasonable inputs with an error that increases with time. The error usually oscillates. For continuous, linear systems, an error that oscillates with constant amplitude is considered to be unstable. However, some discontinuous systems are inherently oscillatory, an oscillating error is considered stable as long as the amplitude does not increase with time. A second concern of control system analysis is the question: How quickly does the error diminish? Control system engineers must answer both questions in the process of designing a control system that is stable and minimizes errors resulting from disturbances, changes in setpoints, or changes in load. The methods for analysing the control systems are: Bode Diagrams, Nyquist Diagrams, and Root-Locus. The Root-Locus technique is a graphical method of determining the stability or in stability of a control system. The Root-Locus consists of a plot of the roots of a certain characterictic equation of the control system as the gain varies from zero to infinity. From the root-locus plot, the designer can observe the relationship between the system gain and the stability of the system. The root-locus is particularly useful to the designer in determining the effect of gain changes on the stability and performance of the control system. Introduction: The basic characteristic of the transient response of a closed-loop system is closely related to the location of closed-loop poles. If the system has a variable loop gain, then the location of the closed-loop poles depends on the value of the loop gain chosen. It is important, therefore, that the designer know how the closed-loop poles move in the s-plane as the loop gain is varied. From the design viewpoint, in some systems simple gain adjustment may move the closed loop poles to desired locations.then the design problem may become the selection of an appropriate gain value. If the gain adjustment alone does not yield a desired result, addition of a compensator to the system will be necessary.

3 The closed-loop poles are the roots of the characteristic equation. Finding the roots of the characteristic equation of degree higher than 3 is laborious and will need computer solution. However, just finding the roots of the characteristic equation may be of limited value, because as the gain of the open-loop transfer function varies the characteristic equation changes and the computations must be repeated. A simple method for finding the roots of the characteristic equation has been developed by W. R. Evans and used extensively in control engineering. This method, called root-locus method, is one in which the roots of the characteristic equation are plotted for all values of a system parameter. The roots corresponding to a particular value of this parameter can then be located on the resulting graph. Note that the parameter is usually the gain, but any other variable of the open-loop transfer function may be used. Unless otherwise stated, we shall assume that the gain of the open-loop transfer function is the parameter to be varied through all values, from zero to infinity. By using the root-locus method the designer can predict the effects on the location of the closed loop poles of varying the gain value or adding open-loop poles and/or open-loop zeros. Therefore, it is desired that the designer have a good understanding of the method for generating the root loci of the closed-loop system, both by hand and by use of a computer software like MATLAB. Root-Locus Method: The basic idea behind the root-locus method is that the values of s that make the transfer function around the loop equal 1 must satisfy the characteristic equation of the system. The locus of roots of the characteristic equation of the closed loop system as the gain is varied from zero to infinity gives the method its name. Such a plot clearly shows the contributions of each open-loop pole or zero to the locations of the closed-loop poles. In designing a control system, we find that the root-locus method proves quite useful since it indicates the manner in which the open-loop poles and zeros should be modified so that the response meets system performance specificiations. This method is particularly suited to obtaining approximate results very quickly. Some control systems may involve more than one parameter to be adjusted. The root-locus diagram for a system having multiple parameters may be constructed by varying one parameter at a time. The root-loci for such a case is called the root-contour. By using the root-locus method, it is possible to determine the value of the loop gain K that will make the damping ratio of the dominant closed-loop poles as prescribed. If the location of an open-loop pole or zero is a system variable, then the root-locus method suggests the way to choose the location of an open-loop pole or zero.

4 Description: One condition for stability of a closed-loop control system is that all roots of the following equation must have negative real parts: 1 + G (s) H (s) = 0 (1) Equation (1) is reffered to as the closed-loop characteristic equation. Recall that G (s) H (s) is the open-loop transfer function and equation (1) is the denominator of the closed-loop transfer function. The root-locus is a graphical method of analysis in which all possible roots of the closed-loop characteristic equation are plotted as the controller gain K is varied from 0 to infinity. This graph of all possible roots gives the designer considerable insight into the relationship between gain K and the stability of the system. The root-locus method not only answers the question of stability, but also gives designer information such as the damping ratio and damping constant of the system. We begin with three simple examples to show how the roots of the characteristic equation change with changes in gain K. We will use s o to represent any value of s that satisfies the characteristic equation. Our first example has the following open-loop transfer function: The characteristic equation is: G (s) H (s) = K / (s+2) Solving for s o : 1 + [ K /(s o +2) ] = 0 S o = -(K + 2) Thus the root-locus begins at s o = -2 when K = 0 and moves to the left on the real axis as K increases. Figure 1-a shows the root locus of example 1. Our second example has the following open-loop transfer function, characteristic equation, and solution: G (s) H (s) = K / (s+2) [ K / (s o +2) 2 ] = 0 s o = -2 ± jk

5 Thus the root-locus begins at s o = -2 when K = 0 and moves both up and down along a vertical line as K increases. Figure 1-b shows the root-locus of example 2. Figure1: Root locus of 1 + G (s) H (s) = 0 for three simple systems.

6 Our third example has the following open-loop transfer function, characteristic equation, and solution: G (s) H (s) = [ K(s + 6) / (s+2) ] 1 + [ K(s o + 6) / (s o + 2) ] = 0 s o K(s o + 6) = 0 s o (K + 1) + 6K + 2 = 0 s o = - [ (6K + 2) / (K + 1) ] The root-locus is the plot of s o as K varies from 0 to infinity. Our first observation is that s o will always be a negative real number, and magnitude of s o will increase as K increases. We really only need the two endpoints of the locus, the values of s o when K = 0 and K =. When K = 0, When K =, s o = - [ (6K + 2) / (K + 1) ] K = 0 = -2 s o = - [ (6K + 2) / (K + 1) ] K = The last expression will be easier to evaluate if we divide the numarator and denominator of the fraction by K. s o = - [ (6 + 2 / K) / (1 + 1 / K) ] K = = -6 Figure 1-c shows the root-locus diagram of this system. The following table of additional values of s o shows how s o varies as K increases. K 1 / s o

7 Examples: Example 1: Determine the root-locus of a control system with the following open-loop transfer function. G (s) H (s) = K / [s(s+10)] Solution: 1 + G (s) H (s) = 1 + K / [s o (s o +10)] = [s o (s o +10) + K] / [s o (s o +10)] The characteristic equation is s o (s o +10) + K = s 2 o + 10s o + K = 0 The roots are given by the formula: s 1 s 2 = K = K The following table shows how s varies as K increases. K s 1 s j5-5+j j5-5-j10

8 Figure 2: Root locus of 1 + G (s) H (s) = 0 for the system with G (s) H (s) = K / [s(s+10)] Figure 3 illustrates the relationship between the roots of the characteristic eqn., the damping ratio (), and the damping coefficient (). The follwing symbols and terminology will be used in the following discussion. = damping coefficient, second -1 o = resonant frequency, radian / second d = damped resonant frequency, radian / second = damping ratio = operating point angle When the roots of the characteristic equation are imaginary, they always come in conjugate pairs given by the following equations. s 1 = - + j d s 2 = - - j d Notice in Figure 3 that o is the hypotenuse of a right triangle, and d and are the two legs of the right triangle. Thus, by the Pythagorean theorem, o = 2 + d 2

9 Figure 3: Root locus diagram showing the relationship between the roots of the characteristic equation, the damping ratio, and the damping coefficient. The damping ratio is given by = / o Therefore = cos -1 (3) and = d 2 = 2 ( d ) 2 d = [ 2 (1-2 )] / 2 d = [ 1-2 / ] (4) Equation 4 can be used to determine the value of K that will result in a specified damping ratio.

10 Example 2: Find the value of gain K that will give a damping ratio of 0.5 in the control system in example 1. Solution: Equation 4 gives the required value of d. From example 1, = 0.5. d = [ 1-2 ] / = [ ] / 0.5 = 8.66 In example 1, when (25 - K) is negative, the roots are imaginary and d = K d = K - 25 = 8.66 K - 25 = K = = 100 Thus the control system will have a damping ratio of 0.5 when the gain K is equal to 100. The methods used in the previous examples are not adequate for the more realistic example that follows. Let us now consider the following general form of the open-loop transfer function: G (s) H (s) = K { [ (s-z 1 )(s-z 2 )...(s-z m ) ] / [ (s-p 1 )(s-p 2 )...(s-p n ) ] } (5) Where K>0 z 1, z 2,...z m are the zeros of G (s) H (s) (zeros are values of s for which G (s) H (s) = 0) p 1, p 2,...p n are the poles of G (s) H (s) (poles are values of s for which G (s) H (s) = ) Note that m is the number of zeros in G (s) H (s) and n is the number of poles. We will be using those two numbers to compute certain characteristics of the root-locus plot. Consider also the following form of the closed-loop characteristic equation: G (s) H (s) = -1 = o (6) Any point s o that makes the angle of G (s ) H (s ) equal to an odd multiple of ± 180 o will satisfy equation 6. This is commonly called angle condition of the closed-loop characteristic equation. All points that satisfy the following angle condition also satisfy the closed loop characteristic equation and lie on the root-locus.

11 Angle Condition of the Closed-loop Characteristic Equation Angle of [G (s ) H (s ) ] = ±N(180) (7) Where N = 1, 3, 5, 7, 9,...= { set of odd integers } If s o is a point on the root-locus, then the value of K can be computed by the following magnitude condition. Magnitude Condition of the Closed-loop Characteristic Equation K = magnitude of [(s o -p 1 )(s o -p 2 )... (s o -p n ) / (s o -z 1 )(s o -z 2 )... (s o -z m )] (8) A detailed study of the angle condition has resulted in the developement of the following rules for construction of root-locus plots. Theories: Root Locus Rules G (s) H (s) = K [ (s-z 1 )(s-z 2 )... (s-z m ) / (s-p 1 )(s-p 2 )... (s-p n ) ] m= the number of poles n= the number of zeros Begin the root locus plot by drawing an s-plane diagram with equal scales for the real and imaginary axes. Then plot each zero and pole of G (s) H (s) on the s-plane diagram. Use an X to mark the location of each pole and O to mark the location of each zero. Then apply the following rules to plot the root locus of 1 + G (s) H (s). 1) The root locus will have n branches that begin at the n poles with the value K = 0. 2) The root locus will have m branches that will terminate on the m finite zeros as K approaches infinity. 3) The remaning n-m branches will go to infinity along asymptotes as K approaches infinity. 4) The asymptotes meet at a point called the centroid, C, which is computed as follows:

12 C= C= poles - zeros (number of poles) (number of zeros) (9) (p 1 + p 2 + p p n ) (z 1 + z z m ) (number of poles) (number of zeros) The n-m asymptote angles are given by: N = ± N(180 o ) / (n-m) ; N = 1, 3, 5, 7, 9,... (10) For each N, 2 angles are computed. Use enough values of N to compute the required n-m angles. 5) Begin at the right and number the zeros and poles that lie on the real axis without regard to whether the points are a zero or a pole. A typical numbering will appear as follows: If there is an even number of points on the real axis, then the root locus will lie on the part of the real axis between each odd point and the following even point and the following even point as illustrated below If there is an odd number of points on the real axis, then the root locus will extend to infinity from the highest numbered point as illustrated below ) If two zeros are connected, there must be a breaking point between them. If two poles are connected, there must be a breakout point between them.

13 If a pole and a zero are connected, they usually are a full branch, starting at the pole with K = 0 and ending at the zero as K ) The break points occur where the derivative of K with respect to s is equal to zero. Computation of the breakpoints is demonstrated in the next example. 8) The points where the branches cross the imaginary axis are determined by replacing s by j in the characteristic equation and then solving for. 9) The angle condition may be used to determine the angle of departure from complex poles or the angle of arrival at complex zeros. Draw a line from each of the other zeros or poles to the zero or pole in question. Measure the angle formed by the positive real axis and the line to each of the other zeros and poles. The departure or arrival angle is equal to 180 o plus the sum of all the zero angles minus the sum of all the pole angles. 10) The operating point for a given damping ratio, zeta (), can be determined as follows: First, determine the operating point angle, = cos -1 (). Then draw a line from the origin of the s-plane that forms an angle of with the negative real axis. This line intersects the root locus at the operating point. The length of the line from the origin to the operating point represents the natural frequency, o. 11) The value of K at the operating point can be determined as follows: First, draw lines from each pole and zero to the operating point. Measure the length of each line and determine the product of the lengths of the lines from the poles and the product of the lenghts of the lines from the zeros. If there are no zeros (or poles) then use 1 as the product. The value of K at the operating point is equal to the poles product divided by the zeros product. Example 3: Determine the roots locus of a control system with the following open-loop transfer function: G (s) H (s) = K / s(s + 3)(s + 9)

14 Then determine the operating point, s o, for a damping ratio of 0.5 ( = 0.5). Also find the gain, K, and the resonant frequency, o, at the operating point. Solution: 1) The open-loop transfer function has no zeros and three poles at S = 0, s = -3, and s = -9. The root locus will have three branches that will all go to infinity along three asymptotes (Rules 1, 2, 3). 2) The asymptotes meet at the centroid, C, determined by equation 9 (Rule 4). C= poles - zeros (number of poles) (number of zeros) [ 0 + (-3) + (-9) ] [0] C= = The asymptote angles are determined by equation = ± 1(180) / (3 0) = ± 60 o = ± 3(180) / (3-0) = ± 180 o Therefore, the asymptote angles are +60 o, -60 o, and +180 o. 3) The root locus lies on the portion of the real axis between s = 0 and s = -3, and the portion from s = -9 to s = - (Rule 5). 4) There is a break out point oon the real axis between s = 0 and s = -3 (Rule 6). 5) The break out point occurs where the derivative of K with respect to s is equal to zero (dk / ds = 0). To obtain this derivative, we first solve the characterictic equation for K: 1 + [K / s(s+3)(s+9)] = 0 K = -(s s s) The power rule may be used to obtain dk / ds. dk / ds = -(3s s + 27) Now set dk / ds = 0 to obtain the break out point. dk / ds = -(3s s + 27) = 0 Dividing both sides of the equation by 3 and then applying the quadratic formula gives us two candidates for the desired breakout point. s 2 + 8s + 9 = 0

15 s = -8 ± s = and The break out point is between 0 and 3, so s = is the desired breakout point (Rule7). 6) The points where the branches cross the imaginary axis are obtained from the characteristic equation with s replaced by j. s s s + K = 0 (j) (j) (j) + K = 0 -j j + K = 0 (K ) + j(27-3 ) = 0 Set the imaginary part equal to zero and solve for K = 0 2 = 27 = ± 27 = ± 3 3 Set the real part equal to zero and solve for K. K = 12 2 = 12(27) = 324 Therefore, the branches cross the imaginary axis at = ± 3 3 with K = 324 (Rule 8). 7) The root-locus plot is shown in figure 4. The characteristic equation, s s s + K = 0 involves the solution of a cubic equation, which is considerably more involved than a quadratic equation. However, the roots when K = 0 are easy to find. They are 0, -3, and, -9. The next step is to use the fact that one of the three roots will always be real. Assume that the real root is a, and divide the characteristic equation by the factor ( s + a ) to get a general equation for the remaning two roots. ( s s s + K ) / (s + a ) = s 2 + (12- a)s + 27 a(12-a) The remainder of the division is K 27a + 12a 2 a 3. We can now express the characteric equation as follows: ( s+a ) [ s 2 + (12-a)s + 27 a(12-a) ] = 0 (11)

16 The value of K must be such that the remainder is 0. K = 27a 12a 2 + a 3 (12) Figure 4: Root locus of 1 + G (s) H (s) = 0 for the system with G (s) H (s) = K / [s(s+3)(s+9)]. A calculator or computer program may be used to obtain exact points on the root locus using Equations 11 and 12 and the quadratic formula. A trial and error procedure may also be employed to obtain the exact operating point. However, the following graphical analysis provided quick results with sufficient accuracy for most applications. 8) The expanded view of the root locus in Figure 5 shows the determination of the operating point for a damping ratio of 0.5 (Rule 10). The line of constant damping ratio makes an angle of as given by Equation 3. = cos = 60 o The operating point is at the intersection of the damping ratio () = 0.5 line and the root locus. From the graph, we obtain the approximate value j1.95 for the coordinates of the operating point. Using a calculator, the more exact value of the coordinate is j Operating point = j = o The resonant frequency is equal to the length of the line from the origin to the operating point. From the graph, the distance is estimated to be 2.25 (the same as the calculated value).

17 Resonant frequency = 2.25 radians / second 9) The value of K at the operating point is determined by the poles product divided by zeros product (Rule11). From the graph we obtain the following lengths: LP o = length from the pole at (0,0) =2.25 LP 3 = length from the pole at (-3,0) = 2.70 LP 9 = length from the pole at (-9,0) = 8.11 Poles product = (2.25)(2.70)(8.81) = 49.3 Zeros product = 1 (There are no zeros.) K = 49.3 / 1 = 49.3 Figure 5: Expanded view of the operating point region of the root locus shown in Figure 4. At the operating point the system has a damping ratio of 0.5 and a resonant frequency of 2.25 radians / second.

18 Conclusions: The relative stability and the transient response performance of a closed-loop control system are directly related to the location of the closed-loop roots of the characteristic equation. Therefore, we have investigated the movement of the characteristic roots on the s-plane as the system parameters are varied by utilizing the root locus method. The root locus method, a graphical technique, can be used to obtain an approximate sketch in order to analyze the initial design of a system and determine suitable alterations of the system structure and values. A computer is commonly used to calculate several accurate roots at important points on the locus.

19 References: 1) Kuo B. b C., Otomatik Kontrol Sistemleri, Literatür YayTnlarT, pp , ) Bateson R. b N., Introduction to Control System Technology, Macmillian Publishing Company, pp , pp , ) Dorf R. b C., and Bishop R. b H., Modern Control Systems, Addison-Wesley Publishing Company, pp , p. 386, ) Vegte J. b V. D., Feedback Control Systems, Prentice-Hall, pp , ) Ogata K., Modern Control Engineering, Prentice-Hall, pp , ) ÖzdaW M. b N., Dinibütün A. b T., and Kuzucu A., Otomatik Kontrol Temelleri, Birsen YayTnevi, pp , 1998.

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)