CHAPTER # 9 ROOT LOCUS ANALYSES

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1 F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closed-loop system is closely related to the location of the 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. 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 the 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. 1

2 F K א By using the root-locus method the control Engineer can predict the effects on the location of the closed-loop poles with varying the gain value. 1. Root Locus Method The root locus is the locus of roots of the characteristic equation of the closed-loop system as a specific parameter (usually, gain K) is varied from zero to infinity, giving 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. By using the root-locus method, it is possible to determine the value of the 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. 1.1 Angle and magnitude conditions Consider the control system shown in Fig. 1, whose closed loop T.F. is; Fig.1 closed loop control system The characteristic equation of this system is; The quantity G(s)H(s) is called loop T.F. or open-loop T.F. Assuming that the loop T.F. is a rational function including a gain K, this gives Or = 1 (1) 2

3 F K א Since G(s)H(s) is a complex quantity, eqn. (1) can be split into two equations by equating the angles and magnitudes of both sides, respectively, to obtain the following: Angle condition: = ±180 (2) Magnitude condition: (3) The values of s that fulfill both the angle and magnitude conditions are the roots of the characteristic equation, or the closed-loop poles. Then the root loci for the system are the loci of the closed-loop poles as the gain K is varied from zero to infinity. 2.2 Root Locus Sketch To begin sketching the root locus of a system by the root-locus method we must know the location of the poles and zeros of G(s) H(s) Step #1 K=0 points are located at the open-loop poles Step #2 K= points are located at the open-loop zeros The poles and zeros referred above include those at infinity, if any. EX: Consider the characteristic equation Dividing both sides by the terms that do not contain K, we get When K = 0, the 3 poles are at s = 0, s = 2, and s = 3 as shown in Fig. 2. 3

4 E F K א =0 = 2 =3 = 3 When K is, the 3 zeros are at s = 1, s = and as shown in Fig. 2. = 1 =1 Fig. 2, K=0 and K= points of the root locus Number of branches on the root loci We must know that, the number of branches of root locus plot equals the number of poles. In the previous example, there are 3 poles. So that the total number of root locus branches is THREE. Also for the control system that has 3 poles and shown in Fig. 3, it has three root locus branches. Symmetry of The root loci Root locus is symmetrical w.r.t. the real axis of the s-plane as shown in Fig. 3. Fig. 3, three pole system gives three root locus branches; also the root locus is symmetrical around the real axis 4

5 F K א Step #3 Number of Asymptotes if there are zeros located at, there are asymptotes equal to those zeros at. Simply we can calculate the number of asymptotes by: Number of asymptotes = For the previous example, since there are TWO zeros at infinity OR n m = 2, there are TWO asymptotes. Step #4 Angle of Asymptotes we can calculate the angles of asymptotes by = Where k = 0, 1, 2,., 1 Substituting k=0 we get the angle of 1 st asymptote Substituting k=1 we get the angle of 2 st asymptote, etc. In case of number of asymptotes =2, therefore the angles are θ 0 = 90 and θ 1 = 270 (as shown in Fig. 4-a) In case of number of asymptotes =3, therefore the angles are θ 0 = 60, θ 1 = 180 and θ 2 = 300 (as shown in Fig. 4-b) In case of number of asymptotes =4, therefore the angles are θ 0 = 45, θ 1 = 135, θ 2 = 225 and θ 3 = 315 (as shown in Fig. 4-c) (a) (b) (c) Fig. 4, angle of asymptotes 5

6 F K א Step #5 Intersection of Asymptotes with Real Axis The point of intersection of asymptotes of the root locus lies on the real axis of the s-plane, at σ, where = The point of intersection of the asymptotes (σ) represents the center of gravity of the root locus, and is always a real number. Since the poles and zeros of G(s)H(s) are either real or in complex-conjugate pairs, the imaginary parts in the numerator of σ equation always cancel each other out. Thus, the summation terms may be replaced by the real parts of the poles and zeros of G(s)H(s), respectively. That is, = Example: suppose we have a control system = The point of intersection of asymptotes with real axis is Step #6 = = 5 3 = 1.67 Root Locus on Real Axis On a given section of the real axis, root locus are found in this section only if the total number of poles and zeros of G(s)H(s) to the right of the section is odd. On another explanation, for s1 to be a point on the root locus, there must be an odd number of poles and zeros of G(s)H(s) to the right of that point. We can explain this by the following pole-zero configurations shown in Fig. 5. 6

7 F K א Fig. 5, root locus on real axis Where the dotted line shows Inverse Root Locus (IRL) where the system gain K changes from to 0 (i.e. K<0). On the other hand, the solid line shows Root Locus (RL) where the system gain K changes from 0 to (i.e. K>0) Step #7 Angles of Departure for complex poles OR And Angles of Arrival for complex zeros The angle of departure or arrival of a root locus at a pole or zero, respectively, of G(s)H(s) denotes the angle of the tangent to the locus near the point. The angle of departure is defined as the angle at which the root locus leaves the pole. The angle of arrival is defined as the angle at which the root locus moves toward the zero. We can explain how to calculate the angle of departure by the following example: Consider the characteristic equation of a control system S(S+3)(S 2 +2S+2) + K(S+1) = 0 The angle of departure of the root locus at (s +1 j) is represented by θ 2, measured with respect to the real axis. Let us assign s 1 to be a point on the RL leaving the 7

8 F K א pole at (s +1 j) and is very close to the pole as shown in Fig. 6. Then, s1 must satisfy Eqn. (2). Thus, Fig. 6, calculation of angle of departure Based on eqn. 2, since no zeros = 180 = 180 = = 71.6 Step #8 Intersection of the Root Locus with the Imaginary Axis The points where the root locus intersect the imaginary axis of the s-plane, and the corresponding values of K, may be determined by means of the Routh-Hurwitz criterion explained in the previous lecture. 8

9 F K א Step #9 Breakaway Points Breakaway points on the root locus of an equation correspond to multiple-order roots of the equation. Figure (7-a) illustrates a case in which two branches of the root locus meet at the breakaway point on the real axis and then depart from the axis in opposite directions. In this case, the breakaway point represents a double root of the equation when the value of K is assigned the value corresponding to the point. Fig. 7, break away point Figure (7-b) shows another common situation when two complex-conjugate root locus approach the real axis, meet at the breakaway point, and then depart in opposite directions along the real axis. In general, a breakaway point may involve 9

10 F K א more than two root locus. Figure (7-c) illustrates a situation when the breakaway point represents a fourth-order root. A root-locus diagram can have, of course, more than one breakaway point. Moreover, the breakaway points need not always be on the real axis. Because of the conjugate symmetry of the root loci, the breakaway points not on the real axis must be in complex conjugate pairs. At the breakaway points, the following properties must be satisfied: 1) (3) where G 1 (s)h 1 (s) = K G(s)H(s) setting K=1 2) All real solutions of Eqn. (3) are breakaway points on the root locus for all values of K, since the entire real axis of the s-plane is occupied by the root locus. 3) The complex-conjugate solutions of Eqn. (3) are breakaway points only if they satisfy the characteristic equation or are points on the root locus. It is important to point out that the condition for the breakaway point given in Eqn. (3) is necessary but not sufficient. The angles at which the root locus arrive or depart from a breakaway point depend on the number of branches that are involved at the point. For example, the root locus shown in Figs. (7-a) and (7-b) all arrive and break away at 180 apart, whereas in Fig. (7-c), the four root loci arrive and depart with angles 90 apart. In general, n root locus branches arrive or depart a breakaway point at 180/n degrees apart. Example: Consider the second-order equation The breakaway points on the root loci must satisfy 10

11 F K א Or We find the two breakaway points of the root locus at s = and Fig. 8 Breakaway points of the example Figure 8, shows that the two breakaway points are all on the root locus for positive K. this mean both points are used as breakaway points. Another example: Consider the equation Fig. 9 Breakaway points of the example 11

12 F K א The above derivative can be reduced to The solution of this equation gives the breakaway point as s 1 = and s 2 = The root locus shown in Fig. 9, indicates that, the point S 1 is located in IRL portion, this means that it can't be a proper breakaway point. Another example: The solutions of the last equation are s 1 = 2, s 2 = 2 + j2.45, and s 3 = 2 j2.45. As shown in Fig. 10 that the two complex points S 2 and S 3 are considered as breakaway points. Fig. 10 Breakaway points of the example 12

13 F K א Another example: In this example, we shall show that not all the solutions of Eqn. (3) are breakaway points on the root locus. Consider the root loci of the equation The roots of the above equation is s 1 = j0.471 and s 2 = j From the root locus shown in Fig. 11, these two roots are not breakaway points on the root loci. Fig. 11 Breakaway points of the example 2.3 Calculation of K from the Root Locus Once the root locus is constructed, the values of K at any point (as s 1 for example) on the locus can be determined by using the angle condition described in eqn. (3). Graphically, the magnitude of K can be written as: 13

14 F K א For the root locus shown in Fig. 12, the value of K at point s 1 is given by where A and B are the lengths of the vectors drawn from the poles of G(s)H(s) to the point s 1, and C is the length of the vector drawn from the zero of G(s)H(s) to s 1. In this case, s 1 is on the locus where K is positive Fig. 12 Calculation of K using magnitude condition Example (1): Consider the control system whose characteristic equation is Dividing both sides of the last equation by the terms that do not contain K, we have K = 0 points: at s = 0, s = 5, and s = 6 and S= 1+j and S= 1 j. 14

15 E F K א =0 = 5 = 6 = 1 = 1+ =5 K = points, at s = 3. = 3 =1 Number of asymptotes = = 5 1 = 4 Angle of asymptotes θ 0 = 45, θ 1 = 135, θ 2 = 225 and θ 3 = 315 Intersection of asymptotes with real axis: Angle of departure: from the angle condition, θ D = 180 so that θ D = 43.8 Intersection of the root locus with the imaginary axis is determined using Routh s tabulation. We can obtain the value of K to satisfy the stability as: 15

16 F K א So that 0 < K < 35 At K = 35, the auxiliary equation A(s) is Substituting K = 35 in A(s) we find Then S = ± j 1.34 Break away points: Taking the derivative after making K = 1 we can obtain Since there is only one breakaway expected, only one root of the last equation is the correct solution of the breakaway point. The five roots of the above equation are: Clearly, the breakaway point is at The other four solutions do not satisfy the requirements. Fig. 13, Root locus of example 1. 16

17 F K א Example (2): For the control system described by the block diagram shown in Fig. 14, draw the root locus. Fig. 14, Block diagram for example 2. 17

18 F K א The root locus of this example is shown in Fig. 15. Fig. 15, Root locus of example 2. Example (3) Draw the root locus of the control system shown in Fig. 16. Fig. 16, control system of example 3. 18

19 F K א Using Routh, we can obtain the intersection of root locus with imaginary axis. From Routh array, K = 150 at S = ± j5.0 The root locus of this example is shown in Fig

20 F K א Fig. 17, Root locus of example 3. Example (4) Draw the root locus for the control system shown in Fig. 18. There are 3 asymptotes Fig. 18, Block diagram for example 4. 20

21 F K א Fig. 19, Root locus of example 4. Example (5) Draw the root locus for the control system whose T.F. is given below, = a) Find the max value of K to make a sustained oscillation system and find the frequency of this oscillation. 21

22 F K א b) Calculate the value of K at the breakaway point. Hint: please don t waste your valuable time in calculating the breakaway points. There is only ONE breakaway point at S =

23 F K א 23

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26 F K א Fig. 20 Root locus of example 5. 26

27 F K א Matlab code to draw the root locus is as follows: 27

28 F K א Example (6) Consider the characteristic equation S 2 (S + a) + K(S + b) = 0 Draw root locus of that system when b = 1 and a = 10, 9, and 8 = +1 + At a = 10, there are TWO breakaway points at S= -2.5 and at S = -4.0 as shown in Fig. 21(a) At a = 9, there are ONE breakaway point at S= -3.0 as shown in Fig. 21(b) 28

29 F K א At a = 8, there is NO breakaway points as shown in Fig. 21(b) Fig. 21, Root locus of example 6. 29

30 F K א Example (7) Consider the control system shown in Fig. 22. Plot the root locus for this system by applying the general rules and procedure for constructing root loci, and write the MATLAB code to get root-locus plots. Fig. 22, Control system of example 7. MATLAB Program SHOWN IN Fig. 23 will plot the root-locus diagram for the system. The plot is shown in Figure 24. Fig. 23, Matlab code for example 7. It can be seen from the root-locus plot of Fig. 24 that this system is stable only for limited ranges of the value of K-that is, 0 < K < 12 and 73 < K < 154. The system becomes unstable for 12 < K < 73 and 154 < K. 30

31 F K א Fig. 24, Matlab root locus obtained from m-code. 31

32 F K א Example 8 A simplified form of the open-loop transfer function of an airplane with an autopilot in the longitudinal mode is Sketch the root locus when a = b = 1, damping ratio = 0.5, and ω n = 4. Find the range of gain K for stability. The open loop TF is There are 3 Asymptotes at angles of 60 & 180 & -60 σ = -2/3 = Breakaway points at S = 0.45 and at S = But the points S = ± J 2.16 do not represent break away points To get intersection of RL with imaginary axis, the characteristic equation is Routh array is The values of K that make the s 1 term in the first column equal zero are K = 35.7 and K = The crossing points on the imaginary axis can be found by solving the auxiliary equation obtained from the s 2 row, that is, by solving the following equation for s: 32

33 F K א The results are The angle of departure (θ D ) of the root locus for the complex poles is calculated as θ D = = -54.5" 33

34 E F K א Example 9 The characteristic equation of a control system is S + 12 S + 64 S S +=0 a) Sketch the system root locus and write the corresponding MATLAB code. b) Find the value of K to give critical-stable system, c) Find K that gives a critically-damped system. From the system characteristic equation, we can obtain the system open-loop transfer function by dividing by the term doesn't include K = S + 12 S + 64 S S = SS + 12 S + 64 S = SS+4S+4+j4S+4 j4 Step#1 K = 0 points: at s = 0, s = 4, and S= 4+j4 and S= 4 j4. =0 = 4 = 4 4 = 4+4 =4 Step#2 K = points: at,, and. =0 Step#3 Number of asymptotes = = 4 0 = 4 Step#4 Angle of asymptotes θ 0 = 45, θ 1 = 135, θ 2 = 225 and θ 3 = 315 Step#5 Intersection of asymptotes with real axis: = = Step#6 Root locus on real axis is as shown in figure below. 34

35 F K א Step#7 Break away points: 11= 1 S + 12 S + 64 S S = S + 12 S + 64 S S = =0 Solving the above equation for beark away points, we get: S1 = (Accepted) S2 = j2.553 (Rejected) S3 = j2.553 (Rejected) Step#8 Angle of departure for complex poles: Step#9 Intersection of root locus with imaginary axis: From Routh based on the system characteristic equation: S + 12 S + 64 S S +=0 S K S S K S S 0.. K From the first column of the above Routh array; K > K > 0 K < < K < Max. value of K for satiability K= (Critically stable system) Auxiliary equation S = 0 S = ± j (frequency of sustained oscillation) Matlab Code to draw the root locus of this problem is: 35

36 F K א The value of K for critically damped system is at break away point = حاصلضربأطوالاألقطاب = حاصلضربأطوالاألصفار 1 L1 = L2 = = L3=L4 = = = =

37 E F K א Example 10: For the control system shown in below, a) Sketch the root locus & write the Matlab code, b) Find the maximum value of K to make all poles real, c) Find the value of K to give a damping ratio of R(S) + _ C(S) Characteristic equation is given as = =0 +1 = +2+4 K = 0 points: at s = 0, s = 2, and s = 4 K = points, at s = 1. =0 = 2 =3 = 4 = 1 = =1 37

38 F K א Number of asymptotes = = 3 1 = 2 Angle of asymptotes θ 0 = 90 and θ 1 = 270 Intersection of asymptotes with real axis: = = Root Locus on Real axis: As shown in graphics Intersection of R.L. with imaginary axis: First obtain the characteristic equation =0 Using Routh, 1 K+8 6 K K K>0 5K+48>0, this means K>-9.6 This means from both conditions, K>0 The system is stable for all values of K and there is no intersection of R.L with Imagenary axis Breakaway points: = =0 S 1 = (accepted as a breakaway point) S 2 = 0.79 ± J0.86 (rejected) Root locus of this system is shown below. The Matlab code is: 38

39 F K א The maximum value of K to make all poles real is at breakaway point = =

40 E F K א Example 11: For the control system shown below, draw the root locus and find the value of K to give a damping ratio of 0.4, write the Matlab code. R(S) + _ C(S) K = 0 points: at s = 0, s = 0, and s = 4 =0 = 0 =3 = 4 K = points, at s = 2+j2 and s = 2+j2. = 2+2 =2 2 =2 Number of asymptotes = = 3 2 = 1 Angle of asymptotes θ 0 = 180 Intersection of asymptotes with real axis: No Need, since the asymptote is coincident on real axis Angle of arrival of complex zeros: = 180 =45 Intersection of R.L. with imaginary axis: First obtain the characteristic equation =0 Using Routh, 40

41 F K א Then K > 0 also K > -4 and K > -2 Then we find that K > 0 This means that the system is stable for all values of K This indicate that there no intersection with imaginary axis Root locus of this system is shown below. Since ξ = 0.4, then β = cos -1 (0.4) = we draw a line at angle β intersect with RL as shown above. Applying the angle condition, Taking tan for both sides: =180 =

42 F K א = = = =0 Solving this equation for x, we find that x 1 = x 2 = j x 3 = j (accepted) (rejected) (rejected) After that we obtain the lengths of poles and zero to that point = = = = = = = = Matlab code is given below: = =

43 F K א Example 12: For the closed-loop control system shown below, a) Using block diagram algebra, find the system transfer function C(S)/R(S). b) Obtain the system characteristic equation. c) Draw the root locus as the system gain T change from 0 to, and write the Matlab code, d) Find the minimum value of T such that all poles are real, e) Find the value of T that gives a damping ratio of R(S) + _ C(S) The blocks 12 & 5/(S+8) are cascade + + R(S) C(S) 3 20 The blocks 60/(S+8) & 3/20 are canonical R(S) C(S) The blocks 60/(S+17) cascaded with 2/(S+T) and the result is canonical with 3/40 43

44 F K א R(S) + _ C(S) R(S) + _ C(S) = The system characteristic equation is Rearrange the above equation to be: = = =0 To obtain the open loop T.F. divide by the term that doesn't include T = = = Steps to draw Root Locus: Step#1 T = 0 points: at s = J3, s = J = 3 =2 = 3 Step#2 T = points: at S=0,, =0 =1 44

45 F K א Step#3 Number of asymptotes = = 2 1 = 1 Step#4 θ 0 = 180 Angle of asymptotes Step#5 Intersection of asymptotes with real axis: Step#6 Root locus on real axis is as shown in figure below. (No need) Step#7 Break away points: 11= S 2 = 0 S 2 = 9 S 1 = -3 (accepted) and S 2 = 3 (rejected) Step#8 Angle of departure (θ d ) : 90 - (90 + θ d ) = 180 θ d = -180 Step#9 Intersection of root locus with imaginary axis (No need) 45

46 F K א The minimum value of T for all poles are real is at break away point. L3 = 3 1=2=3 +3 = = =6 3 the value of T that gives a damping ratio of (θ=45) Angle condition at the point of intersection of the line of 45 with root locus = = = = 9 2 =1 = 4.5= = = = = = =3 = = Example 13: A control system with unity feedback has a forward open loop transfer function: = a) Draw the root locus for this system b) Write the Matlab code. b) Find the value of K to make all poles equal c) Find the value of K to give a damping ratio of and find the location of roots 46

47 F K א K = 0 points: at s = 0, s = 2 =0 =2 = 2 K = points, at s = 1+j1 and s=1 j =2 1+1 Number of asymptotes = = 2 2 = 0 (no asymptotes) Angle of asymptotes None Intersection of asymptotes with real axis: None Root Locus on real axis : As shown in next page Angle of Arrival for complex zeros θ A : from the angle condition, (90 + θ A ) - (45 + tan -1 1/3) = 180 θ A = Intersection of Root Locus with Imaj. Axis: Characteristic equation is S 2 (K+1) + S(2-2K) + 2K = 0 Using Routh Array: S 2 K+1 2K S 1 2-2K 0 S 0 2K from which we can obtain that 2K>0 K>0 2-2K>0 K<1 K+1>0 K>-1 the range of stability is 0 < K < 1 at K =1 the system is critically stable auxiliary equation A(S) = 2S = 0 S = ± J1 Break away points 1 =

48 F K א S 2 - S - 1 = 0 S = 1.6 (Rejected) and S = (Accepted) b) The roots are equal at break away point (S=-0.61) From the magnitude condition: = = c) To get damping ratio of 0.707, draw a line with angle 45 intersect with the root locus at point A, that is located at x distance on the X-axis from the angle condition, taking tan for both sides: = = = 2 4x = 2 x = 0.5 from the magnitude condition, = =0.333 The location of roots is at S = J 0.5 and S = J 0.5 ξ = K=1 & S= J A x +1 K=1 & S= -J 1 48

49 F K א Example 14: For the control system shown below, a) Calculate the transfer function C(s)/R(s), b) Plot the root locus as the gain K changes from 0 to, and type the Matlab code c) Determine the value of K that necessary to place the real root at S= -5 and then determine the damping ratio for the conjugate complex roots. d) Determine the maximum value of K to give critically damped system, then find the location of roots. R(S) + _ K C(S) The block diagram can be rearranged as: R(S) + _ K C(S) R(S) C(S)

50 E F K א R(S) + _ C(S) Therefore, the closed loop T.F. is: = The system characteristic equation is given as: = = 0 = Let 0.3K = k k = 0 points: at s = 0, s = 1, s = 3 =0 = 1 =3 = 3 k = points, at s = j1.82 and s= j = Number of asymptotes = = 3 2 = 1 (one asymptote) Angle of asymptote θ = 180 Intersection of asymptotes with real axis: Since one asymptote with angle 180, no meaning for intersection Root Locus on real axis : As shown in next page Angle of Arrival for complex zeros θ A : from the angle condition, (90 + θ A ) ( tan / tan / tan /1.834 ) = 180 θ A =

51 F K א Intersection of Root Locus with Imaj. Axis: No Intersection Break away points 1 = =0 S = c) To locate a pole at (S=-5), connect all poles and zeros to that point From the magnitude condition: = =3 0.3 K = 3 K = 3/0.3 = 10 ### Substitute the value of K in the characteristic equation Then S 3 + 7S S + 20 But there is a pole at S = -5 divide the characteristic equation by S+5, we get S 2 + 2S + 4 that gives the equation of the other two poles S 1 = -1+J 3 S 2 = -1 -J 3 To get the damping ratio of these conjugate complex poles, calculate the angle β β = tan -1 3/1 = 60 the damping ratio ξ = cos(60) = 0.5 #### d) To get critically damped system, ξ = 1, This is happened at break away point. From the magnitude condition: = =

52 F K א 0.3 K = K = 0.123/0.3 = 0.41 ### Substitute the value of K in the characteristic equation Then S S S at break away point there are two equal poles with equation (S+0.515) 2 = S S by division we obtain the location of third pole that is at S = J J 1.82 Example 15: The closed-loop T.F. of a control system is = Sketch the system root locus and write the corresponding MATLAB code then find a) The value of K to give continuous-oscillated system and find the location of poles. b) The value of K to give a critically-damped system and find the location of poles. c) The value of K to give a damping ratio of system and find the location of poles. d) The value of K to locate one pole at S = 5 and find the location of other pole. 52

53 F K א The closed-loop T.F. of a control system is = The system characteristic equation is + +=0 + += = + And the system block diagram is as shown below K=0 points At S = 0 At S = 3 K = points At S = -1 There is only ONE asymptotes at angle = 180 No need for the intersection of asymptotes with real axis Root locus on real axis is as shown in Fig. 53

54 F K א Break away points = + = + =0 المقام S 2 + 2S -3 = 0 (S+3)(S-1) = 0 S 1 = 1 (accepted as break away point) S 2 = -3 (accepted as break in point) Intersection of root locus with the imaginary axis The system characteristic equation is + +=0 Using routh S 2 1 K S 1 k-3 S 0 K K > 0 (1) K 3 > 0 K > 3 (2) From (1) and (2) K > 3 Auxiliary equation is S = 0 S = ± j 3 a) The value of K to give continuous-oscillated system and find the location of poles. Is obtained from Routh array K = 3, the location of roots are at S 1,2 = ± j 3 b) The value of K to give a critically-damped system and find the location of poles. Is obtained at the break in point K = 6 3/2 = 9. And the location of roots at S 1,2 = -3 c) The value of K to give a damping ratio of system and find the location of poles. By drawing a line with 45 as shown in Fig. the intersection of this line with the root locus is located at x Applying the angle condition =

55 F K א Taking tan for both sides 2x 2 + 2x 3 = 4x 2x 2 2x 3 = 0 x 1 = (accepted) x 2 = (rejected at it is ve value) 3+ 1 = = = =6.646 To find the location of poles at this value of K, the characteristic equation is S 1,2 = ± j =0 d) The value of K to locate one pole at S = 5 and find the location of other pole. To locate pole at S = -5, the value of K = 5 8/4 = 10 To find the location of other pole at this value of K, the characteristic equation is S 1 = -5 and S 1 = =0 Example 15: The characteristic equation of a control system is S + 5 S + 17 S + 13 S +=0 a) Sketch the system root locus and write the corresponding MATLAB code b) Find The value of K to give critical-stable system and find the location of poles c) Find K that gives a system with a damping ratio = 1, and find the location of poles. First we obtain the loop T.F. GH(S) as shown in figure below. 55

56 F K א 56

57 F K א Using Routh S K S S K S 1 S 0. K. K > 0 (1) K > 0 K < (2) 0 < K < Auxiliary equation A(S) = 14.4 S = 0 S = ± J The root locus of the control system is shown below 57

58 F K א a) The value of K to give critical-stable system and find the location of poles. From Routh array, K = There are two poles at S = ± J i.e. its equation is S To know the other two poles, substitute the value of K in the characteristic equation S S S S = 0 By long division Then the other two roots at S = ± J S + 5 S + 17 S + 13 S = b)the value of K to give a system with a damping ratio = 1, and find the location of poles. This is at break away point Substitute this value of S in the characteristic equation K= =0 K = There are two poles at S = there equation is = S S To know the other two poles, substitute the value of K in the characteristic equation S S S S = 0 By long division S + 5 S + 17 S + 13 S Then the other two roots at S = -2.1 ± J2.99 =

59 F K א Salman Bin Abdul Aziz University College of Engineering Department of Electrical Engineering EE3511 Automatic Control Systems Sheet 7 (Root Locus) 1) Sketch the root locus diagram for the following control systems: = = = = ˆèˆÃÖ]< fâ<àe<á^û <íãú^q< í ß]<íé Ò< íéñ^e ãóö]<í ß]<ÜŠÎ< Œ^ŠÖ]<ëçjŠ¹] = = = = = = ) Given the following characteristic equations; S 3 + 5S 2 + (6+K)S + 8K = 0 S 3 + 3S 2 + (K+2)S + 10K = 0 Sketch the root locus then find - The value of K for all roots to be real, and find the location of roots - The value of K to give a critically damped system, and find the location of roots - The value of K to make ζ=0.707, and find the location of roots - Find the maximum value of K for stability 3) A control system has a forward and feedback T.F. as K G(s) =, H(s) = 1 2 S(S + 4S + 13) Sketch the root locus then find 59

60 F K א - The value of K to give a damping ratio of Find the location of the roots at that damping ratio - The value of K that gives a marginally stable system 4) A unity feedback control system whose forward T.F is 2 K (S + 14S + 74) G(s) = (S + 3)(S + 5)(S + 10) Sketch the root locus then find - Maximum value of K that make all roots real, then find the location of roots - The value of K that make the real root at S= 14, then find the corresponding damping ratio - The value of K that gives minimum damping ratio, then find the location of roots 5) For the control system shown below - Plot the root locus - Determine the value of K to place the real root at S=-5, then determine the damping ratio - Determine the maximum value of K so that ζ=1, then find the location of roots R(S) + _ K C(S) ) A control system has a closed loop T.F. with unity feedback as =

61 F K א Draw the R.L. and design the system to have a damping ratio of 0.866, then find the location of roots. 7) The root locus of the two poles P1 and P2 is shown in Fig. 2. It is found that at K=10, the damping ratio is Find the two roots P1 and P2 - Find the value of K to give ζ = 0.5, then find the location of roots jω ξ = K=10-1 -P1 P2 σ 7) Draw the root locus for the control system whose T.F. is given below, = c) Find the max value of K to make a sustained oscillation system and find the frequency of this oscillation. d) Calculate the value of K at the breakaway point. Hint: There is only ONE breakaway point at S =

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)

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(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications: 1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.

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Problems -X-O («) s-plane. s-plane *~8 -X -5. id) X s-plane. s-plane. -* Xtg) FIGURE P8.1. j-plane. JO) k JO) Problems 1. For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4] *~8 -X-O

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