GEE338 Linear Control Systems Chapter 3 Stability of Control Systems in Laplace Doman Prof. Fawzy Ibrahim and Eng. Medhat Toubar Electronics and Communication Department Misr International University (MIU) 1 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim Chapter Contents 3.1 Stability Concept 3.2 Stability Analysis Using Routh-Hurwitz Criterion 3.3 Root Locus Techniques 3.3.1 Definition of Root Locus 3.3.2 Root Locus Construction Procedure 2 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.1 Stability Concept Stability: A system is stable if every bounded input yields a bounded output. We call this statement the Bounded-Input, Bounded-Output (BIBO) definition of stability, so: • A system is stable if every Bounded Input yields a Bounded Output (BIBO). • A system is unstable if any bounded input yields an unbounded output. • If the closed-loop system poles are in the left half of the plane and hence have a negative real part, the system is stable. That is, stable systems have closedloop transfer functions with poles only in the Left Half-Plane (LHP). Example 3.1 Check the stability of a control system described by closed-loop transfer functions, T(s) as: C (s) 32 T (s) 2 R ( s ) s 12s 32 The impulse response function g(t) can be described as follows: • Knowing the system transfer function, T(s) and the input, R(s) then output, C(s) can be determined: C ( s ) R ( s )T ( s ) • If the input is unit impulse, r(t) = (t) its Laplace transform R(s) = 1, therefore: 32 32 C ( s ) R ( s )T ( s ) 1x 2 s 12 s 32 ( s 4)( s 8) 3 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.1 Stability Concept • Impulse response function g(t) is the inverse Laplace transform of T(s), using partial fraction, we get: 32 32 A B T (s) 2 s 12s 32 ( s 4)( s 8) ( s 4) ( s 8) A ( s 4)T ( s) s 4 32 ( s 8) B ( s 8)T ( s) s 8 T (s) s 4 32 ( s 4) 32 8 4 s 8 32 8 4 8 8 ( s 4) ( s 8) g (t ) 8(e 4t e 8t )u (t ) • • • The system is stable because g(t) 0 for t or every Bounded Input yields a Bounded Output (BIBO). Note that all the closed-loop system poles are in the Left Half-Plane (LHP). A system is stable if the natural response approaches zero as time approaches infinity. 4 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.1 Stability Concept • • • A system is unstable if the natural response approaches infinity as time approaches infinity. A system is unstable if any bounded input yields an unbounded output. If some of the closed-loop system poles are in the right half of the s-plane and hence have a positive real part, the system is unstable. Example 3.2 Check the stability of a control system described by closed-loop transfer functions, T(s) as: T (s) C (s) 5 2 R( s) s s 6 The impulse response function g(t) can be described as follows: • Knowing the system transfer function, T(s) and the input, R(s) then output, C(s) can be determined: • If the input is unit impulse, r(t) = (t) its Laplace transform R(s) = 1, therefore: C ( s ) R ( s )T ( s ) C ( s ) R ( s )T ( s ) 1x 5 of 40 5 5 s 2 s 6 ( s 3)( s 2) GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.1 Stability Concept • Impulse response function g(t) is the inverse Laplace transform of T(s), using partial fraction, we get: 5 5 A B T (s) 2 s s 6 ( s 3)( s 2) ( s 3) ( s 2) A ( s 3)T ( s) s 3 5 ( s 2) B ( s 2)T ( s) s 2 T (s) • • • • s 3 5 ( s 3) 1 1 ( s 2) ( s 3) 5 1 5 s 2 5 1 5 g (t ) (e 2t e 3t )u (t ) The system is unstable because g(t) for t or every bounded input yields a unbounded output. Note that some of the closed-loop system poles are in the Right Half-Plane (RHP). A system is stable if the natural response approaches infinity as time approaches infinity. If the closed-loop system poles are in the right half of the s-plane and hence have a positive real part, the system is unstable. 6 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion • • • A stable system is a dynamic system with a bounded response to a bounded input. Consider the closed loop transfer function of a system as: bm s m bm1 s m1 b0 N ( s) T ( s) n n 1 a n s a n 1 s a0 ( s ) Where N(s) is the numerator of the transfer function and ∆(s) or Q(s) is denominator or the characteristic equation or polynomial of the system which is given by: (s) Q( s) an s n an1s n1 a0 • • • For the system described by T(s) to be stable, the root of the characteristic equation must lie in the Left Half Plane (LHP). The Routh-Hurwitz criteria or test is a numerical procedure for determining the number of Right Half-Plane (RHP) and imaginary axis roots of the characteristic polynomial. The method requires two steps: (1) Generate a data table called a Routh table and (2) interpret the Routh table to tell how many closed-loop system poles are in the left half-plane, the right half-plane, and on the j-axis. 7 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Step #1: Generate a date table called a Routh table as follows: • Consider the characteristic equation which is given by: (s) Q(s) a4 s 4 a3 s 3 a2 s 2 a1s a0 Rules for Routh table creation • Any row of the Routh table can be multiplied by a positive constant without changing the values of the rows below. • To avoid the division by zero, an epsilon is assigned to replace the zero in the first column. Table 3.1 Initial layout for Routh table 8 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Step #1: Generate a date table called a Routh table as follows: • Further rows of the schedule are then completed as follows: Table 3.2 Completed Routh table 9 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Step #2: Interpret the Routh table to tell how many closed-loop system poles are in the: - left half-plane - right half-plane. • The number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column. • Notes: 1. If the coefficients of the characteristic equation have differing algebraic, there is at least one RHP root. For Example: (s) Q( s) 7s 5 5s 4 3s 3 2s 2 s 10 Has definitely one or more RHP roots. 2. If one or more of the coefficients of the characteristic equation have zero value, there are imaginary or RHP roots or both. For Example: (s) Q( s) s 6 3s 5 2s 4 8s 2 3s 17 has imaginary axis roots indicated by missing s3 term. 10 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Example 3.3 For the system shown in Fig. 3.1, determine T(s) and then apply the Routh-Hurwitz Method to check the its Stability. Solution: The system closed loop transfer function T(s) is calculated as: 1000 G(s) 1000 ( s 2)( s 3)( s 5) T (s) 3 2 1000 1 G(s) H (s) 1 s 10 s 31s 1030 ( s 2)( s 3)( s 5) Therefore, the characteristic equation is given by: ( s ) s 3 10s 2 31s 1030 (a) (b) Fig. 3.1 (a) Feedback system for; (b) equivalent closed loop system. 11 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion ( s ) s 10s Example 3.3 Solution: Step #1: Generate the Routh table as follows: 3 2 31s 1030 Table 3.3 The Completed Routh table for Example 3.3 Divide by 10 Note: Any row of the Routh table can be multiplied by a positive constant without changing the values of the rows below. Step #2: Interpret the Routh table: Since there are two sign changes in the left column (+ + - +). Therefore, the system is unstable because the characteristic equation, (s) has two roots in the right half plane so two poles of T(s) are in the right half plane. 12 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Example 3.4 Apply the Routh-Hurwitz Method to determine the stability of a the closed loop system whose transfer function is given by: 5 4 3 2 Q ( s ) ( s ) 3 s 5 s 6 s 3 s 2s 1 Solution 10 T (s) 5 3s 5s 4 6s 3 3s 2 2s 1 Step #1: Generate the Routh table as follows: Table 3.4 The Completed Routh table Step #2: Interpret the Routh table: Since there are two sign changes in the left column (+ + - +). Therefore, the system is unstable because the characteristic equation, (s) has two roots in the right half plane so two poles of T(s) are in the right half plane. 13 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Example 3.5 Apply the Routh-Hurwitz Method to check the stability of a system whose characteristic equation is given by: ( s) s5 2 s 4 3 s3 6 s 2 5s 3 Solution Step #1: Generate the Routh table as follows: Table 3.5 (a) The Completed Routh table for Example 3.5 Note: To avoid the division by zero, an epsilon () is assigned to replace the zero in the first column. 14 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Example 3.5 Solution Step #2: Interpret the Routh table to tell how many closed-loop system poles are in the right half-plane. To begin the interpretation, we must assume a sign, positive or negative, for the quantity (ε) as illustrated in Routh table. Table 3.5(b) The Completed Routh table for Example 3.5 • Since, there are two sign changes in the left column (+ + - +); Therefore, the system is unstable because the characteristic equation, (s) has two roots in the right half plane so two poles of T(s) are in the right half plane. 15 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.2 Stability Analysis Using Routh-Hurwitz Criterion Example 3.5 Apply the Routh-Hurwitz Method to determine the values of K that make the system stable whose characteristic equation is given by. s 5 13s 4 54s 3 82s 2 (60 K ) s 3K 0 Solution Step #1: Generate the Routh table as follows: Table 3.5(b) The Completed Routh table for Example 3.5 s5 1 54 60 K s4 13 82 3K s3 47.7 60 0.769K 0 3K 0 0 0 65.6 0.212K s2 s s 1 3940 105K 0.163K 2 65.6 0.212K 0 3K 0 0 65.6 0.212K 0 K 309 3940 105K 0.163K 2 0 K0 K 35 Step #2: Interpret the Routh table. Then the values of k that makes the system stable are: 0 K 35 16 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.1 Definition of Root Locus • Consider the feedback control system shown in Fig. 3.2, transfer function is given by: C (s) k G( s) R( s) 1 k G ( s) H ( s) T ( s) • the closed loop Where k is a constant gain parameter. The poles of the transfer function are the roots of the characteristic equation given by: 1 k G( s) H ( s) 0 • The root locus is the locus of the characteristic equation of the closed-loop system as a specific parameter (usually, gain K) is varied from zero to infinity. For 0 ≤ k < ∞. R (s ) G (s ) k C (s ) H (s ) Fig. 3.2 feedback control system. 17 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.1 Definition of Root Locus • If the roots of the characteristic equation given by: 1 k G ( s ) H ( s ) 0 kG( s ) H ( s ) 1 or G ( s ) H ( s ) 1 k • Then • For positive k, this means that a point, s which is a point on locus must satisfy the magnitude and angle and conditions: • The magnitude condition: | kG( s ) H ( s ) | 1 or | G ( s ) H ( s ) | • 1 k The angle condition: G ( s ) H ( s ) (2i 1)180o where 18 of 40 i 0, 1, 2, 3, .... GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Sketch the root locus for the system shown in Fig.3.3. Solution Step 1: Write the open loop transfer function or the characteristic equation of the system and find the open loop poles and zeros as follows: kG( s ) H ( s ) k ( s 3) s ( s 1)( s 2)( s 4) Open loop Poles = {0, -1, -2, -4} # of poles = n = 4. Open loop Zeros= {-3} # of zeros = m = 1. So, n – m = 3. Root locus properties : 1) Symmetry As all roots are either real or complex conjugate pairs so that the root locus is symmetrical about the real axis. 2) Number of branches of the root locus is equal to the number of poles of the open-loop transfer function. Fig. 3.3 feedback control system. 19 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Solution 3) Locus start and end points: The locus starting points (k=0) are at the openloop poles and the locus ending points (k = ∞) are at the open-loop zeros. n -m branches end at infinity. Step 2: Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis that are root loci. The root locus on the real axis always lies in a section of real axis to the left of an odd number of poles and zeros as shown in Fig. 3.4(a). Root locus segments exists after the poles at 0, -2 and -4 Fig. 3.4(a) pole-zero plot of kG(s)H(s). 20 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Solution Step 3: Draw the asymptotes as follows: Asymptotic Angles: As k approaches +∞, the branches of the locus become asymptotic to straight lines with angles: (2i 1)1800 (2i 1)1800 A where i 0, 1, 2, 3 ,.... nm 3 The number of asymptotes are three [n - m = 3]. for i = 0 1 = 60o, for i = 1 2 = 180o, and for i = -1 3 = -60o. Center of Asymptotes: The starting point of the asymptotes. These linear asymptotes are centered at a point on the real axis at: pole values of GH zero values of GH A nm 1 2 4 3 1.33 3 • These asymptotes are shown in Fig. 3.4(b). 21 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Solution Step 3: Draw the asymptotes as follows: Fig. 3.4(b) The asymptotes of the root locus. 22 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Solution Step 4. Breakaway and break-in or entry points: The point where the locus leaves the real axis is called the breakaway point and the point where the locus returns to the real axis is called the break-in point. The loci leave or return to the real axis at the maximum gain k. A pair of locus segments leave or enter the real axis at angles ±90 degree and k is calculated by trial and error from the 1 1 s( s 1)( s 2)( s 4) equation: | G( so ) H ( so ) | and k | | k | G( s) H ( s) | ( s 3) Therefore, the break away point lies between 0 and -1 at maximum value for k which is calculated as follows: s -0.3 -0.4 -0.45 -0.5 k 0.489 0.532 0.534 0.525 It is clear that the breakaway point is at s = -0.45. This point is shown in Fig. 4.4(c). 1 s( s 1)( s 2)( s 4) s 4 7 s 3 14s 2 8s 1 KG ( s) H ( s) 0 and k G( s) H ( s) ( s 3) ( s 3) • For maximum K, dK/ds = 0, we get: dk ( s 3)(4s 3 31s 2 28s 8) ( s 4 7 s 3 14s 2 8s) 0 2 ds ( s 3) 23 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Solution Step 4. Breakaway and break-in or entry points: 4 s 4 26s 3 77s 2 84s 24 0 • Solve for s, we get the K is at s = - 0.434 and K = 0.5346. Fig. 3.4(c) The complete root locus. 24 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.6 Solution Step 5. j axis crossing: The closed loop transfer function is given by: T ( s) The c/s equation is: KG ( s ) k ( s 3) 4 1 KG ( s ) H ( s ) s 7 s 3 14s 2 (8 k ) s 3k 1 KG (s) H (s) s 4 7s 3 14s 2 (8 k )s 3k The Routh-table for the characteristic equation is shown The coefficient of s1 = 0 k 2 65k 720 0 Then k = 9.65 The intersection points are: (90 k ) s 2 21k 0 80.35s 2 202.7 0 thus s1, 2 j1.59 • These points are shown in Fig. 3.4(c). We also conclude that the system is stable for 0 < k < 9.65. 25 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.7 Sketch the root locus for the system shown in Fig.3.5. Solution Step 1: Write the open loop transfer function or the characteristic equation of the system and find the open loop poles and zeros as follows: kG( s ) H ( s ) k ( s 3)( s 5) ( s 1)( s 2) Open loop Poles = {-1, -2} # of poles = n = 2. Open loop Zeros= {3, 5} # of zeros = m = 2. So, n – m = 0. Step 2 Real Axis Segments: Draw the pole zero plot of kG(s)H(s), then locate the segments of real axis that are root loci. The root locus on the real axis always lies in a section of real axis to the left of an odd number of poles and zeros as shown in Fig. 3.6(a) R(s ) k ( s 3)( s 5) ( s 1)( s 2) C (s ) Fig. 3.5 feedback control system. 26 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.7 Solution Step 3: Draw the asymptotes, center of Asymptotes are not applied for this example. Root locus segments exists after the zero at 5 and after the pole at -1. Fig. 3.6(a) pole-zero plot of kG(s)H(s). 27 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.7 Solution Step 4. Breakaway and break-in or entry points: A pair of locus segments leave or enter the real axis at angles ±90 degree and k is calculated by trial and error from the equation: 1 1 ( s 1)( s 2) | G( so ) H ( so ) | k and k | G( s) H ( s) | ( s 3)( s 5) Therefore, the breakaway point lies between -1 and -2 and the break in point lies between 3 and 4 at maximum value for k which is calculated as listed in Table 3.6(a). Table 3.6(a). It is clear that the breakaway and break-in points are at: • at s = -1.45 breakaway and • at s = -3.82 break-in points 28 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.7 Solution Step 5: Angles of departure and approach is not applied for this example. Step 6 j axis crossing: The closed loop transfer function is given by: G( s) k ( s 2 8s 15) T ( s) 1 G ( s ) H ( s ) (1 k ) s 2 (3 8k ) s ( 2 15k ) The c/s equation is: 1 G( s) H ( s) (1 k ) s 2 (3 8k ) s (2 15k ) The Routh-table for the characteristic equation is shown in Table 3.6 (b) The coefficient of s1 = 0 3 8k 0 Table 3.6(b). Then 3/8 = 0.375 The intersection points are: (1 k ) s 2 (2 15k ) 0 1.375s 2 7.625 0 • thus s1, 2 j 2.35 These points are shown in Fig. 3.6(b). We also conclude that the system is stable for 0 < k < 0.375. 29 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.7 Solution Fig. 3.6(c) The complete root locus. 30 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.8 Sketch the root locus for the system shown in Fig.3.7. Solution Step 1: Write the open loop transfer function or the characteristic equation of the system and find the open loop poles and zeros as follows: k ( s 2 4 s 20) kG( s ) H ( s ) ( s 2)( s 4) Open loop Poles = {-2, -4} # of poles = n = 2. Open loop Zeros= {(2 + j4), (2 – j4)} # of zeros = m = 2. So, n – m = 0. Step 2 Real Axis Segments: Draw the pole zero plot of kG(s)H(s), then locate the segments of real axis that are root loci. The root locus on the real axis always lies in a section of real axis to the left of an odd number of poles and zeros as shown in Fig. 3.6(a) Fig. 3.7 feedback control system. 31 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.8 Solution Step 3: Draw the asymptotes, center of Asymptotes are not applied for this example. Root locus segments exists after the pole at -2. Fig. 3.8(a) pole-zero plot of kG(s)H(s). 32 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.8 Solution Step 4. Breakaway and break-in or entry points: A pair of locus segments leave or enter the real axis at angles ±90 degree and k is calculated by trial and error from the equation: 1 1 ( s 2)( s 4) | G( so ) H ( so ) | k and k | G( s) H ( s) | ( s 2 4s 20) Therefore, the breakaway point lies between -2 and -4. The breakaway point lies at s = -2.88 at maximum value for k = 0.0248. Step 5 Angles of departure and approach: The angle of departure p of a locus from a complex pole is given by: p dep 180 poles zeros p = 180 – [(sum of the other GH pole angles to the pole under consideration) - (sum of the GH zero angles to the pole )] [It is applied only when there are two complex poles] The angle of approach z of a locus to a complex zero is given by: z arv 180 zeros poles z = [(sum of the other GH pole angles to the zero under consideration) - (sum of the GH zero angles to the zero )] -180 z = 180 + [1 + 2] - [3] = 180 + [26.6 + 45] – [90] = 161.6 degree These angles are shown in Fig.3.8(b). [if there are two complex zeros] 33 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.8 Solution Step 5 Angles of departure (p) and approach (z): Fig. 3.8(a) Angles of departure (p) and approach (z) of kG(s)H(s). 34 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.8 Solution Step 6 j axis crossing: The closed loop transfer function is given by: G( s) k ( s 2 4 s 20) T ( s) 1 G ( s ) H ( s ) (1 k ) s 2 (6 4k ) s (8 20k ) The c/s equation is: 1 G( s) H ( s ) (1 k ) s 2 (6 4k ) s (8 20k ) The coefficient of s1 = 0 6 4k 0 Then k = 1.5 The intersection points are: (1 k ) s 2 (8 20k ) 0 2.5s 2 38 0 thus s1, 2 j 3.89 • These points are shown in Fig. 3.8(c). We also conclude that the system is stable for 0 < k < 1.5. 35 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.8 Solution Notes: a) The angle of the desired poles is at = cos-1(0.45) =63.3 °. A line is drawn from the origin to intersect the root locus at the point 3.4 116.7° with a gain k = 0.417. b)Therefore, this system is stable for 0 < k < 1.5. Fig. 3.8(c) The complete root locus. 36 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.9 Sketch the root locus for the system shown in Fig.3.9 Solution Step 1: Write the open loop transfer function or the characteristic equation of the system as follows: kG( s ) H ( s ) R(s ) k ( s 2) ( s 3)( s 2 2 s 2) k ( s 2) ( s 3)( s 2 2 s 2) C (s ) Fig. 3.9 Step 2. Real Axis Segments: Draw the pole zero plot of kG(s)H(s), then locate the segments of real axis that are root loci. The root locus on the real axis always lies in a section of real axis to the left of an odd number of poles and zeros as shown in Fig. 3.10 37 of 21 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.9 Solution Root locus segments exists after the zero at -2. The are two complex poles -1± j1 A Fig. 3.10 The root locus of Example 3.9. 38 of 21 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.3 Root Locus Techniques 3.3.2 Root Locus Construction Procedure Example 3.9 Solution Step 3. Asymptotic Angles: is not applied for this example. Step 5. Breakaway and break-in or entry points: is not applied for this example. Step 6. Angles of departure and approach: The angle of departure p of a locus from a complex pole is given by: p = 180 – [(sum of the other GH pole angles to the pole under consideration) - (sum of the GH zero angles to the pole)] p = 180o – [(2 + 4) - 3] = 180o - [90o + 26.6o - 45o] =108.4o This point is shown in Fig. 3.10 39 of 21 GEE338 Ch.3 Stability Prof Fawzy Ibrahim 3.4 Design of Compensators by Root-Locus Method Example 3.11 Solution Note: The solution to such a problem is not unique, there are infinitely many solutions. • One possible solution is shown in Fig. 3.13 and Kc is calculated as before. Gc ( s ) K c s 1/ s 1.5 1.03 s 1 / s 3.6 Fig. 3.13 Another possible pole and zero selection of compensator. 40 of 40 GEE338 Ch.3 Stability Prof Fawzy Ibrahim