Advanced Control Engineering - Chapter 5 pps

//SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 110 ± [110±144/35] 9.8.2001 2:30PM 5 Classical design in the s-plane 5.1 Stability of dynamic systems The response of a linear system to a stimulus has two components: (a) steady-state terms which are directly related to the input (b) transient terms which are either exponential, or oscillatory with an envelope of exponential form. If the exponential terms decay as time increases, then the system is said to be stable.If the exponential terms increase with increasing time, the system is considered unstable. Examples of stable and unstable systems are shown in Figure 5.1. The motions shown in Figure 5.1 are given graphically in Figure 5.2. (Note that (b) in Figure 5.2 does not represent (b) in Figure 5.1.) The time responses shown in Figure 5.2 can be expressed mathematically as: For (a) (Stable) x o (t)  Ae Àt sin(!t  )(5:1) For (b) (Unstable) x o (t)  Ae t sin(!t  )(5:2) For (c) (Stable) x o (t)  Ae Àt (5:3) For (d) (Unstable) x o (t)  Ae t (5:4) From equations (5.1)±(5.4), it can be seen that the stability of a dynamic system depends upon the sign of the exponential index in the time response function, which is in fact a real root of the characteristic equation as explained in section 5.1.1. //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 111 ± [110±144/35] 9.8.2001 2:30PM mg xt o () N (c) Stable (b) Unstable xt o () (d) Unstable mg N xt o () (a) Stable xt o () Fig. 5.1 Stable and unstable systems. A A A t A (a) t t t (b) (c) (d) xt () o xt () o xt () o xt () o Fig. 5.2 Graphical representation of stable and unstable time responses. Classical design in the s-plane 111 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 112 ± [110±144/35] 9.8.2001 2:30PM 5.1.1 Stability and roots of the characteristic equation The characteristic equation was defined in section 3.6.2 for a second-order system as as 2  bs c  0(5:5) The roots of the characteristic equation given in equation (5.5) were shown in section 3.6.2. to be s 1 , s 2  Àb Æ  b 2 À 4ac p 2a (5:6) These roots determine the transient response of the system and for a second-order system can be written as (a) Overdamping s 1 À 1 s 2 À 2 (5:7) (b) Critical damping s 1  s 2 À (5:8) (c) Underdamping s 1 , s 2 À Æ j! (5:9) If the coefficient b in equation (5.5) were to be negative, then the roots would be s 1 , s 2  Æ j! (5:10) The roots given in equation (5.9) provide a stable response of the form given in Figure 5.2(a) and equation (5.1), whereas the roots in equation (5.10) give an unstable response as represented by Figure 5.2(b) and equation (5.2). The only difference between the roots given in equation (5.9) and those in equation (5.10) is the sign of the real part. If the real part  is negative then the system is stable, but if it is positive, the system will be unstable. This holds true for systems of any order, so in general it can be stated: `If any of the roots of the characteristic equation have positive real parts, then the system will be unstable'. 5.2 The Routh±Hurwitz stability criterion The work of Routh (1905) and Hurwitz (1875) gives a method of indicating the presence and number of unstable roots, but not their value. Consider the character- istic equation a n s n  a nÀ1 s nÀ1 ÁÁÁa 1 s a 0  0(5:11) 112 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 113 ± [110±144/35] 9.8.2001 2:30PM The Routh±Hurwitz stability criterion states: (a) For there to be no roots with positive real parts then there is a necessary, but not sufficient, condition that all coefficients in the characteristic equation have the same sign and that none are zero. If (a) above is satisfied, then the necessary and sufficient condition for stability is either (b) all the Hurwitz determinants of the polynomial are positive, or alternatively (c) all coefficients of the first column of Routh's array have the same sign. The number of sign changes indicate the number of unstable roots. The Hurwitz determinants are D 1  a 1 D 2  a 1 a 3 a 0 a 2         D 3  a 1 a 3 a 5 a 0 a 2 a 4 a 1 a 3               D 4  a 1 a 3 a 5 a 7 a 0 a 2 a 4 a 6 a 1 a 3 a 5 a 2 a 4                       etc: (5:12) Routh's array can be written in the form shown in Figure 5.3. In Routh's array Figure 5.3 b 1  1 a nÀ1 a nÀ1 a nÀ3 a n a nÀ2         b 2  1 a nÀ1 a nÀ1 a nÀ5 a n a nÀ4         etc: (5:13) c 1  1 b 1 b 1 b 2 a nÀ1 a nÀ3         c 2  1 b 1 b 1 b 3 a nÀ1 a nÀ5         etc: (5:14) Routh's method is easy to apply and is usually used in preference to the Hurwitz technique. Note that the array can also be expressed in the reverse order, commen- cing with row s n . · · q 1 p 1 · · s 0 s 1 s n –3 s n s n –2 s n –1 c 1 b 1 a n –1 a n c 2 b 2 a n –3 a n –2 c 3 b 3 a n –5 a n –4 ·· Fig. 5.3 Routh's array. Classical design in the s-plane 113 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 114 ± [110±144/35] 9.8.2001 2:30PM Example 5.1 (See also Appendix 1, A1.5) Check the stability of the system which has the following characteristic equation s 4  2s 3  s 2  4s 2  0(5:15) Test 1: All coefficients are present and have the same sign. Proceed to Test 2, i.e. Routh's array s 0 2 s 1 8 s 2 À12 s 3 24 s 4 112 (5:16) The bottom two rows of the array in (5.16) are obtained from the characteristic equation. The remaining coefficients are given by b 1  1 2 24 11          1 2 (2 À 4) À1(5:17) b 2  1 2 20 12          1 2 (4 À 0)  2(5:18) b 3  0(5:19) c 1 À1 À12 24         À1(À4 À 4)  8(5:20) c 2  0(5:21) d 1  1 8 80 À12          1 8 (16 À 0)  2(5:22) In the array given in (5.16) there are two sign changes in the column therefore there are two roots with positive real parts. Hence the system is unstable. 5.2.1 Maximum value of the open-loop gain constant for the stability of a closed-loop system The closed-loop transfer function for a control system is given by equation (4.4) C R (s)  G(s) 1  G(s)H(s) (5:23) In general, the characteristic equation is most easily formed by equating the denomi- nator of the transfer function to zero. Hence, from equation (5.23), the characteristic equation for a closed-loop control system is 1  G(s)H(s)  0(5:24) 114 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 115 ± [110±144/35] 9.8.2001 2:30PM Example 5.2 (See also Appendix 1, examp52.m) Find the value of the proportional controller gain K 1 to make the control system shown in Figure 5.4 just unstable. Solution The open-loop transfer function is G(s)H(s)  8K 1 s(s 2  s 2) (5:25) The open-loop gain constant is K  8K 1 (5:26) giving G(s)H(s)  K s(s 2  s 2) (5:27) From equation (5.24) the characteristic equation is 1  K s(s 2  s 2)  0(5:28) or s(s 2  s 2)  K  0(5:29) which can be expressed as s 3  s 2  2s K  0(5:30) The characteristic equation can also be found from the closed-loop transfer function. Using equation (4.4) C R (s)  G(s) 1  G(s)H(s) Given the open-loop transfer function in equation (5.27), where H(s) is unity, then C R (s)  K s(s 2 s2) 1  K s(s 2 s2) (5:31) Cs () Rs () Plant Control Valve Proportional Controller + – K 1 4 s 2 (++2) ss 2 Fig. 5.4 Closed-loop control system. Classical design in the s-plane 115 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 116 ± [110±144/35] 9.8.2001 2:30PM Multiplying numerator and denominator by s(s 2  s 2) C R (s)  K s(s 2  s 2)  K (5:32) C R (s)  K s 3  s 2  2s K (5:33) Equating the denominator of the closed-loop transfer function to zero s 3  s 2  2s K  0(5:34) Equations (5.30) and (5.34) are identical, and both are the characteristic equation. It will be noted that all terms are present and have the same sign (Routh's first condition). Proceeding straight to Routh's array s 0 K s 1 (2 À K) s 2 1 K s 3 12 (5:35) where b 1  1 1 K 12          (2 À K) b 2  0 c 1  K To produce a sign change in the first column, K ! 2(5:36) Hence, from equation (5.26), to make the system just unstable K 1  0:25 Inserting (5.36) into (5.30) gives s 3  s 2  2s 2  0 factorizing gives (s 2  2)(s 1)  0 hence the roots of the characteristic equation are s À1 s  0 Æj  2 p and the transient response is c(t)  Ae Àt  B sin(  2t p  )(5:37) From equation (5.37) it can be seen that when the proportional controller gain K 1 is set to 0.25, the system will oscillate continuously at a frequency of  2 p rad/s. 116 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 117 ± [110±144/35] 9.8.2001 2:30PM 5.2.2 Special cases of the Routh array Case 1: A zero in the first column If there is a zero in the first column, then further calculation cannot normally proceed since it will involve dividing by zero. The problem is solved by replacing the zero with a small number " which can be assumed to be either positive or negative. When the array is complete, the signs of the elements in the first column are evaluated by allowing " to approach zero. Example 5.3 s 4  2s 3  2s 2  4s 3  0 s 0 3 s 1 4 À 6/" s 2 " 3 s 3 24 s 4 123 (5:38) Irrespective of whether " is a small positive or negative number in array (5.38), there will be two sign changes in the first column. Case 2: All elements in a row are zero If all the elements of a particular row are zero, then they are replaced by the derivatives of an auxiliary polynomial, formed from the elements of the previous row. Example 5.4 s 5  2s 4  6s 3  12s 2  8s 16  0 s 0 16 s 1 8/3 s 2 616 s 3 824 s 4 21216 s 5 168 (5:39) The elements of the s 3 row are zero in array (5.39). An auxiliary polynomial P(s)is therefore formed from the elements of the previous row (s 4 ). i.e. P(s)  2s 4  12s 2  16 dP(s) ds  8s 3  24s (5:40) The coefficients of equation (5.40) become the elements of the s 3 row, allowing the array to be completed. Classical design in the s-plane 117 //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 118 ± [110±144/35] 9.8.2001 2:30PM 5.3 Root-locus analysis 5.3.1 System poles and zeros The closed-loop transfer function for any feedback control system may be written in the factored form given in equation (5.41) C R (s)  G(s) 1  G(s)H(s)  K c (s À z c1 )(s À z c2 ) FFF(s Àz cn ) (s À p c1 )(s À p c2 ) FFF(s Àp cn ) (5:41) where s  p c1 , p c2 , FFF, p cn are closed-loop poles, so called since their values make equation (5.41) infinite (Note that they are also the roots of the characteristic equation) and s  z c1 , z c2 , FFF, z cn are closed-loop zeros, since their values make equation (5.41) zero. The position of the closed-loop poles in the s-plane determine the nature of the transient behaviour of the system as can be seen in Figure 5.5. Also, the open-loop transfer function may be expressed as G(s)H(s)  K(s À z 01 )(s À z 02 ) FFF(s Àz 0n ) (s À p 01 )(s À p 02 ) FFF(s Àp 0n ) (5:42) where z 01 , z 02 , FFF, z 0n are open-loop zeros and p 01 , p 02 , FFF, p 0n are open-loop poles. X X X X X X X X XXX σ X X X X jω Fig. 5.5 Effect of closed-loop pole position in the s-plane on system transient response. 118 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC05.3D ± 119 ± [110±144/35] 9.8.2001 2:30PM 5.3.2 The root locus method This is a control system design technique developed by W.R. Evans (1948) that determines the roots of the characteristic equation (closed-loop poles) when the open-loop gain-constant K is increased from zero to infinity. The locus of the roots, or closed-loop poles are plotted in the s-plane. This is a complex plane, since s   Æj!. It is important to remember that the real part  is the index in the exponential term of the time response, and if positive will make the system unstable. Hence, any locus in the right-hand side of the plane represents an unstable system. The imaginary part ! is the frequency of transient oscillation. When a locus crosses the imaginary axis,   0. This is the condition of marginal stability, i.e. the control system is on the verge of instability, where transient oscilla- tions neither increase, nor decay, but remain at a constant value. The design method requires the closed-loop poles to be plotted in the s-plane as K is varied from zero to infinity, and then a value of K selected to provide the necessary transient response as required by the performance specification. The loci always commence at open-loop poles (denoted by x) and terminate at open-loop zeros (denoted by o) when they exist. Example 5.5 Construct the root-locus diagram for the first-order control system shown in Figure 5.6. Solution Open-loop transfer function G(s)H(s)  K Ts (5:43) Open-loop poles s  0 Open-loop zeros: none Characteristic equation 1  G(s)H(s)  0 Substituting equation (5.3) gives 1  K Ts  0 i.e. Ts  K  0(5:44) Rs () Cs () K Ts + – Fig. 5.6 First-order control system. Classical design in the s-plane 119 [...]... +1)(s + 0.7s + 2) Imaginary Axis 1 0 .5 0 σa –0 .5 –1 –1 .5 –2 –3 .5 –3 –2 –2 .5 –1 .5 –1 –0 .5 Real Axis 0 0 .5 1 1 .5 0 0 .5 1 1 .5 Fig 5. 22 Proportional control, ship roll stabilization system 2 1 .5 2 1 G (s)H (s) = K(s + 4s + 8) 2 Imaginary Axis s (s + 1)(s + 0.7s + 2) 0 .5 0 –0 .5 –1 –1 .5 –2 –3 .5 –3 –2 .5 –2 –1 .5 Fig 5. 23 PID control, ship roll stabilization system –1 –0 .5 Real Axis ... calculated using the magnitude criterion as shown in Figure 5. 15 From Figure 5. 15 x(x ‡ 2)(x ‡ 5) ˆ 11: 35 (5: 84) Substituting x ˆ 0:73 (i.e s1 ˆ 5: 73) in equation (5. 84) provides a solution Hence the closed-loop poles for K ˆ 11: 35 are s ˆ 5: 73, À0:7 Æ j1: 25 (5: 85) Example 5. 9 (See also Appendix 1, examp59.m) The open-loop transfer function for a control system is G(s)H(s) ˆ K s(s2 ‡ 4s ‡ 13) Find the... rad/s (5: 93) Equating real parts À4!2 ‡ K ˆ 0 K ˆ 52 (5: 94) Angle of departure (Rule 10): If angle of departure is d , then from Figure 5. 16 a ‡ b ‡ d ˆ 180 d ˆ 180 À a À b d ˆ 180 À 123 À 90 ˆ À33 (5: 95) Locate point that corresponds to  ˆ 0: 25 From equation (5. 52) ˆ cosÀ1 (0: 25) ˆ 75: 5 (5: 96) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 132 ± [110±144/ 35] 9.8.2001 2:30PM 132 Advanced. .. Asymptote angles (Rule 5) (1 ‡ 0)  ˆ ˆ 60 , k ˆ 0 3À0 3 (1 ‡ 2) ˆ  ˆ 180 , k ˆ 1 2 ˆ 3À0 (1 ‡ 4) 5 ˆ ˆ 300 , k ˆ 2 3 ˆ 3À0 3 1 ˆ (5: 106) (5: 107) (5: 108) Asymptote intersection (Rule 6) a ˆ f(À1) ‡ (À0: 35 ‡ j1:37) ‡ (À0: 35 À j1:37)g À 0 3À0 a ˆ À0 :57 (5: 109) (5: 110) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 138 ± [110±144/ 35] 9.8.2001 2:30PM 138 Advanced Control Engineering Characteristic... //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 130 ± [110±144/ 35] 9.8.2001 2:30PM 130 Advanced Control Engineering jω x +5 x+2 x X –2 X 5 s1 σ Fig 5. 15 Determination of real closed-loop pole Closed-loop poles (For K ˆ 11: 35) : Since the closed-loop system is third-order, there are three closed-loop poles Two of them are given in equation (5. 81) The third lies on the real locus that extends from 5 to ÀI Its value is calculated... signals and control elements saturate Example 5. 6 Construct the root-locus diagram for the second-order control system shown in Figure 5. 8 Open-loop transfer function K G(s)H(s) ˆ (5: 46) s(s ‡ 4) Open-loop poles s ˆ 0, À4 Open-loop zeros: none Characteristic equation 1 ‡ G(s)H(s) ˆ 0 R (s) + K s (s + 4) – Fig 5. 8 Second-order control system C (s) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 121...//SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 120 ± [110±144/ 35] 9.8.2001 2:30PM 120 Advanced Control Engineering jω –∞ X σ Fig 5. 7 Root-locus diagram for a first-order system Roots of characteristic equation sˆÀ K T (5: 45) When K is varied from zero to infinity the locus commences at the open-loop pole s ˆ 0 and terminates at minus infinity on the real axis as shown in Figure 5. 7 From Figure 5. 7 it can be seen... 180 (5: 55) jG(s)H(s)j ˆ 1 (5: 56) The angle criterion Equation (5. 55) may be interpreted as `For a point s1 to lie on the locus, the sum of all angles for vectors between open-loop poles (positive angles) and zeros (negative angles) to point s1 must equal 180 ' In general, this statement can be expressed as Æ Pole Angles À Æ Zero Angles ˆ 180 (5: 57) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D... k ˆ 1 3À0 (5: 64) (1 ‡ 4) 5 ˆ ˆ 300 (À60 ), 3À0 3 k ˆ 2, i:e: n À m À 1 (5: 65) Asymptote intersection (Rule 6) a ˆ f(0) ‡ (À2) ‡ ( 5) g À 0 3À0 a ˆ À2:33 (5: 66) (5: 67) Characteristic equation: From equation (5. 24) 1‡ K ˆ0 s(s ‡ 2)(s ‡ 5) (5: 68) or s(s ‡ 2)(s ‡ 5) ‡ K ˆ 0 giving s3 ‡ 7s2 ‡ 10s ‡ K ˆ 0 (5: 69) Breakaway points (Rule 8) Method (a): Re-arrange the characteristic equation (5. 69) to make... 7s2 À 10s (5: 70) dK ˆ À3s2 À 14s À 10 ˆ 0 ds (5: 71) 3s2 ‡ 14s ‡ 10 ˆ 0 (5: 72) Multiplying through by ±1 s 1 , s 2 ˆ b ˆ À14 Æ p 142 À 120 6 b ˆ À3:79, À0:884 (5: 73) //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 128 ± [110±144/ 35] 9.8.2001 2:30PM 128 Advanced Control Engineering Method (b) 1 1 1 ‡ ˆ0 ‡ b b ‡ 2 b ‡ 5 (5: 74) Multiplying through by, b (b ‡ 2)(b ‡ 5) (b ‡ . À1 (5: 54) Since equation (5. 54) is a vector quantity, it can be represented in terms of angle and magnitude as = G(s)H(s)  180  (5: 55) G(s)H(s) jj  1 (5: 56) The angle criterion Equation (5. 55) . Engineering //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 1 15 ± [110±144/ 35] 9.8.2001 2:30PM Example 5. 2 (See also Appendix 1, examp52.m) Find the value of the proportional controller gain K 1 to make the control system shown. position in the s-plane on system transient response. 118 Advanced Control Engineering //SYS21/D:/B&H3B2/ACE/REVISES(0 8-0 8-0 1)/ACEC 05. 3D ± 119 ± [110±144/ 35] 9.8.2001 2:30PM 5. 3.2 The root