Chapter4 Time Domain Analysis of Control System Routh stability - PDF document

� � Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Chapter4 Time Domain Analysis of Control System � Routh stability criterion � Steady state errors � Transient response of the first-order system � Transient response of the second-order system Time domain performance specifications The relationship between the performance specifications and system parameters � Transient response of higher-order systems 1 Definition of characteristic equation + − + + + m m 1 C ( s ) b s b s L b s b = = − m m 1 1 0 G ( s ) + − + + + 1 n n R ( s ) a s a s a s a L − n n 1 1 0 The characteristic equation of the system is defined as + − + + + = n n 1 a s a s L a s a 0 − n n 1 1 0 2 Automatic Control System 1

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Transfer function: m ∏ + k ( s z ) i C ( s ) = = = i 1 G ( s ) q r R ( s ) ∏ ∏ + + σ + ω + σ − ω ( s p ) [ s ( j )][( s ( j )] j k k k k = = j 1 k 1 z i : zeros of the closed loop − − σ ± ω =− ζ ω ± ω − ζ 2 p , j j 1 poles of the closed loop j k k k n n k 3 m ∏ + k ( s z ) i 1 Step response: = ⋅ = i 1 C ( s ) q r ∏ ∏ s + + σ + ω + σ − ω ( s p ) [ s ( j )][( s ( j )] j k k k k = = j 1 k 1 α + β q a r a s ∑ ∑ = + + j 0 k k C s ( ) + s s p     = = + ζ ω + ω − ζ + ζω − ω − ζ 2 2 j 1 k 1 j s ( j 1 ) s ( j 1 )     k n n k n n k q r ∑ ∑ − = + + − ζ ω ω − ζ + ω − ζ p t t 2 2 C ( t ) a a e j e ( B cos 1 t C sin 1 t ) k k 0 j k k k k k k = = j 1 k 1 q r ∑ ∑ − − ζ ω = + p t + ω − ζ + ϕ 2 t a a e j D e k k sin( 1 t ) 0 j k k k k = = j 1 i 1 4 Automatic Control System 2

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University 4.1 Routh stability criterion Consider that the characteristic equation of a LTI system = + − + + + = n n 1 F ( s ) a s a s a s a 0 L − n n 1 1 0 Where all the coefficients are real numbers. In order that there be no roots of the above equation with positive real parts, it is necessary but not sufficient that 1. All the coefficients of the polynomial have the same sign. 2. None of the coefficients vanishes. 5 The Routh tabulation = + − + − + + + = n n 1 n 2 F ( s ) a s a s a s a s a 0 L − 0 1 2 n 1 n for n = 6 6 s a a a a 0 2 4 6 5 s a a a 0 1 3 5 − − − × a a a a a a a a a a a 0 = = = 4 1 2 0 3 1 4 0 5 1 6 0 s A B a 0 6 a a a 1 1 1 − − × − × Aa a B Aa a a A 0 a 0 = = = 3 3 1 5 1 6 1 s C D 0 0 A A A − − × × − × BC AD Ca A 0 C 0 A 0 = = = 2 6 s E a 0 0 6 C C C − ED Ca = 1 6 s F 0 0 0 E − × Fa E 0 0 6 s 0 0 0 F 6 Automatic Control System 3

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Rout-Hurwitz criterion The roots of the polynomial are all in the left half of the s-plane if all the elements of the first column of the Routh Array are of the same sign. If there are changes of signs in the elements of the first column, the number of sign changes indicates the number of roots with positive real parts. The necessary and sufficient condition for the stability of a system 7 Example: The characteristic equation of a system is + + + = 3 2 s 4 s 10 s 50 0 Determine the stability of the system using Routh criterion. （ 1 ） . Check the necessary condition Solution: （ 2). Routh array is 3 s 1 10 2 s 4 50 − 1 s 2 . 5 0 0 s 50 0 The system has two roots located in the right half of the s-plane. 8 Automatic Control System 4

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Special case 1 The first element in any one row of the Routh Array is zero but the other elements are not. We can replace the zero element in the Routh tabulation by an arbitrary small positive number ε and then proceed with the Routh test. 9 Example: The characteristic equation of a system is + + + + + = 5 4 3 2 s s 5 s 5 s 2 s 1 0 Determine the stability of the system using Routh criterion. Solution: Routh array is 5 1 5 2 s 4 s 1 5 1 ε 3 s 0 ( ) 1 0 ε − 5 1 2 s 1 0 ε ε − − ε 2 5 1 1 s 0 0 ε − 5 1 0 s 1 0 0 There are two sign changes in the first column of the tabulation, the 10 system has two roots located in the right half of the s-plane. Automatic Control System 5

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University Special case 2 The elements in one row of the Routh Array are all zero. 11 Example: The characteristic equation of the system is: + + + = 3 2 16 16 0 s s s The Routh array is : 3 s 1 16 2 s 1 16 3 s 1 16 1 s 0 0 2 s 1 16 0 s 16 0 = s + = 2 1 The auxiliary equation is: A ( s ) 16 0 s 2 0 0 s 16 0 The sign of the elements in the first column does not change, the system is stable . 12 Automatic Control System 6

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University + + R s ( ) C s ( ) K s ( 1) Example: + + s Ts ( 1)(2 s 1) − Determine the value of K and T to make the closed loop be stable. Solution: the characteristic equation: + + + + + = 3 2 2 Ts ( 2 T ) s ( K 1 ) s K 0 + 3 s 2 T K 1 Routh array: + 2 s 2 T K + + − ( 2 T )( K 1 ) 2 TK 1 s 0 + 2 T 0 s K 0 13 > > > K 0 , 2 T 0 , T 0 + T > 2 0 + + − > ( 2 T )( K 1 ) 2 TK 0 The condition for the stability is: + T 2 < < 0 K − T 2 14 Automatic Control System 7

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University K 6 + T 2 < K 4 − T 2 2 Stable 0 6 8 4 T 2 + T 2 < < 0 K − T 2 15 The characteristic equation of the system is Example: + + + = 3 2 s 8 s 10 s 2 0 Determine the stability of the system. Analyze how many roots = − lie between the imaginary axis and the line s 1 . 3 s 1 10 Solution: Routh array: 2 s 8 2 1 s 9 . 75 0 the system is stable. 0 s 2 0 = s 1 − Let s 1 Im The characteristic equation becomes: − 1 0 Re + − − = 3 2 s 5 s 3 s 1 0 1 1 1 16 Automatic Control System 8

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University + − − = 3 2 s 5 s 3 s 1 0 1 1 1 − 3 s 1 3 1 − Routh array for the above equation: 2 s 5 1 1 − 1 s 2 . 8 0 1 − 0 s 1 0 1 there is one root on the right side of the Im line s = -1. − 0 Re 1 17 4.6 Steady state errors ∞ = ∞ − ∞ （ ） e c ( ) r ( ) 18 Automatic Control System 9

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University + R s ( ) E s ( ) C s ( ) G s ( ) − B s ( ) H s ( ) R ( s ) = E ( s ) + 1 G ( s ) H ( s ) 1 = = = e lim e t ( ) lim s E ( s ) lim s R ( s ) + s s → ∞ → → 1 G ( s H ) ( s ) t s 0 s 0 19 the types of the control systems m ∏ + K ( s z ) + + + r i K ( s z )( s z ) ( s z ) L = = = r 1 2 m i 1 G ( s ) ν + + + q s ( s p )( s p ) L ( s p ) ∏ ν + 1 2 q ( ) s s p j = + j v 1 ν = 0, type 0 system ν = 1, type 1 system ν = 2, type 2 system 20 Automatic Control System 10

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University a. position error constant = ⋅ For a step input r ( t ) R u ( t ) R ( s ) = = = e lim e ( t ) lim sE ( s ) lim s + ss → ∞ → → 1 G ( s ) H ( s ) t s 0 s 0 R / s R = = lim s lim + + → → 1 G ( s ) H ( s ) 1 G ( s ) H ( s ) s 0 s 0 R R = = lim1 e + + ss → G s H s ( ) ( ) 1 lim G s H s ( ) ( ) s 0 → s 0 = lim G s H s ( ) ( ) K P → s 0 21 b. velocity error constant = For a ramp input: r ( t ) Rt R ( s ) = = = e lim e ( t ) lim sE ( s ) lim s + ss → ∞ → → 1 G ( s ) H ( s ) t s 0 s 0 2 / R s R = = lim s lim + → → 1 G ( s ) H ( s ) sG ( s ) H ( s ) s 0 s 0 R = = K lim sG s H s ( ) ( ) e ss v K → s 0 v 22 Automatic Control System 11

Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University c. Acceleration error constant 1 = for acceleration input 2 r t ( ) Rt 2 R ( s ) = = = e lim e ( t ) lim sE ( s ) lim s + ss → ∞ → → 1 G ( s ) H ( s ) t s 0 s 0 3 R / s R = = lim s lim + 2 → → 1 G ( s ) H ( s ) s G ( s ) H ( s ) s 0 s 0 R = = 2 K lim s G ( s ) H ( s ) e a ss → K s 0 a 23 Steady state error 24 Automatic Control System 12