Chapter 7. Steady-State Errors Things to know - The steady-state error for a unity feedback system - A system s steady-state error performance - The steady-state error for disturbance inputs - The steady-state error for nonunity feedback system - Designing system parameters to meet steady-state error performance specifications - The steady-state error for systems represented in state space Chapter 7. Steady-State Errors /9
Chapter 7. Steady State Errors Definition and Test Inputs: t s0 e lim e( t) lim se( s) Table 7. Test waveforms for evaluating steadystate errors of position control systems Chapter 7. Steady-State Errors /9
Figure 7. Test inputs for steady-state error analysis and design vary with target type Figure 7. Steady-state error: a. step input; b. ramp input Chapter 7. Steady-State Errors 3/9
Figure 7.3 Closed-loop control system error: a. general representation; b. representation for unity feedback systems e lim e( t) lim se( s) t s0 Figure 7.4 System with: a. finite steady-state error for a step input; b. zero steady-state error for step input e e t se s t s0 s0 s lim ( ) lim ( ) lim 0 s Chapter 7. Steady-State Errors 4/9
7. Steady- State Error for Unity Feedback Systems 폐루프제어시스템의오차신호 E(s) 는 E( s) R( s) C( s) R( s) T( s) R( s) Gs ( ) 폐루프제어시스템이안정하다는가정아래서최종값정리 (final value theorem) 를이용하여 e e t se s ss sr() s lim ( ) lim ( ) lim t s0 s0 Gs ( ) Gs () m ( s z ) i n N s ( s p ) j i j 시스템형태에따른정상상태오차 Chapter 7. Steady-State Errors 5/9
Ex. 7.3 Find the steady-state errors for inputs of 5 u( t), 5 tu( t), 5 t u( t) 5 u( t), 5 tu( t), 5 t u( t) Figure 7.6 Feedback control system for Example 7.3 ) For a step input 5 ut ( ) sr( s) 5 5 estep lim e step( t) lim se( s) lim = = =0 t s0 s0 G( s) lim G( s) ) For a ramp input 5 tu( t) sr( s) 5 5 eramp lim e ramp ( t) lim se( s) lim = = = t s0 s0 G( s) lim sg( s) 00 0 3) For a parabolic input 5 t u( t) sr( s) 0 0 eparabola lim e parabola ( t) lim se( s) lim = = = t s0 s0 G( s) lim s G( s) 0 s0 s0 s0 Chapter 7. Steady-State Errors 6/9
7.3 Static Error Constants and System Type ) For a step input, ) For a ramp input, Rs () s Gs () s0 m ( s z ) i n N s ( s p ) j sr( s) estep lim e step ( t) lim se( s) lim = t s0 s0 G( s) lim G( s) r( t) tu( t), R( s) s sr( s) eramp lim e ramp ( t) lim se( s) lim = t s0 s0 G( s) lim sg( s) s0 v p i j 3) For a parabolic input, r( t) t u( t), R( s) 3 s sr( s) eparab lim e parab ( t) lim se( s) lim = t s0 s0 G( s) lim s G( s) s0 Chapter 7. Steady-State Errors 7/9 a
System Type Figure 7.8 Feedback control system for defining system type sr() s ess lim e( t) lim se( s) lim, t s0 s0 Gs ( ) Gs () m ( s z ) s n i n j ( s p ) i j Chapter 7. Steady-State Errors 8/9
Ex. 7.4 Find the steady -state errors for inputs ) Type 0 (n=0) e e e 5.08 lim sg( s) 0 s0 lim s G( s) 0 s0 500 5 p lim G( s) s0 v a step ramp parab 8 0 ( ) 0.6 ( ) ( ) v a p Figure 7.7 Feedback control systems for Example 7.4 Chapter 7. Steady-State Errors 9/9
) Type (n=) e e e p v a step lim G( s) s0 500 5 6 lim sg( s) 3.5 s0 80 ramp parab lim s G( s) 0 s0 ( ) 0 ( ) 0.03 3.5 ( ) v a p Chapter 7. Steady-State Errors 0/9
) Type (n=) p v a lim G( s) s0 lim sg( s) s0 500 4 5 6 7 80 lim s G( s) 875 s0 e e e step ramp parab ( ) 0 p ( ) 0 v ( ).40 875 a 3 Chapter 7. Steady-State Errors /9
Table 7. Relationships between input, system type, static error constants, and steady-state errors Chapter 7. Steady-State Errors /9
7.4 Steady-State Error Specifications Figure 7.9 A robot used in the manufacturing of semiconductor random-access memories (RAMs) similar to those in personal computers. Steady-state error is an important design consideration for assembly-line robots. Westlight/ Charles O Rear. Chapter 7. Steady-State Errors 3/9
Ex. 7.6 Find the value of so that there is 0% error in the steady state for ramp input Figure 7.0 Feedback control system for Example 7.6 Sol) Since the system is Type I e ramp ( ) 0. v 5 v 0 lim sg( s) s0 67 6 7 8 Chapter 7. Steady-State Errors 4/9
7.5 Steady-State Error for Disturbances Figure 7.Feedback control system showing disturbance C( s) E( s) G ( s) G ( s) D( s) G ( s) But E( s) R( s) C( s) C( s) R( s) E( s) E( s) G ( s) G ( s) R( s) D( s) G ( s) G ( s) E s R s D s G ( s) G ( s) G ( s) G ( s) ( ) ( ) ( ) Chapter 7. Steady-State Errors 5/9
Find Steady-State Errors sr( s) sg ( s) lim e( t) lim se( s) lim lim D( s) t s0 s0 G 0 ( s) G( s) s G ( s) G( s) e ( ) e ( ) R D In case of a step disturbance D( s) / s e D( ) lim D( s) s0 G( s) G( s) sg ( s) lim lim G ( s) G () s s0 s0 Figure 7. Figure 7. system rearranged to show disturbance as input and error as output, with R(s) = 0 Chapter 7. Steady-State Errors 6/9
Example 7.7 Find the steady-state error component due to a step disturbance for the system of Fig. 7.3 Sol) Fig. 7.3 Feedback control system for Example 7.7 e sg ( s) ( ) ( ) lim lim G 0 000 000 ( s) G () s D( ) lim D( s) s0 G s G s s0 s0 Chapter 7. Steady-State Errors 7/9
7.6 Steady-State Error for Nonunity Feedback Systems Figure 7.5 Forming an equivalent unity feedback system from a general nonunity feedback system Chapter 7. Steady-State Errors 8/9
Example 7.8 Find the system type and steady state errors Sol) Figure 7.6 Nonunity feedback control system G () s e G( s) 00( s 5) 3 G( s) H( s) G( s) s 5s 50s 400 00 5 5 p lim Ge( s) s0 400 4 e( ) 4 5/ 4 p Chapter 7. Steady-State Errors 9/9
( 제어시스템설계 3 장참조 ) 특성방정식과안정도 Lyapunov 방법 : 선형및비선형시스템에대한안정도조사 선형시스템의경우 : 특성방정식의근을직접조사하거나특성방정식의근중에서양의실수부를갖는근이존재하는지를판정하는방법을이용하여시스템의안정도를조사하고있다. 시스템의안정도를판별하기위하여전달함수 G(s) 를고려 y( s) N( s) ( s z) ( s zm) Gs () r( s) D( s) ( s p ) ( s p ) 단위스텝입력 (r(s) = /s) 을가했을때의출력 y(s) 를생각하기로한다 ( s z ) ( s z ) m y( s) G( s) r( s), pi i j i s( s p) ( s pn) 또한, 식 (3.8) 을다음과같이부분분수로전개한다 ys c c c c 0 n () s s p s p s p n n (3.80) (3.8) (3.8) Chapter 7. Steady-State Errors 0/9
위식을역 Laplace 변환하면출력 y(t) 는다음과같다. y t c0 ce ce cne us t pt pt n ( ) pt ( ) 여기서, u () s t 는단위스텝함수이다. (3.83) 그림 3.3 σ 값에따른과도응답 시스템의안정도 : 특성방정식 D(s) 의근의상태에따라결정 n D( s) a s a s a s a 0 n n n 0 시스템이안정특성방정식의근의실수부가모두음수. (3.84) Chapter 7. Steady-State Errors /9
( 제어시스템설계 3 장참조 ) 3.8 MATLAB 을이용한제어시스템성능및안정도평가 MATLAB과 SIMULIN를이용하여시스템의성능및안정도를평가하고또한시스템의응답특성을컴퓨터시뮬레이션하는방법 [ 예제 3.9] 다음과같은전달함수 G(s) 로표현되는시스템의안정도와시간역성능을평가하기로한다. 이시스템은 Gs () s 5 s0 p, j3 실수부가음수이므로시스템은안정하다. 에극점이있다. 시스템의모든극점의 Chapter 7. Steady-State Errors /9
num = [5]; den = [ 0]; pole = roots(den) pole = -.0000 + 3.0000i -.0000-3.0000i MATLAB 프로그램 3. step(num, den) grid xlabel('time(sec)') ylabel('output') title('unit step response of G(s)=5/s^+s+0') 그림 3.4 Gs () s 5 s0 의 단위스텝응답 Chapter 7. Steady-State Errors 3/9
MATLAB 프로그램 3. 는시스템의 시간역성능을나타내는시간응답에 관한지수인최대값시간 ( t p ), 퍼센트오버슈트 ( P.O ), 상승시간 ( t r ), % 정착시간 ( t s ), 정상상태오차 ( e ss ) 를 구하는프로그램이다. MATLAB 프로그램 3. num = [5]; den = [ 0]; % Fv is finalvalue Fv = polyval(num, 0) / polyval(den, 0); [y, x, t] = step(num, den); [Y, k] = max(y); % Tp is peak time Tp = t(k); % PO is percent-overshoot PO = 00*(Y-Fv)/Fv; % compute rising time n= ; while y(n)<fv, n=n+; end % Tr is rising time Tr = t(n); % compute settling time l = length(t); while (y( ㅣ ) > 0.98 * Fv) & (y( ㅣ ) <.0 * Fv) l = l- end % Ts is settling time Ts = t(l); Ess = -Fv [Tp PO Tr Ts Ess] ans =.0606 35.0607 0.4545 5.0000 0.500 Chapter 7. Steady-State Errors 4/9
[ 예제 3.] 그림 3.6 은 /4- 차량시스템모델의개략도이다. /4- 차량시스템모델은자동차의네바퀴중한바퀴에대한모델이다. 시스템파라미터들의값은다음과같다. m m k k s u s t : ( 현가장치위의질량 ) 40( kg) : ( 현가장치아래의질량) 36( kg) : ( 코일스프링상수) 6,000( N / m) : ( 타이어탄성의스프링상수) 60,000( N / m) c : ( 현가장치의감쇠계수) 500( N sec/ m) 그림 3.6 /4- 차량시스템의모델의개략도 Chapter 7. Steady-State Errors 5/9
Newton 의제 법칙을이용하여 /4- 차량시스템의운동방정식을유도 m z c( z z ) k ( z z ) 0 s s s u s s u m z c( z z ) k ( z z ) k ( z z ) 0 u u u s s u s t u r 이시스템은 4 차시스템이므로다음과같이상태변수를선정 x z z ( 현가장치의변위), x z x z z ( 타이어의상대변위), x z s u s 3 u r 4 u 이때, /4- 차량시스템의상태방정식은다음과같이유도된다. x x x4 c ks x ( x x4) x ms ms x3 x4 zr c k k x ( x x ) x x s t 4 4 3 ms mu mu Chapter 7. Steady-State Errors 6/9
혹은, X AX Bu 여기서, A 0 0 0 ks / ms c / ms 0 c / m s 0 B 0 0 0 ks / mu c / mu kt / mu c / mu 0 그리고, u z r 이다. 시스템행렬 A에주어진시스템파라미터들의값을대입하면, A 0 0 66.67.083 0.083 0 0 0 444.4 3.89 4444 3.89 Chapter 7. Steady-State Errors 7/9
MATLAB 프로그램 3.4 A = [0 0 -; -66.67 -.083 0.083; 0 0 0 ; 444.4 3.89-4444 -3.89]; B = [0 ; 0; -; 0]; C = [ 0 0 0]; D = [0]; [num, den] = sstf(a, B, C, D) num =.0e+003 * 0 0.0000 0.0000-4.4444 0.0000 den =.0e+005 * 0.0000 0.000 0.0496 0.096.9630 t = (0 : 0.0 : 5); impulse(num, den, t) grid ylabel( 'output (m)' ) title ('impulse response of suspension system') 그림 3.7 MATLAB 을이용한 /4- 차량시스템의임펄스응답선도 Chapter 7. Steady-State Errors 8/9
4 th edition Home Work #5 (Due date: one weeks from today). Solve Problem () on page 357(405) in the text book.. Solve Problem () on page 358(407) in the text book. 3. Solve Problems 30(9) on page 360(40) in the text book. 4. Solve Problem 45(4) on page 36(44) in the text book. Chapter 7. Steady-State Errors 9/9