Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax:
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1 Control Sytem Engineering ( Chapter 7. Steady-State Error Prof. Kwang-Chun Ho kwangho@hanung.ac.kr Tel: Fax:
2 Introduction In thi leon, you will learn the following : How to find the teady-tate error for a unity feedback ytem How to pecify a ytem teady-tate error performance How to deign ytem parameter to meet teady-tate error performance pecification
3 Introduction In chapter 1, we learnt about 3 requirement needed when deigning a control ytem Tranient repone Stability Steady-tate error (SSE Up until now, we only covered until tranient repone and tability We learned in chapter 4, there are 4 type of tranient repone for a nd -order ytem Overdamped, Underdamped, Undamped, Critically damped 3
4 Introduction An example of elevator repone The tranient repone for elevator can be conidered a overdamped The ytem i table but ha teady-tate error! 4
5 Introduction What i teady-tate error? Steady-tate error i the difference between the input and output for a certain tet input a time approache infinity( t The concept of tead-tate error i limited to ytem that are table In thee ytem, the natural repone approache zero a time approache infinity Tet input ued for teady-tate error analyi and deign are Step, Ramp, Parabola lim et ( lim ( rt ( ct ( t t 5
6 Introduction Tet waveform for evaluating teady-tate error of poition control ytem Tet input for teady-tate error analyi and deign vary with target type 6
7 Introduction Example of ytem teted uing the tet ignal Targeting ytem: Targeting a tatic target. (e.g. a topping car We tet the ytem uing tep input becaue the poition of the car i in contant poition Targeting a car moving with contant velocity We tet the ytem uing ramp input becaue the car i moving in contant velocity Targeting an accelerating car We tet the ytem uing parabola input becaue the car i accelerating 7
8 SSE for unity feedback ytem Unity feedback ytem can be repreented a Steady tate error can be calculated from a ytem cloed-loop tranfer function, T(, or the open-loop tranfer function, G(, for unity feedback ytem SSE in term of T( Conidering figure (a, we find the SSE E( between the input, R( and the output C( E ( R ( C ( E( R([1 T(] C( R( T( 8
9 SSE for unity feedback ytem We can find final value of the error, in term of T( uing Math Ref.: We can only ue thi equation if T( i table, E( ha no pole in the right-half plane or pole on the imaginary axi other than the origin Initial Value Theorem t 0 0 e( e( lim e( t lim E( lim R( [1 T( ] If the function x(t and it firt derivative are Laplace tranformable and x(t Ha the Laplace tranform X(, and the lim X( exit, then lim X( lim x( t x(0 t0 Thi i particularly ueful in circuit and ytem to find out the initial condition in the time domain 9
10 SSE for unity feedback ytem Final Value Theorem Proof: lim X ( lim x( t x( 0 t Thi i particularly ueful in circuit and ytem to find out the the final value of x(t in the time domain 10
11 SSE for unity feedback ytem Example: Find the teady tate-error for a unity feedback ytem that ha T( = 5/( and the input i a unit tep Solution: R( =unit tep = 1/ T( = 5/( +7+10, we mut check the tability of T( uing Routh table or pole If T( i table, E( i then 11
12 SSE for unity feedback ytem E ( Before calculating the final value of the error, we mut check the poition of E( pole The pole for E( are at (0,0, (-,0 and (-5,0 Since all the pole are not on the right half plane or the imaginary axi we can ue the equation to calculate final error value in term of T( Finally, SSE i given by 1 5 elim R1 T lim (imilar to Output of Fig. 7.(a
13 SSE for unity feedback ytem SSE in term of G( Conidering figure (b, we find the SSE E( between the input, R( and the output C( E ( R ( C ( R ( E( C( E( G( 1 G ( Applying the final value theorem and verifying that the ytem i table, we have e R( ( lim E( lim G ( 13
14 SSE for Tet Input We are going to ue three type of input R( : tep, ramp and parabola So, the final value of the error for thi type of input can be decribed a e (1/ 1 ( lim 0 1 G ( 1 lim G ( 0 e( 0 (1/ 1 lim 0 1 G( lim G( 3 (1/ 1 e( lim 0 1 G( lim G( 0 (Step Input (Ramp Input (Parabolic Input 14
15 SSE for Tet Input Example: Find the teady-tate error for input of 5u(t, 5tu(t, and 5t u(t to the ytem with no integration below Solution: ut ( : etep ( 1lim G( tu( t : e ( ramp lim G( 0 tut e lim G( 0 5 ( 3 parabola ( Can get the ame reult by T(!! (imilar to Output of Fig. 7.(a (imilar to Output 3 of Fig. 7.(b 15
16 SSE for Tet Input Example: Find the teady-tate error for input of 5u(t, 5tu(t, and 5t u(t to the ytem with one integration below Solution: ut ( : etep ( 0 1lim G( tu( t : e ( ramp lim G( tut e lim G( 0 5 ( 3 parabola ( (imilar to Output 1 of Fig. 7.(a (imilar to Output of Fig. 7.(b Improvement for a ramp input! (No improvement for a parabolic input! 16
17 SSE for Tet Input Now let define the tatic error contant e e e ( lim 0 1 G ( 1 lim G ( 1 Kp 0 Example: (1 / 1 1 (1 / 1 1 ( lim 0 1 G ( lim G ( Kv 0 3 (1 / 1 1 ( lim 0 1 ( lim G G ( Ka 0 Find SSE via tatic error contant Solution: (Poition Contant (Velocity Contant (Acceleration Contant 17
18 SSE for Tet Input 18
19 SSE for Tet Input Firt tep i to calculate the tatic error contant K K K p v a 500( ( 5( 6 500(0 (0 5(0 6 limglim ( 8( 10( 1 (0 8(0 10(0 1 (500( ( 5( 6 lim Glim ( 8( 10( 1 lim Glim 0 0 ( Next tep i to calculate the final error value (500( ( 5( 6 8( 10( etep ( 0.161, eramp (, eparabola ( 1 K K K p v a 19
20 SSE for Tet Input SSE via ytem type We are till focuing on unity negative feedback ytem Below i a feedback control ytem for defining ytem type We define the ytem type to be the value of n in the denominator Type 0 when n = 0 Type 1 when n = 1 Type when n = 0
21 SSE for Tet Input Relationhip between input, ytem type, tatic error contant, and teady-tate error can be ummarized a (n=0 (n=1 (n= 1
22 SSE Specification We can ue the tatic error contant to repreent the teady-tate error characteritic of our ytem Example: Gain deign to meet a teady-tate error pecification (Find the value of K o that there i 10% error in the teady tate (The ytem i of Type 1, ince only Type 1 have K v that are finite contant Solution: 1 K 5 e( 0.1 Kv 10 lim G( K 0 v K 67
23 3 SSE for Diturbance Feedback control ytem are ued to compenate for diturbance (or unwanted input that enter a ytem Conider feedback control ytem howing diturbance SSE i Uing ( ( ( ( ( ( 1 G D G G E C ( ( ( C R E ( ( ( 1 ( ( ( ( 1 1 ( 1 1 D G G G R G G E
24 SSE for Diturbance e( lim E( lim R( 0 01 G ( G ( 1 G ( lim D ( e ( ( 0 R e D 1 G1( G( Auming a tep diturbance D(=1/ e D ( 1 1 lim lim G1 ( G ( 0 0 The teady-tate error produced by a tep diturbance can be reduced by increaed the dc gain of G 1 ( or decreaing the dc gain of G ( 4
25 SSE with State Space Uing the final value theorem, one can find the teady-tate error for ingle-input, ingle-output ytem in tate-pace form Conider the cloed-loop ytem repreented in tate pace The Laplace tranform of the error i uing x=ax+b r, Applying the final value theorem, we have y =Cx -1 E ( R ( Y ( = R (1 C( IA B -1 Y( R( C( I A B -1 e( lim E( lim R( 1C( I A B 0 0 5
26 SSE with State Space Example: Evaluate the SSE for the ytem Solution: Uing -1 e( lim E( lim R( 1C( I A B 0 0 For the unit-tep, R(=1/, e( 4/5 For the unit-ramp, R(=1/, e( Notice that the ytem behave like a Type 0 ytem 6
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