Basic Control Systems

Chapter 5: Time Response CONTENTS PREFACE .................................................................................................................. iii CHAPTER 1 ......................................................................................................................... 1 1.1 1.2 1.3 1.4 INTRODUCTION ADVANTAGES OF CONTROL SYSTEM OPEN-LOOP AND CLOSED-LOOP SYSTEM DESIGN OBJECTIVES PROBLEMS AND SOLUTIONS .................................................................................. 3 CHAPTER 2 ............................................................................................................... 9 2.1 2.2 2.3 2.4 INTRODUCTION LAPLACE TRANSFORM THEOREM INVERSE LAPLACE TRANSFORM LAPLACE TRANSFORM TABLE AND THEOREMS PROBLEMS AND SOLUTIONS ................................................................................ 12 CHAPTER 3 ....................................................................................................................... 23 3.1 3.2 3.3 3.4 3.5 INTRODUCTION ELECTRICAL NETWORK 3.2.1 Loop or Mesh Analysis 3.2.2 Nodal Analysis MECHANICAL SYSTEM MECHANICAL SYSTEM: SYSTEM WITH GEARS ELECTROMECHANICAL SYSTEM PROBLEMS AND SOLUTIONS ................................................................................ 32 CHAPTER 6 6.1 6.2 INTRODUCTION ROUTH-HURWITZ CRITERION 72 Chapter 5: Time Response CHAPTER 5 TIME RESPONSE CHAPTER OUTCOMES Upon completion of this chapter, student should be able to: 1. 2. 3. 4. 5. 5.1 Define poles and zeros of a transfer function Determine the system response characteristic by observing its poles and zeros Describe quantitatively the characteristic response of first order and second order system Distinguish the characteristic response between first order and second order system Differentiate between overdamped, underdamped, undamped and critically damped responses INTRODUCTION Time response is an evaluation of state and output response with respect to time. The time response of a control system consists of transient response and steady-state response. Transient response is defined as the part of the time response that goes to zero as time increases. It is a function of systems dynamic and is independent on the input function. The steady-state response is simply the part of the total response that remains after the transient has died out[4]. This response is a function of both the system dynamics and input function. In this chapter, the response of first order and second order systems will be discussed. 5.2 POLES AND ZEROS Poles of a transfer function are defined as any roots of the denominator of the transfer function that are common to roots of the numerator[1]. And zeros of a transfer function are defined as any roots of the numerator of the transfer function that are common to roots of the denominator [1]. The output response of a system is the sum of two responses: the forced response and the natural response. A pole of the input function generates the form of the forced response whereas a pole of the transfer function generates the form of the natural response. The poles and zeros generate the amplitudes for both the forced and natural responses. 73 Chapter 5: Time Response 5.3 FIRST ORDER SYSTEM There is only one pole in a first order system. Figure 5.1 describes a block diagram of first order system without zero and its plot on s-plane. jω R(s) = 1/s G(s) = C(s) a σ sa x -a Figure 5.1 C ( s) a 1  . For a unit step input which R( s)  , the R( s ) s  a s a output response of C (s) is G( s) R( s)  . Taking the inverse Laplace Transform, the s ( s  a) output response is given by c(t ) 1  e  at . This is illustrated in Figure 5.2. Referring to Figure 5.1, G( s)  Figure 5.2 Parameter a is the important part in order to analyze first order system response. When 1 t , a c(t ) t  1  1  e 1  1  0.37  0.63 …(5.1) a 74 Chapter 5: Time Response Three transient response performance specifications for first-order system are:  Time Constant Based on Figure 5.2, time constant is the time for the step response to reach 63% of 1 its final value. In other word, is a time constant of the response. The unit of a 1 reciprocal of time constant is  frequency . Since the pole of the transfer sec onds function is at –a, we can say that the pole is located at the reciprocal of the time constant, and the farther the pole from the imaginary axis, the faster the transient response.  Rise Time, Tr It is defined as the time for waveform to go from 0.1 to 0.9 of its final value [1]. It also can be obtained from 2.2 …(5.2) Tr  a  Settling Time, Ts The time for the response to reach and stay within 2% or 5% of its final value is called as settling time. It also can be calculated using 4 …(5.3) Ts  a 5.4 SECOND ORDER SYSTEM A system which has two poles is called a second-order system. The second order system can display characteristics much like a first-order system or, depending on component values, display damped or pure oscillations for its transient response. A general equation for second order system with no zero is shown in Figure 5.3 as below: R(s) b s  as  b C(s) 2 Figure 5.3 Table 5.1 shows the numerical examples of the second-order system responses which are overdamped, critically damped, underdamped and undamped, by assigning appropriate values to parameters a and b. 75 Chapter 5: Time Response Table 5.1: Pole zero location and output response of second order systems (overdamped, critically damped, underdamped and undamped). System Pole zero on s-plane Output snse, c(t) jω R(s) 4 2 s  8s  4 C(s) σ Overdamped x x -7.46 -0.54 jω R(s) 4 2 s  2s  4 C(s) x x σ Critically damped -2 jω R(s) 4 s  2s  4 -1 + j1.73 x -1 - j1.73 x C(s) σ 2 Underdamped jω R(s) 4 2 s 4 x j2 C(s) σ x -j2 Undamped 76 Chapter 5: Time Response Two physically meaningful specifications for second-order system are defined:Natural Frequency, wn the frequency of oscillation of the system without damping.  Damping Ratio, damping ration is defined as:exponentia l decay frequency natural frequency (rad/s) 1 natural period  2 exponentia l time constant Consider the general system b G( s )  2 s  as  b This general second-order transfer function equal to: 2 wn G( s )  2 2 s  2wn s  wn Solving the poles location of the transfer function of G(s) yields:   s1,2  wn  jwn  2  1 …(5.4) …(5.5) …(5.6) The various cases of second-order response are a function of  and are summarized in Table 5.2. Table 5.2: Relationship between various cases of second order system and damping ratio,  . Second order System Overdamped Critically damped Underdamped Undamped Damping ratio,   1  1 0  1  0 77 Chapter 5: Time Response 5.5 UNDERDAMPED SECOND ORDER SYSTEM The common model for physical problem of second-order system is Underdamped. The transient response performance specifications for underdamped system are percentage overshoot (%OS), peak time (Tp), settling time (Ts) and rise time (Tr). These specifications are defined as follow (refer Figure 5.4): Figure 5.4 Peak time, Tp : The time required to reach the first, or maximum peak or Tp can be calculated using Eq (5.7): TP   …(5.7) wn 1   2 Percent overshoot, %OS : The amount that the waveform overshoots the steady state, or final value at the peak time: %OS  Cmax  C final C final  100  e (  / where, Cmax  1  e (  / 1 2 ) 1 2 )  100 …(5.8) and for a unit step C final  1 The inverse of Eq (5.8) allows in solving for  :    ln( %OS ) 100  2  ln 2 ( %OS 100 ) …(5.9) 78 Chapter 5: Time Response Settling time, Ts : The time required for the transient’s damped oscillations to reach and stay within +2% (or +5%) of the steady state value: For criteria between +2% 4 Ts  wn For criteria between +5% 3 Ts  wn …(5.10) …(5.11) Rise time, Tr : The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value. Or we can use the formula: ( 1.76 3  0.417 2  1.039  1 ) Tr  wn …(5.12) 79 Chapter 5: Time Response PROBLEMS AND SOLUTIONS 1. For each of the transfer functions shown below, find the location of the poles and zeros, plot/sketch them on s-plane, and then determine the system order. 5 ( s  3)(s  6) (s 2) (b) T ( s )  s( s 2  2 s  5 ) 20 (c) T ( s )  4 3 s  15 s  73s 2  129 s  70 (a) T ( s)  Solution: (a) T ( s)  numerator 5 ( s  3)(s  6) denominator Zero : No zero because no variable ‘s’ in numerator. Pole : ( s  3 )( s  5 )  0 ( s  3 )( s  5 )  0 (s  3)  0 s  3 (s5)0 s  5 Therefore the system has no zero and 2 poles located at -3 and -5. System is a second order system (It has two poles: order system depends on number of poles) Imaginary axis, jw s-plane j5 j4 j3 j2 j1 Real axis, σ X -7 -6 -5 X -4 -3 -2 -1 0 80 Chapter 5: Time Response (b) T( s )  Zero Pole (s2) s( s 2  2 s  5 ) You can find the answer using: 2 1) Use the formula: as + bs + c = 0 -b± √b2-4ac 2a : (s 2) 0 s  2 2) Scientific calculator How? Depend on the type of calculator. Basically using: Mode EQN. Please refer the manual of calculator. : s( s 2  2s  5 )  0 s 0 s 2  2s  5  0 s  -1+j2; -1 –j2 3) Matlab (software) % write this command in %MATLAB roots([1 2 5]); Therefore the system has 1 zero at -2 and 3 poles located at 0, -1+j1 and -1–j1. System is a third order system (It has three poles: order system depends on number of poles) Imaginary axis, jw s-plane j5 j4 j3 X j2 j1 X O Real axis, σ -7 -6 -5 -4 -3 -2 -1 0 1 -j1 X 2 3 4 5 6 7 -j2 -j3 -j4 -j5 (c) T( s )  20 s  15 s  73s 2  129 s  70 4 3 Zero : No zero because no variable ‘s’ in numerator Pole : s 4  15s 3  73s 2  129 s  70  0 s  1,2,5,7 You can find the answer using: Matlab % write this command in MATLAB roots([1 2 5]); Therefore the system has no zero 4 poles located at -1, -2, -5 and -7. System is a fourth order system (It has four poles: order system depends on number of poles) 81 Chapter 5: Time Response Imaginary axis, jw s-plane j5 j4 j3 j2 X Real axis, σ -7 -6 j1 X X X -5 -4 -3 -2 -1 0 1 -j1 2 3 4 5 6 7 -j2 -j3 -j4 -j5 2. A first-order system shown in Figure Q5-2, do the following: i) Find the value of a if the settling time of the system, Ts = 1.6 s. ii) Calculate the time constant, Tc and rise time, Tr. iii) Then find the step response, c(t) of the system. iv) Sketch the graph of the response and show (label) the Ts, Tc and Tr. R(s) = 1/s G(s) = C(s) a sa Figure Q5-2 Solution: i) Ts  4 T herefore, a  ; a ii) 4 Ts 1 T ime constant,   4   2.5 1.6  0.4 s 2.5 Tr  2.2  0.88 s 2.5 iii) G(s)  2.5 C(s)  R(s) s  2.5 for step response, R(s)  C(s)  A s A  1 s , C(s)  G(s)  R(s)  2.5 s  2.5  1 s B s  2.5 2.5  1, s  2.5 s0 B 2.5  1 s s2.5 Taking inverse Laplace Transform, c(t)  1  e  t 82 Chapter 5: Time Response Step Response 1 Ts = 1.6 s 0.9 Tr = 1.6 s 0.8 0.7 0.63 τ = 0.4 s Amplitude 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 2 2.5 Tim e (sec) 3. Determine the output step response, c(t) for the open loop system with the transfer function, G(s)  3 s4 and then calculate the time, t when the c(t)=0.6. Solution: G(s)  C(s) 3  R(s) s4 for step response, R(s)  1 3 1 , C(s)  G(s)  R(s)   s s4 s A B  s s4 3 3 A  , s  4 s0 4 C(s)  C(s)  3 B 3 s  s 4 3 4 3 4 4 s s4 T aking inverse Laplace T ransform, c(t)  3 3 4t 3  e  (1  e  4 t ) 4 4 4 T he time t when c(t)  0.6. 3 (1  e  4 t ) 4 4 e  4t  1 - (  0.6)  1 - 0.8  0.2 3 - 4t  ln 0.2  -1.609 0.6  t  0.402s 83 Chapter 5: Time Response 4. Find the value of natural frequency, wn for the unity feedback system when the dominant pole is -0.833 + j1.44 with   0.5 . Solution: The generalequationof locationpole on s - planeis -  n  jn 2 Therefore, n  0.833 0.833 When   0.5, n   1.666 0.5 5. Find the output step response, c(t) for the closed-loop system with the transfer 5 function G  s   , and then determine the time constant, rise time and settling s5 time. (Prove your answer using matlab simulation.) Solution: G s  C s  5 s5 and G s R s 1 G s H s H  s  1 5 s  5   5  1   1  s5 5 s  5  s 55 s5  .: C s R s 1  10     2  s  10  1 1  10  for step response, R ( s )  , C  s     s 2  s ( s  10)  1 A B      2  s s  10  A 10 ( s  10) 1 , B s 0 10  1 s s 10 taking inverse LaplaceTransform, c(t )   1 1  e 10t 2  from equation 4.4, we can conclude that , a  10 .: Time cons tan t ,  .: Rise Time, Tr  1  0.1s , 10 .: Settling Time, Ts  4  0.4 s, 10 2.2  0.22 s 10 84 Chapter 5: Time Response % first-order system num=[5] ; % numerator den=[1 5];% denumerator G=tf(num,den); % tf Gc=feedback(G,[1]); % [c,t]=step(Gc,t); % t=[0:0.01:2]; % plot(t,c)  6. Given the transfer function G( s)  100 , find Tp ,% OS andTs . s  25s  100 2 Solution: n 100 , the value of damping ratio,   2 2 s  15s  100 s  2 n s   n 2 and natural frequency,  n are 0.75 and 10 respectively. Thus,  Tp   0.475 n 1   2 2 Comparing G( s)  %OS  e Ts  7. 4  n   / 1 2     100  2.84%  0.533s Referring to Figure Q5-7, determine the step response for the following transfer functions, G(s) : R(s) G(s) C (s) Figure Q5-7 G( s)  900 s  90s  900 2 85 Chapter 5: Time Response Solution: 1 900 Given R( s)  , C ( s)  G( s) R( s)  2 s s s  90s  900 900 900 Factorize C ( s)   2 s( s  90s  900) s( s  78.54)(s  11.46)  i) ii) By partial fraction, C ( s)  iii)  A B C   s s  78.54 s  11.46 A 1 B  0.171 C  1.171 1 0.171 1.171   s s  78.54 s  11.46 iv) Thus, C ( s)  v) Taking inverse Laplace Transform, c(t )  u(t )  0.171e 78.54t  1.171e 11.46t 8. C (s) 20  2 , find its time response specifications R( s) s  5s  16 and sketch the output response. For a system that having Solution:  n 2 16   n  4 rad / sec 2 n  5   d  n 1   2 5 2 n  0.625   d  4 1  (0.625) 2  3.12 rad / sec Time response specification: TS  Tr  4 1.6 s (4)(0.625) 2.16  0.6 n 2.16(0.625)  0.6  4  0.488 s TP   3.12 1.01 s %OS 100  e   1 2  100  e  0.625 1 0.6252  8.08% 86 Chapter 5: Time Response Step Response 1.4 C(t) 1.2 1 Amplitude 0.8 0.6 0.4 1.01 0.2 0 0 0.5 0.488 9. 1.5 1.6 1 2 2.5 Time (sec) t (sec) A closed-loop control system with negative unity feedback has a plant with a transfer function, G s   4 . s 2  3s  2 a) Determine the closed-loop transfer function, C s  R s  . b) Find  , wn, Tp, %OS, Ts and Tr c) Using Matlab program, plot the output response of the system to a unit step input and compare the results Solution: C s G s   R s 1 G s H s 4  s  3s  2 4 R s 1 1 s 2  3s  2 4 2 s  3 s2  2 s  3s  2  4 s 2  3s  2 4  2 s  3s  6 C s 2 Comparing the result from (a) to equation (4.9), we get; n  6 , 2n  3  3  2  6   0.612 87 Chapter 5: Time Response TP   n 1   2  %OS  100 e (  2.45 1  0.6122 1 2 ) ,  1.62s Ts  4  2.67s , n Tr  2.16  0.6  0.79s n  8.8% (c) 0.063 % Pole-zero plot num=[4] ; den=[1 3 2]; G=tf(num,den); Gc=feedback(G,[1]); t=[0:0.01:5]; [c,t]=step(Gc,t); xlabel('Time (sec)'),ylabel('c1(t)') plot(t,c,'--'),grid 0.33 1.12 1.62 2.67 From simulation, we get ; Ts  2.67s, Tp  1.62, Tr  1.12  0.33  0.79 and %OS  10. 0.063 100%  8.75% 0.72 Figure Q5-10 shows a negative-unity feedback system with G(s) and H (s) are given as follow: 1 H (s)  5(s  1)(s  2) G( s)  2 s ( s  12) Find out the output of the closed-loop system for a unit-step function input. Then, determine  and wn. Then conclude the type of system (overdamped, critically damped, underdamped or undamped). 88 Chapter 5: Time Response R(s) Y (s) + G(s) H(s) Figure Q5-10 Solution: i) The closed-loop transfer function of the system is Y ( s) G( s) 1 0.2    R( s) 1  G( s) H ( s) 5( s  1)(s  2) ( s  1)(s  2) M ( s)  ii) iii) 1 Given, R( s)  , output of the system s 0.2 Y ( s )  M ( s ) R( s )  s( s  1)(s  2) By partial fraction, A B C   s s 1 s  2 Y ( s)  iv) From calculation, A  0.1 B  0.2 C  0.1 v) Thus, Y ( s)  vi) vii) 0.1 0.2 0.1   s s 1 s  2 Referring to Laplace Transform table, find the inverse Laplace transform and the output time response is given by: y(t )  0.1u(t )  0.2e t  0.1e 2t Determine the value of  and wn. Compare the characteristic equation (denominator) of closed-loop transfer function with a general equation of second order-system. s 2  2 n s  n  s 2  3s  2 2 89 Chapter 5: Time Response 2 n  3 n  2 2   n  2  1.414  3  1.061 2  1.414 Because of   1 , system is overdamped. All the formulas to calculate TP , Ts ,%OS and Tr is not valid. These formulas are only applicable for underdamped system. 11. A system of Figure Q5-11 is to have the following specifications Peak time, T p = 0.3627 s and  = 0.5. Design the values of K1 and K 2 required for the specifications of the system to be met. R(s) + + K1 _ _ 10 s( s  1) C(s) K2s Figure Q5-11 Solution: 10K G(s)  1 2 s  (1  K )s 2 T ( s)  10K 1 s 2  (1  10K )s  10K 2 1  n 2 s 2  2 ns  n 2 Therefore; 10K  n 1 2 and 1  10K 2  2 n Given Tp  0.3627 and   0.5  Tp  n 1 -  2  n    Tp 1 -  2  10 0.3627 1  (0.5) 2 2 10K  n  100 1 100 K   10 1 10 1  10K 2  2(0.5)(10)  10 10  1 K   0.9 2 10 90 Chapter 5: Time Response 12. Figure Q5-12 (a) and (b) show the unit step responses of system A and system B, respectively. For each system: Figure Q5-12(a) a) b) c) d) Figure Q5-12(b) Find percent overshoot, %OS and peak time, Tp. C ( s) Find the transfer function, T ( s)  , of the systems R( s ) Calculate settling time. You have to design a system which operate with these following specifications: Time settling < 5 s Percent overshoot < 30% Which system you will choose, system A or system B? Solution: a) ReferringFigure below : For System A %OS  Cmax  Cfinal Cfinal x100  1.25  1 x100  25% 1 Tp  1.3 s 91 Chapter 5: Time Response Cmax = 1.25 Cfinal = 1 Tp = 1.3 ReferingFigure below : For System B %OS  C max Cfinal x100  Cfinal 1.67  1 x100  67% 1 Tp  0.3 s Cmax = 1.67 Cfinal = 1 Tp =0.3 b) A general Transfer function for second order system is given as: System A T (s )  n 2 s 2  2 ns  n 2  n  Tp 1 -  2    2.642 1.3 1  (0.404) 2 For System A,   ln(%OS)  0.404  2  ln 2 (% OS) Refer to the figure above,Tp  1.3 Tp   Therefore T(s)   2.6422 s 2  2(0.404)(2.642)s  2.6422 6.980 s 2  2.135s 6.980 n 1 -  2 92 Chapter 5: Time Response System B T(s)  n 2 s 2  2 ns  n 2  n  Tp 1 -  2    10.556 0.3 1  (0.126) 2 For System B,    ln(%OS )  0.126  2  ln 2 (%OS ) Refer to the figure above,Tp  0.3 Tp   Therefore T(s)   10.5562 s 2  2(0.126)(10.556)s  10.5562 111.429 s 2  2.66s  111.429 n 1 -  2 4 4   3.748 s  n 0.404  2.642 4 4 For System B, Ts    3.007 s  n 0.126  10.556 c) For System A, Ts  d) System A 93 View publication stats