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Time Response Analysis Of Control System Presentation

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Published in: Programming Technology
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Variation of the response of the system with respect to time is called time response analysis.

Aswathi K / Abu Dhabi

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  1. Digital and Non-Linear Control Time Domain Analysis
  2. Introduction ' In time-domain analysis the response of a dynamic system to an input is expressed as a function of time. ' It is possible to compute the time response of a system if the nature of input and the mathematical model of the system are known. Usually, the input signals to control systems are not known fully ahead of time. ' It is therefore difficult to express the actual input signals mathematically by simple equations.
  3. Standard Test Signals The characteristics of actual input signals are a sudden shock, a sudden change, a constant velocity, and constant acceleration. The dynamic behavior of a system is therefore judged and compared under application of standard test signals an impulse, a step, a constant velocity, and constant acceleration. The other standard signal of great importance is a sinusoidal signal.
  4. Standard Test Signals ' Impulse signal — The impulse signal imitate the sudden shock characteristic of actual input signal. —If A=l, the impulse signal is called unit impulse signal. t
  5. Standard Test Signals ' Step signal — The step signal imitate the sudden change characteristic of actual input signal. u(t) — If A=l, the step signal is called unit step signal u(t) t
  6. Standard Test Signals ' Ramp signal —The ramp signal imitate the constant velocity characteristic of actual input signal. r(t) — If A=l, the is called signal r(t) r(t) ramp signal with slope A r(t) t ramp signal unit ramp unit ramp signal o 12t
  7. Standard Test Signals Parabolic signal - The parabolic signal the imitate constant acceleration characteristic of actual input signal. 2 2 p(t) t 4SA parabolic signal with slope A OSA the parabolic signal called unit is parabolic signal. Unit parabolic signal
  8. Relation between standard Test Signals ' Impulse Step e Ramp ' Parabolic ö(t) — u(t) r(t)
  9. Laplace Transform of Test Signals ' Impulse ' Step u(t)
  10. Laplace Transform of Test Signals Ramp ' Parabolic r(t) = IR(s) 2 2 2 ID(s) =
  11. Time Response of Control Systems ' Time response of a dynamic system response to an input expressed as a function of time. System ' The time response of any system has two components ' Transient response ' Steady-state response.
  12. Time Response of Control Systems ' When the response of the system is changed from equilibrium it takes some time to settle down. ' This is called transient response. ' The response of the system after the transient response is called steady state response. 6 5 4 3 2 1 0 0 Step Input Response Transient Response 2 4 6 16 o Q.) Q.) 18 8 10 12 Tiræ (sec) 14 20
  13. Time Response of Control Systems ' Transient response depend upon the system poles only and not on the type of input. ' It is therefore sufficient to analyze the transient response using a step input. ' The steady-state response depends on system dynamics and the input quantity. ' It is then examined using different test signals by final value theorem.
  14. Introduction The first order system has only one pole. c(s) K IR(s) Ts+l Where K is the D.C gain and Tis the time constant of the system. ' Time constant is a measure of how quickly a 1st order system responds to a unit step input. D.C Gain of the system is ratio between the input signal and the steady state value of output.
  15. Introduction The first order system given below. 10 3S-kl D.C gain is 10 and time constant is 3 seconds. ' For the following system 3 D.C Gain of the system is seconds. 3/5 1/5S-l-1 3/5 and time constant is 1/5
  16. Impulse Response of 1st Order System ' Consider the following 1st order system IR(s) R(s) c(s) c(s) ö(s)
  17. Impulse Response of 1st Order System c(s) ' Re-arrange following equation as c(s) ' In order to compute the response of the system in time domain we need to compute inverse Laplace transform of the above equation. -1 at C(t)
  18. Impulse Response of 1st Order System T 2s then c(t) 3 and 1.5 0.5 10 Time
  19. Step Response of 1st Order System ' Consider the following 1st order system IR(s) IR(s) c(s) c(s) 1 s(Ts +1) ' In order to find out the inverse Laplace of the above equation, we need to break it into partial fraction expansion (page 867 in the Textbook) K ICT c(s)
  20. Step Response of 1st Order System 1 c(s) = K ' Taking Inverse Laplace of above equation c(t) = Where u(t)=l When t=T (time constant) 1)_ - 0.632K
  21. Step Response of 1st Order System K 10 and then T=1.5s -exp(-t/T)) 11 10 D.C Gain = K Step Response steady state output Input Unit Step Input Time 10 10
  22. Step Response of 1st order System System takes five time constants to reach its final value. c(t) 1 0.632 1 Slope -wry co—I —e
  23. Step Response of 1st Order System K 10 and -exp(-t/T)) 10 T=ls T=3s T=7s 10 Time 15
  24. Step Response of 1st Order System and 3, 5, 10 10 -exp(-t/T)) K=IO 10 Time
  25. Relation Between Step and impulse response The step response of the first order system is Differentiating c(t) with respect to t yields dc(t) d dt dt dc(t)
  26. Analysis of Simple RC Circuit VT(t) R • i (t) + v (t) = VT (t) C v(t) i(t) dt dv(t) + v(t) dt state variable dv(t) C dt Input waveform
  27. Analysis of Simple RC Circuit Step-input response: vou(t) dv(t) RC + v(t) = vou(t) dt + vou(t) match initial state: output response for step-input: v(t) = vo (1 —e
  28. RC Circuit -t/RC) waveform v(t) = vo(l -e under step input vou(t) t = 0.69RC — i.e., delay = 0.69RC (50% delay) t = O. IRC t = 2.3RC — i.e., rise time = 2.2RC (if defined as time from 10% to 90% of Vdd) For simplicity, industry uses Elmore delay
  29. Elmore Delay Delay l. 2. 500/0-500/0 point delay Delay-0.69 RC
  30. Example 1 ' Impulse response of a 1st order system is given below. -O.5t c(t) = 3e Find out — Time constant T - D.C Gain K — Transfer Function — Step Response
  31. Example 1 The Laplace Transform of Impulse response of a system is actually the transfer function of the system. Therefore taking Laplace Transform of the impulse response given by following equation. -O.5t c(t) = 3e 3 c(s) 3 x ö(s) S +0.5 S +0.5 c(s) c(s) 3 ö(s) — R(s) S + 0.5 c(s) 6
  32. Example 1 ' Impulse response of a 1st order system is given below. -O.5t c(t) = 3e Find out — Time constant - D.C Gain c(s) — Transfer Function R(s) — Step Response 6 2S+1
  33. Example 1 ' For step response integrate impulse response -O.5t c(t) = 3e = 31 e 05tdt -O.5t —6e We can find out C if initial condition is known e.g. 0 — —6e -O.5t c s (t) = 6— 6e
  34. Example 1 ' If initial conditions are not known then partial fraction expansion is a better choice c(s) 6 1 since R(s) is a step input, R(s) 6 c(s) s(2s + l) = 6— 6e 6 s(2s -k l) 6 s(2s -k l) c(t) S 6 6 O.5t
  35. Ramp Response of 1st Order System ' Consider the following 1st order system IR(s) 1 IR(s) 2 c(s) S2(TS -k l) ' The ramp response is given as c(t) =Kt-T+Te c(s)
  36. Parabolic Response of 1st Order System ' Consider the following 1st order system IR(s) 1 R(s) Therefore, 3 c(s) c(s) S3(TS -k l)
  37. Practical Determination of Transfer Function of 1st Order Systems Often it is not possible or practical to obtain a system's transfer function analytically. ' Perhaps the system is closed, and the component parts are not easily identifiable. The system's step response can lead to a representation even though the inner construction is not known. With a step input, we can measure the time constant and the steady-state value, from which the transfer function can be calculated.
  38. Practical Determination of Transfer Function of 1st Order Systems ' If we can identify T and K empirically we can obtain the transfer function of the system. c(s) IR(s) Ts+l
  39. Practical Determination of Transfer Function of 1st Order Systems For example, assume the unit step response given in figure. From the response, we can measure the time constant, that is, the time for the amplitude to reach 63% of its final value. Since the final value is about 0.72 the time constant is evaluated where the curve reaches 0.63 x 0.72 = 0.45, or about 0 13 second. K is simply steady state value. 0.8 0.7 0.6 0.3 0.2 0.1 0.1 Thus 1
  40. First Order System with a Zero l/a and pole at ' Zero of the system lie at Step response of the system would be: c(s) = s(Ts + 1) c(s)
  41. First Order System With Delays ' Following transfer function is the generic representation of 1st order system with time lag. c(s) K —std IR(s) Ts+l t d is the delay time.
  42. First Order System With Delays c(s) K —std IR(s) Ts+l 1 Unit Step Step Response t
  43. First Order System With Delays Step Response 10 10 S(3S -k 1) L-l[e OSF(s)] = 10 -10 -1/30-2) C(s) IR(s) 1
  44. Second Order System We have already discussed the affect of location of poles and zeros on the transient response of 1st order systems. Compared to the simplicity of a first-order system, a second-order system exhibits a wide range of responses that must be analyzed and described. Varying a first-order system's parameter (T, K) simply changes the speed and offset of the response Whereas, changes in the parameters of a second-order system can change the form of the response. A 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. 44
  45. Introduction A general second-order system is characterized following transfer function. IR(s) 2 by the CCs) c(s) R(s) 2 n S2 -k + On 2 un-damped natural frequency of the second order system, which is the frequency of oscillation of the system without damping. damping ratio of the second order system, which is a measure of the degree of resistance to change in the system output. 45
  46. Example 2 ' Determine the un-damped natural frequency and damping ratio of the following second order system. c(s) 4 ' Compare the numerator and denominator of the given transfer function with the general 2nd order transfer function. c(s) 2 S2 -k + On 2 On = 2 240 n S = 2S 0.5 46
  47. c(s) Introduction -1 -1 2 R(s) S2 + -k On 2 ' Two poles of the system are — On; -k On on; — on 42 42 47
  48. Introduction — On; -k On 42 — I According the value of a second-order system can be set into one of the four categories (page 169 in the textbook): 1. Overdamped - when the system has two real distinct poles (4 >1). 6 -b 48
  49. Introduction — On; -k On 42 — I According the value of a second-order system can be set into one of the four categories (page 169 in the textbook): Underdamped - when the system has two complex conjugate poles (0
  50. Introduction — On; -k On 42 — I According the value of a second-order system can be set into one of the four categories (page 169 in the textbook): Undamped - when the system has two imaginary poles ( = O). 3. 6 -b 50
  51. Introduction — On; -k On 42 — I According the value of a second-order system can be set into one of the four categories (page 169 in the textbook): Critically damped - when the system has two real but equal poles (4 = 1). 4. 6 -b 51
  52. Underdamped System For 4- and the 2nd order system's response due to a unit step input is as follows. Important timing characteristics: delay time, rise time, peak time, maximum overshoot, and settling time. c(t) 1 0.5 0 Allowable tolerance 0.05 or 0.02 52
  53. Delay Time ' The delay (t d) time is the time required for the response to reach half the final value the very first time. c(t) 1 0.5 O
  54. Rise Time ' The rise time is the time required for the response to rise from 10% to 90%, 5% to 95%, or 0% to 100% of its final value. ' For underdamped second order systems, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time is commonly used. c(t) 1 0.5 0
  55. peak Time ' The peak time is the time required for the response to reach the first peak of the overshoot. c(t) 1 0.5 0
  56. Maximum Overshoot The maximum overshoot is the maximum peak value of the response curve measured from unity. If the final steady-state value of the response differs from unity, then it is common to use the maximum percent overshoot. It is defined by c(t ) P COO) x 100% Maximum percent overshoot c 00) The amount of the maximum (percent) overshoot directly indicates the relative stability of the system. 56
  57. Settling Time The settling time is the time required for the response curve to reach and stay within a range about the final value of size specified by absolute percentage of the final value (usually 2% c(t) Allowable tolerance 0.05 or 0.02 0.5
  58. Step Response of underdamped System c(s) 2 2 Step Response R(s) S2 -k -k On 2 2 S S -k -k On The partial fraction expansion of above equation is given as c(s) - 1 c(s) - (S -k 24a-)n ) 2 c(s) - 1 S S2 -k Von S + On 2 + 240 n + von S -k Ct)n2 1 58
  59. Step Response of underdamped System 1 c(s) S -k 24a.)n (s + ) 2 -k (Dn2 1 — 42 Above equation can be written as S + 24a.)n 1 c(s) - (S -k CCD n + 2 Where = on 1 -4 , is the frequency of transient oscillations and is called damped natural frequency. ' The inverse Laplace transform of above equation can be obtained easily if is written in the following form: 1 c(s) Con 59
  60. Step Response of underdamped System 1 cG) - 1 cG) 1 cG) S + 4CDn (S + CCD n + Cl)} S + 4Ct)n (S + 4CDn + S + CCI)n c(t) COS ct)dt — 4ct)n (S + CCI) n + on 1—42 1-42 (S + + cz)2d sin o d t 1-42 60
  61. Step Response of underdamped System Cont c(t) = I—e COS Odt — Cont sin Odt c(t) 1-42 Cont ' When sin Odt COS O dt -k 1-42 on 1—42 n c(t) = I — cosont 61
  62. Step Response of underdamped System c(t) Cont COS O dt -k if 4=0.1 and on = 3 1.8 1.6 1.4 1.2 0.8 0.6 0.4 0.2 sin Odt 62 10
  63. Step Response of underdamped System c(t) Cont COS O dt -k if = 0.5 and on 1.4 1.2 0.8 0.6 0.4 0.2 sin Odt 63 10
  64. Step Response of underdamped System c(t) Cont COS O dt -k if = 0.9 and on 1.4 1.2 0.8 0.6 0.4 0.2 sin Odt 64 10
  65. Step Response of underdamped System Cont c(t) = I—e COS O d t + 120 10.0 8.0 6.0 sin t 14.0 4+0 2.0 0.0 2.0 0.5 0.0 0.4 0.8 2.0 14.0 12.0 10.0 8.0 6.0 4.0 20 65
  66. — On 4 -k On 42 S-PIane (Underdamped System) -1 2-1 Since 02
  67. Analytical Solution Cont ' Page 171 in the textbook c(t) sin Odt COS O d t + 1-42 Tt—ß Rise time: set c(t)=l, we have tr = 2 dc(t) peak t inn e: set = 0, we have tp dt Maximum overshoot: Mp = c(tp) (for unity output) ' Settling time: the time for the outputs always 4 within 2% of the final value is approximately —
  68. Empirical Solution Using MATLAB ' Page 242 in the textbook
  69. Steady State Error ' If the output of a control system at steady state does not exactly match with the input, the system is said to have steady state error Any physical control system inherently suffers steady-state error in response to certain types of inputs. Page 219 in the textbook A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to a ramp input.
  70. Classification of Control Systems ' Control systems may be classified according to their ability to follow step inputs, ramp inputs, parabolic inputs, and so on. The magnitudes of the steady-state errors due to these individual inputs are indicative of the goodness of the system.
  71. Classification of Control Systems ' Consider the unity-feedback control system with the following open-loop transfer function SN(TIS + + 1) (Tps + 1) ' It involves the term in the denominator, representing N poles at the origin. ' A system is called type 0, type 1, type 2, respectively.
  72. Classification of Control Systems As the type number is increased, accuracy is improved. increasing the type However, aggravates the stability problem. number A compromise between steady-state accuracy and relative stability is always necessary.
  73. Steady State Error of Unity Feedback Systems ' Consider the system shown in following figure. GCS) The closed-loop transfer function is 1 + G(s) GCS) SN(TIS + + 1)
  74. Steady State Error of Unity Feedback Systems Steady state error is defined as the error between the input signal and the output signal when t* 00. The transfer function between the error signal 1 input signal R IR(s) 1 + G(s) and the The final-value theorem provides a convenient way to find the steady-state performance of a stable system. Since E(s) is The steady state error is lim e(t) 1 1 + G(s) limsE(s) SR(s) lim
  75. Static Error Constants The static error constants are figures of merit of control systems. The higher the constants, the smaller the steady-state error. ' In a given system, the output may be the position, velocity, pressure, temperature, or the like. Therefore, in what follows, we shall call the output position," the rate of change of the output "velocity," and so on. This means that in a temperature control system position" represents the output temperature, "velocity" represents the rate of change of the output temperature, and so on.
  76. Static Position Error Constant ( The steady-state error of the system for a unit-step input is — lim 1 The static position error constant K is defined by p = limG(s) = G(o) Thus, the steady-state error in terms of the static position K is given by error constant p 1 ess
  77. Static Position Error Constant ( Type O , For a system K(Tas + + 1 ) = lim (TIS + + 1) or higher order systems Type 1 For K(Tas + + K = lim s-+0 SN(TIS + + l) 00, for N > 1 ' For a unit step input the steady state error e is 1 O, for type 0 systems for type 1 or higher systems
  78. Static Velocity Error Constant ( The steady-state error of the system for a unit-ramp input is 1 — lim ess — 2 1 — lim s-+() SG(S) The static velocity error constant IS is defined by = lim sG(s) Thus, the steady-state error in terms of the static velocity IS is given by error constant 1 ess —
  79. Static Velocity Error Constant ( Type O , For a system SK(Tas + + 1 ) = lim (TIS + -k 1) • • • Type 1 For systems SK(Tas + + 1) Kv = lim s(TIS + + 1 ) ' For type 2 or higher order systems SK(Tas + + 1 ) Kv = lim 00, for N 2 SN(TIS + + 1)
  80. Static Velocity Error Constant (Kv ' For a ramp input the steady state error e is 00, ss for type 0 systems for type 1 systems for type 2 or higher systems
  81. Static Acceleration Error Constant ( The steady-state error of the system for parabolic input is ess — ess 1 — lim + G(s) s3 1 lim S2G(S) The static acceleration error constant Ka is defined by Ka = lims2G(s) Thus, the steady-state error in terms of the static acceleration Ka is given by error constant 1
  82. Type O system For Type 1 systems For type 2 systems Static Acceleration Error Constant ( Ka lim ICa = lim s2K(Tas + + 1) (TIS -k '+ 1) s2K(Tas + + '1) s(TlS + + 1) s2K(Tas + + 1) = lim s-»o e(TlS + + 1) type 3 or higher order systems For s2K(Tas + + Ka = lim S->0 SN(TIS + + 1) m 00, -K for N 3
  83. Static Acceleration Error Constant ( ' For a parabolic input the steady state error e is 1 O, for type 0 and type 1 systems for type 2 systems for type 3 or higher systems

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