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- 1. Feedback Control Systems (FCS) Lecture-13-14 Time Domain Analysis of 1st Order Systems Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/
- 2. Introduction • The first order system has only one pole. C(s) R( s ) K Ts 1 • Where K is the D.C gain and T is 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.
- 3. Introduction • For the first order system given below 10 G(s) 3s 1 • D.C gain is 10 and time constant is 3 seconds. • And for following system 3/5 3 G(s) s 5 1 / 5s 1 • D.C Gain of the system is 3/5 and time constant is 1/5 seconds.
- 4. Impulse Response of 1st Order System • Consider the following 1st order system δ(t) 1 K R( s ) 0 C(s) 1 Ts t R( s ) (s) K C(s) Ts 1 1
- 5. Impulse Response of 1st Order System K C(s) 1 Ts • Re-arrange above equation as K /T C(s) s 1/T • In order to represent the response of the system in time domain we need to compute inverse Laplace transform of the above equation. L A 1 s Ae a at c( t ) K T e t /T
- 6. Impulse Response of 1st Order System K • If K=3 and T=2s then c ( t ) e t /T T K/T*exp(-t/T) 1.5 c(t) 1 0.5 0 0 2 4 6 Time 8 10
- 7. Step Response of 1st Order System • Consider the following 1st order system K R( s ) Ts R( s ) C(s) 1 1 U (s) s K C(s) s Ts 1 • In order to find out the inverse Laplace of the above equation, we need to break it into partial fraction expansion Forced Response C(s) K s Natural Response KT Ts 1
- 8. Step Response of 1st Order System C(s) K 1 T s 1 Ts • Taking Inverse Laplace of above equation c( t ) K u(t ) e • Where u(t)=1 c( t ) K 1 e t /T t /T • When t=T c( t ) K 1 e 1 0 . 632 K
- 9. Step Response of 1st Order System • If K=10 and T=1.5s then c( t ) K 1 e t /T K*(1-exp(-t/T)) 11 10 Step Response 9 8 K 10 Input D .C Gain steady state output 1 c(t) 7 63 % 6 5 4 3 2 Unit Step Input 1 0 0 1 2 3 4 5 Time 6 7 8 9 10
- 10. Step Response of 1st Order System • If K=10 and T=1, 3, 5, 7 c( t ) K 1 e t /T K*(1-exp(-t/T)) 11 10 T=1s 9 8 T=3s c(t) 7 T=5s 6 T=7s 5 4 3 2 1 0 0 5 10 Time 15
- 11. Step Response of 1st order System • System takes five time constants to reach its final value.
- 12. Step Response of 1st Order System • If K=1, 3, 5, 10 and T=1 c( t ) K 1 e t /T K*(1-exp(-t/T)) 11 10 K=10 9 8 c(t) 7 6 K=5 5 4 K=3 3 2 K=1 1 0 0 5 10 Time 15
- 13. Relation Between Step and impulse response • The step response of the first order system is c( t ) K 1 t /T e K Ke t /T • Differentiating c(t) with respect to t yields dc ( t ) d dt dt K Ke dc ( t ) K dt T e t /T t /T (impulse response)
- 14. Example#1 • Impulse response of a 1st order system is given below. c(t ) • Find out – – – – Time constant T D.C Gain K Transfer Function Step Response 3e 0 .5 t
- 15. 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. c(t ) 3e 3 C(s) 3 1 0 .5 S 0 .5 t 0 .5 S C(s) C(s) (s) R( s ) C(s) R( s ) 3 S 6 2S 1 0 .5 (s)
- 16. Example#1 • Impulse response of a 1st order system is given below. c(t ) 3e 0 .5 t • Find out – – – – – Time constant T=2 D.C Gain K=6 6 Transfer Function C ( s ) R( s ) 2S 1 Step Response Also Draw the Step response on your notebook
- 17. Example#1 • For step response integrate impulse response c(t ) 0 .5 t 3e c ( t )dt 0 .5 t 3 e c s (t ) dt 0 .5 t 6e C • We can find out C if initial condition is known e.g. cs(0)=0 0 6e C c s (t ) 0 .5 0 C 6 6 6e 0 .5 t
- 18. Example#1 • If initial Conditions are not known then partial fraction expansion is a better choice C(s) R( s ) 6 2S since R ( s ) is a step input , R ( s ) 1 1 s 6 C(s) s 2S 6 s 2S A 1 6 s 2S 1 s B 2s 6 1 c( t ) s 6 1 6 s 0 .5 6e 0 .5 t
- 19. Partial Fraction Expansion in Matlab • If you want to expand a polynomial into partial fractions use residue command. y( s ) x( s ) r1 s Y=[y1 y2 .... yn]; X=[x1 x2 .... xn]; [r p k]=residue(Y, X) r2 p1 s rn p2 s k pn
- 20. Partial Fraction Expansion in Matlab • If we want to expand following polynomial into partial fractions 4s Y=[-4 8]; X=[1 6 8]; [r p k]=residue(Y, X) r =[-12 p =[-4 k = [] 8] -2] s 2 8 6s 8 4s s 2 8 r1 6s 4s s 2 8 8 6s s r2 p1 s 12 8 s 4 p2 8 s 2
- 21. Partial Fraction Expansion in Matlab • If you want to expand a polynomial into partial fractions use residue command. C(s) Y=6; X=[2 1 0]; [r p k]=residue(Y, X) r =[ -6 p =[-0.5 k = [] 6] 0] 6 s 2S 1 6 s 2s 6 1 s 6 0 .5 s
- 22. Ramp Response of 1st Order System • Consider the following 1st order system K R( s ) C(s) 1 Ts 1 R( s ) s 2 K C(s) s 2 1 Ts • The ramp response is given as c( t ) K t T Te t /T
- 23. Ramp Response of 1st Order System • If K=1 and T=1 c( t ) K t T Unit Ramp Response 10 Unit Ramp Ramp Response c(t) 8 6 4 error 2 0 0 5 10 Time 15 Te t /T
- 24. Ramp Response of 1st Order System • If K=1 and T=3 c( t ) K t T Unit Ramp Response 10 Unit Ramp Ramp Response c(t) 8 6 4 error 2 0 0 5 10 Time 15 Te t /T
- 25. Parabolic Response of 1st Order System • Consider the following 1st order system K R( s ) R(s) Ts 1 s • Do it yourself 3 Therefore, C(s) 1 C(s) K 3 s Ts 1
- 26. 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.
- 27. Practical Determination of Transfer Function of 1st Order Systems • If we can identify T and K from laboratory testing we can obtain the transfer function of the system. C(s) R( s ) K Ts 1
- 28. Practical Determination of Transfer Function of 1st Order Systems • For example, assume the unit step response given in figure. K=0.72 • 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. C(s) 5 R( s ) s 7 T=0.13s • Thus transfer function is obtained as: C(s) R( s ) 0 . 72 0 . 13 s 5 .5 1 s 7 .7
- 29. 1st Order System with a Zero C(s) K (1 s) R( s ) Ts 1 • Zero of the system lie at -1/α and pole at -1/T. • Step response of the system would be: K (1 C(s) s) 1 s Ts K( K C(s) s c( t ) K Ts K T ( T) Partial Fractions 1 T )e t /T Inverse Laplace
- 30. 1st Order System with & W/O Zero (Comparison) C(s) R( s ) c( t ) K 1 C(s) Ts e 1 t /T K (1 R( s ) K Ts c( t ) K K ( s) 1 T )e t /T T • If T>α the shape of the step response is approximately same (with offset added by zero) c( t ) K K ( n )e t /T T c (t ) K 1 n T e t /T
- 31. 1st Order System with & W/O Zero • If T>α the response of the system would look like Unit Step Response 10 10 (1 R( s ) 3s 2s) 9 1 c(t) C(s) 9.5 8.5 8 7.5 c( t ) 10 10 3 (2 3 )e t/3 7 6.5 offset 0 5 10 Time 15
- 32. 1st Order System with & W/O Zero • If T<α the response of the system would look like Unit Step Response of 1st Order Systems with Zeros 14 10 (1 R( s ) c(t ) 1 .5 s 10 10 1 .5 2s) 13 1 (2 Unit Step Response C(s) 1)e t / 1 .5 12 11 10 9 0 5 10 Time 15
- 33. 1st Order System with a Zero Unit Step Response of 1st Order Systems with Zeros 14 Unit Step Response 13 12 11 T 10 T 9 8 7 6 0 5 10 Time 15
- 34. 1st Order System with & W/O Zero Unit Step Response of 1st Order Systems 14 Unit Step Response 12 T 10 T 8 6 1st Order System Without Zero 4 2 0 0 2 4 6 Time 8 10
- 35. Home Work • Find out the impulse, ramp and parabolic response of the system given below. C(s) K (1 R( s ) Ts s) 1
- 36. Example#2 • A thermometer requires 1 min to indicate 98% of the response to a step input. Assuming the thermometer to be a first-order system, find the time constant. • If the thermometer is placed in a bath, the temperature of which is changing linearly at a rate of 10°C/min, how much error does the thermometer show?
- 37. PZ-map and Step Response jω C(s) R( s ) C(s) R( s ) K Ts 1 10 s 1s T 1 -3 -2 -1 δ
- 38. PZ-map and Step Response jω C(s) R( s ) K C(s) R( s ) 1 Ts T 10 s 2 C(s) 5 R( s ) 0 .5 s -3 1 0 .5 s -2 -1 δ
- 39. PZ-map and Step Response jω C(s) R( s ) K C(s) R( s ) 1 Ts T 10 s 3 C(s) 0 . 33 s -3 3 .3 R( s ) 0 . 33 s 1 -2 -1 δ
- 40. Comparison C(s) C(s) 1 R( s ) R( s ) 1 s Step Response s 10 Step Response 1 0.1 0.08 0.6 0.06 Amplitude 0.8 Amplitude 1 0.4 0.2 0 0.04 0.02 0 1 2 3 Time (sec) 4 5 6 0 0 0.1 0.2 0.3 Time (sec) 0.4 0.5 0.6
- 41. First Order System With Delays • Following transfer function represents the 1st order system with time lag. C(s) R( s ) K Ts 1 • Where td is the delay time. e st d
- 42. First Order System With Delays C(s) R( s ) K Ts 1 st d e 1 Unit Step Step Response td t
- 43. First Order System With Delays Step Response 10 K 10 C(s) R( s ) 10 3s 1 e 2s Amplitude 8 6 4 2 td 0 0 2s T 3s 5 10 Time (sec) 15
- 44. Examples of First Order Systems • Armature Controlled D.C Motor (La=0) Ra La B u ia Ω(s) U(s) eb T K t Ra Js B K t K b Ra J
- 45. Examples of First Order Systems • Electrical System Eo (s) 1 Ei (s) RCs 1
- 46. Examples of First Order Systems • Mechanical System X o(s) X i(s) 1 b k s 1
- 47. Examples of First Order Systems • Cruise Control of vehicle V (s) U (s) 1 ms b
- 48. To download this lecture visit http://imtiazhussainkalwar.weebly.com/ END OF LECTURES-13-14

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