time domain analysis, Rise Time, Delay time, Damping Ratio, Overshoot, Settling Time Calculations

Time Response- Transient, Steady State
Standard Test Signals- U(t), S(t), R(t)
Analysis of First order system - for Step input
Analysis of second order system -for Step input
Time Response Specifications- Rise Time, Delay time, Damping Ratio, Overshoot, Settling Time
Calculations

1. Time domain Analysis Time Response- Transient, Steady State Standard Test Signals- U(t), S(t), R(t) Analysis of First order system - for Step input Analysis of second order system - for Step input Time Response Specifications- Rise Time, Delay time, Damping Ratio, Overshoot, Settling Time Calculations

2. Generally speaking, the response of any system thus has two parts Transient Response Steady State Response Time Response . . 2

3. i.e. As the nam e su g g e s ts t h a t t r ans i e n t r es p onse r emains onl y f o r some ti m e f r o m i n it i al s t a t e to final state. t  Transient Response That part of the time response that goes to zero as time becomes very large is called as “ Transient Response ” L c (t)  . . 3

4. Steady State Response That part of the response that remains after the transients have died out is called “ Steady State Response ”. From the steady state we can know; How long it took before steady state was reached. Wh e ther the r e is a n y e r r o r b e t w e e n the desi r ed and actual values. Whether this error is constant, zero or infinite i.e. unable to track the input. . . 4

5. Steady State Response . . 5

6. Steady State Response . . 6

7. Standard Test Signal Step Input Mathematical Representations Graphical Representations r(t) = R. u(t) = t>0 t<0 This signal signifies a sudden change in the reference input r(t) at time t=0 Laplace Representations s L {Ru(t)}  R . . 7

8. Ramp Input Mathematical Representations Graphical Representations r(t) = R.t = t>0 t<0 Signal have constant velocity i.e. constant change in it’s value w.r.t. time Laplace Representations L {Rt}  R s 2 Standard Test Signal . . 8

9. Impulse Input Graphical Representations Mathematical Representations r( t) =  (t) =1 = t>0 t<0 Laplace Representations L {  (t)}  1 The function has a unit value only for t=0. In practical cases, a pulse whose time approaches zero is taken as an impulse function. Standard Test Signal . . 9

10. Analysis of first order system for Step input Consider a first order system as shown; + - R(s) C(s) 1 Ts He r e G (s)  1 Ts G and H(s)  1 C ( s) R ( s) 1  GH   1 1 1    Ts Ts 1 1  Ts  . . 10

11. r(t) = u(t) = t>0 t<0 For step input; Taking Laplace transform; R (s)  L {Ru(t)}  1 s but C (s)  1 R (s) 1  Ts 1  C (s)  R ( s ) 1  Ts Analysis of first order system for Step input . . 11

12. 1  1 1  Ts s  C (s)  B s Using partial fraction;  C (s)  A  T s  1 Solving;  A  s . C (s) | s   1   1 T s  1 T  B  ( s  1 ) C (s) | Analysis of first order system for Step input . . 12

13.  C (s)  1  s 1 T s  1 Taking Inverse Laplace transform; 1 1 } 1  1  c (t)  L  1 { C (s)}  L { }  L  1 { s s  T 1 t T  c (t)  1  e Analysis of first order system for Step input . . 13

14. Plot c(t) vs t; Sr. No. t C(t) 1 T 0.632 2 2T 0.86 3 3T 0.95 4 4T 0.982 5 5T 0.993 6  1 Analysis of first order system for Step input . . 14

15. Time Constant (T) The value of c(t)=1 only at t=∞. P r act i c al l y the v a l u e of c(t) i s with i n 5 % of f i nal value at t=3T and within 2% at t=4T. In practice t=3T or 4T may be taken as steady state. Ho w qu i ckly the v alue r ea c h e s st eady st a t e i s a function of the time constant of the system. Hence smaller T indicates quicker response. . . 15

16. Damping E v e r y s y st em h a s a t endency t o oppo s e the oscillatory behavior of the system which is known as “Damping” . . . 16

17. Damping Factor (  ) Th e d a mping i n a n y s y st em i s measu r ed b y a factor or ratio which is known as damping ratio. It is denoted by ( Z e t a)  . . 17

18. Analysis of second order system for Step input Consider a second order system as shown; - R(s) + C(s)  n 2 s (s  2  n ) He r e  n 2 s (s  2  n ) G (s)  and H(s)  1  C(s)  G R ( s) 1  GH  n 2 s (s  2  n )  n 2 s (s  2  n ) 1   s 2  n 2  2  n s   n 2  . . 18

19. This is the standard form of the closed loop transfer function These poles of transfer function are given by; s 2  n 2  2  n s   n 2 C (s)  R ( s)  2  n s   n 2  s 2 2  2  n  ( 2   n ) 2  4 (  n ) 2  s   2  n 2   n 2    n   2     n   n  1 Analysis of second order system for Step input . . 19

20. The poles are; (i) Real and Unequal if i.e. They lie on real axis and distinct  2  1    1 (ii) Real and equal if i.e. They are repeated on real axis  2  1    1 (iii) Complex if i.e. Poles are in second and third quadrant  2  1    1 Analysis of second order system for Step input . . 20

21. Relation between  and pole locations (i)    1 Under damped Pole Location Step Response c(t) . . 21

22. (ii) Critically damped   1 Pole Location Step Response c(t) R el a tion b e t w ee n  . . 22 and pole locations

23. ( i ii) over damped   1 Pole Location Step Response c(t) R el a tion b e t w ee n  . . 23 and pole locations

24. ( i v)   Pole Location Step Response c(t) R el a tion b e t w ee n  . . 24 and pole locations

25. ( v )     1 Pole Location Step Response c(t) R el a tion b e t w ee n  . . 25 and pole locations

26. ( v i)    1 Pole Location Step Response c(t) R el a tion b e t w ee n  . . 26 and pole locations

27. ( v ii)    1 Pole Location Step Response c(t) Relation between  and pole locations . . 27

28. Time Response Specifications . . 28

29. Time Response Specifications Delay Time (t d ) : I t i s time r equi r ed f o r the r es p ons e t o r each 50% of the final value in the first attempt. t d  1  0.7   n . . 29

30. Rise Time (t r ) : It is time required for the response to rise from 10% to 90% of the final value for overdamped systems. (It is 0 to 100% for under damped systems) t r     whe r e, and  d  1   tan  1   2  d   n 1   2 Time Response Specifications . . 30

31.  { 1   2 } % M p  e  100  Time Response Specifications Peak Overshoot (M p ): The maximum overshoot is the maximum peak value of the response curve measured from unity. It is therefore largest error between input and output during the transient period. . . 31

32. T p    d Time Response Specifications Peak Time (t p ): I t i s the time r equi r ed f o r the r espon s e to reach the first peak. . . 32

33. 4 T s  4 T   n Time Response Specifications Settling Time (t s ): It is the time required for the response curve to reach and stay within a specified percentage (usually 2% or 5%) of the final value. . . 33

34. Example 6 A unity feedback system has 16 G ( s )  s ( s  5) If a step input is given calculate Damping Ratio Overshoot Settling Time Solution: 16 G ( s )  s ( s  5) H ( s )  1 s 2 . . 34 Determine the closed loop transfer function 16 C ( s )  G s ( s  5) 16 R ( s ) 1  GH  5 s  16   1  16 s ( s  5)

35. E x ample 6 c o n t …… Compare closed loop TF with standard form of second order system s 2  n 2  2  n s   n 2 s 2 16  5 s  16  Compare denominators of both Natural Frequency;  n 2  16 Damping Ratio; 2  n s  5 s   n  4 rad / sec 5 5  0.625     2   n 2  4 Settling Time; T s  4 . . 35 4  1.6 sec   n (0.625)  (4)

36. Example 6 c o n t …… Overshoot  {  } 1   2 % M p  e  100 1  ( 0.625) 2 . . 36  { ( 0.625)  } % M p  e  100 % M p  8.08%