Chapter 2 Laplace Transform

5,815 views

Published on

Published in: Education, Technology
1 Comment
22 Likes
Statistics
Notes
No Downloads
Views
Total views
5,815
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
511
Comments
1
Likes
22
Embeds 0
No embeds

No notes for slide

Chapter 2 Laplace Transform

  1. 1. CHAPTER 2 LAPLACE TRANSFORM AND TRANSFER FUNCTION MAR JKE
  2. 2.      2.1 Explain the concept of Laplace Transform 2.2 Understand the concept of transfer function 2.3 Understand block diagram representation 2.4 Explain Signal Flow Graph representation Identify the Mason’s gain formula LAPLACE TRANSFORM AND TRANSFER FUNCTION MAR JKE
  3. 3.  The transform method is used to solve certain problems, that are difficult to solve directly.  In this method the original problems is first transformed and solved.  Laplace transform is one of the tools for solving ordinary linear differential equations. Definition of Laplace Transform MAR JKE
  4. 4.  First : Convert the given differential equation from time domain to complex frequency domain by taking Laplace transform of the equation  From this equation, determine the Laplace transform of the unknown variable  Finally, convert this expression into time domain by taking inverse Laplace transform MAR JKE
  5. 5.  Laplace transform method of solving differential equations offers two distinct advantages over classical method of problem solving  From this equation, determine the Laplace transform of the unknown variable  Finally, convert this expression into time domain by taking inverse Laplace transform MAR JKE
  6. 6.  The Laplace transform is defined as below: Let f(t) be a real function of a real variable t defined for t>0, then f (t) .e dt F(s) L f(t) - st 0 Where F(s) is called Laplace transform of f(t). And the variable ‘s’ which appears in F(s) is frequency dependent complex variable  It is given by, s σ jω where = Real part of complex variable ‘s’ = Imaginary part of complex variable ‘s’  MAR JKE
  7. 7. Example 1  Find the Laplace transform of e-at and 1 for t ≥ 0 MAR JKE
  8. 8.  Find the Laplace transform of e-at and 1 for t ≥ 0  Solution : i) f(t) = e-at  ii) f(t) = 1 MAR JKE
  9. 9.  The operation of finding out time domain function f(t) from Laplace transform F(s) is called inverse Laplace transform and denoted as L-1 -1 L F(s) -1 L L (f(t)) f(t)  Thus,  The time function f(t) and its Laplace transform F(s) is called transform pair Inverse Laplace Transform MAR JKE
  10. 10. The properties of Laplace transform enable us to find out Laplace transform without having to compute them directly from the definition.  The properties are given:  A) The Linear Property  The Laplace transformation is a linear operation – for functions f(t) and g(t), whose Laplace transforms exists, and constant a and b, the equation isb:g(t) L a f(t) aLf(t) bLg(t)  Properties of Laplace Transforms MAR JKE
  11. 11. B) Differentiation df(t) L sLf(t) - f(0)  According to this property, dt  It means that inverse Laplace transform of a Laplace transform multiplied by s will give derivative of the function if initial conditions are zero.  C) n-fold differentiation n  According to this property, d f(t)  n L dt n s Lf(t) - s n -1 f(0) - s n -2 ' f (0) - ... f n -1 MAR JKE (0)
  12. 12.  D) Integration Property t L 1 f ( ) s In general, the Laplace transform of an order n is L 1 s  Lf(t) s 0  1 f (0 ) n .... Lf(t) f(t)dt f n -1 s n (0) n f n -2 s (0) n -1 n ... f (0) s Laplace transform exists if f(t) does not grow too t fast as MAR JKE
  13. 13.  E) Time Shift  The Laplace transform of f(t) delayed by time T is equal to the Laplace transform of f(t) multiplied by e-sT ; that is L[f ( t – T ) u( t – T )] = e-sT F(s), where u (t – T) denotes the unit step function, which is shifted to the right in time by T. MAR JKE
  14. 14. F) Convolution Integral  The Laplace transform of the product of two functions F1(s) and F2(s) is given by the convolution integrals t  L -1 F1 ( s)F 2 ( s) f 1 ( t) f 2 ( t - )d 0 t f 1 (t - )f 2 ( ) d 0 where L-1F1(s) = f1(t) and L-1F2(s) = f2(t) MAR JKE
  15. 15.  G) Product Transformation  The Laplace transform of the product of two functions f1(t) and f2(t) is given by the complex convolution integral 1 c L -1 f 1 ( t)f 2 j ( t) 2 j F1 ( )F2 ( )d c- j  H) Frequency Scaling  Thes inverse Laplace transform of the functions F af(at), where -1 L F(s) f(t) a MAR JKE
  16. 16.  I) Time Scaling  The L f Laplace transform of a functions t aF(as) where F(s) Lf(t) a MAR JKE
  17. 17. J) Complex Translation  If F(s) is the Laplace transform of f(t) then by the complex translation property,  MAR JKE
  18. 18.  K) Initial Value Theorem  The Laplace transform is very useful to find the initial value of the time function f(t). Thus if F(s) is the Laplace transform of f(t) then,  The only restriction is that f(t) must be continuous or at the most , a step discontinuity at t=0. MAR JKE
  19. 19.  L) Final Value Theorem  Similar to the initial value, the Laplace transform is also useful to find the final value of the time function f(t). Thus if F(s) is the Laplace transform of f(t) then the final value theorem states that,   The only restriction is that the roots of the denominator polynomial of F(s) i.e poles of F(s) have negative or zero real parts MAR JKE
  20. 20. Example 2  Find the Laplace transform of sin t MAR JKE
  21. 21.  Find the Laplace transform of sin t MAR JKE
  22. 22. MAR JKE
  23. 23. f(t) F(s) 1 1/s Constant K K/s K f(t), K is constant K F(s) t 1/s2 tn n /sn+1 e-at 1/s+a eat 1/s-a e-at tn n /((s+a)n+1 ) sin t /(s2 + 2) cos t s/(s2 + 2) e-at sin t /((s+a)2 + 2) Table of Laplace Transforms: Table 1 : Standard Laplace Transform pairs MAR JKE
  24. 24. f(t) e-at cos t F(s) (s+a)/((s+a)2 + sinh t /(s2 - 2) cosh t s/(s2 - 2) t e-at 1/(s+a)2 1 - e-at a/s(s+a) Table 1 Contd…. MAR JKE 2)
  25. 25. Function f(t) Laplace Transform F(s) Unit step = u(t) 1/s A u(t) A/s Delayed unit step = u(t-T) e-Ts/s A u(t-T) Ae-Ts /s Unit ramp = r(t) = t u(t) 1/s2 At u(t) A/s2 Delayed unit ramp = r(t-T) = (t-T) u(tT) e-Ts /s2 A(t-T) u(t-T) Ae-Ts /s 2 Unit impulse = (t) 1 Delayed unit impulse = (t-T) e-Ts Impulse of strength K i.e K (t) K Table 2 : Laplace transforms of standard time functions MAR JKE
  26. 26.  Let F(s) is the Laplace transform of f(t) then the inverse Laplace transform is denoted as, f(t)  -1 L F(s) The F(s), in partial fraction method, is written in the form as, F(s) N(s) D(s)   Where N(s) = Numerator polynomial in s D(s) = Denominator polynomial in s Inverse Laplace Transform MAR JKE
  27. 27. The roots of D(s) are simple and real  The function F(s) can be expressed as,  N(s) N(s) D(s) F(s) (s - a)(s - b)(s - c)... where a, b, c… are the simple and real roots of D(s).  The degree of N(s) should be always less than D(s) Simple and Real Roots MAR JKE
  28. 28.  This can be further expressed as, N(s) K1 K2 K3 (s - a)(s - b)(s - c)... F(s) (s - a) (s - b) (s - c) .... where K1, K2, K3 … are called partial fraction coefficients  The values of K1, K2, K3 … can be obtained as, K1 (s - a). F(s) K (s - b). F(s) 2 K3 (s - c). F(s) s s s a b c MAR JKE
  29. 29.  In general, K  Where sn = nth root of D(s)  n L e (s - s n ). F(s) at s and so on sn 1 (s  a) Is standard Laplace transform pair.  Once F(s) is expressed in terms of partial fractions, with coefficients K1, K2 … Kn, the inverse Laplace transform can be easily obtained -1 at bt ct  f(t) L F(s) K 1e K 2e K 3e MAR JKE ...
  30. 30. Example 3  Find the inverse Laplace transform of given F(s) (s F(s) s(s 2) 3)(s 4) MAR JKE
  31. 31.  Solution : The degree of N(s) is less than D(s). Hence F(s) can be expressed as, F(s) K1 s where (s K1 K2 K3 1 F(s) K3 K2 6 s 3) s. F(s) (s (s (s s 0 3). F(s) 4). F(s) 4) (s s. s(s s s 3 4 2) 3)(s (s 2 4) 0 (s 3). s(s (s s s(s 3x4 6 2) 3)(s (s 4 ). 1 (-3 4) s 3 (-3)x(-3 2) 3)(s (-4 4) s 4 2) (-4)x(-4 1 1 3 2 (s 3) ( s 4 ) MAR JKE 1 4) 2) 3 - 3) 1 2
  32. 32.  Taking inverse Laplace transform, f(t) 1 1 6 3 e - 3t - 1 e - 4t 2 MAR JKE
  33. 33.  The given function is of the form, N(s) F(s) (s - a) n D(s) There is multiple root of the order ‘n’ existing at s=a.  The method of writing the partial fraction expansion for suchKmultiple roots K is, N' (s) K K  F(s) 0 (s - a) 1 n (s - a) 2 n -1 (s - a) n -2 ... n -1 (s - a) D' (s) N' (s) D' (s) where represents remaining terms of the expansion of F(s) Multiple Roots MAR JKE
  34. 34. A separate coefficient is assumed for each power of repetative root, starting from its highest power n to 1  For ease of solving simultaneous equations, find the coefficient -K0 nby the same method for simple K 0 (s a) . F(s) s a roots   Finding the Laplace inverse transform of expanded F(s) refer to standard transform pairs, MAR JKE
  35. 35. Example 4  Obtain the inverse Laplace transform of given f(s) MAR JKE
  36. 36. MAR JKE
  37. 37. MAR JKE
  38. 38. MAR JKE
  39. 39.  If there exists a quadratic term in D(s) of F(s) whose roots are complex conjugates then the F(s) is expressed with a first order polynomial in s in the numerator as, As B N' (s) F(s) (s 2 s ) D' (s)  N' (s) Where (s2 + s + ) is the quadratic whose roots D' (s) are complex conjugates while represents remaining terms of the expansion.  The A and B are partial fraction coefficients. Complex Conjugate Roots MAR JKE
  40. 40. The method of finding the coefficients is same as the multiple roots.  Once A and B are known, then use the following method for calculating inverse Laplace As B transform.(s) F (s s )  Consider A and B are know.  1 2 2  Now complete the square in the denominator by L.T. 4(F.T.) calculating last term as,  Where L.T = Last term term F.T = First term (M.T.) M.T = Middle MAR JKE
  41. 41. 2 L.T. 4 As F1 ( s) B 2 s 2 As 2 s 4 4 B 2 2 s 2 2 where 4  Now adjust the numerator As + B in such a way that it is of the form MAR JKE
  42. 42. Example 5  Find the inverse Laplace transform of MAR JKE
  43. 43. MAR JKE
  44. 44. MAR JKE
  45. 45. MAR JKE
  46. 46. MAR JKE
  47. 47. MAR JKE
  48. 48. The control system can be classified as electrical, mechanical, hydraulic, thermal and so on.  All system can be described by integrodifferential equations of various orders  While the o/p of such systems for any i/p can be obtained by solving such integrodifferential equations  Mathematically, it is very difficult to solve such equations in time domain  Application of Laplace Transform in Control System MAR JKE
  49. 49.  The Laplace transform of such integrodifferential equations converts them into simple algebraic equations  All the complicated computations then can be easily performed in s domain as the equations to be handled are algebraic in nature.  Such transformed equations are known as equations in frequency domain MAR JKE
  50. 50.  By eliminating unwanted variable, the required variable in s domain can be obtained  By using technique of Laplace inverse, time domain function for the required variable can be obtained  Hence making the computations easy by converting the integrodifferential equations into algebraic is the main essence of the Laplace transform MAR JKE

×