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# Applications laplace transform

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### Applications laplace transform

1. 1. 1 CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
2. 2. 2 METHODOLOGY Examples of nonlinear circuits: logic circuits, digital circuits, or any circuits where the output is not linearly proportional to the input. Examples of linear circuits: amplifiers, lots of OPM circuits, circuits made of passive components (RLCs). If the circuit is a linear circuit Laplace transform of the sources of excitation: s(t) → S(s) Laplace transform of the all the elements in the circuit Find the output O(s) in the Laplace freq. domain Obtain the time response O(t) by taking the inverse Laplace transform Stop or approximate the circuit into a linear circuit and continue NO YES
3. 3. 3 THE s-DOMAIN CIRCUITS  Equation of circuit analysis: integrodifferential equations.  Convert to phasor circuits for AC steady state.  Convert to s-domain using Laplace transform.  KVL, KCL, Thevenin,etc.
4. 4. 4 KIRCHHOFF’S VOLTAGE LAW  Consider the KVL in time domain:  Apply the Laplace transform: 0)()()()( 4321 =++++ tvtvtvtv 0)()()()( 4321 =++++ sVsVsVsV
5. 5. 5 KIRCHHOFF’S CURRENT LAW  Consider the KCL in time domain:  Apply the Laplace transform: 0)()()()( 4321 =++++ tItItItI 0)()()()( 4321 =++++ titititi
6. 6. 6 OHM’S LAW  Consider the Ohm’s Law in time domain   Apply the Laplace transform  RsIsV RR )()( = Rtitv RR )()( =
7. 7. 7 INDUCTOR  Inductor’s voltage – In the time domain: – In the s-domain: dt di LtvL =)( )]0()([)( − −= LLL issILsV
8. 8. 8 INDUCTOR  Inductor’s current – Rearrange VL(s) equation: s i sL sV sI L L )0()( )( − +=
9. 9. 9 CAPACITOR  Capacitor’s current – In the time domain: – In the s-domain: dt dv Ctic =)( )]0()([ − −= ccc vssVC(s)I
10. 10. 10 CAPACITOR  Capacitor’s voltage – Rearranged IC(s) equation: ( ) ( ) )(v s (s)I sC (s)V ccc − += 011
11. 11. 11 RLC VOLTAGE  The voltage across the RLC elements in the s-domain is the sum of a term proportional to its current I(s) and a term that depends on its initial condition. )]0()([)( − −= LLL issILsV ( ) ( ) )(v s (s)I sC (s)V ccc − += 011
12. 12. 12 CIRCUIT ANALYSIS FOR ZERO INITIAL CONDITIONS (ICs = 0)
13. 13. 13 IMPEDANCE  If we set all initial conditions to zero, the impedance is defined as: [all initial conditions=0] )( )()( sI sVsZ =
14. 14. 14 IMPEDANCE & ADMITANCE  The impedances in the s-domain are  The admittance is defined as: sC sZ sLsZ RsZ C L R 1 )( )( )( = = = sCsY sL sY R sY C L R = = = )( 1 )( 1 )(
15. 15. 15 Ex.  Find vc(t), t>0 H1 F5.0 Ω3 )(tu −+ )(tvc + − )(tvL − + )(tvR
16. 16. 16 Obtain s-Domain Circuit  All ICs are zero since there is no source for t<0 s s 2 3 s 1 −+ )(sVc + − )(sVL − + )(sVR )(sI
17. 17. 17 Convert to voltage sourced s-Domain Circuit s s 2 3 s 3 −+ )(sVc + − )(sVL −+ )(sVR )(sI − +
18. 18. 18 23 3 )( 0 3 )(3 2 2 ++ − =⇒ =+      ++ ss sI s sI s s:KVLBy Find I(s)
19. 19. 19 Find Capacitor’s Voltage  The capacitor’s voltage:  Rewritten: )23( 6 )( 2 )( 2 ++ − =⋅= sss sI s sVc )2)(1( 6 )23( 6 )( 2 ++ − = ++ − = ssssss sVc
20. 20. 20 Using PFE  Expanding Vc(s) using PFE:  Solved for K1, K2, and K3: 21)2)(1( 6 )( 321 + − + += ++ − = s K s K s K sss sVc 2 3 1 63 )2)(1( 6 )( + − + + − = ++ − = ssssss sVc
21. 21. 21 Find v(t)  Using look up table: 2 3 1 63 )2)(1( 6 )( + − + + − = ++ − = ssssss sVc ( ) )(363)( 2 tueetv tt c −− −+−=
22. 22. 22 Ex.  Find the Thevenin and Norton equivalent circuit at the terminal of the inductor. 1 H 0 .5 F 3 Ω u ( t )
23. 23. 23 Obtain s-domain circuit s 2 / s 3 1 / s
24. 24. 24 Find ZTH 2 / s 3 s ZTH 2 3+=
25. 25. 25 Find VTH or Voc 2 / s 3 1 / s + V T H - ss VTH 31 3 =⋅=
26. 26. 26 Draw The Thevenin Circuit  Using ZTH and VTH: 2 / s 3 + - 3 / s
27. 27. 27 Obtain The Norton Circuit  The norton current is: 2 / s 3 3 / ( 3 s + 2 ) 23 3 23 3 + = + == s s s Z V I TH TH N
28. 28. 28 Ex.  Find v0(t) for t>0.
29. 29. 29 s-Domain Circuit Elements Laplace transform all circuit’s elements ssC s F ssLH tu 31 3 1 1 1 )( =→ =→ →
30. 30. 30 s-Domain Circuit
31. 31. 31 Apply Mesh-Current Analysis 21 3 5 3 1 1 I s I s −      +=Loop 1 Loop 2 ( ) 2 2 1 21 35 3 1 3 5 3 0 IssI I s sI s ++=∴       +++−=
32. 32. 32 Substitute I1 into eqn loop 1 ( ) ( ) sss I Isss I s Iss s 188 3 1883 3 35 3 1 5 3 1 1 232 2 23 22 2 ++ =∴ ++=       −++      +=
33. 33. 33 Find V0(s) 22 2 20 )2()4( 2 2 3 188 3 )( ++ = ++ = = s ss sIsV
34. 34. 34 Obtain v0(t) tetv t 2sin 2 3 )( 4 0 − = 22 )2()4( 2 2 3 )( ++ = s sVo
35. 35. 35 Ex.  The input, is(t) for the circuit below is shown as in Fig.(b). Find i0(t) 1 0 2 t(s) is(t) (b) )(tis Ω1H1 )(tio (a)
36. 36. 36 s-Domain Circuit )(sIs 1s )(sIo
37. 37. 37  Using current divider: )1()( 1 )( →      + = sI s s sIo
38. 38. 38 1 0 2 t(sec) is(t) Derive Input signal, Is 0 t is1(t) 0 2 t is2(t) −
39. 39. 39 Obtain Is(t) and Is(s)  Expression for is(t):  Laplace transform of is(t): )2()()( −−= tututis ( ) )2(1 111 )( 22 →−=−= −− ss s e ss e s sI
40. 40. 40  Substitute eqn. (2) into (1): 11 1 )1( )1( )( 2 2 0 + − + = + − = − − s e s ss es sI s s
41. 41. 41 )2()()( )2( −−= −−− tuetueti tt o Inverse Laplace transform
42. 42. 42 CIRCUIT ANALYSIS FOR NON-ZERO INITIAL CONDITION (ICs ≠ 0)
43. 43. 43 TIME DOMAIN TO s-DOMAIN CIRCUITS  s replaced t in the unknown currents and voltages.  Independent source functions are replaced by their s-domain transform pair.  The initial condition serves as a second element, the initial condition generator.
44. 44. 44 THE ELEMENTS LAW OF s- DOMAIN )0( 1 )( 1 )( )0()()( )()( − − +=⇒ −=⇒ =⇒ CCC LLL RR v sC sI sC sV LissLIsV sRIsV
45. 45. 45 THE ELEMENTS LAW OF s- DOMAIN )0()()( )0()( )( )( )( − − −=⇒ +=⇒ =⇒ CC LL L R R CvssCVsI s i sL sV sI R sV sI
46. 46. 46 TRANSFORM OF CIRCUITS- RESISTOR  In the time domain:  In the s-domain: i ( t ) + v ( t ) - R v ( t ) = i ( t ) R I ( s ) + V ( s ) - R V ( s ) = I ( s ) R
47. 47. 47 TRANSFORM OF CIRCUITS- INDUCTOR  In the time domain:
48. 48. 48 TRANSFORM OF CIRCUITS- INDUCTOR  Inductor’s voltage:  Inductor’s current:
49. 49. 49 TRANSFORM OF CIRCUITS- CAPACITOR  In the time domain:
50. 50. 50 TRANSFORM OF CIRCUITS- INDUCTOR  Capacitor’s voltage:  Capacitor’s current:
51. 51. 51 Ex.  Find v0(t) if the initial voltage is given as v0(0- )=5 V
52. 52. 52 s-Domain Circuit
53. 53. 53 Apply nodal analysis method 5.2 1 1 )2( 10 1 5.2 10101 1 10 5.02 1010 0 10 )1( 10 0 + + =+⇒ =++ + −⇒ +=++ − + s sV sVV s V VVV o oo s oos
54. 54. 54 Cont’d )2)(1( 3525 25 1 10 )2( 0 ++ + =∴ + + =+ ss s V s sVo
55. 55. 55 Using PFE  Rewrite V0(s) using PFE:  Solved for K1 and K2: 21)2)(1( 3525 21 + + + = ++ + = s K s K ss s Vo 15;10 21 == KK
56. 56. 56 Obtain V0(s) and v0(t)  Calculate V0(s):  Obtain V0(t) using look up table: 2 15 1 10 )( + + + = ss sVo )()1510()( 2 tueetv tt o −− +=