Applications laplace transform
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This document describes circuit analysis using Laplace transforms. It discusses analyzing both linear and nonlinear circuits in the Laplace domain. Key steps include taking the Laplace transform of circuit elements and sources, setting up equations for elements like resistors, inductors and capacitors in the s-domain, and using circuit analysis techniques like KVL, KCL to solve for output variables. It also addresses analyzing circuits with both zero and non-zero initial conditions. Examples are provided to illustrate the process.
![7
INDUCTOR
Inductor’s voltage
– In the time domain:
– In the s-domain:
dt
di
LtvL =)(
)]0()([)( −
−= LLL issILsV](https://image.slidesharecdn.com/applicationsofthelaplacetransform-150504063102-conversion-gate01/85/Applications-laplace-transform-7-320.jpg)
![9
CAPACITOR
Capacitor’s current
– In the time domain:
– In the s-domain:
dt
dv
Ctic =)(
)]0()([ −
−= ccc vssVC(s)I](https://image.slidesharecdn.com/applicationsofthelaplacetransform-150504063102-conversion-gate01/85/Applications-laplace-transform-9-320.jpg)
![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](https://image.slidesharecdn.com/applicationsofthelaplacetransform-150504063102-conversion-gate01/85/Applications-laplace-transform-11-320.jpg)
![13
IMPEDANCE
If we set all initial conditions to zero,
the impedance is defined as:
[all initial conditions=0]
)(
)()(
sI
sVsZ =](https://image.slidesharecdn.com/applicationsofthelaplacetransform-150504063102-conversion-gate01/85/Applications-laplace-transform-13-320.jpg)
Applications laplace transform
- 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 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 KIRCHHOFF’S VOLTAGE LAW Consider the KVL in time domain: Apply the Laplace transform: 0)()()()( 4321 =++++ tvtvtvtv 0)()()()( 4321 =++++ sVsVsVsV
- 5. 5 KIRCHHOFF’S CURRENT LAW Consider the KCL in time domain: Apply the Laplace transform: 0)()()()( 4321 =++++ tItItItI 0)()()()( 4321 =++++ titititi
- 6. 6 OHM’S LAW Consider the Ohm’s Law in time domain Apply the Laplace transform RsIsV RR )()( = Rtitv RR )()( =
- 7. 7 INDUCTOR Inductor’s voltage – In the time domain: – In the s-domain: dt di LtvL =)( )]0()([)( − −= LLL issILsV
- 8. 8 INDUCTOR Inductor’s current – Rearrange VL(s) equation: s i sL sV sI L L )0()( )( − +=
- 9. 9 CAPACITOR Capacitor’s current – In the time domain: – In the s-domain: dt dv Ctic =)( )]0()([ − −= ccc vssVC(s)I
- 10. 10 CAPACITOR Capacitor’s voltage – Rearranged IC(s) equation: ( ) ( ) )(v s (s)I sC (s)V ccc − += 011
- 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 CIRCUIT ANALYSIS FOR ZERO INITIAL CONDITIONS (ICs = 0)
- 13. 13 IMPEDANCE If we set all initial conditions to zero, the impedance is defined as: [all initial conditions=0] )( )()( sI sVsZ =
- 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 Ex. Find vc(t), t>0 H1 F5.0 Ω3 )(tu −+ )(tvc + − )(tvL − + )(tvR
- 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 Convert to voltage sourced s-Domain Circuit s s 2 3 s 3 −+ )(sVc + − )(sVL −+ )(sVR )(sI − +
- 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 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 Find v(t) Using look up table: 2 3 1 63 )2)(1( 6 )( + − + + − = ++ − = ssssss sVc ( ) )(363)( 2 tueetv tt c −− −+−=
- 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 Obtain s-domain circuit s 2 / s 3 1 / s
- 24. 24 Find ZTH 2 / s 3 s ZTH 2 3+=
- 25. 25 Find VTH or Voc 2 / s 3 1 / s + V T H - ss VTH 31 3 =⋅=
- 26. 26 Draw The Thevenin Circuit Using ZTH and VTH: 2 / s 3 + - 3 / s
- 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 Ex. Find v0(t) for t>0.
- 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 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 ++ =∴ ++= −++ +=
- 34. 34 Obtain v0(t) tetv t 2sin 2 3 )( 4 0 − = 22 )2()4( 2 2 3 )( ++ = s sVo
- 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)
- 37. 37 Using current divider: )1()( 1 )( → + = sI s s sIo
- 38. 38 1 0 2 t(sec) is(t) Derive Input signal, Is 0 t is1(t) 0 2 t is2(t) −
- 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 Substitute eqn. (2) into (1): 11 1 )1( )1( )( 2 2 0 + − + = + − = − − s e s ss es sI s s
- 41. 41 )2()()( )2( −−= −−− tuetueti tt o Inverse Laplace transform
- 42. 42 CIRCUIT ANALYSIS FOR NON-ZERO INITIAL CONDITION (ICs ≠ 0)
- 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 THE ELEMENTS LAW OF s- DOMAIN )0( 1 )( 1 )( )0()()( )()( − − +=⇒ −=⇒ =⇒ CCC LLL RR v sC sI sC sV LissLIsV sRIsV
- 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 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 TRANSFORM OF CIRCUITS- INDUCTOR In the time domain:
- 48. 48 TRANSFORM OF CIRCUITS- INDUCTOR Inductor’s voltage: Inductor’s current:
- 49. 49 TRANSFORM OF CIRCUITS- CAPACITOR In the time domain:
- 50. 50 TRANSFORM OF CIRCUITS- INDUCTOR Capacitor’s voltage: Capacitor’s current:
- 51. 51 Ex. Find v0(t) if the initial voltage is given as v0(0- )=5 V
- 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
- 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 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 −− +=