Laplace Transformation - CAE Users

1 Laplace Transformation Laplace Transformation Definition: . f (t ) L{ f (t )} = F ( s ) = f (t )e st dt 0. time domain frequency domain Usefulness: differential algebraic equations equations Analogy: a log a a b log a + log b Circuit Analysis Using Laplace Transforms Time domain Complex frequency (t domain) domain (s domain). Linear Circuit Differential Laplace transform Algebraic equation L equation Classical Algebraic techniques techniques Response Inverse transform Response waveform L-1 transform Basic Laplace transform Pairs Laplace transform of Some Basic Functions . L{ (t )} = (t )e st dt 0. 0+. = (t )e dt st 0. 0+. = (t )dt 0. =1. Laplace transform of Some Basic Functions . L{u (t )} = u (t )e dt st 0.

2 = e st dt 0.. 1 st . = e . s 0. 1. =. s Laplace transform of Some Basic Functions . L{t u (t )} = t u (t )e st dt 0.. = te dt st 0.. 1 st 1 st = t ( e ) + e dt s 0 s 0. 1. = 2. s Laplace transform of Some Basic Functions . L{e } = e e dt t t st 0.. = e ( s + )t dt 0.. 1 ( s + )t . = e . s + 0. 1. =. s + . Basic Laplace Transformation Properties Proofs of Basic Laplace Transformation Properties F ( s). L{ f ( )d } =. t 0 s .. Proof: L{ 0 f ( )d } = 0 0 f ( )d e dt t t st .. e st e st . f ( )d f (t ) . t = dt s 0. 0 0. s . 1 F (s). = f (t )e dt =. st s 0 s Proofs of Basic Laplace Transformation Properties df (t ). L{ } = sF ( s ) f (0 ). dt df (t ) df (t ).. st Proof: L{ } = . e dt dt 0 dt [.]

3 = f (t )e ] . st . 0 .. 0. ( ). f (t ) se st dt . = f (0 ) + s f (t )e dt = sF ( s ) f (0 ). st . 0. Poles and Zeros of F(s). bm s m + bm 1s m 1 + K + b1s + b0. F (s) =. an s n + an 1s n 1 + K + a1s + a0. ( s z1 )( s z 2 ) K ( s z m ). F (s) = K. ( s p1 )( s p2 ) K ( s pn ). Scale factor: K = bm/an Poles: s = pk (k = 1, 2, .., n). Critical frequencies Zeros: s = zk (k = 1, 2, .., m). Pole-Zero Diagrams pole location zero location j j j . s-plane s-plane s-plane j .. 1 . j . 1 A( s + ) 1. F (s) = F (s) = F (s) =. s +1 (s + )2 + 2 s pole : s = 1 zero : s = pole : s = 0 + j 0. poles : s = j . Poles and Waveforms If poles in right-plane, waveform increases without bound as time approaches infinity Complex poles come in pairs that produce oscillatory waveforms Real poles produce exponential waveforms If poles in left-plane, waveform decays to zero as time approaches infinity If poles on j-axis, waveform neither decays nor grows Inverse Laplace Transforms ( s z1 )( s z 2 ) K ( s z m ).

4 F (s) = K. ( s p1 )( s p2 ) K ( s pn ). m < n, , F(s) is a proper rational function k1 k2 kn F (s) = + +K+ ki : residues ( s p1 ) ( s p2 ) ( s pn ). ( ). f (t ) = k1e p1t + k 2 e p2t + K + k n e pnt u (t ). How to determine k's? Distinct poles: ki = ( s pi ) F ( s ) s = p i Repeated poles of multiplicity r: r associated residues ki0 = ( s pi ) r F ( s ) s = p i dq k = q ( s pi ) r F ( s ) s = p i q q = 1L r 1. ds i Example of Inverse Laplace Transforms ( s z1 ) k1 k2 k3. F (s) = K = + +. ( s p1 )( s p2 )( s p3 ) s p1 s p2 s p3. Multiply with (s-p1). ( s z1 ) k ( s p1 ) k3 ( s p1 ). ( s p1 ) F ( s ) = K = k1 + 2 +. ( s p2 )( s p3 ) s p2 s p3. Set s = p1. ( p1 z1 ). ( s p1 ) F ( s ) s = p =K = k1.

5 1. ( p1 p2 )( p1 p3 ). Similarly, k 2 = ( s p2 ) F ( s ) s = p 2. k3 = ( s p3 ) F ( s ) s = p 3. Example of Inverse Laplace Transforms ( s z1 ) k1 k 20 k 21. F (s) = K = + +. ( s p1 )( s p2 ) 2. s p1 ( s p2 ) 2. s p2. Multiply with (s-p2)2. ( s z1 ) k1 ( s p2 ) 2. ( s p2 ) F ( s ) = K. 2. = + k 20 + k 21 ( s p2 ). ( s p1 ) s p1. Set s = p2. ( p2 z1 ). ( s p2 ) F ( s ). 2. =K = k 20. s = p2 ( p2 p1 ). Differentiate s d ( s p2 ) 2 F ( s ) = k 21. ds s = p2. Inverse Laplace Transforms m n, , F(s) is an improper rational function For example: s 3 + 6 s 2 + 12 s + 8. F (s) =. s 2 + 4s + 3. s+2 1 1. = s+2+ 2 = s+2+ 2. + 2. s + 4s + 3 s +1 s + 3. Recall: L{ (t) }= 1. d (t ) . L = s L { }. (t) ( 0.)

6 =s dt . Inverse transform : f (t ) =. d (t ). dt + 2 (t ) + [1 t 2. e + 2. e ]u (t ). 1 3t Pathologic waveforms Pathological waveforms usually do not occur in real circuits Improper rational functions may occur during manipulation Additional Laplace transform Properties and Pairs Proofs of Some Additional Properties t translation: If L{ f (t )} = F ( s ), then L{ f (t )u (t )} = e s F ( s ) for > 0.. Proof: L{ f (t )u (t )} = f (t )u (t )e st dt 0.. = f (t ' )e s (t '+ ) dt ' Let t ' = t . 0.. s =e f (t ' )e st dt ' = e s F ( s ). 0. Proofs of Some Additional Properties Scaling: If L{ f (t )} = F ( s ), 1 s then L{ f ( t )} = F ( ) for > 0.. Proof: L{ f ( t )} = f ( t )e st dt 0. t'. = f (t ' )e ( s )t '.

7 D( ) Let t ' = t 0 . 1 1 s . ( s )t '. = f (t ' )e dt ' = F ( ). 0 . Proofs of Some Additional Properties Initial/Final Values: lim f (t) = lim sF ( s ). t 0 + s . lim f (t) = lim sF ( s ). t s 0.. df (t ) st Proof: sF ( s ) f (0 ) = 0 .. e dt dt 0 + df (t ) df (t ). lim( sF ( s ) f (0 )) = lim .. e dt + lim +. st e st dt s s 0 dt s 0 dt 0 + df (t ). = lim e 0 dt + 0 = f (0 + ) f (0 ). s 0 dt lim sF ( s ) = f (0 + ). s . Proofs of Some Additional Properties Proof of Initial/Final Values: df (t ) st lim( sF ( s ) f (0 )) = lim .. e dt s 0 s 0 0 dt df (t ). = lim e 0 dt = f ( ) f (0 ). s 0 0 dt lim sF ( s ) = lim f (t ). s 0 t . Initial-value property holds when F(s) is a proper rational function (f(0+) is well defined).

8 Final-value property holds when poles of sF(s) are in the left- half plane (except for a simple pole at the origin). Proofs of Some Additional Properties Convolution Property: L {F1 ( s ) F2 ( s )} = f1 ( ) f 2 (t )d . t 1. 0. = f 2 ( ) f1 (t )d . t 0.. Proof: F1 ( s) = f1 ( )e s d . 0.. ( ). F1 ( s) F2 ( s ) = f1 ( ) F2 ( s )e s d ). 0. = f1 ( ) f 2 (t )u (t )e st dt d ).. 0 0 . = f1 ( ) f 2 (t )d e st dt . t 0 0 . Example of Convolution Property 1 . 1 t Show: L 2 . = te (s + ) . 1 1 t ( t ). = 0 e e 1. Proof: L d . (s + ) (s + ) . = e t d = te t t 0.