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Unformatted text preview: Review: Laplace Transform Laplace Transform of Signals: Signals are represented as func tions of time t . For a signal f ( t ), its onesided Laplace transform is defined by F ( s ) := L [ f ( t )] = Z  f ( t ) e st dt. where s = + j is a complex number and is sufficiently large so that the integral converges. can be simply replaced by 0 if f ( t ) does not contain impulse at t = 0. Inverse Laplace Transform: Given F ( s ), f ( t ) can be obtained as f ( t ) = 1 2 j Z c + j c j F ( s ) e st ds where c is large enough to include all p k , poles of F ( s ), on the left of straight line s = c . Thus f ( t ) can be computed using residue theorem: f ( t ) = X Res p k F ( s ) e st = X Res p k { F ( s ) } e p k t . 1 Example L1: Find the Laplace Transform of the step function a 1( t ) and ramp function bt 1( t ). Solution: By the Laplace transform integral L [ a 1( t )] = Z a 1( t ) e st dt = ae st s fl fl fl fl fl fl fl = a s Note that e st 0 as t for s = + j with > 0. Similarly, L [ bt 1( t )] = Z bt 1( t ) e st dt =  bte st s be st s 2 = b s 2 Example L2: Find the Laplace Transform of the unitimpulse func tion ( t ). Solution: By the Laplace transform integral L [ ( t )] = Z  ( t ) e st dt = Z + ( t ) e st dt = 1 . Example L3: Find the Laplace Transform of the sinusoid function. Solution: By the Laplace transform integral L [sin t ] = Z  sin t e st dt = Z  e jt e jt 2 j e st dt = Z  e ( j s ) t e ( j s ) t 2 j dt = 1 2 j 1 s j 1 s + j = s 2 + 2 2 Properties of Laplace Transforms 1. Superposition: L [ 1 f 1 ( t ) + 2 f 2 ( t )] = 1 F 1 ( s ) + 2 F 2 ( s ). 2. Time delay: L [ f ( t T )] = e sT F ( s ). L [ f ( t T )] = Z  f ( t T ) e st dt = Z  T f ( ) e s ( + T ) d = e sT Z  f ( ) e s d = e sT F ( s ) 3. Timescaling: L [ f ( at )] = 1 a F ( s a ) , a > 0. L [ f ( at )] = Z  f ( at ) e st dt = 1 a Z  f ( at ) e ( s/a )( at )) d ( at ) = F ( s/a ) a 4. Shift in frequency: L [ e at f ( t )] = F ( s + a ). L [ e at f ( t )] = Z  e at f ( t ) e st dt = Z  f ( t ) e ( s + a ) t dt = F ( s + a ) 5. Differentiation: L [ f ( t )] = sF ( s ) f (0 ), and L f ( m ) ( t ) = s m F ( s ) s m 1 f (0 ) s m 2 f (0 )  f ( m 1) (0 ) . L [ f ( t )] = Z  f ( t ) e st dt = f ( t ) e st fl fl fl fl  + Z  f ( t ) se st dt = f (0 ) + sF ( s ) 3 6. Integration: L [ R t f ( ) d ] = F ( s ) s . L [ Z t f ( ) d ] = Z  Z t f ( ) de st dt = 1 s Z  Z t f ( ) dde st = 1 s " Z t f ( ) de st fl fl fl fl fl  Z  f ( t ) e st dt # = 1 s F ( s ) 7. Convolution: L [ f 1 ( t ) * f 2 ( t )] = F 1 ( s ) F 2 ( s )....
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 Fall '07
 Chen

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