Notes-1B

Notes-1B - Review: Laplace Transform Laplace Transform of...

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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 one-sided 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 unit-impulse 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. Time-scaling: 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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Notes-1B - Review: Laplace Transform Laplace Transform of...

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