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

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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 0 - f ( t ) e - st dt. where s = σ + is a complex number and σ is sufficiently large so that the integral converges. 0 - 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 πj Z σ c + j σ c - j F ( s ) e st ds where σ c is large enough to include all p k , poles of F ( s ), on the left of straight line s = σ 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
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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 0 a 1( t ) e - st dt = - ae - st s fl fl fl fl fl fl fl 0 = a s Note that e - st 0 as t → ∞ for s = σ + with σ > 0. Similarly, L [ bt 1( t )] = Z 0 bt 1( t ) e - st dt = - bte - st s - be - st s 2 0 = b s 2 Example L2: Find the Laplace Transform of the unit-impulse func- tion δ ( t ). Solution: By the Laplace transform integral L [ δ ( t )] = Z 0 - δ ( t ) e - st dt = Z 0 + 0 - δ ( t ) e - st dt = 1 . Example L3: Find the Laplace Transform of the sinusoid function. Solution: By the Laplace transform integral L [sin ωt ] = Z 0 - sin ωt e - st dt = Z 0 - e jωt - e - jωt 2 j e - st dt = Z 0 - e ( - s ) t - e ( - - s ) t 2 j dt = 1 2 j 1 s - - 1 s + = ω s 2 + ω 2 2
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Properties of Laplace Transforms 1. Superposition: L [ α 1 f 1 ( t ) + α 2 f 2 ( t )] = α 1 F 1 ( s ) + α 2 F 2 ( s ). 2. Time delay: L [ f ( t - T )] = e - sT F ( s ). L [ f ( t - T )] = Z 0 - f ( t - T ) e - st dt = Z - T f ( τ ) e - s ( τ + T ) = e - sT Z 0 - f ( τ ) e - = e - sT F ( s ) 3. Time-scaling: L [ f ( at )] = 1 a F ( s a ) , a > 0. L [ f ( at )] = Z 0 - f ( at ) e - st dt = 1 a Z 0 - 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 0 - e - at f ( t ) e - st dt = Z 0 - 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 0 - ˙ f ( t ) e - st dt = f ( t ) e - st fl fl fl fl 0 - + Z 0 - f ( t ) se - st dt = - f (0 - ) + sF ( s ) 3
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6. Integration: L [ R t 0 f ( τ ) ] = F ( s ) s . L [ Z t 0 f ( τ ) ] = Z 0 - Z t 0 f ( τ ) dτe - st dt = - 1 s Z 0 - Z t 0 f ( τ ) dτde - st = - 1 s " Z t 0 f ( τ ) dτe - st fl fl fl fl fl 0 - - Z 0 - f ( t ) e - st dt # = 1 s F ( s ) 7. Convolution: L [ f 1 ( t ) * f 2 ( t )] = F 1 ( s ) F 2 ( s ). L [ f 1 ( t ) * f 2 ( t )] = Z 0 - •Z 0 f 1 ( τ ) f 2 ( t - τ ) e - st dt = Z 0 f 1 ( τ ) •Z 0 - f 2 ( t - τ ) e - st dt = Z 0 f 1 ( τ ) e - •Z 0 - f 2 ( t - τ ) e - s ( t - τ ) dt = F 2 ( s ) Z 0 f 1 ( τ ) e - = F 1 ( s ) F 2 ( s ) 8. Time product: L [ f 1 ( t ) f 2 ( t )] = 1 2 πj R σ + j σ - j F 1 ( x ) F 2 ( s - x ) dx . L [ f 1 ( t ) f 2 ( t )] = Z 0 - f 1 ( t ) f 2 ( t ) e - st dt = Z 0 - 1 2 πj Z σ c + j σ c - j F 1 ( x ) e xt dx f 2 ( t ) e - st dt = 1 2 πj Z σ c + j σ c - j F 1 ( x ) •Z 0 - f 2 ( t ) e - ( s - x ) t dt dx = 1 2 πj Z σ c + j σ c - j F 1 ( x ) F 2 ( s - x ) dx =: 1 2 πj F 1 ( s ) * F 2 ( s ) 9. Multiplication by time: L [ tf ( t )] = - F 0 ( s ).
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