Ch06 3
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This document discusses step functions and their Laplace transforms. It defines the unit step function uc(t) and negative step function -uty(c). Translating a function f(t) by c units results in the function g(t)=uc(t)f(t-c). The Laplace transform of uc(t) is 1/s. Translating f(t) corresponds to multiplying F(s) by e-cs in the Laplace domain. Several examples demonstrate finding the Laplace transform and inverse Laplace transform of functions involving step functions.
Ch06 3
- 1. Ch 6.3: Step Functions Some of the most interesting elementary applications of the Laplace Transform method occur in the solution of linear equations with discontinuous or impulsive forcing functions. In this section, we will assume that all functions considered are piecewise continuous and of exponential order, so that their Laplace Transforms all exist, for s large enough.
- 2. Step Function definition Let c ≥ 0. The unit step function, or Heaviside function, is defined by A negative step can be represented by ≥ < = ct ct tuc ,1 ,0 )( ≥ < =−= ct ct tuty c ,0 ,1 )(1)(
- 3. Example 1 Sketch the graph of Solution: Recall that uc(t) is defined by Thus and hence the graph of h(t) is a rectangular pulse. 0),()()( 2 ≥−= ttututh ππ ≥ < = ct ct tuc ,1 ,0 )( ∞<≤ <≤ <≤ = t t t th π ππ π 20 2,1 0,0 )(
- 4. Laplace Transform of Step Function The Laplace Transform of uc(t) is { } s e s e s e e s dte dtedttuetuL cs csbs b b c st b b c st b c st c st c − −− ∞→ − ∞→ − ∞→ ∞ − ∞ − = +−= −== == ∫ ∫∫ lim 1 limlim )()( 0
- 5. Translated Functions Given a function f (t) defined for t ≥ 0, we will often want to consider the related function g(t) = uc(t) f (t - c): Thus g represents a translation of f a distance c in the positive t direction. In the figure below, the graph of f is given on the left, and the graph of g on the right. ≥− < = ctctf ct tg ),( ,0 )(
- 6. Example 2 Sketch the graph of Solution: Recall that uc(t) is defined by Thus and hence the graph of g(t) is a shifted parabola. .0,)(where),()1()( 2 1 ≥=−= tttftutftg ≥ < = ct ct tuc ,1 ,0 )( ≥− <≤ = 1,)1( 10,0 )( 2 tt t tg
- 7. Theorem 6.3.1 If F(s) = L{f (t)} exists for s > a ≥ 0, and if c > 0, then Conversely, if f (t) = L-1 {F(s)}, then Thus the translation of f (t) a distance c in the positive t direction corresponds to a multiplication of F(s) by e-cs . { } { } )()()()( sFetfLectftuL cscs c −− ==− { })()()( 1 sFeLctftu cs c −− =−
- 8. Theorem 6.3.1: Proof Outline We need to show Using the definition of the Laplace Transform, we have { } )( )( )( )( )()()()( 0 0 )( 0 sFe duufee duufe dtctfe dtctftuectftuL cs sucs cus ctu c st c st c − ∞ −− ∞ +− −= ∞ − ∞ − = = = −= −=− ∫ ∫ ∫ ∫ { } )()()( sFectftuL cs c − =−
- 9. Example 3 Find the Laplace transform of Solution: Note that Thus )()1()( 1 2 tuttf −= ≥− <≤ = 1,)1( 10,0 )( 2 tt t tf { } { } { } 3 22 1 2 )1)(()( s e tLettuLtfL s s − − ==−=
- 10. Example 4 Find L{ f (t)}, where f is defined by Note that f (t) = sin(t) + uπ/4(t) cos(t - π/4), and ≥−+ <≤ = 4/),4/cos(sin 4/0,sin )( ππ π ttt tt tf { } { } { } { } { } 1 1 11 1 cossin )4/cos()(sin)( 2 4/ 2 4/ 2 4/ 4/ + + = + + + = += −+= − − − s se s s e s tLetL ttuLtLtfL s s s π π π π π
- 11. Example 5 Find L-1 {F(s)}, where Solution: 4 7 3 )( s e sF s− + = ( )3 7 3 4 71 4 1 4 7 1 4 1 7)( 6 1 2 1 !3 6 1!3 2 1 3 )( −+= ⋅+ = + = −−− − −− ttut s eL s L s e L s Ltf s s
- 12. Theorem 6.3.2 If F(s) = L{f (t)} exists for s > a ≥ 0, and if c is a constant, then Conversely, if f (t) = L-1 {F(s)}, then Thus multiplication f (t) by ect results in translating F(s) a distance c in the positive t direction, and conversely. Proof Outline: { } cascsFtfeL ct +>−= ),()( { })()( 1 csFLtfect −= − { } )()()()( 0 )( 0 csFdttfedttfeetfeL tcsctstct −=== ∫∫ ∞ −− ∞ −
- 13. Example 4 Find the inverse transform of To solve, we first complete the square: Since it follows that 52 1 )( 2 ++ + = ss s sG ( ) ( ) ( ) 41 1 412 1 52 1 )( 222 ++ + = +++ + = ++ + = s s ss s ss s sG { } { } ( )tetfesFLsGL tt 2cos)()1()( 11 −−−− ==+= { } ( )t s s LsFLtf 2cos 4 )()( 2 11 = + == −−