Chapter 4 Lecture Notes-Laplace Transforms 3

Chapter 4 Lecture Notes-Laplace Transforms 3 - McGill...

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McGill University Math 263: Differential Equations and Linear Algebra CHAPTER 4: LAPLACE TRANSFORMS (III) 1 Further Studies of Laplace Transform 1.1 Step Function 1.1.1 Definition u c ( t ) = { 0 t < c, 1 t c. 1.1.2 Some basic operations with the step function Cutting-off head part Generate a new function F ( t ) by “cutting-off the head part” of another function f ( t ) at t = c : F ( t ) = f ( t ) u c ( t ) = { 0 (0 < t < c ) f ( t ) ( t > c ) . 0-0
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Cutting-off tail part Generate a new function F ( t ) by “cutting-off the tail part” of another function f ( t ) at t = c : F ( t ) = f ( t ) [ 1 u c ( t ) ] = { f ( t ) (0 < t < c ) 0 ( t > c ) . Cutting-off both head and tail part Generate a new function F ( t ) by “cutting-off the head part” of another function f ( t ) at t = c 1 and its tail part at t = c 2 > c 1 : F ( t ) = f ( t ) u c 1 ( t ) f ( t ) u c 2 ( t ) = 0 (0 < t < c 1 ) f ( t ) ( c 1 < t < c 2 ) 0 ( t > c 2 ) . (1) It is easy to prove that if 0 < t 1 < t 2 , we may write discontinuous function: f ( t ) = 0 , ( < t < 0) f 1 ( t ) , (0 t < t 1 ) f 2 ( t ) , ( t 1 t < t 2 ) f 3 ( t ) , ( t 2 t < . 0-1
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where f k ( t ) , ( k = 1 , 2 , 3) are known, and defined on ( −∞ , ) into the form f ( t ) = [ f 1 ( t ) u 0 ( t ) f 1 ( t ) u t 1 ( t ) ] + [ f 2 ( t ) u t 1 ( t ) f 2 ( t ) u t 2 ( t ) ] + f 3 ( t ) u t 2 ( t ) 1.1.3 Laplace Transform of Unit Step Function L{ u c ( t ) } = e cs s . One can derive L{ u c ( t ) f ( t c ) } = e cs F ( s ) = e cs L [ f ] . or u c ( t ) f ( t c ) = L 1 { e cs L [ f ] } . Example 1 . Determine L [ f ] , if f ( t ) = 0 , (0 t < 1) t 1 , (1 t < 2) 1 , ( t 2) . 0-2
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Solution: In terms of the unit step function, we can express f ( t ) = ( t 1) u 1 ( t ) ( t 2) u 2 ( t ) . Let g = t . Then we have f ( t ) = g ( t 1) u 1 ( t ) g ( t 2) u 2 ( t ) . It follows that L [ f ] = e s L ( g ) e 2 s L [ g ] = 1 s 2 ( e s e 2 s ) . Example 2 . Determine L 1 [ 2e - s s 2 +4 ] . Solution: We have L [sin 2 t ] = 2 s 2 + 4 . Hence, L 1 [ 2e - s s 2 +4 ] = L 1 { e s L [sin 2 t ] } = u 1 ( t ) sin[2( t 1)] Example 3 . Determine L 1 [ ( s 4)e - 3 s s 2 4 s +5 ] . 0-3
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Solution: Let G ( s ) = ( s 4) s 2 4 s +5 = ( s 4) ( s 2) 2 +1 = [ ( s 2) ( s 2) 2 +1 2 ( s 2) 2 +1 ] .
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