chapter6.page08

chapter6.page08 - ∞ f t is sum of these three products f...

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8 1. LAPLACE TRANSFORM METHODS In general, if b > a, u a ( t ) - u b ( t ) = 0 if 0 < t < a 1 if a < t b 0 if t > b To express a multiple defined function as linear combination of step functions, we need, for each finite interval [a, b] on which f has a different expression, multiply the expression by the difference of step functions u a ( t ) - u b ( t ); for the expression defined on an infinite interval [c, ) we just multiply the expression by u c ( t ) . f ( t ) is the summation of all the products, as illustrated in the following examples, Example 3.1 . Express f ( t ) = t 2 - t + 3 if 0 < t < 2 e t if 2 t < 5 2 t sin( t ) if t 5 as linear combination of step functions. Solution Notice f ( t ) has three different expresses over interval [0, 2), [2, 5) and [5, ), so we have two difference, u ( t ) - u 2 ( t ) and u 2 ( t ) - u 5 ( t ) , and three products, ( t 2 - t + 3)( u ( t ) - u 2 ( t )) for ( t 2 - t + 3) defined on [0,2); e t ( u 2 ( t ) - u 5 ( t )) for e t defined on [2, 5); 2 t sin( t ) u 5 ( t ) for 2 t defined on [5,
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Unformatted text preview: ∞ ). f ( t ) is sum of these three products, f ( t ) = ( t 2-t + 3)( u ( t )-u 2 ( t )) + e t ( u 2 ( t )-u 5 ( t )) + 2 t sin( t ) u 5 ( t ) = ( t 2-t + 3) u ( t ) + ( e t-t 2 + t-3) u 2 ( t ) + (2 t sin( t )-e t ) u 5 ( t ) . a Example 3.2 . Express f ( t ) = t 2 e t if < t < 3 cos( t ) if 3 ≤ t < 6 2 t if t ≥ 6 as linear combination of step functions. Solution Notice f ( t ) has three different expresses over interval [0, 3), [3, 6) and [6, ∞ ), so we need to have two difference, u ( t )-u 3 ( t ) and u 3 ( t )-u 5 ( t ) and f ( t ) = t 2 e t ( u ( t )-u 3 ( t )) + cos( t )( u 3 ( t )-u 6 ( t )) + 2 tu 6 ( t ) = t 2 e t u ( t ) + (cos( t )-t 2 e t ) u 3 ( t ) + (2 t-cos( t )) u 6 ( t ) . a...
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This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.

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