HigherLin4

# 86 inverse transform the process of determining y t

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8.6: Inverse Transform. The process of determining y ( t ) from Y ( s ) is called taking the inverse Laplace transform. It is important to know that this process has a unique result. Indeed, we will use the following theorem. Theorem. Let f ( t ) and g ( t ) be two functions over [0 , ) and α a real number such that f ( t ) and g ( t ) are of exponential order α as t → ∞ , f ( t ) and g ( t ) are piecewise continuous over every [0 , T ], • L [ f ]( s ) = L [ g ]( s ) for every s > α . Then f ( t ) = g ( t ) for every t in [0 , ).

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16 The proof of this result requires tools from complex variables that are beyond the scope of this course. Fortunately, you do not need to know how to prove this result to use it! Its usefulness stems from the fact that solutions y ( t ) to the initial-value problems we are considering lie within the class of functions considered above — namely, they are functions that are of exponential order as t → ∞ and that are piecewise continuous over every [0 , T ]. In fact, they are continuous and piecewise differentiable over every [0 , T ]. This means that if we succeed in finding a function y ( t ) within this class such that L [ y ]( s ) = Y ( s ) then it will be the unique solution of the initial-value problem that we seek. Because the above result states there is a unique f ( t ) that is of exponential order as t → ∞ and is piecewise continuous over every [0 , T ] such that L [ f ]( s ) = F ( s ), we introduce the notation f ( t ) = L 1 [ F ]( t ) . The operator L 1 denotes the inverse Laplace transform . Because it undoes the Laplace transform L , it inherits many properties from L . For example, it is linear. You can also easily read-off from the first and last three entries in table (8.11) of basic forms that L 1 bracketleftbigg n ! s n +1 bracketrightbigg ( t ) = t n , L 1 bracketleftbigg s s 2 + b 2 bracketrightbigg ( t ) = cos( bt ) , L 1 bracketleftbigg b s 2 + b 2 bracketrightbigg ( t ) = sin( bt ) , L 1 bracketleftbigg n ! ( s a ) n +1 bracketrightbigg ( t ) = e at t n , L 1 bracketleftbigg s a ( s a ) 2 + b 2 bracketrightbigg ( t ) = e at cos( bt ) , L 1 bracketleftbigg b ( s a ) 2 + b 2 bracketrightbigg ( t ) = e at sin( bt ) . (8.13) It is also clear from the sixth entry of table (8.11) that L 1 [ e cs F ( s )]( t ) = u ( t c ) f ( t c ) , where f ( t ) = L 1 [ F ]( t ) . (8.14) For us, the process of computing y ( t ) = L 1 [ Y ]( t ) for a given Y ( s ) will be one of expressing Y ( s ) as a sum of terms that will allow us to read off y ( t ) from the basic forms above. To illustrate this process, we will compute y ( t ) = L 1 [ Y ]( t ) for the Y ( s ) found in the examples given in the previous section, thereby completing our solution of the initial-value problems. Example. Find y ( t ) = L 1 [ Y ]( t ) for Y ( s ) = 1 ( s 2)( s 5) + 3 s 2 . By the partial fraction identity 1 ( s 2)( s 5) = 1 3 s 5 + 1 3 s 2 ,
17 you can express Y ( s ) as Y ( s ) = 1 3 1 s 5 + 8 3 1 s 2 . The top right entry of table (8.13) with a = 5 and a = 2 then yields y ( t ) = L 1 [ Y ( s )]( t ) = 1 3 L 1 bracketleftbigg 1 s 5 bracketrightbigg ( t ) + 8 3 L 1 bracketleftbigg 1 s 2 bracketrightbigg ( t ) = 1 3 e 5 t + 8 3 e 2 t .

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