HigherLin4

# Example find y t l 1 y t for y s 3 s 1 s 2 2 s 8 by

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Example. Find y ( t ) = L 1 [ Y ]( t ) for Y ( s ) = 3 s + 1 s 2 2 s 8 . By the partial fraction identity 3 s + 1 s 2 2 s 8 = 3 s + 1 ( s 4)( s + 2) = 13 6 s 4 + 5 6 s + 2 , you can express Y ( s ) as Y ( s ) = 13 6 1 s 4 + 5 6 1 s + 2 . The top right entry of table (8.13) with a = 4 and a = 2 then yields y ( t ) = L 1 [ Y ( s )]( t ) = 13 6 L 1 bracketleftbigg 1 s 4 bracketrightbigg ( t ) + 5 6 L 1 bracketleftbigg 1 s + 2 bracketrightbigg ( t ) = 13 6 e 4 t + 5 6 e 2 t . Example. Find y ( t ) = L 1 [ Y ]( t ) for Y ( s ) = 3 ( s 2 + 4)( s 2 + 9) . By the partial fraction identity 3 ( z + 4)( z + 9) = 3 5 ( z + 4) + 3 5 ( z + 9) , you can express Y ( s ) as Y ( s ) = 3 5 s 2 + 4 3 5 s 2 + 9 = 3 10 2 s 2 + 2 2 1 5 3 s 2 + 3 2 . The bottom left entry of table (8.13) with b = 2 and b = 3 then yields y ( t ) = L 1 [ Y ( s )]( t ) = 3 10 L 1 bracketleftbigg 2 s 2 + 2 2 bracketrightbigg ( t ) 1 5 L 1 bracketleftbigg 3 s 2 + 3 2 bracketrightbigg ( t ) = 3 10 sin(2 t ) 1 5 sin(3 t ) .

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18 Example. Find y ( t ) = L 1 [ Y ]( t ) for Y ( s ) = 7 s + 5 s 2 + 4 + ( 1 e 2 s ) 2 s 3 ( s 2 + 4) ( e 2 s + e 4 s ) 2 s 2 ( s 2 + 4) e 4 s 4 s ( s 2 + 4) . You first derive the partial fraction identities 7 s + 5 s 2 + 4 = 7 s s 2 + 4 + 5 s 2 + 4 , 2 s 3 ( s 2 + 4) = 1 2 s 3 1 8 s + 1 8 s s 2 + 4 , 2 s 2 ( s 2 + 4) = 1 2 s 2 1 2 s 2 + 4 , 4 s ( s 2 + 4) = 1 s s s 2 + 4 . The top left of these is straightforward. The top right identity only involves s 2 , so is simply the identity 2 z ( z + 4) = 1 2 z 1 2 z + 4 , evaluated at z = s 2 . The bottom right identity is simply 2 s times the top right one. Finally, the bottom left identity is obtained by first dividing the top right one by s and then employing the bottom right one divided by 8 to the last term. These partial fraction identites allow you to express Y ( s ) as Y ( s ) = 7 s s 2 + 4 + 5 s 2 + 4 + ( 1 e 2 s ) parenleftbigg 1 2 s 3 1 8 s + 1 8 s s 2 + 4 parenrightbigg ( e 2 s + e 4 s ) parenleftbigg 1 2 s 2 1 2 s 2 + 4 parenrightbigg e 4 s parenleftbigg 1 s s s 2 + 4 parenrightbigg = 7 s s 2 + 2 2 + 5 2 2 s 2 + 2 2 + ( 1 e 2 s ) parenleftbigg 1 4 2 s 3 1 8 1 s + 1 8 s s 2 + 2 2 parenrightbigg ( e 2 s + e 4 s ) parenleftbigg 1 2 1 s 2 1 4 2 s 2 + 2 2 parenrightbigg e 4 s parenleftbigg 1 s s s 2 + 2 2 parenrightbigg . The formulas in the first column of table (8.13) show that L 1 bracketleftbigg 7 s s 2 + 2 2 + 5 2 2 s 2 + 2 2 bracketrightbigg ( t ) = 7 cos(2 t ) + 5 2 cos(2 t ) , L 1 bracketleftbigg 1 4 2 s 3 1 8 1 s + 1 8 s s 2 + 2 2 bracketrightbigg ( t ) = 1 4 t 2 1 8 + 1 8 cos(2 t ) , L 1 bracketleftbigg 1 2 1 s 2 1 4 2 s 2 + 2 2 bracketrightbigg ( t ) = 1 2 t 1 4 sin(2 t ) , L 1 bracketleftbigg 1 s s s 2 + 2 2 bracketrightbigg ( t ) = 1 cos(2 t ) .
19 By combining these facts with formula (8.14), it follows that y ( t ) = 7 cos(2 t ) + 5 2 cos(2 t ) + ( 1 4 t 2 1 8 + 1 8 cos(2 t ) ) u ( t 2) ( 1 4 ( t 2) 2 1 8 + 1 8 cos(2( t 2)) ) u ( t 2) ( 1 2 ( t 2) 1 4 sin(2( t 2)) ) u ( t 4) ( 1 2 ( t 4) 1 4 sin(2( t 4)) ) u ( t 4) ( 1 cos(2( t 4)) ) .

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