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ECE2260F09_HW5p1bsoln

# ECE2260F09_HW5p1bsoln - 2 −[4 s 25[2 s 8 s 4 2 3 2 2 or L...

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2260 F 09 HOMEWORK #5 prob 1b solution E X : Find the Laplace transform of the following waveform: t 2 d dt e 4 t cos(3 t ) [ ] S OL ' N : We start on the inside (of this layered onion) and apply identities to work our way out to the time-domain form given. The innermost term, found in a transform table, is e 4 t cos(3 t ) . L e 4 t cos(3 t ) { } = s + 4 ( s + 4) 2 + 3 2 Now we apply the identity for derivatives. L d dt f ( t ) = sF ( s ) f (0 ) or L d dt e 4 t cos(3 t ) = s s + 4 ( s + 4) 2 + 3 2 e 4 t cos(3 t ) t = 0 or L d dt e 4 t cos(3 t ) = s ( s + 4) ( s + 4) 2 + 3 2 1 Now we apply the identity for multiplication by t n . L t n f ( t ) { } = ( 1) n d n F ( s ) ds n or L t 2 d dt e 4 t cos(3 t ) = ( 1) 2 d 2 ds 2 s ( s + 4) ( s + 4) 2 + 3 2 1 or L t 2 d dt e 4 t cos(3 t ) = ( 1) 2 d 2 ds 2 s ( s + 4) ( s + 4) 2 + 3 2 ( s + 4) 2 + 3 2 ( s + 4) 2 + 3 2 or

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L t 2 d dt e 4 t cos(3 t ) = d 2 ds 2 4 s + 25 ( s + 4) 2 + 3 2 or L t 2 d dt e 4 t cos(3 t ) = d ds 4 ( s + 4) 2 + 3 2 4 s + 25 [( s + 4) 2 + 3 2 ] 2 (2 s + 8) or L t 2 d dt e 4 t cos(3 t ) = d ds 4[( s + 4) 2 + 3 2 ]
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Unformatted text preview: 2 ] − [4 s + 25][2 s + 8] [( s + 4) 2 + 3 2 ] 2 or L t 2 d dt e − 4 t cos(3 t ) = − d ds − 4 s 2 − 50 s − 100 [( s + 4) 2 + 3 2 ] 2 or L t 2 d dt e − 4 t cos(3 t ) = − − 8 s − 50 [( s + 4) 2 + 3 2 ] 2 − 2 − 4 s 2 − 50 s − 100 [( s + 4) 2 + 3 2 ] 3 (2 s + 8) or, (with assistance from Matlab ® symbolic computations), L t 2 d dt e − 4 t cos(3 t ) = − 2 4 s 3 + 75 s 2 + 300 s + 175 [( s + 4) 2 + 3 2 ] 3...
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ECE2260F09_HW5p1bsoln - 2 −[4 s 25[2 s 8 s 4 2 3 2 2 or L...

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