This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 2 = 0 ) K 2 = 1 6 K 1 + K 3 = 0 ) K 3 = 6 Which gives us: F ( s ) = 1 s s +6 s 2 +6 s +25 = 1 s s +3+3 ( s 2 +3) 2 +4 2 = 1 s s +3 ( s 2 +3) 2 +4 2 3 ( s 2 +3) 2 +4 2 We know that: L [ Ae at cos !t ] = A ( s + a ) ( s + a ) 2 + ! 2 and L [ Be at sin !t ] = B! ( s + a ) 2 + ! 2 from the constant multiplication and frequency shift properties. SO: ) f ( t ) = 1 e 3 t cos 4 t 3 4 e 3 t sin 4 t b) s +4 s 2 +8 s +160 = s +4 s 2 +8 s +16+144 = s +4 ( s +4) 2 +12 2 Glancing at the above problem, we can write the solution by inspection: ) f ( t ) = e 4 t cos 12 t Page 4 of 4...
View
Full
Document
This note was uploaded on 03/20/2010 for the course ME 450 taught by Professor Liburdy during the Spring '08 term at Oregon State.
 Spring '08
 Liburdy

Click to edit the document details