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solns4.nb 1 Chapter 4 (Laplace transforms): Solutions

Chapter 4 ( laplace transforms): Solutions (The table of laplace transforms is used throughout.) Solution (a) HsinH4 tLcos H2 tLL= ikjj1 2sinH4 tLy{zz=1 2 HsinH4 tLL=1 24 s2+16=2 s2+16. Solution (b) Hcosh2 HtL L= ikjjjjikjj1 2 H t- -tLy{zz2y{zzzz= ikjjjj 2 t 4+1 2+ -2 t 4y{zzzz=1 4 1 s-2+1 2 1 s+1 4 1 s+2=s2-2 sHs2-4L. Solution (c) Hcos Ha tLsinh Ha tLL= ikjj1 2 H a t- -a tL cos Ha tLy{zz=1 2 H a tcos Ha tL- -a tcos Ha tLL=1 2ikjjjjs-a Hs-aL2+a2-s+a Hs+aL2+a2y{zzzz=a s2-2 a3 s4+4 a4. Solution (d) Ht2 -3 tL= H -3 tt2L=2 Hs+ Solution (a)Use partial fractions, set s Hs+3L Hs+5L=A Hs+3L+B Hs+5L s=A Hs+5L+B Hs+3 LUsing the cover up method for example we see thatA= -3 2, B=5 Hs+3L Hs+5L= -3 2 1 Hs+3L+5 2 1 Hs+5L= ikjj-3 2 -3 ty{zz+ ikjj5 2 -5 ty{ so the inverseLaplacetransform -1 ikjjjs Hs+3L Hs+5Ly{zzz=-3 2 -3 t+5 2 -5 t.}}}}}}}}}

Chapter 4 (Laplace transforms): Solutions (The table of Laplace transforms is used throughout.) Solution 4.1(a) ¸ HsinH4tL cos H2tLL = …

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