Question

a Apply given an inverse franction Laplace and find transform resulting to f (t) F 6) = 300s (5+1)(8 + 25 +26) and explain how Please clearly Show all work to get to this answer. Answers! F#) = -126 61.2 e cos (5t +78.79)

Please clearly show all work and explain how they computed this answer.

Answer 1

By using partial fractions, inverse Laplace transform and response of second order control system for exponential input we get the solution. There is a small correction in the final answer.

Sol'r 30o 5 FCs) BS+ C FeS) A S+ ACsT25t26)CBs+c) (S+) 300 S As t2As+26A + Bst BS+cstC 3ooS Compare Coeficients both sides on A十B =0 24+Btcニ 300 26 A t C = O A =- B put it in 3 B tc = 300 B t26 (-Bt c= o -B+ C/= 300 6B O ニ 300 25 B A B -12 2A+BtCニ 300 2(-1ノ+ 12tc= 300 ー2 Cニ32

12-513(2 FCs) 12 St s'42st6 E+1+ 25 S+25+ 1t -2 2S+2 S that Kno We t eu ct) L.7 Co 5utf St S+15 5 e sin5tut «Li Inverse Laplace Tansfoym to pply Now -t -t -1Et+ s (5t + 60 e sin(st) -t 1et12e st5sinl5 From above it y observed that 5 d /sec

3C wn- 5 26 25= sine= if Let -t L Cat 5 sin(yt A 12 e sinst A 12 e t in(st) 2 A tos (18- + sin(18-7Sin(sy A = fI-19 e C5t - T8 10 61.14 e A = wo

Now - 12 e61 T (5t-78-70) 19 e simi lar to Note The Oyder System esponse Second the Cortrol System tor exponential in input 12 e -12 e -t CaCt 17 3て)- + sin Cut) Wdwn -