Question

Using the equation in the figure using the inverse Laplace transform Figure it out.

F(s) 11s+28 (s+2)2 (s+5)

Answer 1

A (5775 +10) + B (5+5) + C (5²+ 45+4) (6 given, Í (S) IIS + 28 (5+2)2 (5+5) F (S) B IIS +39 (5+2)2 (5+6 A S+2 sts (5+2) IS + 28 (5+2)² (5+5) A (5+2) (5+5) + B(5+5) + 6(3+2)² (5+2)? (5+5) 115 +28 = A(8+75+10) + B (5+5) + (8 +215 + 4) 2 os tlls +28 os²+ 115 +28 = 52(A+C) + s (A+B+40) + 10 A +58 +4 0 on comparing co-efficient را Atc A = -c > (11) S 7 A + B + 4C 7A+B - 4 A = "=3 A+B constant 28 - 10 A +5B +4c 28 10 A +5B - 4 A 6A +5B

Now, 3 A + B = 11 6 A+ SB = 28 x 2 then subtract by ② 649 + 2B = 22 GA +5B = 28 -3B B put B = 2 in ea 3 A + 2 = 11 9 Now C .A c = -3 so, A = 3, B = 2; C = -3 45+5 į Ls) Sie + 6)*-s () (5+2) 2 (sts) taking ILT Both sides, st5 ['? Foy & 3 Lt 2 st +22" }ets-y-4 -st 3 e -2+ -2t 2 е t le *}-30-1

we know è at 11 e sta flt) -at f(t) sniting property (2) if Liff (t)} Lt & I (sta)} : 너로화 ) 4 ny 2ni so, L' ? Fis)} - 2t = 3e + 2e .-2t -st Be الا