Question

Find the inverse (unilateral) Laplace transforms of the following functions:

(a) $\frac{2s+5}{s^{2} +5s+6}$

(b) $\frac{3s +5 }{s^{2} +4s+13}$

(c) $\frac{(s+1)^{2}}{s^{2}-s-6}$

(d)$\frac{5}{s^{2}(s+2)}$

(e) $\frac{2s+1}{(s +1)(s^{2} +2s +2)}$

(f) $\frac{s+2}{s(s +1)^{2}}$

(g) $\frac{1}{(s +1)(s +2)^{4}}$

(h) $\frac{s +1}{s(s +2)^{2}(s^{2} +4s+5)}$

(i) $\frac{s^{2}}{(s +1)^{2}(s^{2} +2s+5)}$

Answer 1

(a) Given 25+5 $455+6 25+ 5 (8+2) (5+3). ... Invense Laplace transforms of 25+5 le. Ilh 25+5 ? 3275346 1 st 55+6s ( 5+55+6 ( - ta'}} } ezt te 37

- Given 35+5 8745+13 = 3(3+2)-1

sem to = 324 Cos (34) - et sin (3+). Given (5+1) 7 (5+1) 2. e) (s²_8-6) .) -575+6 = 1- 347) (5+5+6) (35+7) (3-3) (3+2) = 1+ Now, let 353 Tsites - Compaining - A(5+2) + B (S-3) (5-3) (3+2) = S(A+B) + 2A-3B (5-3) (SP2) numenatons nominations, we have A+B=3 & 2A-3 B=7 > A=3-B = 2(3-3) - B 27 = 340 – 5B=1 = 16 1 » B=-YS

S- > otest2 (88) at St +16 et 2 e24 where s(t) is Dirac delta function e

d Given 5 I 5(512) Let tes = 4+ siz this is the form of partial fraction decomposition. Then 5 = 5(3+2) A+ (5+2) Bt se s(s+2) = 5 (Ate) + s (2A+B) + 2B - 82(3+2) Compaining the denominat numerator's We have Atezo, ZA+B 20, ZB=5 >e=5 & B = 5 7 2 = 2 16 > 2A=-B A = -2 m + L

= = } 동동은라