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Question Number 48877 by absa last updated on 29/Nov/18

Commented by absa last updated on 29/Nov/18

$f i n d t h e i n v e r s e l a p l a c e t r a n s f o r m a t i o n ?$
	Answered by Abdulhafeez Abu qatada last updated on 30/Nov/18
	$((s^2 −3)/(s(s^2 +a))) = (A/s) + ((Bs+C)/(s^2 +a)) s^2 −3 = A(s^2 +a) + (Bs+C)s s^2 −3 = (A+B)s^2 + Cs + Aa ....where a is a constant By comparing coefficients, A + B = 1 C = 0 Aa = −3 C=0, A = ((−3)/a), B = 1 − (((−3)/a)) = 1 + (3/a) = ((a+3)/a) ((s^2 −3)/(s(s^2 +a))) = ((−3)/(as)) + (((((a+3)/a))s)/(s^2 +a)) L^(−1) {((s^2 −3)/(s(s^2 +a)))} = L^(−1) {((−3)/(as)) + (((((a+3)/a))s)/(s^2 +a))} L^(−1) {((s^2 −3)/(s(s^2 +a)))} = ((−3)/a) + (((a+3)/a))cos((√a)t)$
	$\frac{s^{2} - 3}{s (s^{2} + a)} = \frac{A}{s} + \frac{B s + C}{s^{2} + a}$ $s^{2} - 3 = A (s^{2} + a) + (B s + C) s$ $s^{2} - 3 = (A + B) s^{2} + C s + A a . . . . w h e r e a i s a c o n s t a n t$ $B y c o m p a r i n g c o e f f i c i e n t s,$ $A + B = 1$ $C = 0$ $A a = - 3$ $C = 0, A = \frac{- 3}{a}, B = 1 - (\frac{- 3}{a}) = 1 + \frac{3}{a} = \frac{a + 3}{a}$ $\frac{s^{2} - 3}{s (s^{2} + a)} = \frac{- 3}{a s} + \frac{(\frac{a + 3}{a}) s}{s^{2} + a}$ $L^{- 1} {\frac{s^{2} - 3}{s (s^{2} + a)}} = L^{- 1} {\frac{- 3}{a s} + \frac{(\frac{a + 3}{a}) s}{s^{2} + a}}$ $L^{- 1} {\frac{s^{2} - 3}{s (s^{2} + a)}} = \frac{- 3}{a} + (\frac{a + 3}{a}) c o s (\sqrt{a} t)$
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