Question

1. (a) Using the Tables of Laplace transforms, along with the operational theorems, de- termine the inverse Laplace transform of 3s 7 82 -2s + 10 (b) Hence determine the inverse Laplace transform of 3s +7 -2s S2-2s10

Answer 1

3s+7 F(s) s2 2s10

3s+7 F(s) (s 1)9 2

F(s)3(s-1) +10 (s 1)9 2

s1 F(s) 3 10 (s 19 2 (s 1)9

take inverse laplace

s 1 f (t) L3 -1)910 -1)9 J

s 1 f (t)3L 1 (s 1)2+9 10L (s 1)29

use inverse laplace rule

LF (S-a}= e" f (t)

so here

S 1 (s-1)2+9e'L s2 9

and  

1 (s 1)9 e'L 29J

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f (t) 3e' L 1 2910e L-1 s2 9

$f(t)=3e^tL^{-1}\left\{\frac{s}{s^2+9}\right\}+10\cdot \frac{1}{3} \cdot e^tL^{-1}\left\{\frac{ 3}{s^2+9}\right\}$

f(t) 3e cos (3t)10esin (3t) 3

${\color{Red} f(t)=3e^t\cos \left(3t\right)+10e^t\frac{1}{3}\sin \left(3t\right)}$

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$F(s)=\frac{3s+7}{s^2-2s+10}e^{-2s}$

$F(s)=\frac{e^{-2s}\left(3s+7\right)}{s^2-2s+10}$

take inverse laplace

$f(t)= L^{-1}\left\{\frac{e^{-2s}\left(3s+7\right)}{s^2-2s+10}\right\}$

apply laplace rule $L^{-1}\left\{e^{-as}F\left(s\right)\right\}=u\left(t-a\right)f\left(t-a\right)$

$\mathrm{For\:}\frac{e^{-2s}\left(3s+7\right)}{s^2-2s+10}:\quad F\left(s\right)=\frac{\left(3s+7\right)}{s^2-2s+10},\:\quad \:a=2$

here we have already find inverse laplace of $F\left(s\right)=\frac{\left(3s+7\right)}{s^2-2s+10}$

and that is $f(t)=3e^t\cos \left(3t\right)+\frac{10}{3}e^t\sin \left(3t\right)$

but as explained above here a=2, so inverse laplace is

${\color{Red} f(t)=\text{u}\left(t-2\right)\left(3e^{t-2}\cos \left(3\left(t-2\right)\right)+\frac{10}{3}e^{t-2}\sin \left(3\left(t-2\right)\right)\right)}$