Inverse Laplace problem using partition fraction

Question

Hello I am solving inverse Laplace transform using partial fraction. The question is:

$$ X(s) = \frac{10(s+1)}{s(s^2+4s+8)} => \frac{10(s+1)}{s((s+2)^2+4)} $$

$$ \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$ for finding C1 $$ \frac{10(s+1)}{s((s+2)^2+4)} = \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$ Multiplying both side by $s$ and set $s=0$, we get $C1=5/4$ for finding C3 $$ \frac{10(s+1)}{s((s+2)^2+4)} = \frac {C1} {s} + \frac{C2}{((s+2)+4)} + \frac{C3}{((s+2)^2+4)} $$ Multiplying both side by $((s+2)^2+4)$ and set $s=-2$, we get $C3=5$
For C2 I will take first derivative $X(s)$and Multiplying both side by $((s+2)^2+4)$ $$ \frac{d}{ds}[((s+2)^2+4) X(s)] = \frac{d}{ds} \frac{10s}{s}$$ $$ \frac {s(10) - (10s + 10)*1} {s^2} $$ We set $s=-2$
I got $C2=-5/2$...

but the solution I have they said $C2 = -5/4$ Please can you tell where I did mistake?

Answer 1

Notice, for factorization of denominator you should find the roots of quadratic equation: $s^2+4s+8=0$ as $$s=\frac{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)}=-2\pm 2i$$ $$\implies s^2+4s+8=(s+2+2i)(s+2-2i)$$

Now, one should have the following partial fractions $$X(s)=\frac{10(s+1)}{s(s^2+4s+8)}=\frac{10(s+1)}{s(s+2+2i)(s+2-2i)}$$

$$\frac{10(s+1)}{s(s+2+2i)(s+2-2i)}=\frac{A}{s}+\frac{B}{s+2+2i}+\frac{C}{s+2-2i}$$

Now, you can solve for $A, B$ & $C$ & then apply formula for inverse Laplace for each partial fraction $$L^{-1}\left[\frac{1}{s+a}\right]=e^{-at}$$

Answer 2

(BIG) HINT:

$$\mathcal{L}_{s}^{-1}\left[\frac{10(s+1)}{s(s^2+4s+8)}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{5}{4s}-\frac{5(s-4)}{4(s^2+4s+8)}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{5}{4s}\right]_{(t)}-\mathcal{L}_{s}^{-1}\left[\frac{5(s-4)}{4(s^2+4s+8)}\right]_{(t)}=$$ $$\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{1}{s}\right]_{(t)}-\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{s-4}{s^2+4s+8}\right]_{(t)}=$$ $$\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{1}{s}\right]_{(t)}-\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{s-4}{(s+2)^2+4}\right]_{(t)}=$$ $$\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{1}{s}\right]_{(t)}-\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{s}{(s+2)^2+4}-\frac{4}{(s+2)^2+4}\right]_{(t)}=$$ $$\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{1}{s}\right]_{(t)}-\frac{5}{4}\left(\mathcal{L}_{s}^{-1}\left[\frac{s}{(s+2)^2+4}\right]_{(t)}-\mathcal{L}_{s}^{-1}\left[\frac{4}{(s+2)^2+4}\right]_{(t)}\right)=$$ $$\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{1}{s}\right]_{(t)}-\frac{5}{4}\left(\mathcal{L}_{s}^{-1}\left[\frac{s}{(s+2)^2+2^2}\right]_{(t)}-\mathcal{L}_{s}^{-1}\left[\frac{2\cdot2}{(s+2)^2+2^2}\right]_{(t)}\right)=$$ $$\frac{5}{4}\mathcal{L}_{s}^{-1}\left[\frac{1}{s}\right]_{(t)}-\frac{5}{4}\left(\mathcal{L}_{s}^{-1}\left[\frac{s}{(s+2)^2+2^2}\right]_{(t)}-2\mathcal{L}_{s}^{-1}\left[\frac{2}{(s+2)^2+2^2}\right]_{(t)}\right)$$

Now use:

• $F\left(s\right)\to f(t)\Longrightarrow \frac{1}{s}\to1$
• $F\left(s\right)\to f(t)\Longrightarrow F(s-a)\to e^{at}f(t)$
• $F\left(s\right)\to f(t)\Longrightarrow F(s+a)\to e^{-at}f(t)$
• $F\left(s\right)\to f(t)\Longrightarrow \frac{\omega}{s^2+\omega^2}\to\sin(\omega t)$
• $F\left(s\right)\to f(t)\Longrightarrow \frac{s}{s^2+\omega^2}\to\cos(\omega t)$