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I've understood that, to find the transfer function of a system, given its weight function, Laplace transform should be used.
I.e. if $$w(t)=50\,\cdot(e^{-5t}\,-\,e^{-10t})\,\cdot\,u(t)$$ then
$$W(s) = \int\limits_0^\infty w(t)\,e^{-st}\:dt$$
How is this solved further? I understand that w(t) is inserted into the formula for the transfer function W(s), but cannot obtain the correct answer through sequential math steps.
The final answer should be:
$$ W(s) = \frac{5}{(1+0.2s)(1+0.1s)}$$
More often than not, one should not literally apply the definition of Laplace transform to compute the L-equivalent of a time domain function.
Let us just use linearity: $$ \mathscr{L}[a\cdot f(t) + b\cdot g(t)]=a\cdot\mathscr{L}[f(t)]+b\cdot \mathscr{L}[g(t)] $$ Where \$a,b\$ are scalar quantities.
Let's rewrite your \$w(t)\$: $$ w(t)= 50\cdot(e^{-5t}-e^{-10t})\cdot u(t)= 50\cdot\Big(e^{-5t}\cdot u(t) - e^{-10t}\cdot u(t) \Big) $$ Now we apply \$\mathscr{L}[\ \cdot\ ]\$ and use linearity: $$ W(s)=\mathscr{L}[w(t)]= 50\Big[\mathscr{L}[e^{-5t}\cdot u(t)]-\mathscr{L}[e^{-10t}\cdot u(t)]\Big] $$
Can you take it from here?
