# Find the transfer function of the system given by its weight function

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.

$$W(s) = \frac{5}{(1+0.2s)(1+0.1s)}$$

• You can solve the integral... But should you? Try to use the linearity of the Laplace transform, and also $\mathscr{L}[e^{-\alpha t}\cdot u(t)]=\frac{1}{s+\alpha}$ Aug 20, 2017 at 17:00
• Is the w(t) equation correct? Should there be a convolution operation? Is u(t) meant to be the input signal or the Heaviside function?
– Chu
Aug 21, 2017 at 6:47

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: $$W(s)=\mathscr{L}[w(t)]= 50\Big[\mathscr{L}[e^{-5t}\cdot u(t)]-\mathscr{L}[e^{-10t}\cdot u(t)]\Big]$$