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:

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

  • \$\begingroup\$ 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}\$ \$\endgroup\$ Commented Aug 20, 2017 at 17:00
  • \$\begingroup\$ 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? \$\endgroup\$
    – Chu
    Commented Aug 21, 2017 at 6:47

1 Answer 1


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?


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