0
\$\begingroup\$

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:

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

\$\endgroup\$
  • \$\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\$ – Vladimir Cravero Aug 20 '17 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 Aug 21 '17 at 6:47
0
\$\begingroup\$

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?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.