Please help-me in this question, first I calculate the I(s) and I get
$$ I(s) = \dfrac{ \dfrac{1}{s+1} }{ s^2+\frac{R}{L}s+\frac{1}{LC} } $$
How I proceed to find Vo(t)?
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\$\begingroup\$ How doyou express the input voltage and impedence of the components using the Laplace transform? Note that exercises are fine here, but some effort in solving is expected of the poster. \$\endgroup\$– clabacchioCommented Dec 7, 2014 at 17:44
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\$\begingroup\$ I(s) = (1/s+1)/((s^2+(R/L)*s+(1/LC)) \$\endgroup\$– Diego CardosoCommented Dec 7, 2014 at 17:48
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\$\begingroup\$ You know \$I(s)\$ through capacitor. Calculate \$V_o(s)\$ and get \$V_o(t)\$ by taking its Laplace inverse \$\endgroup\$– nidhinCommented Dec 7, 2014 at 18:24
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\$\begingroup\$ Can u give-me a example, please? Vo(s) = (RSC+1)I(s) ? \$\endgroup\$– Diego CardosoCommented Dec 7, 2014 at 19:10
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2 Answers
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First, basic thing:
$$Vo(s) = I(s) \cdot X_C(s)$$
Replace Xc with its Laplace equivalent:
$$X_C = \dfrac{1}{sC}$$
Replace the formulas the first equation and you get the voltage function. You might need to massage the equation to simplify it.
Then you need to anti-transform: you can do it using the rules and tables.
answered Dec 7, 2014 at 20:30
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Thanks,
to resolve this problem I did the following steps,
transform the circuit in S-domain
then, resolve inverse of \$I(s) \cdot X_C \rightarrow I(s) = \dfrac{1\over s+1}{6+s+10\over s} \$ and Xc = 10/s
\$ \mathcal{L}^-1(I(s) \cdot X_C) \$
and got the correct answer:
\$ V_O (t) = \dfrac{2}{e^t} - \dfrac{2(cos(t) + 2sin(t))}{e^{3t}} \$