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# Find G(s) such that the Laplace transform of the charge on the capacitor, Q(s) = L{q(t)}, can be expressed as Q(s) = G(s)E(s), where E(s) = L{ e(t)} is the Laplace transform of the impressed voltage. Assume that at time t = 0 the charge on the capacitor is zero and the currents $i_{1} \text { and } i_{2}$ are zero.

Solution

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Applying Kirchoff's second law in the given circuit, we obtain the following equations.

$Ri_1+L_1\frac{di_1}{dt}+L_2\frac{di_2}{dt}=e(t) \tag1$

and

$L_1\frac{di_1}{dt}+Ri_1+\frac{1}{C}\int idt=e(t) \tag2$

The current $i$ at time $t$ is defined by

$i=q'(t)=\frac{dq}{dt} \tag3$

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