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Consider a circuit with \$L_{1}\$ and \$L_{2}\$ as inductors and \$C_{1}\$ and \$C_{2}\$ as the capacitors. \$I\$ and \$V\$ are the manifest variables

Manifest Variables

I want a single differential equation without the latent variables that links V(t) and I(t) (i.e. describes the behaviour)

Thus, taking the Laplace transform, we get

$$I=V\left(\frac{sC_{1}}{s^{2}L_{1}C_{1}+1}+\frac{sC_{2}}{s^{2}L_{2}C_{2}+1}\right)$$

Do I want to take the inverse Laplace transform here, or must I apply Laplace transforms on the equations derived by Kirchoff's laws:

I write \$I_{L_{1}}\$, \$I_{L_{2}}\$, \$I_{C_{1}}\$, \$I_{C_{2}}\$, \$V_{L_{1}}\$, \$V_{L_{2}}\$, \$V_{C_{1}}\$, \$V_{C_{2}}\$ as the latent variables.

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Then I derive

\begin{equation}\begin{cases}I=I_{L_{1}}+I_{L_{2}}\\ I_{L_{1}}=I_{C_{1}}\\ I_{L_{2}}=I_{C_{2}}\\ I_{C_{1}}+I_{C_{2}}=I\end{cases}\end{equation}

\begin{equation}\begin{cases}V=V_{L_{1}}+V_{C_{1}}\\ V=V_{L_{2}}+V_{C_{2}}\\ V_{L_{1}}+V_{C_{1}}=V_{L_{2}}+V_{C_{2}}\end{cases}\end{equation}

\begin{equation}\begin{cases}L_{1}\frac{dI_{L_{1}}}{dt}=V_{L_{1}}\\ L_{2}\frac{dI_{L_{2}}}{dt}=V_{L_{2}}\\ C_{1}\frac{dV_{C_{1}}}{dt}=I_{C_{1}}\\ C_{2}\frac{dV_{C_{2}}}{dt}=I_{C_{2}}\end{cases}\end{equation}

After some elimination, I end up with

\begin{equation}\begin{cases} I=I_{L_{1}}+I_{L_{2}} \\ I_{L_{1}}=C_{1}\frac{dV_{C_{1}}}{dt} \\ I_{L_{2}}=C_{2}\frac{dV_{C_{2}}}{dt}\end{cases} \end{equation}

And \begin{equation} \begin{cases} V={L_{1}}\frac{dI_{L_{1}}}{dt}+V_{C_{1}} \\ V=L_{2}\frac{dI_{L_{2}}}{dt}+V_{C_{2}} \end{cases} \end{equation}

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Taking a particular V(s) and then performing the inverse LT will give you i(t), but that will be in transcendental form, and not a differential equation. You can obtain a DE from your equation using the property: \$sX(s)\rightarrow\frac{dx(t)}{dt}\$. The resultant DE is not particularly user-friendly, though!

***Added in response to comment:

Write your original equation in TF form and add the two fractions:

\$\frac{I(s)}{V(s)}= \frac{As^3 +Bs}{Cs^4 +Ds^2+1}\$

Cross multiply:

\$Cs^4 I(s) + Ds^2 I(s) + I(s)=As^3 V(s)+Bs V(s)\$

Inverse LT:

\$C\frac{d^4I(t)}{dt^4}+D\frac{d^2I(t)}{dt^2}+I(t)= A\frac{d^3V(t)}{dt^3}+B\frac{dV(t)}{dt}\$

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  • \$\begingroup\$ It doesn't have to be user-friendly since I don't need to work with it. But since I'm not so familiar with electrical engineering, I'm going to have to ask when does \$sX(s)\$ go to \$\frac{dx(t)}{dt}\$? \$\endgroup\$ – Jason Born Sep 19 '15 at 23:39
  • \$\begingroup\$ It's a property of the Laplace Transform. Given the LT of a signal, say, \$x(t)\rightarrow X(s)\$, then the LT of \$\frac{dx(t)}{dt}\$ is \$sX(s)\$. For example, given \$\frac{Y(s)}{X(s)}=\frac{2}{3s+1}\$, cross-multiply: \$3sY(s)+Y(s)=2X(s)\$, now inverse LT: \$3\frac{dy(t)}{dt}+y(t)=2x(t)\$. Also, dividing a LT by s is equivalent to integrating the function of time. \$\endgroup\$ – Chu Sep 20 '15 at 0:08
  • \$\begingroup\$ I can reduce the problem to \$I(t)=\mathcal{L}^{-1}\left(sV(s)\cdot\frac{C_{1}}{s^{2}L_{1}C_{1}+1}+sV(s) \cdot \frac{C_{2}}{s^{2}L_{2}C_{2}+1}\right)\$. I know \$sV(s)\to\frac{dV(t)}{dt}\$, but how do I deal with the \$\frac{C_{1}}{s^{2}L_{1}C_{1}+1}\$ part of the equation? It's quite problematic. \$\endgroup\$ – Jason Born Sep 20 '15 at 16:53
  • \$\begingroup\$ I'm not sure what you mean by 'TF form'... \$I(s)=\frac{V(s)s^{3}(C_{1}C_{2}L_{2}+C_{1}C_{2}L_{1})+V(s)s(C_{1}+C_{2})}{s^{4}C_{1}C_{2}L_{1}L_{2}+s^{2}(L_{1}C_{1}+L_{2}C_{2})+1}\$ Is this what you meant? \$\endgroup\$ – Jason Born Sep 20 '15 at 18:45
  • \$\begingroup\$ See addition to original answer \$\endgroup\$ – Chu Sep 20 '15 at 18:50
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Simplify your first equation and then use Laplace properties to transform it back to a differential equation.

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  • \$\begingroup\$ The OP is adding admittances, so their equation is correct. \$\endgroup\$ – Chu Sep 19 '15 at 8:34
  • \$\begingroup\$ @Chu I see what you mean, i stand corrected. \$\endgroup\$ – vini_i Sep 19 '15 at 12:31

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