# Writing a single differential equation that describes the behaviour (electrical circuit)

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

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.

Then I derive

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

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

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

After some elimination, I end up with

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

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

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!

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:

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

• 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 $\frac{dx(t)}{dt}$? – Jason Born Sep 19 '15 at 23:39
• 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. – Chu Sep 20 '15 at 0:08
• 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. – Jason Born Sep 20 '15 at 16:53
• 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? – Jason Born Sep 20 '15 at 18:45
• See addition to original answer – Chu Sep 20 '15 at 18:50

Simplify your first equation and then use Laplace properties to transform it back to a differential equation.

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