# Differential Equation of a LC Low Pass Filter

I'm trying to learn about different applications of RLC circuits and have been presented with a weird looking circuit...

I've only ever dealt with circuits that can be modeled with a 2nd-order differential equation, so I'm unsure how to approach this. I've researched the circuit and found this article on low pass filters comprised of LC components. This is a lot different than the RC low pass filters I've studied.

Is the circuit above a low pass filter as well? Does the resistor merely a load resistance or does it influence the behavior of the following LC components? The values aren't provided in the graphic, but C1 and C2 have different capacitance values. The article shows the capacitors both being equal to C/2.

How would I relate the output voltage y(t) to the input voltage x(t) using a differential equation? I tried to create a differential equation using mesh analysis but quickly got mixed and couldn't figure out how to manipulate the equations I got into a differential equation. This article shows a similar circuit being solved but I'm confused by how everything relates.

Here's what I've done thus far...

EDIT: I think I figured it out...

• The $L$ forms a voltage divider with $C_2$, the whole of which is itself then in parallel with $C_1$. This result then forms a voltage divider with $R$. That's about it.
– jonk
Sep 19, 2018 at 23:01
• I assume you want to solve this, generally, not specifically. You mentioned differential equations, so that is why I assumed this. (Phasors are more about when you know $\omega$.) You can use the approach I mentioned with nodal equations (if you like) or as complex number fractions, then simplify and put in standard form for analysis.
– jonk
Sep 19, 2018 at 23:09
• This could be a pi-filter, where R is the source resistance. What exactly do you want to do with it - time domain or frequency domain analysis? Either of these could be done via the Laplace transform. Or is it just an exercise in differential equation analysis?
– Chu
Sep 19, 2018 at 23:28
• If you could tidy-up the jumble, we might be able to help.
– Chu
Sep 19, 2018 at 23:34
• Assigning $V_2$ as the output node at the top of $C_2$ and $V_1$ as the node at the top of $C_1$, then using nodal I get something like: :\begin{align*} \tau_1&=\sqrt{L\:C_2}\\\\ \tau_2&=\frac{L}{R}\\\\ \tau_3&=R\:C_1\\\\ \frac{\text{d}^2\, V_2}{\text{d}\, t^2}-\frac{V_2}{\tau_1^2}&=-\frac{V_1}{\tau_1^2}\\\\ \tau_3\cdot\frac{\text{d}^2\, V_1}{\text{d}\, t^2}+\frac{\text{d}\,V_1}{\text{d}\,t}-\frac{V_1}{\tau_2}&=\frac{\text{d}\,x_t}{\text{d}\,t}-\frac{V_2}{\tau_2} \end{align*}as a start. Have at it from there.
– jonk
Sep 20, 2018 at 6:49