Hacker News new | ask | show | jobs
by Dylan16807 3146 days ago
So if users had a small low-pass filter installed at their fuse box it wouldn't be a problem?
1 comments

A low pass filter that can pass 50 kW isn't exactly a trivial device or small in any sense of the word.
Low pass filters don't have the power go through them. It's a capacitor going across the power supply, with a resistor for damping.
Except unless you have some sort of an impedance follower you'll probably just end up attenuating your powerline networking to the point of not working.
Most of the useful signal in the system is going directly from one circuit to another. The signal headed out toward the power meter is mostly dead, and you might even improve the networking by getting rid of those extra echoes.

Unless I'm mistaken?

I'd be really surprised if it works that way. I'm not a power systems engineer but I've worked a fair bit on the low voltage side of things.

My guess is that the connection to your house to the service wires in incredibly low impedance, otherwise you'd see all sorts of voltage sags when you used a large appliance with an inductive load(or power tool). So any change large enough to affect the service line is going to go straight into your house. Circuits aren't directional(unless you have a diode or voltage/impedance follower) hence why it would impact the quality of power-line networking.

Additionally any sort of low pass filter(RC, LC or RLC) involves putting a inductor(P) or resistor(R) in series with the circuit(AKA your transmission wires) which isn't simple or cheap.

The impedance going to all your neighbors is pretty low too, but I don't think the system relies on signals going out that far and echoing back. High frequencies have more of a transmission line model, which is pretty directional. There's free impedance everywhere.

My understanding is that you mostly care about the minor reflections caused by wire junctions in a system like this, and that line headed outside is more or less a signal sink no matter what. It's possible a capacitor would need to be a bit further away than directly at the breaker box, but by the time you reach the pole I'm reasonably confident that absorbing the entire signal with $.50 of components and no inductor would be fine.

Only one way to find out... and I would love to hear about your results :)