Electronic – Final voltage of two parallel (switched) capacitors

Sparing the maths, an easy way to think about EMI is a high frequency voltage will take the path of lowest impedance back to its source. In short, the filter direction will make a difference, in the real world but not so much in the ideal case.

If the filter is placed with caps to the noise source, the preferred current path will be through the caps. You could imagine L as a large impedance placed after a small impedance, the high frequency currents preferred path back to the source is through the smaller resistor, of course you have to make sure the inductance of the filter is significantly higher than the inductance of the cap (which is going to be in the nH range).

^{simulate this circuit – Schematic created using CircuitLab}

The problem with the second configuration is there is no place for the HF current to go after it is blocked from the filter inductor. Any voltage RF that gets through the inductor will be seen by both the load and the caps equally. It may also be attenuated, however, there may be another preferred path an that is through mutual inductance between cables and capacitive coupling through the air to other physical things to get back to the source. It is extremely difficult to come up with a circuit model of exactly what would happen in a situation like this. There are too many parasitics and cables and terminals start functioning like antennas. If you have ever studied antenna theory, it is more of an art. If the inductor could do the blocking on its own, then why would you need caps in the first place? Ideally they look similar, once you add the parasitics in and account for other effects they are not.

^{simulate this circuit}

How do I know this? I actually encountered this very problem with a product this week. I had an issue with a switching load. The load was producing RF, which was causing interference on the PCB, even to the extent of causing a micro controller to 'drop out' and have a communications failure. I placed a ferrite on the cable which would be just like adding a filter with only HF inductance\impedance. The problem was attenuated. Then I added capacitors the problem was gone, and it worked better when the caps were pointed toward the load.

Remember the rule: HF will take the lowest impedance pathway, if you provide a better low inductance pathway, you can attenuate the problem. Oh, if your building or selecting a filter, make sure it has low inductance between the caps by placing components close together.

For control purposes you need to linearize the system around a single operating point. Considering the synchronous DQ frame, the grid voltage is DC and therefore can be neglected.

Do not replace the grid voltage with a resistor. You are mixing the power delivery circuit with the filter circuit.

The resistor will cause huge damping where you don't want the damping to be. And your inductances are way too small.

First, you need to select the maximum current ripple. This will set your inductance. Then you need to select the shunt capacitor value based on maximum current harmonic distortion that is allowed into the utility.

For stability purposes you need to add passive (resistor) or active (measurement + controls) damping into the system.

Inductance L2 could be partly physical and partly the inherent grid inductance. So you need to make sure the design and controls are stable for any grid output inductance (which varies based on the grid loading and time of day).

So, to answer your question:

You need to control the current so calculate the inverter voltage to grid current (L2) transfer function. No reason to calculate voltage to voltage transfer function since the grid voltage is established by the utility.

\$ \frac{I_{L2}}{V_{inverter}} = ... \$

Btw. this is a simple sophomore level calculation. But the implications, the control and other dependencies make it a grad-level problem. Good luck!