In another thread I came across this regarding WinISD and EQ:

It was my understanding that to correctly model EQ power requirements in WinISD, you utilise the fact that for every 3dB increase, twice the power is required. Hence for every 3dB of EQ boost you would halve the signal power you enter into WinISD. (E.g. For 500w and 6dB EQ boost you would enter 125w.)The real power handling requirements actually shows up on the VA plot...

I've never fully understood exactly how the VA plot relates to amplifier requirements. The VA represents the "apparent load" not the "real load". As I understand it, the "apparent load" can never be less than the "real load". I'm assuming you can calculate the "real load" utilising the impedance phase. :nerd:

To put this in practical questions assuming:

- I have an amp that is capable of 500w RMS.
- I boost 6dB with an SOS high pass filter.
- Specifying 125w RMS as the source input in WinISD, the VA plot peaks at 278VA
- The impedance phase is +42 degrees at the VA peak.

Questions:

- Is 278VA really the limit of my amp here? Or would the limit be 500VA?
- Or would it be higher than 500VA because of the Impedance Phase since it is only the "apparent load"? So with a +42 degree impedance angle, this would be 672VA [500*(1/Cos(42))]?
- Or is all this irrelevant since the amp is actually voltage limited? Hence why 500w was dropped to 125w since the voltage will be a quarter.

The ultimate reason for all these questions is that the answers will significantly effect what HPF cutoff to use when modelling. I probably just don't get it... :scratch:

It would be nice to know how to understand how WinISD works in this area as it is a bit unintuitive.

Can anyone shine any light on this?