IIP3 calculation: Pin or Pout?

Hi,
I've seen a few sources post the following and needed some clarification. Thanks in advance:
1. IIP3=Pin+(Pout-IM3)/2
2. IIP3=Pin+(Pin-IM3)/2
Is #2 valid only if the gain is unity (Pin=Pout)?
Or is there something else I'm missing?

-Robert

OIP3(dBm)=Pout(dBm)+[Pout(dBm)-IM3(dBm)]/2
IIP3(dBm)=OIP3(dBm)-Gain(dB)
where Gain is the gain of the amplifier

IIP3 is calculated by #1.
I've never seen #2.

Hi Pancho,
Here's an example of #2 from Razvi:

Thanks, Robert

Topic of the quoted lection is different, digital receiver not amplifier. There's neither a gain nor Pout in this case.

In #2 they may be discussing this IM3 at the input.

For example,
OIP3 = Pout + (Pout - Po,IM3)/2
Pout = G + Pin
Po,IM3 = G + Pi,IM3
OIP3 = IIP3 +G

therefore,
IIP3 + G = Pin + G + (Pin + G - P,IM3 - G)/2
IIP3 = Pin + (Pin - Pi,IM3)/2

where (Pin - Pi,IM3) = (Pout - Po,IM3). Theoretically, this is because the input and output powers are just differentiated by the gain (or loss) of the system but you can't measure intermods at the input I don't think. Normally there's enough isolation that you don't have measurable IMDs seen at the input.

Hi Niccoh,
Thanks for the help and it makes sense. In the Razavi ex, the IM3 is designated at the input.

Here's another explanation/scenario where G is unity so Pout=Pin:
1. http://www.microwavejournal.com/arti...r-nonlinearity.

Thanks, Robert