Hi,

first I assume you are trying to calculate the required IIP3, not the IIP3 of some circuit block. You don't need the blockers to be in the N+2 and N+4 channel, N+1 and N+2 will do. The 3rd order output will be

y₃ = Bx³
10log(y₃) = 30log(x) + 10logB

the first order is

y₁ = Ax
10log(y₁) = 10log(x) + 10log(A)

From here its pretty straight forward. IIP3 is the value of x where y₁ and y₃ are equal. The gain is of course equal to 10log(A). you can substitute the powers in the N+1 and N+2 channels in y₃.


cheers,
Aaron