I think the underlying point is that IMDs add together as voltages
not powers so you have to work back to the levels in Volts, not
power, in order to can see the relative contributions and the
total overall level. IMD levels in dB are powers; power=20log(V)
and V=10^(power/20) which is where the 20 comes from.
Steve
> Hi Gary,
>
> B = 20Log(1+10^(-A/20))
>
> Let's say you measured the input IMD3 at -44 dB and the output IMD3 at
> -32 dB. The difference is 12 dB and that is A. Divide A by 20 which
> equals 0.6. Negate that and raise 10 to that power which is 0.251. Add
> 1 to that which is 1.251 and take the Log of that which is 0.973.
> Multiply by 20 which is 1.945 which is ~2 dB. This is B which is the
> contribution to the output IMD3 due to the input IMD3. Therefore the
> output IMD3 is really -34 dB if there was no contribution from the input.
>
> I hope that explains it. If not get back to me and I'll give it another
> try. If I confused you, I am sure I confused someone else.
>
> 73, Tom W0IVJ
>
> On 5/6/2012 1:49 PM, Gary Schafer wrote:
>> Hi Tom,
>>
>> I can't seem to make your math work for finding input IMD contribution?
>> Where does the divide by 20 come from in finding B?
>>
>> 73
>> Gary K4FMX
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