EMRFD Message Archive 13206

Message Date From Subject
13206 2016-10-12 18:42:29 Jim Leslie 3rd order Input Intercept calculation
In Feb 1993 QST there is an article by Jacob Makhinson "A high dynamic range MF/HF receiver front end". At the beginning of the article in
the spectrum analyzer picture, he states "...the 3rd order intermod products at 8970 and 9010kHz are down 64dB relative to either
test signal. This equates to a 3rd order input intercept of +42dBm".
I've spent a bit of time on this & I'm hoping this is not too trivial, but I'd like to understand the math involved to arrive at the above result.
thx & 73
Jim Leslie
VE6JF
13207 2016-10-12 20:44:12 Bill Carver Re: 3rd order Input Intercept calculation
Rather than pop out some formula, since you want to understand what's involved, I came up with this rationale. It may, or may not, give some insight into the IMD/intercept math. Here goes:

Given the answer of +42 third order intercept, and the IMD being 64 dB below the incident tones, we can work backwards.
What +42 intercept means, is the third order IMD and the incident signals were projected to have the same amplitude, to meet, at +42 dBm. IE, with inputs = +42 dBm the third order IMD will also be +42 dBm.
As the incident signal amplitudes are reduced the third order IMD drops three times as fast. So if the incident signal drops 1 dB to +41, IMD will drop by 3 dB to +39. The 1 dB drop in incident produced a 2 dB difference between either of the incident tones and either third order IMD product. Jacob says in his test that difference was 64 dB. To get 64 dB difference the incident tones would have to drop 32 dB. So if equal at +42, the incident tones must be +10 dBm. And the third order IMD would drop by three times that, by 96 dB to (+42 - 96 = -54 dBm).

So the applied tones with +10 dBm, the IMD was -54 dBm. Going full circle now, increasing the +10 to +42 is a 32 dB increase. The IMD must increase by 96 dB from -54 to +42 dBm.

You're told you can draw a picture of the 1:1 slope of incident signal, and 1:3 slope of the IMD, intersecting at +42.
The graphical equivalent of my verbal description is the reverse: START at the +42 meeting point, drawing two lines down and to the left at 1:1 and 1:3 slopes until the difference between them is 64 dB.

Once you get the concept Jim, you don't need to know a formula.

Good 'ol Jake. Good engineer. Met him for lunch a few times, many decades ago. Wonder what he's doing these days? Probably retired by now.

Bill - W7AAZ

13208 2016-10-13 06:58:17 Jim Leslie Re: 3rd order Input Intercept calculation
Thanks for that great explanation Bill! I don't get it yet, but after sitting down with morning coffee and reading it a dozen times and actually
going through the exercise of drawing it all out on paper, I will. :)
73 Jim

13209 2016-10-13 22:43:38 Jim Leslie Re: 3rd order Input Intercept calculation
I think I may have it. After some pondering and also re-reading the section in EMRFD p2-22 a couple more times, it is starting to make sense. So: if I see for example the 3rd
order products down 52 dB it would correspond to IIP3 of 36dBm right?
jim

13210 2016-10-13 23:37:41 Bill Carver Re: 3rd order Input Intercept calculation
13211 2016-10-14 09:56:14 vasilyivanenko Re: 3rd order Input Intercept calculation
Thanks Bill for the super earlier explanation --  and greetings all..

Some people I've helped with this found easy web-based "simulation tools" helpful. For example
IP3 Calculator

Changpuak -- his entire website offers great learning + proves interesting for those who code, solder and measure in the RF hobby

Respectfully, T/V
13212 2016-10-14 11:15:16 Jim Leslie Re: 3rd order Input Intercept calculation