Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Third - order Intercept Point (IP3) Equation derivation

Status
Not open for further replies.

promach

Advanced Member level 4
Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,202
Helped
2
Reputation
4
Reaction score
5
Trophy points
1,318
Activity points
11,643
Could anyone elaborate on how to derive the following IP3 equation in the screenshot ?

Screenshot from 2017-12-23 23-24-00.png

Note: Extracted from https://www.keysight.com/upload/cmc_upload/All/ENA_IMD_Measurement_Summary.pdf#page=4
 

IP3out is a fictious output level where 3rd order products reach to the fundamental's level. 3rd order products rise 3 times faster than fundamentals(as the slops indicate) while you increase the input level.
D distance between two runners having velocities v and 3v closes in D/2v sec. In that time, the fundamental having velocity v takes D/2. As a result, output level(named as IP3out) becomes initial level plus D/2 as the formula states.
 

Hi,
You just have to use the equations of a straight line knowing one point and it's slope.

Find out the equations of the line that has slope one, and the equation of the line that has slope 3.

Use the fact that when the input is IIP3, the value of both lines is OIP3.
 

Why divide delta_P by (N-1) instead of just N ?
 

Just be cautious as an assumption of linearity is made. If the amplifier is not in the linear region actual results will be different.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top