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https://designers-guide.org/forum/YaBB.pl Design >> RF Design >> IIP3 for system https://designers-guide.org/forum/YaBB.pl?num=1168590707 Message started by aaron_do on Jan 12th, 2007, 12:31am |
Title: IIP3 for system Post by aaron_do on Jan 12th, 2007, 12:31am Hi all, I have a system where I have interference in the adjacent channel with C/I = 0 dB and in the alternate channel where C/I=-30 dB, and the signal then proceeds all the way to the CSF. At the output of the CSF, C/I at adjacent is still 0 dBm but C/I for the alternate has been improved. Call the rejection of the alternate channel R in dB. How do I calculate the overall required IIP3 of the system? Should I try and express it in terms of signal plus interference and make sure the interference is less than the noise floor? Without the rejection of the alternate channel, I would normally use the formula IIP3>(3Pinterfer-Psig+SNRmin)/2 where SNRmin is the required SNR at the ADC, Pinterfer is the interferring signal and Psig is the input signal 3 dB above the sensitivity. Secondly, there is an equation Pintermod,in = 3*Psig,in - 2*IIP3 (all in dBm) which gives the input referred intermodulation level. How would I adjust this formula if the intermodulation is 30 dB higher than the signal? Seems like I should add 90 dB right? And to refer it to the output I simply multiply by the gain right? thanks, Aaron |
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