Jan 25, 2008 #1 P passerby Member level 3 Joined Jun 12, 2006 Messages 58 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,694 For 2 tones test with the same amplitude level IIP3(dBm)=Pin(dBm) + Delta(dB)/2 where Delta = Pout(dBm)-IMD(dBm) so we can get IMD=3xPin + G - 2xIIP3 What IMD would be if the 2 tones signals are different amplitude levels? Thanks.
For 2 tones test with the same amplitude level IIP3(dBm)=Pin(dBm) + Delta(dB)/2 where Delta = Pout(dBm)-IMD(dBm) so we can get IMD=3xPin + G - 2xIIP3 What IMD would be if the 2 tones signals are different amplitude levels? Thanks.
Jan 25, 2008 #2 V vfone Advanced Member level 6 Joined Oct 10, 2001 Messages 5,709 Helped 1,616 Reputation 3,237 Reaction score 1,248 Trophy points 1,393 Activity points 36,250 Assuming the picture below we get: PL(dBm) = P1(dBm) - 2*[OIP3(dBm) – P2] PU(dBm) = P2(dBm) - 2*[OIP3(dBm) – P1]
Assuming the picture below we get: PL(dBm) = P1(dBm) - 2*[OIP3(dBm) – P2] PU(dBm) = P2(dBm) - 2*[OIP3(dBm) – P1]
Jan 26, 2008 #3 P passerby Member level 3 Joined Jun 12, 2006 Messages 58 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,694 Could also provide the derivation ? Thanks.
Feb 26, 2008 #4 P passerby Member level 3 Joined Jun 12, 2006 Messages 58 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,694 PL(dBm) = P1(dBm) - 2*[OIP3(dBm) – P2] PU(dBm) = P2(dBm) - 2*[OIP3(dBm) – P1] and OIP3(dBm) = IIP3(dBm)+G(dB) The G here is the gain @ fundamental frequency ? or the gain @ f1 or @ f2? Thanks
PL(dBm) = P1(dBm) - 2*[OIP3(dBm) – P2] PU(dBm) = P2(dBm) - 2*[OIP3(dBm) – P1] and OIP3(dBm) = IIP3(dBm)+G(dB) The G here is the gain @ fundamental frequency ? or the gain @ f1 or @ f2? Thanks
Feb 26, 2008 #5 V vfone Advanced Member level 6 Joined Oct 10, 2001 Messages 5,709 Helped 1,616 Reputation 3,237 Reaction score 1,248 Trophy points 1,393 Activity points 36,250 Yes, G is the gain at fundamental frequency.