Feb 9, 2011 #1 moonnightingale Full Member level 6 Joined Sep 17, 2009 Messages 362 Helped 5 Reputation 10 Reaction score 5 Trophy points 1,298 Activity points 3,832 The question says The interferer is on with probability .25 and off with probability .75. The average transmit power is 10 mW, And solution says 0.75P1 + 0.25P2 = 10 and then he gives the answer P1= 12.25 mW P2=3.25mW How he calculated P1 and P2 and P2 =3.25 mW
The question says The interferer is on with probability .25 and off with probability .75. The average transmit power is 10 mW, And solution says 0.75P1 + 0.25P2 = 10 and then he gives the answer P1= 12.25 mW P2=3.25mW How he calculated P1 and P2 and P2 =3.25 mW
Feb 9, 2011 #2 J JeanPellanda Newbie level 2 Joined Dec 9, 2008 Messages 2 Helped 1 Reputation 2 Reaction score 1 Trophy points 1,283 Activity points 1,288 You should have other formula... From here the unique thing we know is that P2 = 40 - 3P1.