Jan 30, 2006 #1 N neoflash Advanced Member level 1 Joined Jul 2, 2005 Messages 492 Helped 10 Reputation 20 Reaction score 2 Trophy points 1,298 Activity points 4,759 snr requirement Why duo-binary signal coding SNR requirement is 2.5dB more than normal binary coding? According to my understanding, duo-binary coding will have 3 voltage level, binary code will only have two. Thus the SNR ratio between them are 2/3, the difference should not be 2.5dB, it should be 3.5dB. Any inputs?
snr requirement Why duo-binary signal coding SNR requirement is 2.5dB more than normal binary coding? According to my understanding, duo-binary coding will have 3 voltage level, binary code will only have two. Thus the SNR ratio between them are 2/3, the difference should not be 2.5dB, it should be 3.5dB. Any inputs?
Feb 1, 2006 #2 M mehtesham Junior Member level 3 Joined Nov 14, 2005 Messages 31 Helped 1 Reputation 2 Reaction score 0 Trophy points 1,286 Activity points 1,551 2.5dB more transmit powre is needed for duobinary for having the same error probability as binary. as Pe of duobinary and binary are known: Pe(duobinary)=Pe(binary) ---> St(duobinary)~=1.8*St(binary) ---> St(duo,dB)~=2.5dB+St(binary,dB) St is transmit power.
2.5dB more transmit powre is needed for duobinary for having the same error probability as binary. as Pe of duobinary and binary are known: Pe(duobinary)=Pe(binary) ---> St(duobinary)~=1.8*St(binary) ---> St(duo,dB)~=2.5dB+St(binary,dB) St is transmit power.
Feb 1, 2006 #3 N neoflash Advanced Member level 1 Joined Jul 2, 2005 Messages 492 Helped 10 Reputation 20 Reaction score 2 Trophy points 1,298 Activity points 4,759 where is 1.8 coming from?
Feb 1, 2006 #4 M mehtesham Junior Member level 3 Joined Nov 14, 2005 Messages 31 Helped 1 Reputation 2 Reaction score 0 Trophy points 1,286 Activity points 1,551 you know: St(duo)/St(bin)=[(4*Q^-1(2*Pe/3)/pi)^2]/[(Q^-1(Pe))^2] Q^-1(x) is inverse of Q(x) function,pi=3.14 calculating numerically (from table) : for example : Pe=10^-3 then Q^-1(2/3*10^-3)=3.26 , Q^-1(10^-3)=3.09 you have (4*3.26/pi)^2/(3.09)^2~=1.8 for Pe=10^-4,... you will get about 1.8
you know: St(duo)/St(bin)=[(4*Q^-1(2*Pe/3)/pi)^2]/[(Q^-1(Pe))^2] Q^-1(x) is inverse of Q(x) function,pi=3.14 calculating numerically (from table) : for example : Pe=10^-3 then Q^-1(2/3*10^-3)=3.26 , Q^-1(10^-3)=3.09 you have (4*3.26/pi)^2/(3.09)^2~=1.8 for Pe=10^-4,... you will get about 1.8