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I would like to know what is exact and correct definition of $E_b$ (energy per bit) in digital communication. Consider I have a model such as r=a*x+n where a is the channel gain, x is a (+1 or -1) BPSK symbol and n is zero mean AWGN noise. I am defining $E_b$ to be (assuming a=10 and x given:

a=10;
Eb= sum((a*x).^2)/length(x);

However this method seems to be failing since $\sigma$ can be related to $E_b$ using $\sigma=\sqrt(E_b/(2*\text{SNR}))$. So if a is increased the noise power is increased as well and this should not be the case. Therefore should the definition of $E_b$ be like following:

   a=10;
   Eb= sum((x).^2)/length(x);

Meaning that is $E_b$ normalized to the a (channel gain). In this case then $E_b$ is 1 and as a increases the noise power won't be increasing. Can anyone elaborate on this? Should the noise power increases as the amplitude of signal increases? Then in this case performance gets worse and worse as signal amplitude is increased. Is this a correct statement?

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The energy per bit $E_b$ means one thing and one thing only: how many joules is the transmitter spending, on average, per information bit transmitted. Note that the channel has absolutely nothing to do with $E_b$.

However, your question involves the SNR too. In a channel with non-unit gain, you need to decide where you'll measure the SNR.

You can define the SNR at the transmitter: $\text{SNR}=2E_b/N_0$ for BPSK. Obviously, the BER depends on both the SNR and the channel gain.

You can also define the SNR at the matched filter's output: $\text{SNR}=2G^2E_b/N_0$ for BPSK, for a channel with gain $G$. In this case, the BER depends exclusively on the SNR.

When $G$ is random with known probability density, it is common to average the SNR over $G$, in order to obtain the expected BER over many channel realizations.

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    $\begingroup$ I understand but it does not make any sense, assume I measure SNR in transmitter and channel gain is 40 and energy ber pit 1 and add noise with given SNR . then according to $r= ax+n$ power of the signal is always higher than noise so even in SNR -5 dB I should not get any error since vector x is multiplied by "a" that is 40 and it dominates the noise. in this case BER is zero for most SNR even -10 maybe. is there any justification or adjustment that should be made to take the channel gain into the account? $\endgroup$
    – justin
    Aug 25, 2016 at 1:56
  • $\begingroup$ I guess the reasonable question is what if the energy of transmitted signal is not 1 then how do you correctly adjust SNR for that since this is part of transmitter and ha nothing to do with the channel. Should you normalize the signal or what since other wise the signal power will be much higher than noise power? $\endgroup$
    – user59419
    Aug 25, 2016 at 3:53
  • $\begingroup$ @justin It does make sense: if the channel amplifies the signal 40 times and then add noise, the noise will have to be very powerful in order to be able to introduce any errors! In other words, the SNR at the matched filter will be huge. However, in practice the channel gain is always very small. For a wireless channel, the gain can be as small as $10^{-10}$ or less. Please let me know if it still doesn't make sense and I'll add more detail to my answer. $\endgroup$
    – MBaz
    Aug 25, 2016 at 13:01
  • $\begingroup$ @user59419 No, what happens is that over this particular channel, with $G=40$ and a small $N_0$, transmitting with $E_b=1$ results in almost no errors. Of course this is a completely unrealistic channel (especially in wireless applications). $\endgroup$
    – MBaz
    Aug 25, 2016 at 13:02
  • $\begingroup$ @MBaz, thanks I really appreciate it. I was just wondering what if BPSK signal has no- unity energy (lets say Eb=10) and channel gain =G=1 then would this be similar to (G=10 and Eb=1) I am just wondering what justification should be made if the amplitude of signal or its energy is not 1 in the transmitter side, should I normalize it to 1 and then add noise to it, I am asking this since in almost all books Eb is considered to be 1 but what if it is not 1. $\endgroup$
    – justin
    Aug 25, 2016 at 15:09

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