Relation between SNR and Eb/No

Energy per bit (Eb) is defined as the ratio of signal power to
bit-rate,

  Eb = Ps / B [joules / bit],

where Ps is in [watts = joules / sec] and B is in [bits / sec]. The
bit-rate can be expressed in terms of the spectral efficiency b
[bits / sec / Hz] and bandwidth W [Hz] required for the signal as 

  B = b * W [bits / sec].

Therefore

  Eb = Ps / (b * W) [joules / bit].

Now note that if the noise is flat with constant power spectral
density No [watts / Hz = joules], then the noise power Pn in bandwidth
W [Hz = 1 / sec] is

  Pn = No * W [joules / sec = watts].

Thus the signal-to-noise ratio SNR is 

  SNR = Ps / Pn
      = Ps / (No * W),

and Eb / No is related to SNR as

  Eb / No = Ps / (b * W * No)
          = SNR / b [1 / bit].

so, Eb / No is SNR normalized to the spectral efficiency (b).

If we're concerned with energy efficiency, then we have to look no further
than Eb (energy per bit) alone. If we're concerned with spectral efficieny,
then we have to look no further than b (spectral efficiency) alone.
Eb / No is a significant ratio because the performance (i.e., bit-error

rate) of many common digital comm signals is a function of this ratio.

To see why we wouldn't use SNR in performance computations, consider

comparing two signals S1 and S2 with the same bit rate B and energy

per bit Eb but with varying spectral efficiencies b1 = 1 bit / sec / Hz

and b2 = 2 bit / sec / Hz:

  Ps = Eb * B

  W = B / b

  Pn = No * W

  SNR = Ps / Pn

  Qty         Signal 1      Signal 2         Units

  ---         --------      --------         -----

  Ps          Eb * B        Eb * B           [watts]

  W           B             B / 2            [Hz]

  Pn          No * B        No * B / 2       [watts]

  SNR         Eb / No       2 * Eb / No      [none]

  Eb / No     Eb / No       Eb / No          [1 / bit]

The point is that the spectral efficiency caused signal 2 to

have a higher SNR even though the transmitted power was the

same due to the fact that the noise bandwidth is half and

thus the noise power is half. However, this would not have

created any better performance in terms of bit-error rate

since its Eb / No is the same as signal 1.


taken from : http://www.dsprelated.com/showmessage/81112/1.php

Let’s make it more detail :
Eb / No = SNR / b
Eb/No = SNR / (BitRate/BW)
Eb/No(dB) = SNR(dB) – 10*LOG10(BitRate/BW)
Eb/No(dB) = SNR(dB) + 10*LOG10( BW(Hz) / BitRate(Hz) )

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