In digital communications textbooks (e.g., Digital Communications by Bernard Sklar), the relationship between the "received" signal power $P_r$ and the "transmitted" energy per bit $E_b$ is given as follows: $P_r=E_b R_b$ where $R_b$ is the bit rate. What confuses me in this equation is that the channel attenuation is not considered in the equation. Shouldn't the equation be as $P_r=\alpha^2E_b R_b$ instead where $\alpha$ is the channel attenuation?


You are correct that it could be argued that every physical channel will introduce attenuation. In digital communications, however, it is common to define "channel models" that simplify the equations and make the problem easier to understand and to solve.

The most popular model is the "additve white Gaussian noise" (AWGN) model, where the only effect of the channel is to introduce noise. In other words, there is no attenuation. In that model, Sklar's relationship between power and bit energy is correct. This model is very popular because, in many cases, you can ignore the attenuation by assuming an appropriate gain mechanism in the receiver.

Of course, in many other cases the AWGN model is not appropriate. In the wireless channel, for instance, the channel gain is so fundamental that it gets incorporated into the model. If the channel is narrow-band, then the relationship between power and bit energy is similar to what you propose, with $\alpha$ random and possible time-varying.

  • $\begingroup$ What did you mean by narrow-band channel? $\endgroup$
    – ArmenG
    Feb 24 '15 at 2:28
  • $\begingroup$ @ArmenG, a wireless channel is narrowband when its frequency response is approximately flat. This means that the all the transmitted signal's frequency components are attenuated equally. A wide-band (or frequency-selective) channel attenuates different frequencies by different amounts. $\endgroup$
    – MBaz
    Feb 24 '15 at 12:47

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