to give you a short intro of why I am asking here. My Uni is offering a TeleCom Lecture, I love the topic but all they teach is electronics, the exercise sheets are still TeleCom, we get no supporting material and after a few days of Google searching I have reached maximum confusion. Send help ;)

Ok here is the Deal:

We have a BPSK Modulated Signal with spectral efficiency of p = 0.7 bit/s/Hz. We want to Transmit Rb = 2.048 Mbit/s In addition, BER should be 10^-6 or Eb/N0 = 10.5 dB

Task: Find the Bandwith B.

I thought that Rb/p = B will give me the desired result. But here I never used Eb/N0. So I assume I am missing something.

What am I missing? Or is the simple relation I found already correct and Eb/N0 is just there to confuse me?

  • 1
    $\begingroup$ "after a few days of Google searching I have reached maximum confusion" -- I strontly suggest that you stop using Google to substitute for a good textbook. Online learning using Google is severely overrated. $\endgroup$ – MBaz Jun 2 '19 at 17:39
  • $\begingroup$ Thank you! Can you recommend any? I asked my tutors multiple times and they said "yes we will upload a list"... that never happend $\endgroup$ – Clex Jun 2 '19 at 19:04
  • $\begingroup$ Hm... maybe "An Introduction to Analog and Digital Communications" by Haykin. I see it starting at $12 used on Amazon. $\endgroup$ – MBaz Jun 2 '19 at 19:24

Your initial answer is correct -- the bandwidth depends only on the spectral efficiency and the symbol rate. $E_b/N_0$ is a figure of merit largely designed to be independent of the bandwidth.

Let me give you an example of where $E_b/N_0$ might come into play. Say you are required to provide $E_b/N_0 = 10$, and you know that your receiver is rated at $N_0 = 10 ^ {-12}$. Then, $E_b = 10 N_0 = 10^{-11}$. At a rate $R = 2 \times 10^6$, the received power is $(2 \times 10^6)(10^{-11} = 2 \times 10^{-5}$ watts, or $-47$ dB.

If you know the channel attenuation, then you can calculate the transmitted power. From there, you can calculate things like battery life. And so on...

  • $\begingroup$ Thanks for the answer. So I found something else, for BPSK S/N = Eb/N0 and thought well that probably assumes that p = 1 so i went ahead and said S/N = Eb/N0 * p and then went ahead and used Shanon-Hartley to find the Bandwith, the results are diffrent. Is this approach completely of? $\endgroup$ – Clex Jun 2 '19 at 19:04
  • $\begingroup$ Yes, I'd say so :) $\endgroup$ – MBaz Jun 2 '19 at 19:22
  • $\begingroup$ Well I guess then It really just is Rb/p = B, thank you again :) $\endgroup$ – Clex Jun 2 '19 at 19:23
  • $\begingroup$ @Clex There is a way to use a specific form of Shannon's capacity formula here, but I doubt it that it applies to your problem, since it's a quite advanced topic. In any case, please remember to upvote the answer if you found it useful :) $\endgroup$ – MBaz Jun 2 '19 at 20:05
  • $\begingroup$ Just in case, which formula would it be? I tried to upvote but I can't as I don't have enough points in this stack exchange yet. But I am very thankful! $\endgroup$ – Clex Jun 2 '19 at 21:35

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