# Find Bandwith from Spectral Efficiency and Eb/N0

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

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?

• "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.
– MBaz
Jun 2, 2019 at 17:39
• Thank you! Can you recommend any? I asked my tutors multiple times and they said "yes we will upload a list"... that never happend
– Clex
Jun 2, 2019 at 19:04
• Hm... maybe "An Introduction to Analog and Digital Communications" by Haykin. I see it starting at \$12 used on Amazon.
– MBaz
Jun 2, 2019 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...

• 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?
– Clex
Jun 2, 2019 at 19:04
• Yes, I'd say so :)
– MBaz
Jun 2, 2019 at 19:22
• Well I guess then It really just is Rb/p = B, thank you again :)
– Clex
Jun 2, 2019 at 19:23
• @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 :)
– MBaz
Jun 2, 2019 at 20:05
• 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!
– Clex
Jun 2, 2019 at 21:35