Find Bandwidth Efficiency (R/B) given SNR and Eb/N0

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The discussion focuses on calculating the bandwidth efficiency (R/B) of PSK given a bit error rate (BER) of 10^-7 and a signal-to-noise ratio (S/N) of 12 dB. The user attempts to apply the formula Eb/N0 = S/N * B/R but struggles with the conversion of dB values. A key point raised is the need to convert dB to a power ratio correctly, as well as the correct application of the formula R/B (dB) = SNR(dB) - (Eb/N0)(dB). The correct answer for bandwidth efficiency is stated to be 1.2. The discussion highlights the importance of careful arithmetic and understanding of dB conversions in these calculations.
lycaeus
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Thank you in advance for your help!What is the bandwidth efficiency (R/B) of PSK for a bit error rate of 10^-7 on a channel with
an S/N of 12 dB?[/b]
From the graph, for BER= 10^-7, Eb/N0 = 11.2 dB./b]I have applied the formulae Eb/N0 = S/N * B/R
Where, S/N is Signal power over noise power * Bandwidth / Bit Rate

I know for one that I need to convert dB to dBW by applying 10log(dB)

However, I have tried for 5 hours and could not get the answer

The answer to this question is 1.2

Thank you
 
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It looks right to me. Are you making an arithmetical error? Why do you think dBW enters?
You have
R/B (dB) = SNR(dB) - (Eb/No)(dB)
in dB, it is simple to convert to a power ratio.
 

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