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

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SUMMARY

The bandwidth efficiency (R/B) for Phase Shift Keying (PSK) at a bit error rate (BER) of 10^-7 on a channel with a signal-to-noise ratio (S/N) of 12 dB is calculated to be 1.2. The formula used is Eb/N0 = S/N * B/R, where S/N is the signal power over noise power. The conversion from dB to dBW is not necessary for this calculation, as the relationship can be expressed directly in dB without requiring power ratio conversions.

PREREQUISITES
  • Understanding of Phase Shift Keying (PSK)
  • Knowledge of Signal-to-Noise Ratio (S/N) and Bit Error Rate (BER)
  • Familiarity with the Eb/N0 formula
  • Basic arithmetic operations involving logarithms
NEXT STEPS
  • Study the relationship between S/N and Eb/N0 in digital communications
  • Learn how to calculate bandwidth efficiency for different modulation schemes
  • Explore the impact of BER on communication system performance
  • Investigate the use of dB and dBW in signal processing
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Communication engineers, signal processing specialists, and students studying digital communications who need to understand bandwidth efficiency calculations and their implications in system design.

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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