Question
Download Solution PDFA coherent binary phase-shift-keyed (BPSK) transmitter operates at a bit rate of 20 Mbps. For a probability of error P(e) of 10-4 and given carrier-to-noise (C/N) density ratio of 8.8 dB, determine the energy of bit-to-noise (Eb/N0) density ratio for a receiver bandwidth equal to the minimum double-sided Nyquist bandwidth.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The carrier to noise (C/N) density ratio is defined as:
\(\frac{C}{N} = \frac{{{E_b}}}{{{N_0}}}\frac{{{R_b}}}{B}\)
Rb: Bitrate
B: Bandwidth
Calculation:
Given bit rate is Rb = 20Mbps, P(e) = 10-4, C/N = 8.8 dB
Nyquist bandwidth is:
\(B = 2\left( {\frac{{{R_b}}}{2}} \right) = {R_b}\)
Carrier to noise density ratio is:
\(\frac{C}{N} = \frac{{{E_b}}}{{{N_0}}}\frac{{{R_b}}}{{{R_b}}}\)
\(\frac{C}{N} = \frac{{{E_b}}}{{{N_0}}} = 8.8dB\)
Last updated on Jul 2, 2025
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