A 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.

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ESE Electronics 2012 Paper 2: Official Paper
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  1. 23 dB
  2. 2.3 dB
  3. 8.8 dB
  4. 0.88 dB

Answer (Detailed Solution Below)

Option 3 : 8.8 dB
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Detailed Solution

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Concept:

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:

\(\frac{C}{N} = \frac{{{E_b}}}{{{N_0}}}\frac{{{R_b}}}{{{R_b}}}\) 

\(\frac{C}{N} = \frac{{{E_b}}}{{{N_0}}} = 8.8dB\) 

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