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For a given probability of error, binary coherent FSK is inferior to binary coherent PSK by:

This question was previously asked in
UGC NET Paper 2: Electronic Science 29 Oct 2022 Shift 1
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  1. 0 dB
  2. 2 dB
  3. 3 dB
  4. 6 dB

Answer (Detailed Solution Below)

Option 3 : 3 dB
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UGC NET Paper 1: Held on 21st August 2024 Shift 1
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Detailed Solution

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Pe for FSK \(= \frac{1}{2}erfc\sqrt {\frac{{Eb}}{{2{N_o}}}} \)

Pe for PSK \(= \frac{1}{2}erfc\sqrt {\frac{{Eb}}{{{N_o}}}}\)

For a given Pe, PSK is better than FSK by a factor of 2

10log102 = 3 dB
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