Question
Download Solution PDFFor a given probability of error, binary coherent FSK is inferior to binary coherent PSK by:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFPe 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 dBLast updated on Jun 12, 2025
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