GATE EC 2014 Set 1 — Question 61
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Communications → Digital Modulation & Detection → Digital Modulation (ASK, PSK, FSK, QAM)
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Question
Let be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density . The parameter is a function of bit energy and noise power spectral density. A system with two independent and identical AWGN channels with noise power spectral density is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels. 
If the BER of this system is , then the value of is _____.
Correct answer
1.4 to 1.42
Solution
For a standard BPSK system over an AWGN channel with noise power spectral density , the Bit Error Rate (BER) is given by:Given that is a function of bit energy and noise PSD, we identify .In the provided system, the BPSK signal is transmitted through two independent AWGN channels. The output of the summer is:where are independent noise components.
1.Signal Component: The combined signal amplitude is .
2.Noise Component: The combined noise is . Since and are independent, the variance of the sum is the sum of the variances:
The BER for a BPSK system with signal amplitude and noise variance is . Substituting our values:Since , we can write:Comparing this with the given form , we find:The value of is approximately 1.414, which falls within the range [1.4, 1.42].Continue learning with Success Tracker
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