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Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ 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 N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.
If the BER of this system is Q b√γ, then the value of b is _____________.
  • a)
    1.4
  • b)
    1.42
Correct answer is between '1.4,1.42'. Can you explain this answer?
Verified Answer
Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sid...
Bit error rate for BPSK
Function of bit energy and noise
Counterllation diagram of BPSK
Channel is AWGN which implies noise sample as independent
Let 2x + n1 + n2 = x' + n'
where x' = 2x
n′ = n1 + n2
Now Bit error rate
E1 is energy in x2
is PSD of h1
E1 = 4E[as amplitudes are getting doubled]
= N0[ independent and identical channel
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Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ 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 N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.If the BER of this system is Q b√γ, then the value of b is _____________.a)1.4b)1.42Correct answer is between '1.4,1.42'. Can you explain this answer?
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Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ 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 N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.If the BER of this system is Q b√γ, then the value of b is _____________.a)1.4b)1.42Correct answer is between '1.4,1.42'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ 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 N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.If the BER of this system is Q b√γ, then the value of b is _____________.a)1.4b)1.42Correct answer is between '1.4,1.42'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Q√γ be the BER of a BPSK system over an AWGN channel with two-sided noise power spectral density N0/2. The parameter γ 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 N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.If the BER of this system is Q b√γ, then the value of b is _____________.a)1.4b)1.42Correct answer is between '1.4,1.42'. Can you explain this answer?.
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