Question

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 tow 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 _____________.

Answer 1

Bit error rate for BPSK

Function of bit energy and noise P_SD(N₀/2)

Counterllation diagram of BPSK 

Channel is A which implies noise sample as independent

Let 2x + n₁ + n₂ = x¹ + n¹

where x₁ = 2x

n¹ = n₁ + n₂

Now Bit error rate = Q(√(2E¹/N₀¹))

E₁ is energy in x₁

N₀¹ is PSD of h₁

E1 = 4E [as amplitudes are getting doubled]

N₀¹ = N₀ [independent and identical channel]

⇒ Bit error rate = Q(√(4E/N₀)) = Q(√2(2E/N₀)) ⇒ b = √2 or 1.414