In a binary baseband communication system the equiprobable bits are modelled using polar-NRZ scheme with a rectangular basic pulse of amplitude 1 V. These bits are transmitted through an AWGN channel with two sided noise PSD of 10 $^{- 5}$ W/Hz and received by a correlator receiver with optimum threshold detection. If the maximum acceptable bit error rate (BER) is 10 $^{- 4}$ then the maximum allowable bit rate $(R_{b})$ will be ________ kbps.

[Assume that the effect of ISI is negligible and $Q (3.7) = 10^{- 4}$ ]
7.305

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Solution

The correct option is A 7.305
$B E R = Q (\sqrt{\frac{2 E_{b}}{N_{0}}})$

$E_{b} = A^{2} T_{b} = T_{b} ∵ A = 1 f o r b a s i c p u l s e u s e d$

$Q (\sqrt{\frac{2 T_{b}}{N_{0}}}) \leq Q (3.7) = 10^{- 4}$

The above relation will be satisfied when.

$\frac{2 T_{b}}{N_{0}} \geq (3.7)^{2} ∵ Q (x) d e c r e a s e s a s r i n c r e a s e s$

$T_{b} \geq \frac{13.69 N_{0}}{2} s e c$

$T_{b (m i n)} = 13.69 \times 10^{- 5} s e c ∵ \frac{N_{o}}{2} = 10^{- 5} W / H z$

$R_{b (m a x)} = \frac{1}{T_{b (m i n)}} = 7.305 k b p s$

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