Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms $${s_1}\,(t)\, = \,\alpha \,\,\cos \,\,\,2\,\pi {f_1}\,t\,and\,\,{s_{2\,}}(t)\,\, = \,\alpha \,\,\cos \,\,\,2\,\pi {f_2}\,t$$, where $$\,\alpha = 4\,\,\,mV$$. Assume an AWGN channel with two-sided noise power spectral density $$\,{{{N_0}} \over 2} = 0.5\,\, \times \,{10^{ - 12}}$$ W/Hz. Using an optimal receiver and the relation $$Q(v) = {1 \over {\sqrt {2\,\pi } }}\,\int\limits_v^\infty {e{\,^{ - {u^2}/2}}} \,du$$, the bit error probability for a data rate of 500 kbps is

Let U and V be two independent zero mean Gaussian random variables of variances $${{1 \over 4}}$$ and $${{1 \over 9}}$$ respectively. The probability $$P(\,3V\, \ge \,\,2U)$$ is

A BPSK scheme operating over an AWGN channel with noise power spectral density of N₀2, uses equi-probable signals $$${s_1}\left( t \right) = \sqrt {{{2E} \over T}\,\sin \left( {{\omega _c}t} \right)} $$$
and $$${s_2}\left( t \right) = - \sqrt {{{2E} \over T}\,\sin \left( {{\omega _c}t} \right)} $$$

over the symbol interval, $$(0, T)$$. If the local oscillator in a coherent receiver is ahead in phase by 45⁰ with respect to the received signal, the probability of error in the resulting system is

A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X =0) = 9/10, then the probability of error for an optimum receiver will be