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 four phase and an eight-phase signal constellation are shown in the figure below.

For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r₁, and r² of the circles are

A four phase and an eight-phase signal constellation are shown in the figure below.

Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is

Consider a base band binary PAM receiver shown below. The additive channel noise $$n(t)$$ is white with power spectral density $${S_N}\left( f \right) = {N_0}/2 = {10^{ - 20}}$$ $$W/Hz$$. The low-pass filter is ideal with unity gain and cut -off frequency $$1MHz$$. Let $${Y_k}$$ represent the random variable $$y\left( {{t_k}} \right)$$.
$${Y_k} = {N_k}$$ if transmitted bit $${b_k} = 0$$
$${Y_k} = a + {N_k}$$ if transmitted bit $${b_k} = 1$$
Where $${b_k} = 0$$ represents the noise sample value. The noise sample has a probability density function, $${P_{{N_k}}}\left( n \right)\,\,\,\,\,\,\, = 0.5\alpha {e^{ - \alpha \left| n \right|}}$$ (This has mean zero and variance $$2/{\alpha ^2}$$). Assume transmitted bits to be equiprobable and threshold $$z$$ is set to $$a/2 = {10^{ - 6}}V$$.

The value of the parameter $$\alpha $$( in V^-1 ) is