So, energy of S2(t) = (-√(Eb))2 = Eb
Distance between the signaling points, d12 = 2(√(Eb))
Energy per bit for binary '1' and binary '0' (Tb = bit duration)
** For Transmission of binary ‘1’:
$E_b = \int_{0}^{T_b} (A_c \cos(2\Pi f_c t))^2 dt$
$= \int_{0}^{T_b} \frac{(A_c)^2}{2} dt + \int_{0}^{T_b} \frac{(A_c)^2 \cos(4\Pi f_c t)}{2} dt$
$= \int_{0}^{T_b} \frac{(A_c)^2}{2} dt + 0$ (The integral of $\cos(4\Pi f_c t)$ over a complete cycle is zero)
$= \frac{(A_c)^2}{2} \cdot T_b$
From this, $A_c = \sqrt{\frac{2E_b}{T_b}}$
** For Transmission of binary ‘0’:
$E_b = \int_{0}^{T_b} (-A_c \cos(2\Pi f_c t))^2 dt$
$= \int_{0}^{T_b} \frac{(A_c)^2}{2} dt + \int_{0}^{T_b} \frac{(A_c)^2 \cos(4\Pi f_c t)}{2} dt$
$= \int_{0}^{T_b} \frac{(A_c)^2}{2} dt + 0$ (The integral of $\cos(4\Pi f_c t)$ over a complete cycle is zero)
$= \frac{(A_c)^2}{2} \cdot T_b$
From this, $A_c = \sqrt{\frac{2E_b}{T_b}}$
** Constellation Diagram Representation:
For binary '1': $S_1(t) = A_c \cos(2\Pi f_c t) = \sqrt{\frac{2E_b}{T_b}} \cos(2\Pi f_c t) = \sqrt{E_b} \cdot \sqrt{\frac{2}{T_b}} \cos(2\Pi f_c t)$
For binary '0': $S_2(t) = -A_c \cos(2\Pi f_c t) = -\sqrt{\frac{2E_b}{T_b}} \cos(2\Pi f_c t) = -\sqrt{E_b} \cdot \sqrt{\frac{2}{T_b}} \cos(2\Pi f_c t)$
PSK modulation can be visualized using a constellation diagram, where each point represents a symbol. In the presence of noise, points may be away from the original positions, making them harder to distinguish.