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Constellation Diagram of PSK in Detail (with Simulator)


 


 

 

 

 
Fig 1: Constellation Diagram of PSK
 
 In the above figure, the binary bit '1' is represented by S1(t) and the binary bit '0' by S2(t), respectively.

So, energy of S1(t) = (√(Eb))2 = Eb
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': $S_2(t) = -A_c \cos(2\Pi f_c t) = -\sqrt{\frac{2E_b}{T_b}} \cos(2\Pi f_c t) = -\sqrt{E_b} \cdot \sqrt{\frac{2}{T_b}} \cos(2\Pi f_c t)$

 High-order PSK (e.g., 8 PSK, 16 PSK) can transmit more bits per symbol but is more sensitive to noise. Low-order PSK (e.g., BPSK, QPSK) is less susceptible to noise.
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. 

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