Error probability:

• We know that BPSK signal is represented as follows:

Binary 1:$ x _1(t)= \sqrt{2 P_s} $cos$ω_C$ t$

Binary 0: $x _2(t)= -\sqrt{2 P_s} cosω_C t$

Therefore $x _2(t)= - x _1(t)$

• By using the matched filter for detection of BPSK signal.The expression for error probability of an optimum filter is,

• The expression for the signal to noise ration of matched filter is given by,

• Using the Rayleigh’s energy theorem,

• The limits of integration of the last term in equation (3) are 0 to T because x (t) is present only over one bit interval T. Substituting equation (3) into equation (2) we get,

• But x (t)=$ x _1(t)- x_ 2(t) $

and for BPSK $ x _2(t)= - x _1(t)$

• Substituting this value of x (t) into equation (4) we get,

• Substitute this into equation (5) to get

• The value of second term in the RHS in equation (6) is zero

Bandwidth of BPSK:

• From the frequency spectrum of BPSK signal ,it is clear that the bandwidth of a BPSK signal is given by,

BW= Highest frequency – Lowest frequency in main lobe =$(f_c+f_b)- (f_c-f_b)$

∴ BW=2 $f_b$

Where $f_b =1/ T_b$

• Thus the minimum bandwidth of BPSK signal is equal to twice the highest frequency contained in the baseband signal.