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Question 2. (40 Points) Consider a BPSK modulation with 2 signals: 2 where ??(t) and ?1 (t) for t E 10, T,l form a set of orthonormal basis functions. Signals so(t) and si(t) are equiprobable. We suppose that the transmission channel introduces an additive white Gaussian noise n(t). Thus, the received signal is given by r(t) s(t)+n(t), where the additive white Gaussian noise has power spectral density No/2. toT Y-X+N rit) a(t)+n(t) Figure 2: A receiver for the BPSK signals. To detect the BPSK signals, the receiver shown in Figure 2 finds the correlation of r(t) and apo(t) + Boi(t), where a and B are two positive constants satisfying a2 +g2 = 1· The output Y-X + N contains two com and a noise component (a) (5 points) Plot the signal constellation of this BPSK modulation. (b) (5 points) Find the two possible values of X when so(t) and si(t) are sent, respectively. (c) (5 points) What is the mean and variance of N? (d) (10 points) Consider the equivalent channel Y X+N. Derive the maximum likelihood detection rule for this BPSK modulation. (e) (10 points) Derive error probability for the maximum likelihood detection rule in part (d) with respect to ? and ?. (f) (5 points) How do you choose parameters a and B in order to minimize the error probability?

Answer 1

BPSK modulation (Binary Phase Shift Keying) bullet It is a scheme of 2 phase modulation In binary message of BPSK, 0 and 1 are signified by the carrier signal with 2 different phase states bullet i.e for binary 1 theta = 0 degree and for binary 0 theta = 180 degree Part A: Given BPSK signals s_0(t) = + squareroot epsilon/2 phi_0 (t) + squareroot epsilon/2 phi_1 (t) s_1(t) = - squareroot epsilon/2 phi_0 (t) - squareroot epsilon/2 phi_1 (t) Where signals s_0(t) and s_1(t) are equiprobable and phi_1(t) and phi_1(t) are the orthonormal functions Now the expressions for s_0(t) will be, s_0(t) = (squareroot epsilon/2, squareroot epsilon/2) Now the expressions for s_1(t) will be, s_1(t) = (-squareroot epsilon/2, -squareroot epsilon/2) \r\n Graph for signal constellation Part B: To find the values of X Consider the expression for s_0(t), \r\n s_0(t) = (squareroot epsilon_b/2 phi_0(t), squareroot epsilon_b/2 phi_1(t)) [phi_0(t) = phi_1(t) = squareroot 2/T cos omega_c t] Because A_c = squareroot E_b = squareroot A_C^2 T/2 s_0(t) = squareroot A_C^2 T/2/2 squareroot 2/T cos omega_c t + squareroot A_C^2 T/2/2 squareroot 2/T cos omega_c t Substitute squareroot E_b = squareroot A_C^2 T/2 in the above equation, squareroot epsilon_b/2 squareroot 2/T cos omega_c t + squareroot epsilon_b/2 squareroot 2/T cos omega_c t As derived s_0(t), similarly s_1(t) will be s_1(t) = -squareroot epsilon_b/2 squareroot 2/T cos omega_c t - squareroot epsilon_b/2 squareroot 2/T cos omega_c t \r\n Part C: To find the mean and variance of N Consider the equivalent channel, Y = X + N Consider the expression for mean, epsilon(y) = epsilon(-squareroot epsilon_b/2 + n] Mean for s_0 (t), epsilon (y) squareroot epsilon_b/2 Mean for s_0(t), epsilon(y) = -squareroot epsilon_b/2 Now determining the varience epsilon(y^2) = epsilon(- squareroot epsilon_b/2 + n)^2 Bacause (a + b)^2 a^2 + b^2 + 2ab = epsilon(epsilon_b/n + n^2 + 2 (squareroot epsilon_b/2 * n))^2 Bacause epsilon(n^2) = N_0/2 and 2 epsilon(n) = 0 = epsilon_b/2 + N_0/2 \r\n Variance for s_0(t), s_0(t) = epsilon_b/2 + N_0/2 Varience for s_1 (t), s_1(t) = -epsilon_b/2 + N_0/2