Question

A baseband PCM system with a matched filter at the receiver is implemented with \(\pm 5\) V bipolar with a pulse duration of 72\(\mu\) sec. If the noise PSD is \(1.0 \times 10^{-4} V^2/Hz\), the probability of error in this system is given by

• A. \(1.15 \times 10^{-5}\)
• B. \(1.85 \times 10^{-5}\)
• C. \(2.8 \times 10^{-5}\)
• D. \(2.36 \times 10^{-5}\)

Hint:

Pay close attention to whether the noise Power Spectral Density (PSD) is given as single-sided (\(N_0\)) or two-sided (\(N_0/2\)). This is a common source of errors by a factor of 2 in the argument of the Q-function. The formula for \(P_e\) for bipolar signaling is \(Q(\sqrt{2E_b/N_0})\).

Answer 1

The Correct Option is A

Step 1: State the Probability of Error (\(P_e\)) formula for Bipolar Signaling.For bipolar signaling employing a matched filter, the error probability is defined by the Q-function:\[ P_e = Q\left(\sqrt{\frac{2E_b}{N_0}}\right) \]Here, \(E_b\) represents the energy per bit, and \(N_0/2\) signifies the two-sided noise power spectral density. Given a single-sided PSD, \(N_0 = 1.0 \times 10^{-4} V^2/Hz\).
Step 2: Compute the energy per bit, \(E_b\).The signal is characterized by a pulse with amplitude \(A = 5V\) and duration \(T_b = 72 \mu s\).The pulse energy is calculated as:\[ E_b = A^2 T_b = (5V)^2 \times (72 \times 10^{-6} s) \]\[ E_b = 25 \times 72 \times 10^{-6} = 1800 \times 10^{-6} = 1.8 \times 10^{-3} \text{ Joules} \]
Step 3: Determine the Q-function argument.\[ \frac{2E_b}{N_0} = \frac{2 \times (1.8 \times 10^{-3})}{1.0 \times 10^{-4}} = \frac{3.6 \times 10^{-3}}{1 \times 10^{-4}} = 36 \]Thus, we need to find \(P_e = Q(\sqrt{36}) = Q(6)\).
Step 4: Evaluate the Q-function to find \(P_e\).The Q-function, \(Q(x)\), lacks a simple closed-form solution and is generally evaluated using tables or software. \(Q(6)\) yields a very small value.Using approximation or reference tables, \(Q(6) \approx 9.86 \times 10^{-10}\).This result doesn't align with any of the provided options, suggesting a possible error in the problem statement or answer choices. Re-examining the formulas, specifically \(E_b = \int s(t)^2 dt\), confirms its validity. The amplitude is confirmed to be \(\pm 5V\), not peak-to-peak. Considering if the PSD is given as \(N_0/2\)?If \(N_0/2 = 10^{-4}\), then \(N_0 = 2 \times 10^{-4}\).Then \(\frac{2E_b}{N_0} = \frac{2 \times (1.8 \times 10^{-3})}{2 \times 10^{-4}} = 18\).\(P_e = Q(\sqrt{18}) = Q(4.24)\). Consulting tables, \(Q(4.24) \approx 1.15 \times 10^{-5}\). This result aligns with option A.