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: Write the formula for Probability of Error (\(P_e\)) for Bipolar Signaling. For bipolar signaling with a matched filter, the probability of error is given by the Q-function: \[ P_e = Q\left(\sqrt{\frac{2E_b}{N_0}}\right) \] where \(E_b\) is the energy per bit and \(N_0/2\) is the two-sided power spectral density of the noise. The given PSD is single-sided, so \(N_0 = 1.0 \times 10^{-4} V^2/Hz\).
Step 2: Calculate the energy per bit, \(E_b\). The signal is a pulse of amplitude \(A = 5V\) and duration \(T_b = 72 \mu s\). The energy of the pulse is given by: \[ E_b = A^2 T_b = (5V)^2 \times (72 \times 10^{-6} s) \] \[ E_b = 25 \times 72 \times 10^{-6} = 1800 \times 10^{-6} = 1.8 \times 10^{-3} \text{ Joules} \]
Step 3: Calculate the argument of the Q-function. \[ \frac{2E_b}{N_0} = \frac{2 \times (1.8 \times 10^{-3})}{1.0 \times 10^{-4}} = \frac{3.6 \times 10^{-3}}{1 \times 10^{-4}} = 36 \] So, we need to calculate \(P_e = Q(\sqrt{36}) = Q(6)\).
Step 4: Evaluate the Q-function. The Q-function, \(Q(x)\), does not have a simple closed-form expression and is usually evaluated from tables or software. \(Q(6)\) is a very small number. Using a standard approximation or table, \(Q(6) \approx 9.86 \times 10^{-10}\). None of the options match this value. There must be an error in the problem statement or the options. Let's recheck the formulas. Maybe \(E_b = \int s(t)^2 dt\). Yes. Maybe the amplitude is peak-to-peak? No, \(\pm 5V\). Maybe the PSD is \(N_0/2\)? If \(N_0/2 = 10^{-4}\), then \(N_0 = 2 \times 10^{-4}\). Then \(\frac{2E_b}{N_0} = \frac{2 \times (1.8 \times 10^{-3})}{2 \times 10^{-4}} = 18\). \(P_e = Q(\sqrt{18}) = Q(4.24)\). From tables, \(Q(4.24) \approx 1.15 \times 10^{-5}\). This matches option A.