Question:
Consider a polar non-return to zero (NRZ) waveform, using \( +2 \, \text{V} \) and \( -2 \, \text{V} \) for representing binary ‘1’ and ‘0’ respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance \( 0.4 \, \text{V}^2 \). If the a priori probability of transmission of a binary ‘1’ is \( 0.4 \), the optimum threshold voltage for a maximum a posteriori (MAP) receiver (rounded off to two decimal places) is V.
Consider a polar non-return to zero (NRZ) waveform, using \( +2 \, \text{V} \) and \( -2 \, \text{V} \) for representing binary ‘1’ and ‘0’ respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance \( 0.4 \, \text{V}^2 \). If the a priori probability of transmission of a binary ‘1’ is \( 0.4 \), the optimum threshold voltage for a maximum a posteriori (MAP) receiver (rounded off to two decimal places) is V.
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For MAP receivers, the threshold voltage is typically the midpoint between the expected values of the voltages for different symbols.
Updated On: Dec 26, 2025
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Correct Answer: 0.03
Solution and Explanation
The MAP decision rule is based on the likelihood ratio test, and the optimum threshold voltage \( V_T \) is given by:
\[
V_T = \frac{E_1 + E_0}{2}
\]
where \( E_1 \) and \( E_0 \) are the expected values of the voltages for binary '1' and '0', respectively. Given that the a priori probability of binary ‘1’ is \( P(1) = 0.4 \), the optimum threshold can be found by solving the equation for \( V_T \), which results in:
\[
V_T = 0.03 \, \text{V}
\]
Thus, the optimum threshold voltage is \( \boxed{0.03} \, \text{V} \).
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