# Consider a polar non-return to zero (NRZ) waveform, using +2 V and -2 V for representing binary '1' and '0' respectively, is transmitted in the presen

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Consider a polar non-return to zero (NRZ) waveform, using +2 V and -2 V for representing binary '1' and '0' respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance 0.4 V2. 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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Concept:

Optimum threshold voltage

For an Additive White Gaussian Noise function, the optimum threshold value is given by:

${v_{th}} = \frac{{{a_1} + {a_2}}}{2} + \frac{{\sigma _n^2}}{{{a_1} - {a_2}}}\ln \left[ {\frac{{P\left( 0 \right)}}{{P\left( 1 \right)}}} \right]$

P(0): Probability of the error when ‘0’ transmitted

P(1): Probability of the error when ‘1’ transmitted

Calculation:

Given that P(1) = 0.4 then P(0) = 0.6

The variance of the noise is 0.4V2

Given that binary ‘1’ is represented for +2V and binary ‘0’ is represented for -2V

For transmission of binary ‘1’

a1 = E[X + N]

a1 = E[X] + E[N]

Given that mean of noise is zero i.e, E[N] = 0

a1 = E[2] + E[N] = 2 + 0

a1 = 2

For transmission of binary ‘0’

a2 = E[X + N]

a2 = E[- 2] + E[N]

a2 = - 2

Optimum threshold voltage is

${v_{th}} = \frac{{2 - 2}}{2} + \frac{{0.4}}{{2 - \left( { - 2} \right)}}\ln \left[ {\frac{{0.6}}{{0.4}}} \right]$

${v_{th}} = \frac{{0.4}}{{2 - \left( { - 2} \right)}}\ln \left[ {\frac{{0.6}}{{0.4}}} \right]$

Vth = 0.04 V