Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is white with power spectral density S_{N}(f) = N_{0}/2 = 10^{-20} W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1 MHz. Let Y_{t}, represent the random variable y(t_{1})

Y_{t} = N_{s} if transmitted bit b_{k} = 0

Y_{K }= a + N_{k} if transmitted bit b_{k} = 1

where N_{k} represents the noise sample value. The noise sample has a probability density function, P_{N} (n) = 0.5αe (This has mean zero and variance 2/α^{2}), Assume transmitted bits lo be equiprobable and threshold z is set to a/2 = 10^{-6} V

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GATE EC 2010 Official Paper

Option 2 : 10^{7}

__Concept:__

The statistical averages Mean and Variance are called as 'first order moment' and 'second-order moment' respectively.

Variance represents the **total power **of the random signal.

\(mean = E\left[ X \right] = {m_1} = \mathop \smallint \limits_{ - \infty }^\infty x × f\left( x \right)dx\)

\(V[X] = {m_2} = \mathop \smallint \limits_{ - \infty }^\infty x^2 × f\left( x \right)dx\)

Total power can be determined as the area of the curve.

When the input is passed through a system the output is shown by:

S_{N}(f): Input noise PSD

S_{N0}(f): Output noise PSD

H(f): system response

__Calculation:__

The system is defined as shown:

The given mean is 0 and variance is 2/α^{2}

V[X] = E[X^{2}] - (E[X])^{2}

2/α2 = E[X2]

The area under the curve is:

E[X2] = 10^{-20} × 2 × 10^{6}

2/α2 = 10-14 × 2

α2 = 10^{14}

**α = 10 ^{7}**