# GATE Papers >> ECE >> 2019 >> Question No 55

Question No. 55

Let a random process () be described as () = () * () + (), where () is a white noise process with power spectral density SX () = 5 W/Hz. The filter () has a magnitude response given by |()| = 0.5 for −5 $\leq$ f $\leq$ 5, and zero elsewhere. () is a stationary random process, uncorrelated with (), with power spectral density as shown in the figure. The power in (), in watts, is equal to ________ W (rounded off to two decimal places). ##### Answer : 17.40 to 17.60

Solution of Question No 55 of GATE 2019 ECE Paper

y(t) = h(t) $\times$ x(t) + z(t) ... $\boxed1$

x(t) → AWGN   5 $\times$ (f) = 5W / Hz

h(t) = |H(f)| = 0.5 for -5 < f < 5

z(t) → stationary power uncorrelated with x(t) FT of $\boxed1$ is y(f) = |H(f)|.|X(f)| + 2(f)

PSD of y(t) is

sy(F) = |y(F)|2 = |H(f)|2 |x(f)|2 + |z(f)|2 + 2|H(f)|x(f)|z(f)|

$\therefore$ output $=\int\limits_\infty^\infty\mathrm{sy}(5)\mathrm{df}$

$\begin{array}{l}=\int\limits_\infty^\infty\left[\left|\mathrm H\left(\mathrm f\right)\right|^2\left|\mathrm x\left(\mathrm f\right)\right|^2+\left|\mathrm z\left(\mathrm f\right)\right|^2\right]\mathrm{at}\\=\int\limits_\infty^\infty\left(\frac12\right)^25\mathrm{df}+\frac12\times10\times1\\=\frac54\times10+\frac{10}2=17.5\;\mathrm{watts}\end{array}$

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