Let a random process {Y}( {t}) be described as {Y}( {t})= {h}( {t}) * {X}( {t})+ {Z}( {t}), where {X}( {t}) is {a} white noise process with power spectral density {Sx}( {f})=5 {W} {/} {Hz}. The filter {h}( {t}) has a magnitude response given by | {H}( {f})|= 0.5 for -5 \leq f \leq 5, and zero elsewhere. Z(t) is a stationary random process, uncorrelated with {X}( {t}), with power spectral density as shown in the figure. The power in {Y}( {t}), in watts, is equal to ______________ W (rounded off to two decimal places).

GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 56 of 65

Marks: +1, -0

Let a random process

{Y}( {t})

be described as

{Y}( {t})= {h}( {t}) * {X}( {t})+ {Z}( {t})

, where

{X}( {t})

{a}

white noise process with power spectral density

{Sx}( {f})=5 {W} {/} {Hz}

. The filter

{h}( {t})

has a magnitude response given by

| {H}( {f})|=

0.5

for

-5 ≤ f ≤ 5

, and zero elsewhere.

Z(t)

is a stationary random process, uncorrelated with

{X}( {t})

, with power spectral density as shown in the figure. The power in

{Y}( {t})

, in watts, is equal to ______________ W (rounded off to two decimal places).

Solution:

and

y(t)=z(t)+p(t)

∴ {R_{\yy}}(τ)=R_{\zz}(τ)+R_{\pp}(τ)+R_{\px}(τ)+R_{\zp}(τ)

now

{x}( {t}) \& {z}( {t})

are uncorrelated.

∴ R_{\pz}(τ)=R_{\zp}(τ)=0

∴ R_{yy}(τ)=R_{\zz}(τ)+R_{\pp}(τ)

So the power spectral relation can be given by Fourier transform of the above relation.

∴ S_{ {\yy}}( {f})= {S}_{\zz}( {f})+ {S}_{ {\pp}}( {f})

now power of

y(t)=

∫↙{∞}↖{∞} s_{\y y}(f) d f

∴ P=∫↙{∞}↖{∞} s_{\zz}(f) d f+∫↙{∞}↖{∞} S_{\pρ}(f) d f

now

S_{\pp}(f)=|H(w)|^{2} × S_{\xx}(f)

and

∴ {P}=\;{5}/{4} × 10+\;{1}/{2} × 10 × 1

∴ {P}=17.5 {watt}

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