For the system shown in the figure below- the inputXtis a stationary random process with an auto correlation functionRX-4sinc22-If the low-pass filter has a cut-off frequency of2Hz and a passband gain of unity- then the time average power of the output process Ytwill be-W-

(ACF)R_x(t)CTFT→S_x(t)(PSD) sin c(2t)CTFT→12rect ( f2 ) 4sin c^2(2t)CTFT→ [ rect ( f2 ) rect ( f2 ) ] cos(2π t)CTFT→12[δ (f 1)+δ(f+1)] For time average power ...

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Byju's Answer

Standard XII

Physics

Interference Definition

For the syste...

Question

$F o r t h e s y s t e m s h o w n i n t h e f i g u r e b e l o w, t h e i n p u t X (t) i s a s t a t i o n a r y r a n d o m p r o c e s s w i t h a n a u t o c o r r e l a t i o n f u n c t i o n R_{X} (τ) = 4 s i n c^{2} (2 τ) I f t h e l o w - p a s s f i l t e r h a s a c u t - o f f f r e q u e n c y o f 2 H z a n d a p a s s b a n d g a i n o f u n i t y, t h e n t h e t i m e a v e r a g e p o w e r o f t h e o u t p u t p r o c e s s Y (t) w i l l b e . . . . W .$

1.75

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Solution

The correct option is A 1.75
$(A C F) R_{x} (t) C T F T \to S_{x} (t) (P S D) s i n c (2 t) C T F T \to \frac{1}{2} r e c t (\frac{f}{2}) 4 s i n c^{2} (2 t) C T F T \to [r e c t (\frac{f}{2}) r e c t (\frac{f}{2})] c o s (2 π t) C T F T \to \frac{1}{2} [δ (f - 1) + δ (f + 1)]$
$F o r t i m e a v e r a g e p o w e r s, S_{z} (f) = \frac{1}{4} [S_{X} (f - 1) + S_{X} (f + 1)]$

$A f t e r p a s s i n g t h r o u g h a n L P F w i t h f_{c} = 2 H z, w e g e t,$

$P_{y} \int_{- \infty}^{\infty} S_{y} (f) d f = 2 [(1 \times 0.50) + (1 \times 0.25) + (\frac{1}{2} \times 1 \times 0.25)] W = 1.75 W$

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For the system shown in the figure below, the inputX(t)is a stationary random process withan auto correlation functionRX(τ)=4sinc2(2τ)If the low−pass filter has a cut−off frequency of2Hz and a passbandgain of unity, then the time averagepower of the output process Y(t)will be....W. 1.75