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Question

# A real and stationary process X(t) is passed through the system shown below. If the power spectral density (PSD) of the process X(t) is Xx(f), then the PSD of the output process Y(t) will be

A
4SX(f)[1+cos(4πf)]​​​
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B
2SX(f)cos(4πf)
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C
2SX(f)cos2(2πf)
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D
4SX(f)cos2(2πf)
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Solution

## The correct option is D 4SX(f)cos2(2πf)Ans : (c) y(t)=x(t)+x(t−2) H(jω)=Y(jω)X(jω)=1+e−j2ω |H(f)|2=|1+e−j4πf|2=(1+e−j4πf)(1+ej4πf) =1+1+e−j4πf+ej4πf=2+2cos(4πf) =2[1+cos(4πf)]=4cos2(2πf) So,SY(f)=SX(f)|H(f)|2=4SX(f)cos2(2πf)

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