s2Eva_IIT2017_T4 Sy(f) con funcion de transferencia H(f)

2da Evaluación II Término 2017-2018. Febrero 7, 2018

Tema 4.
S_x(f)= \frac{N_0}{2}

H(f) = \frac{1}{1+j2\pi f} S_{YX}(f) = H(f) S_x(f) = \frac{N_0}{2}\frac{1}{1+j2\pi f} R_{YX}(\tau) = F^{-1} [S_{YX}(f)] = \frac{N_0}{2} F^{-1} \Big[\frac{1}{1+j2\pi f} \Big] R_{YX}(\tau) = \frac{N_0}{2} e^{-\tau} , \tau>0

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Autocorrelación de Y(t)

S_{Y}(f) = |H(f)|^2 S_x(f) = \Big|\frac{1}{1+j2\pi f}\Big| ^2 \frac{N_0}{2} = \Big|\frac{1}{1+(2\pi f)^2}\Big| \frac{N_0}{2} S_{Y}(f) = \frac{N_0}{4} \Big|\frac{2}{1+(2\pi f)^2}\Big| R_{Y}(\tau) = F^{-1} [S_{Y}(f)] = F^{-1} \Big[\frac{N_0}{4} \frac{2}{1+(2\pi f)^2}\ \Big] = \frac{N_0}{4} F^{-1} \Big[\frac{2}{1+(2\pi f)^2}\ \Big] R_{Y}(\tau) = \frac{N_0}{4} e^{-|\tau|}

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Potencia promedio

R_{Y}(0) = \frac{N_0}{4} e^{-|0|} = \frac{N_0}{4}