Publicado en por Edison Del Rosario

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

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

Tema 4.
Sx(f)=N02 S_x(f)= \frac{N_0}{2}

H(f)=11+j2πf H(f) = \frac{1}{1+j2\pi f} SYX(f)=H(f)Sx(f)=N0211+j2πf S_{YX}(f) = H(f) S_x(f) = \frac{N_0}{2}\frac{1}{1+j2\pi f} RYX(τ)=F−1[SYX(f)] R_{YX}(\tau) = F^{-1} [S_{YX}(f)] =N02F−1[11+j2πf] = \frac{N_0}{2} F^{-1} \Big[\frac{1}{1+j2\pi f} \Big] RYX(τ)=N02e−τ,τ>0 R_{YX}(\tau) = \frac{N_0}{2} e^{-\tau} , \tau>0
Autocorrelación de Y(t)

SY(f)=∣H(f)∣2Sx(f) S_{Y}(f) = |H(f)|^2 S_x(f) =∣11+j2πf∣2N02 = \Big|\frac{1}{1+j2\pi f}\Big| ^2 \frac{N_0}{2} =∣11+(2πf)2∣N02 = \Big|\frac{1}{1+(2\pi f)^2}\Big| \frac{N_0}{2} SY(f)=N04∣21+(2πf)2∣ S_{Y}(f) = \frac{N_0}{4} \Big|\frac{2}{1+(2\pi f)^2}\Big| RY(τ)=F−1[SY(f)] R_{Y}(\tau) = F^{-1} [S_{Y}(f)] =F−1[N0421+(2πf)2 ] = F^{-1} \Big[\frac{N_0}{4} \frac{2}{1+(2\pi f)^2}\ \Big] =N04F−1[21+(2πf)2 ] = \frac{N_0}{4} F^{-1} \Big[\frac{2}{1+(2\pi f)^2}\ \Big] RY(τ)=N04e−∣τ∣ R_{Y}(\tau) = \frac{N_0}{4} e^{-|\tau|}

Potencia promedio

RY(0)=N04e−∣0∣=N04 R_{Y}(0) = \frac{N_0}{4} e^{-|0|} = \frac{N_0}{4}