2da Evaluación II Término 2017-2018. Febrero 7, 2018
Tema 2 .
θ es una variable aleatoria uniforme, distribuida en el rango [-π π].
f θ ( θ ) = 1 π − ( − π ) ) = 1 2 π f_{\theta} (\theta) = \frac{1}{\pi - (-\pi))} = \frac{1}{2\pi} f θ ( θ ) = π − ( − π ) ) 1 = 2 π 1 valor esperados X(t)
X ( t ) = cos ( ω t + θ ) X(t) = \cos (\omega t + \theta) X ( t ) = cos ( ω t + θ ) E [ X ( t ) ] = E [ cos ( ω t + θ ) ] E[X(t)] = E[ \cos (\omega t + \theta)] E [ X ( t ) ] = E [ cos ( ω t + θ ) ] = ∫ − π π x ( t ) f θ ( θ ) d θ = \int_{-\pi}^{\pi}x(t)f_{\theta}(\theta) d\theta = ∫ − π π x ( t ) f θ ( θ ) d θ = ∫ − π π cos ( ω t + θ ) 1 2 π d θ = \int_{-\pi}^{\pi}\cos (\omega t + \theta) \frac{1}{2\pi} d\theta = ∫ − π π cos ( ω t + θ ) 2 π 1 d θ = 1 2 π sin ( ω t + θ ) ∣ − π π = \frac{1}{2\pi}\sin (\omega t + \theta) \Big|_{-\pi}^{\pi} = 2 π 1 sin ( ω t + θ ) ∣ ∣ ∣ − π π = 1 2 π [ sin ( ω t + π ) − sin ( ω t − π ) ] = \frac{1}{2\pi} [ \sin (\omega t +\pi) -\sin (\omega t -\pi) ] = 2 π 1 [ sin ( ω t + π ) − sin ( ω t − π ) ] E [ X ( t ) ] = 0 E[X(t)] = 0 E [ X ( t ) ] = 0 valor esperado Y(t)
Y ( t ) = sin ( ω t + θ ) Y(t) = \sin (\omega t + \theta) Y ( t ) = sin ( ω t + θ ) E [ Y ( t ) ] = E [ sin ( ω t + θ ) ] E[Y(t)] = E[ \sin(\omega t + \theta)] E [ Y ( t ) ] = E [ sin ( ω t + θ ) ] = ∫ − π π y ( t ) f θ ( θ ) d θ = \int_{-\pi}^{\pi}y(t)f_{\theta}(\theta) d\theta = ∫ − π π y ( t ) f θ ( θ ) d θ = ∫ − π π sin ( ω t + θ ) 1 2 π d θ = \int_{-\pi}^{\pi}\sin(\omega t + \theta) \frac{1}{2\pi} d\theta = ∫ − π π sin ( ω t + θ ) 2 π 1 d θ = 1 2 π [ − cos ( ω t + θ ) ] ∣ − π π = \frac{1}{2\pi} [-\cos(\omega t + \theta)] \Big|_{-\pi}^{\pi} = 2 π 1 [ − cos ( ω t + θ ) ] ∣ ∣ ∣ − π π = 1 2 π [ − cos ( ω t + π ) − ( − cos ( ω t − π ) ) ] = \frac{1}{2\pi} [ -\cos(\omega t +\pi) - (-\cos (\omega t -\pi)) ] = 2 π 1 [ − cos ( ω t + π ) − ( − cos ( ω t − π ) ) ] = 1 2 π [ cos ( ω t − π ) − cos ( ω t + π ) ] = \frac{1}{2\pi} [\cos(\omega t -\pi) - \cos (\omega t +\pi) ] = 2 π 1 [ cos ( ω t − π ) − cos ( ω t + π ) ] E [ Y ( t ) ] = 0 E[Y(t)] = 0 E [ Y ( t ) ] = 0 Correlación X(t) y Y(t)
R X Y [ t , t + τ ] = E [ X ( t ) Y ( t + τ ) ] R_{XY}[t,t+\tau] =E[X(t) Y(t+\tau)] R X Y [ t , t + τ ] = E [ X ( t ) Y ( t + τ ) ] = E [ cos ( ω t + θ ) sin ( ω ( t + τ ) + θ ) ] =E[\cos (\omega t + \theta) \sin (\omega (t+\tau) + \theta)] = E [ cos ( ω t + θ ) sin ( ω ( t + τ ) + θ ) ] = E [ 1 2 [ sin [ ( ω ( t + τ ) + θ ) − ( ω t + θ ) ] + sin [ ( ω ( t + τ ) + θ ) + ( ω t + θ ) ] ] ] =E \Big[ \frac{1}{2}\Big[\sin [(\omega (t+\tau) + \theta) - (\omega t + \theta)] + \sin [(\omega (t+\tau) + \theta) + (\omega t + \theta)] \Big] \Big] = E [ 2 1 [ sin [ ( ω ( t + τ ) + θ ) − ( ω t + θ ) ] + sin [ ( ω ( t + τ ) + θ ) + ( ω t + θ ) ] ] ] = 1 2 E [ sin ( ω τ ) + sin ( 2 ω t + ω τ + 2 θ ) ] =\frac{1}{2}E\Big[\sin (\omega \tau) + \sin (2\omega t+ \omega \tau + 2\theta) \Big] = 2 1 E [ sin ( ω τ ) + sin ( 2 ω t + ω τ + 2 θ ) ] = 1 2 E [ sin ( ω τ ) ] + 1 2 E [ sin ( 2 ω t + ω τ + 2 θ ) ] =\frac{1}{2}E \Big[ \sin (\omega \tau) \Big] + \frac{1}{2}E\Big[\sin (2\omega t+ \omega \tau + 2\theta) \Big] = 2 1 E [ sin ( ω τ ) ] + 2 1 E [ sin ( 2 ω t + ω τ + 2 θ ) ] El primer término no contiene la variable aleatorioa Θ, por lo que se comporta como una constante para el valor esperado.
= sin ( ω τ ) 2 + 1 2 ∫ − π π sin ( 2 ω t + ω τ + 2 θ ) 1 2 π d θ =\frac{\sin (\omega \tau)}{2} + \frac{1}{2}\int_{-\pi}^{\pi}\sin (2\omega t+ \omega \tau + 2\theta) \frac{1}{2\pi} d\theta = 2 sin ( ω τ ) + 2 1 ∫ − π π sin ( 2 ω t + ω τ + 2 θ ) 2 π 1 d θ = sin ( ω τ ) 2 − 1 4 π cos ( 2 ω t + ω τ + 2 θ ) ∣ − π π =\frac{\sin (\omega \tau)}{2} - \frac{1}{4\pi}\cos (2\omega t+ \omega \tau + 2\theta) \Big|_{-\pi}^{\pi} = 2 sin ( ω τ ) − 4 π 1 cos ( 2 ω t + ω τ + 2 θ ) ∣ ∣ ∣ − π π = sin ( ω τ ) 2 − 1 4 π [ cos ( 2 ω t + ω τ + 2 π ) − cos ( 2 ω t + ω τ − 2 π ) ] =\frac{\sin (\omega \tau)}{2} - \frac{1}{4\pi}\Big[ \cos (2\omega t+ \omega \tau + 2\pi) - \cos (2\omega t+ \omega \tau - 2\pi)\Big] = 2 sin ( ω τ ) − 4 π 1 [ cos ( 2 ω t + ω τ + 2 π ) − cos ( 2 ω t + ω τ − 2 π ) ] = sin ( ω τ ) 2 − 0 =\frac{\sin (\omega \tau)}{2} - 0 = 2 sin ( ω τ ) − 0 R X Y [ t , t + τ ] = sin ( ω τ ) 2 R_{XY}[t,t+\tau] =\frac{\sin (\omega \tau)}{2} R X Y [ t , t + τ ] = 2 sin ( ω τ ) C X Y [ t , t + τ ] = R X Y [ t , t + τ ] − E [ X ( t ) ] E [ Y ( t + τ ) ] C_{XY}[t,t+\tau] = R_{XY}[t,t+\tau] - E[X(t)]E[Y(t+\tau)] C X Y [ t , t + τ ] = R X Y [ t , t + τ ] − E [ X ( t ) ] E [ Y ( t + τ ) ] C X Y [ t , t + τ ] = R X Y [ t , t + τ ] − 0 C_{XY}[t,t+\tau] = R_{XY}[t,t+\tau] - 0 C X Y [ t , t + τ ] = R X Y [ t , t + τ ] − 0 C X Y [ t , t + τ ] = sin ( ω τ ) 2 C_{XY}[t,t+\tau] = \frac{\sin (\omega \tau)}{2} C X Y [ t , t + τ ] = 2 sin ( ω τ ) X(t) y Y(t) son procesos con correlación, pues su covarianza cruzada no es igual a cero para todas las selecciones de muestras de tiempo. Sin embargo, X(t1 ) y Y(t2 ) son variables aleatorias no correlacionadas para t1 y t2 dado que ω( t2 – t1 ) = k π, donde k es cualquier número entero.
Los valores de mas medias de X(t) = Y(t) =0 son constantes
R X [ t , t + τ ] = E [ X ( t ) X ( t + τ ) ] R_{X}[t,t+\tau] = E[X(t) X(t+\tau)] R X [ t , t + τ ] = E [ X ( t ) X ( t + τ ) ] = E [ cos ( ω t + θ ) cos ( ω ( t + τ ) + θ ) ] =E[\cos (\omega t + \theta) \cos(\omega (t+\tau) + \theta)] = E [ cos ( ω t + θ ) cos ( ω ( t + τ ) + θ ) ] = E [ 1 2 [ cos [ ( ω t + θ ) − ( ω ( t + τ ) + θ ) ] + cos [ ( ω t + θ ) + ( ω ( t + τ ) + θ ) ] ] ] =E\Big[\frac{1}{2} \Big[ \cos [(\omega t + \theta) -(\omega (t+\tau) + \theta) ] + \cos[(\omega t + \theta)+(\omega (t+\tau) + \theta)] \Big] \Big] = E [ 2 1 [ cos [ ( ω t + θ ) − ( ω ( t + τ ) + θ ) ] + cos [ ( ω t + θ ) + ( ω ( t + τ ) + θ ) ] ] ] = 1 2 E [ cos ( ω τ ) + cos ( 2 ω t + ω τ + 2 θ ) ] ] =\frac{1}{2}E\Big[ \cos (\omega \tau ) + \cos(2\omega t + \omega \tau + 2\theta)] \Big] = 2 1 E [ cos ( ω τ ) + cos ( 2 ω t + ω τ + 2 θ ) ] ] = 1 2 E [ cos ( ω τ ) ] + 1 2 E [ cos ( 2 ω t + ω τ + 2 θ ) ] ] =\frac{1}{2}E\Big[ \cos (\omega \tau )\Big] +\frac{1}{2}E\Big[ \cos(2\omega t + \omega \tau + 2\theta)] \Big] = 2 1 E [ cos ( ω τ ) ] + 2 1 E [ cos ( 2 ω t + ω τ + 2 θ ) ] ] = cos ( ω τ ) 2 cos ( ω τ ) + 0 =\frac{\cos (\omega \tau )}{2} \cos (\omega \tau ) +0 = 2 cos ( ω τ ) cos ( ω τ ) + 0 La autocorrelación depende solo de las diferencias de tiempo τ = t2 -t1
El proceso X(t) clasifica como Estacionario en el sentido amplio.
Tarea
