Let's consider a signal y(t) that is an output of a linear transformation with the addition of noise. $$y ( t ) = Ax ( t − \theta) + m ( t )$$
The hypothesis on the signals x(t), y(t) is that they are both:
- random stationary processes with finite variance
- random processes with zero mean value
I need to find out the power of the signal and noise components separately.
I multiply both members by $x(t-\theta)$ and average, obtaining (E = Expected Value, m(t) is not correlated with x(t), $\sigma^2_x$ is the variance of the input signal) $$E [ x ( t − \theta ) y ( t ) ] = A E [ x ^2 ( t − \theta ) ] + E [ x ( t − \theta ) m ( t ) ] = A \sigma^2_x $$
Hence $$^{(1)}A=\frac{1}{\sigma^2_x}E [ x ( t − \theta ) y ( t ) ]$$
A depends on $\theta$ but since $\theta$ is arbitrary we chose a $\sigma$ that gives the maximum value of A.
$$A_0 = \frac{1}{\sigma^2_x} max_{\theta}E[x(t−\theta)y(t)]= \frac{1}{\sigma^2_x} E[x(t−\theta_0)y(t)] $$
We can therefore evaluate the power of the signal component:
$$^{(2)}S = A_0^2 E[x^2(t−\theta)]=\frac{1}{\sigma^2_x} \{E[x(t−\theta_0)y(t)]\}^2$$
The points $^{(1)}$ and $^{(2)} $ are both confusing to me.
The first one, I'm not sure why the scaling factor A is given by that formula.
The second one I'm not sure why or how the equality holds.
Any help?
Let me break it down a bit more. (Adding a $_\bullet$ after $A$ and $\theta$ as a reminder that they are co-dependent.)
$$\require{enclose}\begin{align}\mathsf E\big[x(t-\theta_\bullet)\,y(t)\big]=&~ \mathsf E\big[A_\bullet\, x^2(t-\theta_\bullet)+m(t)\, x(t-\theta_\bullet)]\big) & \enclose{circle}{1} &~y(t) = A_\bullet\,x(t-\theta_\bullet)-m(t) \\[1ex]=&~ A_\bullet\,\mathsf E\big[x^2(t-\theta_\bullet)\big]+\mathsf E\big[m(t)\, x(t-\theta_\bullet)\big] & \enclose{circle}{2} &~\text{linearity of expectation} \\[1ex] =& A_\bullet\,\mathsf E\big[x^2(t-\theta_\bullet)\big] + 0 & \enclose{circle}{3} &~ \text{x,m are uncorrelated} \\[1ex] = & A_\bullet\,\sigma_x^2 & \enclose{circle}{4} &~\text{x has zero mean} \\[3ex]\therefore~ A_\bullet =&~ \frac 1{\sigma_x^2}\,\mathsf E\big[x(t-\theta_\bullet)\,y(t)\big] &\enclose{circle}{5}&~\text{algebraic rearrangment} \\[3ex]S= &~ A_0^2~\mathsf E\big(x^2(t-\theta_0)\big) & \enclose{circle}6 & ~\text{given} \\[1ex] =&~ A_0 \Big(A_0\mathsf E\big[x^2(t-\theta_0)\big]\Big) & \enclose{circle}7 &~ a^2b = a(ab) \\[1ex] =&~ A_0~\mathsf E\big[x(t-\theta_0)\,y(t)\big] & \enclose{circle}8 &~ \text{refer to }\enclose{circle} 3\text{ above} \\[1ex] =&~ \left(\frac 1{\sigma_x^2}~\mathsf E\big[x(t-\theta_0)\,y(t)\big] \right)\,\mathsf E\big[x(t-\theta_0)\,y(t)\big] & \enclose{circle}9 &~ \text{refer to }\enclose{circle} 5\text{ above} \\[1ex]\therefore~ S =&~ \frac 1{\sigma_x^2}~\left(\mathsf E\big[x(t-\theta_0)\,y(t)\big] \right)^2 & \enclose{circle}{10} &~ (ab)b= ab^2 \end{align}$$
Are any steps not clear enough?