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I'm having trouble showing the following relation:

The autocorrelation of a LTI system with impulse response $h[n]$, input $x[n]$ and output $y[n]=h[n]*x[n]$ is given as:

$r_{yy} = r_{hh}*r_{xx}$

My approach was the following:

$r_{yy} = E(y[n]y[n+m]) = E((h[n]*x[n])\cdot (h[n+m]*x[n+m])) = E((h[n]h[n+m])*(x[n]x[n+m])*(h[n]x[n+m])*(x[n]*h[n+m])) $

but I don't know how to simplify this further...

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  • $\begingroup$ Try expanding the convolutions as sums (beware to use different summing indices). Then, the note that E and Sum can be interchanged if h is BIBO stable and x has finite power. $\endgroup$ – Juancho Mar 3 at 13:11
  • $\begingroup$ I've come so far now: $r_{yy}[m] = E( (\sum_{k=-\infty}^{\infty}h[k]x[n-k])(\sum_{l=-\infty}^{\infty}h[l]x[n+m-l]) ) = \\ \sum_{k=-\infty}^{\infty} \sum_{l=-\infty}^{\infty} E(h[k]h[l]x[n-k]x[n+m-l]) = \sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty} r_{hh}[l-k] r_{xx}[m-(l-k)]$ but does this equal to the convolution? $\endgroup$ – Phinie Mar 3 at 13:33
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    $\begingroup$ Further hint: Set $l-k$ to equal $t$ and calculate the double sum in diagonal stripes rather than an inner sum on $l$ followed by an outer sum on $k$. $\endgroup$ – Dilip Sarwate Mar 3 at 16:16
  • $\begingroup$ That would be a lot easier in the frequency domain. $R_{yy} = Y \cdot Y^*$ and $Y = H \cdot X$ $\endgroup$ – Hilmar Mar 3 at 19:50
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Actually, we have two kinds of autocorrelation functions. One is defined for stochastic signals and the other for deterministic ones.

If $x(t)$ is a stochastic signal, then its autocorrelation function will be $$R_x(t+\tau,t)=\Bbb E\{x(t+\tau)x^*(t)\}$$where $\Bbb E(\cdot)$ denotes the mathematical expectation.

If x(t) is a deterministic signal, then its autocorrelation function will be $$R_x(t)=x(t)*x^*(-t)=\int_{-\infty}^\infty x(\tau)x^*(t+\tau)d\tau$$ where $\cdot *\cdot$ denotes the convolution operation.

Based on the previous category, we have four possible cases:

(1) Both $x(t)$ and $h(t)$ are deterministic.

In this case, $y(t)$ is deterministic and we obtain $$R_y(t){=\int_{-\infty}^\infty y(\tau)y^*(t+\tau)d\tau\\=\int_{-\infty}^\infty \int_{-\infty}^\infty\int_{-\infty}^\infty x(\tau_1)h(\tau-\tau_1)x^*(\tau_2)h^*(t+\tau-\tau_2)d\tau_1d\tau_2d\tau\\=\int_{-\infty}^\infty \int_{-\infty}^\infty x(\tau_1)x^*(\tau_2)\int_{-\infty}^\infty h(\tau-\tau_1)h^*(t+\tau-\tau_2)d\tau d\tau_1d\tau_2\\=\int_{-\infty}^\infty \int_{-\infty}^\infty x(\tau_1)x^*(\tau_2)\int_{-\infty}^\infty h(\tau)h^*(t+\tau+\tau_1-\tau_2)d\tau d\tau_1d\tau_2\\=\int_{-\infty}^\infty \int_{-\infty}^\infty x(\tau_1)x^*(\tau_2)R_h(t+\tau_1-\tau_2) d\tau_1d\tau_2\\=\int_{-\infty}^\infty \int_{-\infty}^\infty x(\tau_1+\tau_2)x^*(\tau_2)R_h(t+\tau_1) d\tau_1d\tau_2\\=\int_{-\infty}^\infty R_h(t+\tau_1) \int_{-\infty}^\infty x(\tau_1+\tau_2)x^*(\tau_2) d\tau_2 d\tau_1\\=\int_{-\infty}^\infty R_h(t+\tau_1) \int_{-\infty}^\infty x(\tau_2)x^*(\tau_2-\tau_1) d\tau_2 d\tau_1\\=\int_{-\infty}^\infty R_h(t+\tau_1) R_x(-\tau_1) d\tau_1\\=R_x(t)*R_h(t) }$$

(2) $x(t)$ is stochastic and $h(t)$ is deterministic.

In this case, $y(t)$ is stochastic and we obtain: $$R_y(t_1,t_2){=\Bbb E\{y(t_1)y^*(t_2)\} \\=\Bbb E\left\{\int_{-\infty}^\infty\int_{-\infty}^\infty x(\tau_1)h(t_1-\tau_1)x^*(\tau_2)h^*(t_2-\tau_2)d\tau_1d\tau_2\right\} \\=\int_{-\infty}^\infty\int_{-\infty}^\infty \Bbb E\{x(\tau_1)x^*(\tau_2)\}h(t_1-\tau_1)h^*(t_2-\tau_2)d\tau_1d\tau_2 \\=\int_{-\infty}^\infty\int_{-\infty}^\infty R_x(\tau_1,\tau_2)h(t_1-\tau_1)h^*(t_2-\tau_2)d\tau_1d\tau_2 \\=\int_{-\infty}^\infty\left[\int_{-\infty}^\infty R_x(\tau_1,\tau_2)h(t_1-\tau_1)d\tau_1\right]h^*(t_2-\tau_2)d\tau_2 \\=\int_{-\infty}^\infty\left[ R_x(t_1,\tau_2)*_{t_1}h(t_1)\right]h^*(t-\tau_2)d\tau_2 \\=\left[ R_x(t_1,t_2)*_{t_1}h(t_1)\right]*_{t_2}h^*(t_2) }$$where $*_u$ means convolution w.r.t. $u$.

(3) $h(t)$ is stochastic and $x(t)$ is deterministic.

This case is very similar to case (2). Since the system is LTI, we can reverse the impulse response with input signal to obtain $$R_y(t_1,t_2)=\left[ R_h(t_1,t_2)*_{t_1}x(t_1)\right]*_{t_2}x^*(t_2)$$

(4) Both $x(t)$ and $h(t)$ are stochastic.

As the statistical sources of $x(t)$ and $h(t)$ are often independent, it is quite reasonable to assume that $x(t)$ and $h(t)$ are also independent. To use this fact, we start by:$$R_y(t_1,t_2){=\Bbb E\{y(t_1)y^*(t_2)\} \\=\Bbb E\left\{\int_{-\infty}^\infty\int_{-\infty}^\infty x(\tau_1)h(t_1-\tau_1)x^*(\tau_2)h^*(t_2-\tau_2)d\tau_1d\tau_2\right\} \\=\Bbb \int_{-\infty}^\infty\int_{-\infty}^\infty \Bbb E\left\{x(\tau_1)h(t_1-\tau_1)x^*(\tau_2)h^*(t_2-\tau_2)\right\}d\tau_1d\tau_2 \\=\Bbb \int_{-\infty}^\infty\int_{-\infty}^\infty \Bbb E\{x(\tau_1)x^*(\tau_2)\}\Bbb E\{ h(t_1-\tau_1)h^*(t_2-\tau_2)\}d\tau_1d\tau_2 \\=\Bbb \int_{-\infty}^\infty R_x(t_1,\tau_2)*_{t_1}R_h(t_1,t_2-\tau_2)d\tau_2 \\=R_x(t_1,t_2)*_{t_1}*_{t_2}R_h(t_1,t_2) }$$ which means that a dual convolution must be performed: one on $t_1$ and one on $t_2$.

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  • $\begingroup$ Thank you a lot for this answer! $\endgroup$ – Phinie Mar 6 at 15:13
  • $\begingroup$ Your welcome. Good luck!! $\endgroup$ – Mostafa Ayaz Mar 6 at 18:07

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