# Autocorrelation of LTI system output

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...

• 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. – Juancho Mar 3 at 13:11
• 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? – Phinie Mar 3 at 13:33
• 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 $k$. – Dilip Sarwate Mar 3 at 16:16
• That would be a lot easier in the frequency domain. $R_{yy} = Y \cdot Y^*$ and $Y = H \cdot X$ – Hilmar Mar 3 at 19:50

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 $$t_2$$.

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