这是一份uwa西澳大学PHYS1100的成功案例

$$
y_{j}(t)=\sum_{i=1}^{n} \int_{-\infty}^{+\infty} h_{i j}(\tau) x_{i}(t-\tau) d \tau
$$
The cross correlation between outputs $y_{j}(t)$ and $y_{k}(t)$ is defined as follows:
$$
C_{j k}(\tau)=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^{+T} y_{j}(t) y_{k}(t+\tau) d t
$$
From the definition of $C_{j k}(\tau)$ it is seen that:
$$
C_{k j}(\tau)=\left\langle y_{k}(t) y_{j}(t+\tau)\right\rangle
$$
where:
$$
\langle()\rangle=\lim {T \rightarrow \infty} \frac{1}{2 T} \int{-T}^{+T}() d t
$$

PHYS1100 COURSE NOTES ：

The next step is true only under the condition that the process is ergodic. In this case, the last equation can be written as:
$$
\begin{aligned}
G_{j k}(\omega)=& \lim {T \rightarrow \infty} \frac{1}{2 T}\left(\int{-T}^{+T} y_{j}(u) e^{i \omega u} d u\right) \
& \times\left(\int_{-T}^{+T} y_{k}(v) e^{-i \omega v} d v\right)
\end{aligned}
$$
This last relation can then be written:
$$
G_{j k}(\omega)=\lim {T \rightarrow \infty} \frac{\bar{y}{j}^{}(T, \omega) \bar{y}{k}(T, \omega)}{2 T} $$ where: $$ \bar{y}{j}(T, \omega)=\int_{-T}^{+T} y_{j}(t) e^{i \omega t} d t
$$
and $y_{j}^{}$ is the complex conjugate of $y_{j}$.
$$
\bar{y}{k}(T, \omega)=\int{-T}^{+T} y_{k}(t) e^{-i \omega t} d t
$$