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Linear Time-Invariant (LTI) Systems:

A linear time-invariant (LTI) system can be represented by its impulse response (Figure 10.6). More specifically, if $X(t)$ is the input signal to the system, the output, $Y(t)$, can be written as \begin{align} \nonumber Y(t)=\int_{-\infty}^{\infty} h(\alpha)X(t-\alpha) \; d\alpha=\int_{-\infty}^{\infty} X(\alpha)h(t-\alpha) \; d\alpha. \end{align} The above integral is called the convolution of $h$ and $X$, and we write \begin{align} \nonumber Y(t)&=h(t)\ast X(t)=X(t) \ast h(t). \end{align} Note that as the name suggests, the impulse response can be obtained if the input to the system is chosen to be the unit impulse function (delta function) $x(t)=\delta(t)$. For discrete-time systems, the output can be written as (Figure 10.6) \begin{align*} Y(n)=h(n)\ast X(n)&=X(n) \ast h(n)\\ &=\sum_{k=-\infty}^{\infty} h(k)X(n-k)=\sum_{k=-\infty}^{\infty} X(k)h(n-k). \end{align*} The discrete-time unit impulse function is defined as \begin{align} \nonumber \delta(n) = \left\{ \begin{array}{l l} 1 & \quad n=0\\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} For the rest of this chapter, we mainly focus on continuous-time signals.

LTI Systems with Random Inputs:

Consider an LTI system with impulse response $h(t)$. Let $X(t)$ be a WSS random process. If $X(t)$ is the input of the system, then the output, $Y(t)$, is also a random process. More specifically, we can write \begin{align*} Y(t)&=h(t)\ast X(t)\\ &=\int_{-\infty}^{\infty} h(\alpha)X(t-\alpha) \; d\alpha. \end{align*} Here, our goal is to show that $X(t)$ and $Y(t)$ are jointly WSS processes. Let's first start by calculating the mean function of $Y(t)$, $\mu_Y(t)$. We have \begin{align*} \mu_Y(t)=E[Y(t)]&=E\left[\int_{-\infty}^{\infty} h(\alpha)X(t-\alpha) \; d\alpha\right]\\ &=\int_{-\infty}^{\infty} h(\alpha)E[X(t-\alpha)] \; d\alpha\\ &=\int_{-\infty}^{\infty} h(\alpha) \mu_X \; d\alpha\\ &=\mu_X \int_{-\infty}^{\infty} h(\alpha) \; d\alpha. \end{align*} We note that $\mu_Y(t)$ is not a function of $t$, so we can write \begin{align*} \mu_Y(t)=\mu_Y=\mu_X \int_{-\infty}^{\infty} h(\alpha) \; d\alpha. \end{align*} Let's next find the cross-correlation function, $R_{XY}(t_1,t_2)$. We have \begin{align*} R_{XY}(t_1,t_2)=E[X(t_1)Y(t_2)]&=E\left[X(t_1) \int_{-\infty}^{\infty} h(\alpha)X(t_2-\alpha) \; d\alpha\right]\\ &=E\left[ \int_{-\infty}^{\infty} h(\alpha)X(t_1)X(t_2-\alpha) \; d\alpha\right]\\ &=\int_{-\infty}^{\infty} h(\alpha)E[X(t_1)X(t_2-\alpha)] \; d\alpha\\ &=\int_{-\infty}^{\infty} h(\alpha) R_X(t_1,t_2-\alpha) \; d\alpha\\ &=\int_{-\infty}^{\infty} h(\alpha) R_X(t_1-t_2+\alpha) \; d\alpha & (\textrm{since $X(t)$ is WSS}). \end{align*} We note that $R_{XY}(t_1,t_2)$ is only a function of $\tau=t_1-t_2$, so we may write \begin{align*} R_{XY}(\tau)&=\int_{-\infty}^{\infty} h(\alpha) R_X(\tau+\alpha) \; d\alpha\\ &=h(\tau)\ast R_X(-\tau)=h(-\tau) \ast R_X(\tau). \end{align*} Similarly, you can show that \begin{align*} R_{Y}(\tau)=h(\tau) \ast h(-\tau) \ast R_X(\tau). \end{align*} This has been shown in the Solved Problems section. From the above results we conclude that $X(t)$ and $Y(t)$ are jointly WSS. The following theorem summarizes the results.

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Frequency Domain Analysis:

Let's now rewrite the statement of Theorem 10.2 in the frequency domain. Let $H(f)$ be the Fourier transform of $h(t)$, \begin{align*} H(f)=\mathcal{F} \{h(t) \}= \int_{-\infty}^{\infty} h(t) e^{-2 j \pi f t} \; dt. \end{align*} $H(f)$ is called the transfer function of the system. We can rewrite \begin{align*} \mu_Y=\mu_X \int_{-\infty}^{\infty} h(\alpha) \; d\alpha \end{align*} as Since $h(t)$ is assumed to be a real signal, we have \begin{align*} \mathcal{F} \{h(-t) \}= H(-f)=H^{*}(f), \end{align*} where $^{*}$ shows the complex conjugate. By taking the Fourier transform from both sides of $R_{XY}(\tau)= R_X(\tau)\ast h(-\tau) $, we conclude Finally, by taking the Fourier transform from both sides of $R_{Y}(\tau)=h(\tau) \ast h(-\tau) \ast R_X(\tau) $, we conclude \begin{align*} S_{Y}(f)&=S_X(f)H^{*}(f)H(f)\\ &=S_X(f)|H(f)|^2. \end{align*}

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Example
Let $X(t)$ be a zero-mean WSS process with $R_X(\tau)=e^{-|\tau|}$. $X(t)$ is input to an LTI system with \begin{align*} |H(f)|=\left\{ \begin{array}{l l} \sqrt{1+4\pi^2 f^2} & \quad |f| \lt 2 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align*} Let $Y(t)$ be the output.

1. Find $\mu_Y(t)=E[Y(t)]$.
2. Find $R_Y(\tau)$.
3. Find $E[Y(t)^2]$.

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