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10.2.5 Solved Problems

Problem

Consider a WSS random process $X(t)$ with \begin{align} \nonumber R_X(\tau) =\left\{ \begin{array}{l l} 1-|\tau| & \quad -1 \leq \tau \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} Find the PSD of $X(t)$, and $E[X(t)^2]$.

  • Solution
    • First, we have \begin{align*} E[X(t)^2] &=R_X(0)=1. \end{align*} We can write triangular function, $R_X(\tau)=\Lambda(\tau)$, as \begin{equation*} R_X(\tau)=\Pi(\tau) \ast \Pi(\tau), \end{equation*} where \begin{align} \nonumber \Pi(\tau) =\left\{ \begin{array}{l l} 1 & \quad -\frac{1}{2} \leq \tau \leq \frac{1}{2} \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} Thus, we conclude \begin{align*} S_X(f) &=\mathcal{F} \{R_X(\tau)\} \\ &=\mathcal{F} \{\Pi(\tau) \ast \Pi(\tau)\}\\ &=\mathcal{F} \{\Pi(\tau)\} \cdot \mathcal{F} \{\Pi(\tau)\}\\ &=\big[\textrm{sinc} (f)\big]^2. \end{align*}

Problem

Let $X(t)$ be a random process with mean function $\mu_X(t)$ and autocorrelation function $R_X(s,t)$ ($X(t)$ is not necessarily a WSS process). Let $Y(t)$ be given by \begin{align*} Y(t)&=h(t)\ast X(t), \end{align*} where $h(t)$ is the impulse response of the system. Show that

  1. $\mu_Y(t)=\mu_X(t) \ast h(t)$.
  2. $R_{XY}(t_1,t_2)=h(t_2) \ast R_X(t_1,t_2)=\int_{-\infty}^{\infty} h(\alpha) R_X(t_1,t_2-\alpha) \; d\alpha$.

  • Solution
      1. 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(t-\alpha) \; d\alpha\\ &=\mu_X(t) \ast h(t). \end{align*}
      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. \end{align*}

Problem

Prove the third part of Theorem 10.2: Let $X(t)$ be a WSS random process and $Y(t)$ be given by \begin{align*} Y(t)&=h(t)\ast X(t), \end{align*} where $h(t)$ is the impulse response of the system. Show that \begin{align*} R_{Y}(s,t)=R_Y(s-t)&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\alpha) h(\beta) R_X(s-t-\alpha+\beta) \; d\alpha d\beta. \end{align*} Also, show that we can rewrite the above integral as $R_{Y}(\tau)=h(\tau) \ast h(-\tau) \ast R_X(\tau)$.

  • Solution
    • \begin{align*} R_{Y}(s,t)&=E[Y(s)Y(t)]\\ &=E\left[ \int_{-\infty}^{\infty} h(\alpha)X(s-\alpha) \; d\alpha \int_{-\infty}^{\infty} h(\beta)X(t-\beta) \; d\beta\right]\\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\alpha) h(\beta) E[X(s-\alpha)X(t-\beta)] \; d\alpha \; d\beta\\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\alpha) h(\beta) R_X(s-t-\alpha+\beta) \; d\alpha \; d\beta. \end{align*} We now compute $h(\tau) \ast h(-\tau) \ast R_X(\tau)$. First, let $g(\tau)=h(\tau) \ast h(-\tau)$. Note that \begin{align*} g(\tau)&=h(\tau) \ast h(-\tau)\\ &= \int_{-\infty}^{\infty} h(\alpha)h(\alpha-\tau) \; d\alpha. \end{align*} Thus, we have \begin{align*} g(\tau) \ast R_X(\tau)&= \int_{-\infty}^{\infty} g(\theta) R_X(\theta-\tau) \; d\theta\\ &=\int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} h(\alpha)h(\alpha-\theta) \; d\alpha \right] R_X(\theta-\tau) \; d\theta\\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\alpha)h(\alpha-\theta) R_X(\theta-\tau) \; d\alpha \; d\theta\\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\alpha)h(\beta) R_X(\alpha-\beta-\tau) \; d\alpha \; d\beta\\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(\alpha)h(\beta) R_X(\tau-\alpha+\beta) \; d\alpha \; d\beta &\big(\textrm{since } R_X(-\tau)=R_X(\tau)\big). \end{align*}

Problem

Let $X(t)$ be a WSS random process. Assuming that $S_X(f)$ is continuous at $f_1$, show that $S_X(f_1) \geq 0$.

  • Solution
    • Let $f_1 \in \mathbb{R}$. Suppose that $X(t)$ goes through an LTI system with the following transfer function \begin{align*} H(f)=\left\{ \begin{array}{l l} 1 & \quad f_1 \lt |f| \lt f_1+\Delta \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align*} where $\Delta$ is chosen to be very small. The PSD of $Y(t)$ is given by \begin{align*} S_Y(f)=S_X(f)|H(f)|^2=\left\{ \begin{array}{l l} S_X(f) & \quad f_1 \lt |f| \lt f_1+\Delta \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align*} Thus, the power in $Y(t)$ is \begin{align*} E[Y(t)^2] & =\int_{-\infty}^{\infty} S_Y(f) \; df \\ &=2\int_{f_1}^{f_1+\Delta} S_X(f) \; df \\ &\approx 2 \Delta S_X(f_1). \end{align*} Since $E[Y(t)^2] \geq 0$, we conclude that $S_X(f_1) \geq 0$.

Problem

Let $X(t)$ be a white Gaussian noise with $S_X(f)=\frac{N_0}{2}$. Assume that $X(t)$ is input to an LTI system with \begin{align*} h(t)=e^{-t}u(t). \end{align*} Let $Y(t)$ be the output.

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

  • Solution
    • First, note that \begin{align*} H(f)&=\mathcal{F} \{h(t)\}\\ &=\frac{1}{1+j2 \pi f}. \end{align*}
      1. To find $S_Y(f)$, we can write \begin{align*} S_Y(f) &=S_X(f)|H(f)|^2 \\ &=\frac{N_0/2}{1+(2 \pi f)^2}. \end{align*}
      2. To find $R_Y(\tau)$, we can write \begin{align*} R_Y(\tau)&=\mathcal{F}^{-1}\{S_Y(f)\} \\ &=\frac{N_0}{4}e^{-|\tau|}. \end{align*}
      3. We have \begin{align*} E[Y(t)^2]&=R_Y(0) \\ &=\frac{N_0}{4}. \end{align*}

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