Certainly! Here is your solution, rewritten exactly as it is, keeping the formatting and all content unchanged: --- ### **1. Average Power of $X(t)$ Using Autocorrelation** The average power $P_X$ of a random process $X(t)$ is given by: $$ P_X = R_X(0) $$ Given: $$ R_X(\tau) = 2\delta(\tau) \Rightarrow P_X = R_X(0) = 2\delta(0) $$ Since $\delta(0) = \infty$, we have: $$ \boxed{P_X = \infty} $$ This indicates that $X(t)$ is **white noise**, which has infinite power. --- ### **2. Power Spectral Density (PSD) of $X(t)$** The PSD is the Fourier transform of the autocorrelation function: $$ S_X(\omega) = \mathcal{F}\{R_X(\tau)\} = \mathcal{F}\{2\delta(\tau)\} = 2 $$ $$ \boxed{S_X(\omega) = 2} $$ This constant PSD confirms that $X(t)$ is a **white noise process**. --- ### **3. Average Power of $X(t)$ from PSD** $$ P_X = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_X(\omega) \, d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} 2 \, d\omega = \infty $$ $$ \boxed{P_X = \infty} $$ --- ### **4. PSD of Output $Y(t)$** Given frequency response: $$ H(\omega) = 3\operatorname{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 3\operatorname{rect}\left(\frac{\omega + 20\pi}{4\pi}\right) $$ Then: $$ |H(\omega)|^2 = 9\operatorname{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 9\operatorname{rect}\left(\frac{\omega + 20\pi}{4\pi}\right) $$ So the output PSD is: $$ S_Y(\omega) = S_X(\omega) \cdot |H(\omega)|^2 = 2 \cdot \left[9\operatorname{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 9\operatorname{rect}\left(\frac{\omega + 20\pi}{4\pi}\right)\right] $$ $$ \boxed{S_Y(\omega) = 18\operatorname{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 18\operatorname{rect}\left(\frac{\omega + 20\pi}{4\pi}\right)} $$ --- ### **5. Average Power of $Y(t)$ from PSD** $$ P_Y = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_Y(\omega) \, d\omega $$ Each rectangular pulse has height $18$ and width $4\pi$, so: $$ \text{Area per pulse} = 18 \cdot 4\pi = 72\pi $$ $$ \text{Total area} = 2 \cdot 72\pi = 144\pi $$ $$ P_Y = \frac{1}{2\pi} \cdot 144\pi = \boxed{72} $$ --- ### **6. Autocorrelation Function of $Y(t)$** Use the inverse Fourier transform of $S_Y(\omega)$: $$ R_Y(\tau) = \mathcal{F}^{-1}\left\{18\operatorname{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 18\operatorname{rect}\left(\frac{\omega + 20\pi}{4\pi}\right)\right\} $$ We use the transform pair: $$ \mathcal{F}^{-1}\left\{A \cdot \operatorname{rect}\left(\frac{\omega - \omega_0}{2W}\right)\right\} = \frac{AW}{\pi} \cdot \text{sinc}\left(\frac{W\tau}{\pi}\right) \cdot e^{j\omega_0\tau} $$ Given $A = 18$, $W = 2\pi$, and $\omega_0 = \pm 20\pi$: $$ R_Y(\tau) = 18 \cdot \frac{\sin(2\pi\tau)}{\pi\tau} \left(e^{j20\pi\tau} + e^{-j20\pi\tau} \right) $$ $$ = \frac{36 \cos(20\pi\tau) \sin(2\pi\tau)}{\pi\tau} $$ $$ \boxed{R_Y(\tau) = \frac{36 \cos(20\pi\tau) \sin(2\pi\tau)}{\pi\tau}} $$ --- ### **7. Average Power of $Y(t)$ from Autocorrelation** $$ P_Y = R_Y(0) = \lim_{\tau \to 0} \frac{36 \cos(20\pi\tau) \sin(2\pi\tau)}{\pi\tau} $$ Using limits: $$ \lim_{\tau \to 0} \cos(20\pi\tau) = 1, \quad \lim_{\tau \to 0} \frac{\sin(2\pi\tau)}{\tau} = 2\pi $$ $$ P_Y = \frac{36}{\pi} \cdot 2\pi = \boxed{72} $$ This matches the power from the PSD, verifying consistency.