Question:

Sure! Here's a clean and concise rewrite of your entire analysis, preserving all essential derivations and explanations: --- ### **1. Average Power of $X(t)$ Using Its Autocorrelation Function** The average power $P_X$ of a random process is given by its autocorrelation function evaluated at zero: $$ P_X = R_X(0) $$ Given: $$ R_X(\tau) = 2\delta(\tau) $$ $$ P_X = 2\delta(0) $$ Since $\delta(0)$ is infinite, the average power is: $$ P_X = \infty $$ This indicates that $X(t)$ is a **white noise process** with infinite power. --- ### **2. Power Spectral Density (PSD) of $X(t)$** The PSD is the Fourier transform of the autocorrelation function: $$ S_X(\omega) = \mathcal{F}\{R_X(\tau)\} = \mathcal{F}\{2\delta(\tau)\} $$ Using the property $\mathcal{F}\{A\delta(t)\} = A$, we get: $$ S_X(\omega) = 2 $$ This flat spectrum confirms that $X(t)$ is white noise. --- ### **3. Average Power of $X(t)$ Using Its PSD** $$ P_X = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_X(\omega) \, d\omega = \frac{1}{2\pi} \int_{-\infty}^{\infty} 2 \, d\omega $$ This integral diverges, confirming: $$ P_X = \infty $$ --- ### **4. PSD of $Y(t)$** For an LTI system, the output PSD is: $$ S_Y(\omega) = S_X(\omega) |H(\omega)|^2 $$ Given: $$ H(\omega) = 3 \, \text{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 3 \, \text{rect}\left(\frac{\omega + 20\pi}{4\pi}\right) $$ Squaring each term (since they don't overlap): $$ |H(\omega)|^2 = 9 \, \text{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 9 \, \text{rect}\left(\frac{\omega + 20\pi}{4\pi}\right) $$ Then: $$ S_Y(\omega) = 2 \cdot |H(\omega)|^2 = 18 \, \text{rect}\left(\frac{\omega - 20\pi}{4\pi}\right) + 18 \, \text{rect}\left(\frac{\omega + 20\pi}{4\pi}\right) $$ --- ### **5. Average Power of $Y(t)$ Using Its 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: $$ \text{Area of one pulse} = 18 \cdot 4\pi = 72\pi $$ $$ \text{Total area} = 2 \cdot 72\pi = 144\pi $$ $$ P_Y = \frac{1}{2\pi}(144\pi) = 72 $$ --- ### **6. Autocorrelation Function of $Y(t)$** $$ R_Y(\tau) = \mathcal{F}^{-1}\{S_Y(\omega)\} $$ Using inverse transform of shifted rect functions: $$ \mathcal{F}^{-1}\left\{18 \, \text{rect}\left(\frac{\omega \pm 20\pi}{4\pi}\right)\right\} = 18 \cdot \frac{\sin(2\pi \tau)}{\pi \tau} e^{\pm j20\pi \tau} $$ So: $$ R_Y(\tau) = \frac{36 \cos(20\pi \tau) \sin(2\pi \tau)}{\pi \tau} $$ --- ### **7. Average Power of $Y(t)$ Using Its Autocorrelation Function** $$ P_Y = R_Y(0) = \lim_{\tau \to 0} \frac{36 \cos(20\pi \tau) \sin(2\pi \tau)}{\pi \tau} $$ Apply known limits: $$ \lim_{\tau \to 0} \cos(20\pi \tau) = 1, \quad \lim_{\tau \to 0} \frac{\sin(2\pi \tau)}{\tau} = 2\pi $$ $$ P_Y = \frac{36}{\pi} \cdot 2\pi = 72 $$ This matches the value found from the PSD. --- rewrite the /human as it is no changes