To solve this problem, we'll address each part step-by-step. ### a) Derive the expression for the output power spectral density. The input process \( X(t) \) has a power spectral density (PSD) \( S_X(\omega) \). For a white noise process, \( S_X(\omega) = \sigma_X^2 \). The output power spectral density \( S_Y(\omega) \) is given by: \[ S_Y(\omega) = |H(\omega)|^2 S_X(\omega) \] For an ideal low-pass filter with bandwidth \( B = 10 \, \text{rad/s} \), the frequency response \( H(\omega) \) is: \[ H(\omega) = \begin{cases} 1, & |\omega| \leq B \\ , & |\omega| > B \end{cases} \] Thus, \[ S_Y(\omega) = \begin{cases} \sigma_X^2, & |\omega| \leq 10 \\ , & |\omega| > 10 \end{cases} \] ### b) Compute the output variance \( \sigma_Y^2 \). The output variance \( \sigma_Y^2 \) is the integral of the output PSD over all frequencies: \[ \sigma_Y^2 = \int_{-\infty}^{\infty} S_Y(\omega) \, d\omega \] Given the expression for \( S_Y(\omega) \): \[ \sigma_Y^2 = \int_{-10}^{10} \sigma_X^2 \, d\omega = \sigma_X^2 \times (10 - (-10)) = \sigma_X^2 \times 20 \] Substituting \( \sigma_X^2 = 3 \): \[ \sigma_Y^2 = 3 \times 20 = 60 \] ### c) If the input mean is , determine the mean of the output. An ideal filter does not change the mean of a stationary process. Therefore, if the input mean is , the output mean will also be . ### d) Discuss the effect of the filter on Gaussianity and stationarity of the process. - **Gaussianity:** The Gaussian nature of the process is preserved through linear filtering. Therefore, the output process \( Y(t) \) remains Gaussian. - **Stationarity:** An ideal low-pass filter applied to a stationary process results in an output that is also stationary. Thus, the output process \( Y(t) \) is stationary. ### Final Answers: a) \( S_Y(\omega) = \begin{cases} 3, & |\omega| \leq 10 \\ , & |\omega| > 10 \end{cases} \) b) \( \sigma_Y^2 = 60 \) c) Mean of output = d) The filter preserves both Gaussianity and stationarity of the process.

# Step-by-Step Solution We will solve the problem regarding the stationary Gaussian random process \( X(t) \) and the ideal low-pass filter. ## Given Data - Variance of the input process: \( \sigma_X^2 = 3 \) - Bandwidth of the filter: \( B = 10 \, \text{rad/s} \) ## a) Derive the Expression for the Output Power Spectral Density The output power spectral density \( S_Y(\omega) \) can be derived from the input power spectral density \( S_X(\omega) \) and the frequency response of the filter \( H(\omega) \). ### Power Spectral Density of Input For a stationary Gaussian random process with white noise: \[ S_X(\omega) = \sigma_X^2 = 3 \] ### Frequency Response of the Filter For an ideal low-pass filter: \[ H(\omega) = \begin{cases} 1, & |\omega| \leq B \\ 0, & |\omega| > B \end{cases} \] ### Output Power Spectral Density Thus, the output power spectral density is: \[ S_Y(\omega) = |H(\omega)|^2 S_X(\omega) = \begin{cases} 3, & |\omega| \leq 10 \\ 0, & |\omega| > 10 \end{cases} \] ## b) Compute the Output Variance \( \sigma_Y^2 \) The output variance \( \sigma_Y^2 \) is calculated by integrating the output PSD over all frequencies: \[ \sigma_Y^2 = \int_{-\infty}^{\infty} S_Y(\omega) \, d\omega \] Substituting the expression for \( S_Y(\omega) \): \[ \sigma_Y^2 = \int_{-10}^{10} 3 \, d\omega = 3 \times (10 - (-10)) = 3 \times 20 = 60 \] ## c) Determine the Mean of the Output If the input mean is \( \mu_X = 0 \), the mean of the output \( \mu_Y \) remains unchanged because an ideal filter does not affect the mean of a stationary process: \[ \mu_Y = \mu_X = 0 \] ## d) Discuss the Effect of the Filter on Gaussianity and Stationarity - **Gaussianity:** - The output process \( Y(t) \) remains Gaussian. Linear filtering of a Gaussian process results in another Gaussian process. - **Stationarity:** - The output process \( Y(t) \) is also stationary. An ideal low-pass filter does not change the stationary nature of the input process. ## Final Answers - **a)** \( S_Y(\omega) = \begin{cases} 3, & |\omega| \leq 10 \\ 0, & |\omega| > 10 \end{cases} \) - **b)** \( \sigma_Y^2 = 60 \) - **c)** Mean of output \( \mu_Y = 0 \) - **d)** The filter preserves both Gaussianity and stationarity of the process.