Let's go through each part of the problem step by step. --- ## 1. Obtain the probability density function (pdf) of a Gaussian random variable with mean \( m = 2 \) and variance \( \sigma^2 = 4 \) The general Gaussian (normal) pdf is: \[ f_X(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left( - \frac{(x - m)^2}{2 \sigma^2} \right) \] Given: - \( m = 2 \) - \( \sigma^2 = 4 \Rightarrow \sigma = 2 \) Substitute into the formula: \[ f_X(x) = \frac{1}{\sqrt{2 \pi \times 4}} \exp \left( - \frac{(x - 2)^2}{2 \times 4} \right) = \frac{1}{\sqrt{8 \pi}} \exp \left( - \frac{(x - 2)^2}{8} \right) \] --- ## 2. Compute \( P( < X < 4) \) This probability is: \[ P( < X < 4) = \Phi \left( \frac{4 - m}{\sigma} \right) - \Phi \left( \frac{ - m}{\sigma} \right) \] where \( \Phi(z) \) is the standard normal cumulative distribution function (CDF). Calculate the standardized variables: \[ z_1 = \frac{4 - 2}{2} = 1 \] \[ z_2 = \frac{ - 2}{2} = -1 \] Using standard normal tables or a calculator: \[ \Phi(1) \approx .8413 \] \[ \Phi(-1) = 1 - \Phi(1) \approx 1 - .8413 = .1587 \] Therefore: \[ P( < X < 4) = .8413 - .1587 = .6826 \] --- ## 3. Determine the value of \( x \) for which \( P(X < x) = .5 \) This is the median of the distribution, which coincides with the mean for a symmetric Gaussian: \[ x = m = 2 \] --- ## 4. Autocorrelation function of a band-limited white noise process Given: \[ S_X(f) = \frac{N_}{2}, \quad |f| \leq B \] The autocorrelation function \( R_X(\tau) \) is related to the PSD via the inverse Fourier transform: \[ R_X(\tau) = \int_{-\infty}^{\infty} S_X(f) e^{j 2 \pi f \tau} df \] Since \( S_X(f) \) is band-limited (rectangular PSD): \[ R_X(\tau) = \int_{-B}^{B} \frac{N_}{2} e^{j 2 \pi f \tau} df \] Compute the integral: \[ R_X(\tau) = \frac{N_}{2} \int_{-B}^{B} e^{j 2 \pi f \tau} df \] This integral evaluates to: \[ R_X(\tau) = \frac{N_}{2} \left[ \frac{e^{j 2 \pi f \tau}}{j 2 \pi \tau} \right]_{-B}^{B} \] which simplifies to: \[ R_X(\tau) = \frac{N_}{2} \times \frac{e^{j 2 \pi B \tau} - e^{-j 2 \pi B \tau}}{j 2 \pi \tau} \] Using Euler's formula: \[ e^{j \theta} - e^{-j \theta} = 2 j \sin \theta \] we get: \[ R_X(\tau) = \frac{N_}{2} \times \frac{2 j \sin(2 \pi B \tau)}{j 2 \pi \tau} = \frac{N_}{2} \times \frac{2 j \sin(2 \pi B \tau)}{j 2 \pi \tau} \] The \( j \) cancels out: \[ R_X(\tau) = \frac{N_}{2} \times \frac{2 \sin(2 \pi B \tau)}{2 \pi \tau} = \frac{N_}{2} \times \frac{\sin(2 \pi B \tau)}{\pi \tau} \] --- ## 5. Compute \( R_X() \) and \( R_X(1/2B) \) ### At \( \tau = \): \[ R_X() = \lim_{\tau \to } \frac{N_}{2} \times \frac{\sin(2 \pi B \tau)}{\pi \tau} \] Using the limit: \[ \lim_{x \to } \frac{\sin x}{x} = 1 \] Set \( x = 2 \pi B \tau \), then: \[ R_X() = \frac{N_}{2} \times \frac{2 \pi B \tau}{\pi \tau} = \frac{N_}{2} \times 2 B = N_ B \] ### At \( \tau = \frac{1}{2B} \): \[ R_X \left( \frac{1}{2B} \right) = \frac{N_}{2} \times \frac{\sin(2 \pi B \times \frac{1}{2B})}{\pi \times \frac{1}{2B}} = \frac{N_}{2} \times \frac{\sin(\pi)}{\pi / (2B)} \] Since \( \sin(\pi) = \): \[ R_X \left( \frac{1}{2B} \right) = \] --- ## 6. Autocorrelation of the system with impulse response \( h(t) = e^{-t} u(t) \) Given: \[ S_X(\omega) = \frac{N_}{2} \] and \[ h(t) = e^{-t} u(t) \] ### Output autocorrelation \( R_y(\tau) \): \[ R_y(\tau) = \frac{N_}{2} \times |H(\omega)|^2 \] where \( H(\omega) \) is the Fourier transform of \( h(t) \): \[ H(\omega) = \int_^{\infty} e^{-t} e^{-j \omega t} dt = \int_^{\infty} e^{-(1 + j \omega) t} dt = \frac{1}{1 + j \omega} \] Therefore: \[ |H(\omega)|^2 = \frac{1}{(1)^2 + \omega^2} = \frac{1}{1 + \omega^2} \] The autocorrelation function: \[ R_y(\tau) = \frac{N_}{2} \times \int_{-\infty}^{\infty} \frac{e^{j \omega \tau}}{1 + \omega^2} \frac{d \omega}{2\pi} \] This is a standard Fourier transform pair, and: \[ R_y(\tau) = \frac{N_}{2} e^{-\left| \tau \right|} \] --- ## **Summary of key results:** - **Gaussian pdf:** \[ f_X(x) = \frac{1}{\sqrt{8 \pi}} \exp \left( - \frac{(x - 2)^2}{8} \right) \] - **Probability \( P( < X < 4) \):** \[ \boxed{.6826} \] - **Median \( x \) where \( P(X < x) = .5 \):** \[ x = 2 \] - **Autocorrelation \( R_X(\tau) \):** \[ R_X(\tau) = \frac{N_}{2} \times \frac{\sin(2 \pi B \tau)}{\pi \tau} \] - **Values:** \[ R_X() = N_ B \] \[ R_X \left( \frac{1}{2B} \right) = \] - **Autocorrelation of the filtered process:** \[ R_y(\tau) = \frac{N_}{2} e^{ - |\tau| } \] --- Let me know if you'd like further clarification!

Let's go through each part of the problem step by step. --- # 1. Probability Density Function of a Gaussian Random Variable The probability density function (pdf) for a Gaussian random variable is given by: \[ f_X(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left( - \frac{(x - m)^2}{2 \sigma^2} \right) \] ### Given: - Mean \( m = 2 \) - Variance \( \sigma^2 = 4 \) (thus \( \sigma = 2 \)) ### Substituting: \[ f_X(x) = \frac{1}{\sqrt{2 \pi \times 4}} \exp\left( - \frac{(x - 2)^2}{2 \times 4} \right) = \frac{1}{\sqrt{8 \pi}} \exp\left( - \frac{(x - 2)^2}{8} \right) \] --- # 2. Compute \( P(0 < X < 4) \) To calculate this probability: \[ P(0 < X < 4) = P(X < 4) - P(X < 0) \] ### Standardize the variables: \[ z_1 = \frac{4 - 2}{2} = 1 \] \[ z_2 = \frac{0 - 2}{2} = -1 \] ### Using the Standard Normal Table: - \( \Phi(1) \approx 0.8413 \) - \( \Phi(-1) \approx 0.1587 \) ### Calculation: \[ P(0 < X < 4) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6826 \] --- # 3. Determine \( x \) for which \( P(X < x) = 0.5 \) The value of \( x \) for which the cumulative probability is 0.5 corresponds to the mean of the distribution: \[ x = m = 2 \] --- # 4. Autocorrelation Function of a Band-Limited White Noise Process ### Given: \[ S_X(f) = \frac{N_0}{2}, \quad |f| < B \] ### Autocorrelation Function: The autocorrelation function \( R_X(\tau) \) is given by the inverse Fourier transform of the power spectral density: \[ R_X(\tau) = \int_{-\infty}^{\infty} S_X(f) e^{j 2 \pi f \tau} df \] ### For Band-Limited White Noise: \[ R_X(\tau) = \int_{-B}^{B} \frac{N_0}{2} e^{j 2 \pi f \tau} df \] ### Evaluating the Integral: \[ R_X(\tau) = \frac{N_0}{2} \left[ \frac{e^{j 2 \pi f \tau}}{j 2 \pi \tau} \right]_{-B}^{B} = \frac{N_0}{2} \times \frac{e^{j 2 \pi B \tau} - e^{-j 2 \pi B \tau}}{j 2 \pi \tau} \] Using Euler's formula: \[ e^{j \theta} - e^{-j \theta} = 2j \sin(\theta) \] ### Resulting Function: \[ R_X(\tau) = \frac{N_0}{2} \times \frac{2 \sin(2 \pi B \tau)}{j 2 \pi \tau} = \frac{N_0}{2} \cdot \frac{\sin(2 \pi B \tau)}{\pi \tau} \] --- # 5. Compute \( R_X(0) \) and \( R_X(1/2B) \) ### At \( \tau = 0 \): Using the limit: \[ R_X(0) = \lim_{\tau \to 0} \frac{N_0}{2} \cdot \frac{\sin(2 \pi B \tau)}{\pi \tau} = \frac{N_0}{2} \cdot 2B = N_0 B \] ### At \( \tau = \frac{1}{2B} \): \[ R_X\left(\frac{1}{2B}\right) = \frac{N_0}{2} \cdot \frac{\sin(2 \pi B \cdot \frac{1}{2B})}{\pi \cdot \frac{1}{2B}} = \frac{N_0}{2} \cdot \frac{\sin(\pi)}{\pi / (2B)} = 0 \] --- # 6. Output Autocorrelation of an LTI System ### Given: - Impulse response: \( h(t) = e^{-t} u(t) \) - Power spectral density: \( S_X(\omega) = \frac{N_0}{2} \) ### Autocorrelation Function: \[ R_Y(\tau) = S_X(\omega) |H(\omega)|^2 \] ### Fourier Transform of \( h(t) \): \[ H(\omega) = \int_0^{\infty} e^{-t} e^{-j \omega t} dt = \frac{1}{1 + j \omega} \] ### Magnitude Squared: \[ |H(\omega)|^2 = \frac{1}{1 + \omega^2} \] ### Final Output Autocorrelation: \[ R_Y(\tau) = \frac{N_0}{2} \cdot \frac{1}{1 + (2 \pi B \tau)^2} \] --- ## Summary of Key Results - **Gaussian pdf**: \( f_X(x) = \frac{1}{\sqrt{8 \pi}} \exp\left( - \frac{(x - 2)^2}{8} \right) \) - **Probability \( P(0 < X < 4) \)**: \( \boxed{0.6826} \) - **Median \( x \) where \( P(X < x) = 0.5 \)**: \( x = 2 \) - **Autocorrelation**: \( R_X(\tau) = \frac{N_0}{2} \cdot \frac{\sin(2 \pi B \tau)}{\pi \tau} \) - **Values**: \( R_X(0) = N_0 B \) \( R_X\left(\frac{1}{2B}\right) = 0 \) - **Output Autocorrelation**: \( R_Y(\tau) = \frac{N_0}{2} e^{-| \tau |} \) --- Let me know if you need further clarification or additional calculations!