Let's go through each part of the step-by-step explanations. --- # 1. **Obtain the probability density function (PDF) of a Gaussian random variable** Given: Mean, \( m = 2 \) - Variance, \( \sigma^2 = 4 \) The PDF of a Gaussian (normal) distribution: \[ f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(x - m)^2}{2\sigma^2}\right) \] Plugging in the values: \[ f_X(x) = \frac{1}{\sqrt{2\pi \times 4}} \exp\left(-\frac{(x - 2)^2}{8}\right) = \frac{1}{2 \sqrt{\pi}} \exp\left(-\frac{(x - 2)^2}{8}\right) \] --- # 2. **Compute \( P( < X < 4) \) using the standard normal table** Transform \( X \) to standard normal \( Z \): \[ Z = \frac{X - m}{\sigma} \] where \( \sigma = \sqrt{4} = 2 \). Calculate the bounds: \[ Z_1 = \frac{ - 2}{2} = -1 \] \[ Z_2 = \frac{4 - 2}{2} = 1 \] So: \[ P( < X < 4) = P(-1 < Z < 1) \] Using standard normal table: \[ P(Z < 1) \approx .8413 \] \[ P(Z < -1) \approx .1587 \] Therefore: \[ P(-1 < Z < 1) = .8413 - .1587 = .6826 \] --- # 3. **Determine \( x \) for which \( P(X < x) = .5 \)** The median of a normal distribution coincides with its mean: \[ x = m = 2 \] since \( P(X < m) = .5 \). **Final answer:** \[ \boxed{ x = 2 } \] --- # 4. **Autocorrelation function of a band-limited white noise process** Given: \[ S_X(f) = \frac{N_}{2}, \quad |f| \leq B \] and the autocorrelation function \( R_X(\tau) \) is 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 and flat over \([-B, B]\): \[ R_X(\tau) = \int_{-B}^{B} \frac{N_}{2} e^{j 2 \pi f \tau} df \] Evaluate the integral: \[ R_X(\tau) = \frac{N_}{2} \int_{-B}^{B} e^{j 2 \pi f \tau} df \] This integral: \[ \int_{-B}^{B} e^{j 2 \pi f \tau} df = \frac{\sin(2 \pi B \tau)}{\pi \tau} \] Thus: \[ \boxed{ R_X(\tau) = \frac{N_}{2} \times \frac{\sin(2 \pi B \tau)}{\pi \tau} } \] --- # 5. **Compute \( R_X() \) and \( R_X(1/2B) \)** - At \( \tau = \): \[ R_X() = \frac{N_}{2} \times \lim_{\tau \to } \frac{\sin(2 \pi B \tau)}{\pi \tau} \] Using \( \lim_{x \to } \frac{\sin x}{x} = 1 \): \[ R_X() = \frac{N_}{2} \times 2 \pi B \times \frac{1}{\pi} = \frac{N_}{2} \times 2 B = N_ B \] - 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 \( \sin(\pi) = \): \[ R_X\left(\frac{1}{2B}\right) = \] --- # **Summary of Results** | Quantity | Final Expression / Value | |---|---| | PDF of \(X\) | \(\displaystyle f_X(x) = \frac{1}{2 \sqrt{\pi}} e^{-\frac{(x - 2)^2}{8}}\) | | \( P( < X < 4) \) | \(\boxed{.6826}\) | | \( x \) for \( P(X < x) = .5 \) | \(\boxed{2}\) | | Autocorrelation \( R_X(\tau) \) | \(\displaystyle R_X(\tau) = \frac{N_}{2} \frac{\sin(2 \pi B \tau)}{\pi \tau} \) | | \( R_X() \) | \(\boxed{N_ B}\) | | \( R_X(1/2B) \) | \(\boxed{}\) | --- Let me know if you'd like further clarification on any step!