Let's analyze and solve each part step-by-step. --- ## 1. Obtain the probability density function (PDF) of a Gaussian random variable with mean \( m = \) and \( \sigma2 = 4 \). ### Solution: The PDF of a Gaussian (normal) distribution is: \[ f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left( - \frac{(x - m)^2}{2\sigma^2} \right) \] Given: - \( m = 2 \) - \( \sigma^2 = 4 \Rightarrow \sigma = 2 \) Plug in the values: \[ f_X(x) = \frac{1}{\sqrt{2\pi \times 4}} \exp\left( - \frac{(x - 2)^2}{2 \times 4} \right) = \frac{1}{2 \sqrt{\pi}} \exp \left( - \frac{(x - 2)^2}{8} \right) \] --- ## 2. Compute \( P( < X < 4) \) using the standard normal table. ### Solution: Standardize the bounds: \[ Z = \frac{X - m}{\sigma} \] - For \( X = \): \[ z_1 = \frac{ - 2}{2} = -1 \] - For \( X = 4 \): \[ z_2 = \frac{4 - 2}{2} = 1 \] Using the standard normal table: \[ P( < X < 4) = P(-1 < Z < 1) = \Phi(1) - \Phi(-1) \] From standard normal distribution table: \[ \Phi(1) \approx .8413, \quad \Phi(-1) = 1 - \Phi(1) = .1587 \] Thus, \[ P( < X < 4) = .8413 - .1587 = .6826 \] --- ## 3. Determine \( x \) such that \( P(X < x) = .5 \). ### Solution: Since \( P(X < x) = .5 \), \( x \) is the median of the normal distribution. For a normal distribution, the median equals the mean: \[ x = m = 2 \] --- ## 4. Obtain the autocorrelation function \( R_X(\tau) \) of a band-limited white noise process with PSD: \[ S_X(f) = \frac{N_}{2} \quad \text{for} \ |f| \leq B \] ### Show that: \[ R_X(\tau) = - \frac{N_ \sin(2 \pi B \tau)}{2 \pi B \tau} \] ### Solution: The autocorrelation function is the inverse Fourier transform of the PSD: \[ R_X(\tau) = \int_{-\infty}^{\infty} S_X(f) e^{j 2 \pi f \tau} df \] Since \( S_X(f) \) is constant over \( [-B, B] \): \[ R_X(\tau) = \int_{-B}^{B} \frac{N_}{2} e^{j 2 \pi f \tau} df \] Compute the integral: \[ R_X(\tau) = \frac{N_}{2} \int_{-B}^{B} e^{j 2 \pi f \tau} df \] \[ = \frac{N_}{2} \left[ \frac{e^{j 2 \pi f \tau}}{j 2 \pi \tau} \right]_{f=-B}^{f=B} \] \[ = \frac{N_}{2} \times \frac{1}{j 2 \pi \tau} [ e^{j 2 \pi B \tau} - e^{-j 2 \pi B \tau} ] \] Recall that: \[ e^{j \theta} - e^{-j \theta} = 2j \sin \theta \] Thus: \[ R_X(\tau) = \frac{N_}{2} \times \frac{1}{j 2 \pi \tau} \times 2j \sin(2 \pi B \tau) \] Simplify: \[ R_X(\tau) = \frac{N_}{2} \times \frac{2j \sin(2 \pi B \tau)}{j 2 \pi \tau} \] \[ = \frac{N_}{2} \times \frac{2j \sin(2 \pi B \tau)}{j 2 \pi \tau} \] Cancel \( j \) and 2: \[ R_X(\tau) = \frac{N_}{2} \times \frac{\sin(2 \pi B \tau)}{\pi \tau} \] Since the original integral is symmetric, the autocorrelation function is real and given by: \[ \boxed{ R_X(\tau) = N_ \frac{\sin(2 \pi B \tau)}{2 \pi \tau} } \] Note the negative sign in the original problem statement appears to be a typo or extra; the standard form is as above, positive. --- ## 5. Compute \( R_X() \) and \( R_X(1/(2B)) \) for \( B = 100\, \text{Hz} \), \( N_ = 1 \). ### Solution: - \( R_X() \): \[ R_X() = N_ \times \frac{\sin()}{2 \pi \times } \] which is indeterminate but using the limit: \[ \lim_{\tau \to } \frac{\sin(2 \pi B \tau)}{2 \pi \tau} = B \] Thus, \[ R_X() = N_ \times B = 1 \times 100 = 100 \] - \( R_X(1/(2B)) \): \[ \tau = \frac{1}{2B} = \frac{1}{2 \times 100} = .0005\, \text{s} \] Calculate numerator: \[ \sin(2 \pi B \tau) = \sin(2 \pi \times 100 \times .0005) = \sin(\pi) = \] Therefore, \[ R_X(1/(2B)) = N_ \times \frac{}{2 \pi \times .0005} = \] --- # **Summary of Final Results:** | Part | Result | |-------|---------| | 1. PDF | \(\displaystyle f_X(x) = \frac{1}{2 \sqrt{\pi}} \exp \left( - \frac{(x - 2)^2}{8} \right)\) | | 2. \(P( < X < 4)\) | \(\boxed{.6826}\) | | 3. Median \(x\) | \(\boxed{2}\) | | 4. Autocorrelation \( R_X(\tau) \) | \(\displaystyle R_X(\tau) = N_ \frac{\sin(2 \pi B \tau)}{2 \pi \tau}\) | | 5a. \( R_X() \) | \(\boxed{100}\) | | 5b. \( R_X(1/(2B)) \) | \(\boxed{}\) | Let me know if you'd like further clarification!