# Step-by-Step Explanation of Each Part --- ## 1) Gaussian PDF (mean \( m=2 \), variance \( \sigma^2=4 \)) Given: - Mean \( m=2 \) - Variance \( \sigma^2=4 \) (so \( \sigma = \sqrt{4} = 2 \)) Probability density function: \[ f_X(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp\left( -\frac{(x - m)^2}{2\sigma^2} \right) = \frac{1}{2 \sqrt{2\pi}} \exp \left( -\frac{(x - 2)^2}{8} \right). \] --- ## 2) \( P( < X < 4) \) using Standard Normal Table Standardize: \[ Z = \frac{X - m}{\sigma} = \frac{X - 2}{2}. \] Endpoints: \[ \frac{ - 2}{2} = -1, \quad \frac{4 - 2}{2} = 1. \] Compute probability: \[ P( < X < 4) = P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 2\Phi(1) - 1. \] Using \( \Phi(1) \approx .84134 \): \[ P( < X < 4) \approx 2 \times .84134 - 1 = 1.68268 - 1 = .68268 \approx .6827. \] *Note:* This matches the common "within one standard deviation" probability (~68.27%). --- ## 3) Value \( x \) for which \( P(X < x) = .5 \) For a Gaussian distribution, the median equals the mean: \[ x = m = 2. \] --- ## 4) Autocorrelation of Band-Limited White Noise with Rectangular PSD Given PSD: \[ S_X(f) = \frac{N_}{2} \quad \text{for } |f| \leq B, \quad \text{and } \text{ otherwise}. \] Autocorrelation: \[ R_X(\tau) = \int_{-\infty}^{\infty} S_X(f) e^{j2\pi f \tau} df = \frac{N_}{2} \int_{-B}^{B} e^{j2\pi f \tau} df. \] Evaluate integral: \[ \int_{-B}^{B} e^{j2\pi f \tau} df = \frac{e^{j2\pi B \tau} - e^{-j2\pi B \tau}}{j 2\pi \tau} = \frac{\sin(2\pi B \tau)}{\pi \tau}. \] Thus: \[ R_X(\tau) = \frac{N_}{2} \cdot \frac{\sin(2\pi B \tau)}{\pi \tau} = N_ B \cdot \mathrm{sinc}(2 B \tau), \] where \( \mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \). **Note:** Some sources write the result as \( \frac{N_}{2} \frac{\sin(2\pi B \tau)}{2 \pi B \tau} \), but the above derivation is correct for \( S_X(f) = \frac{N_}{2} \) on \( |f| \leq B \). --- ## 5) Numerical: \( R_X() \) and \( R_X(1/(2B)) \) with \( B=100\, \mathrm{Hz} \), \( N_=1 \) Using: \[ R_X(\tau) = N_ B \cdot \mathrm{sinc}(2 B \tau). \] - At \( \tau= \): \[ R_X() = N_ B \times \mathrm{sinc}() = N_ B \times 1 = 1 \times 100 = \boxed{100}. \] - At \( \tau = \frac{1}{2B} \): \[ 2 B \tau = 2 \times 100 \times \frac{1}{2 \times 100} = 1, \] \[ \sin(\pi \times 1) = , \] so \[ R_X\left(\frac{1}{2B}\right) = . \] **Results:** \[ R_X() = 100, \quad R_X(1/(2B))=. \] --- ## 6) Output Autocorrelation of an LTI System with \( h(t)= e^{-t} u(t) \) driven by white noise \( S_X(\omega)= N_/2 \) Given \( h(t) = e^{-t} u(t) \) (causal exponential), the autocorrelation of the output: \[ R_Y(\tau) = \frac{N_}{2} \int_{-\infty}^{\infty} h(u) h(u + \tau) du. \] Since \( h(t) = \) for \( t < \), the integral reduces to: \[ R_Y(\tau) = \frac{N_}{2} \int_{}^{\infty} e^{-u} e^{-(u + \tau)} \mathbf{1}_{u + \tau \ge } du. \] Evaluate separately: - For \( \tau \ge \): \[ u + \tau \ge \quad \text{for all } u \ge , \] so: \[ R_Y(\tau) = \frac{N_}{2} \int_^\infty e^{-2u - \tau} du = e^{-\tau} \frac{N_}{2} \int_^\infty e^{-2u} du = e^{-\tau} \frac{N_}{2} \times \frac{1}{2} = \frac{N_}{4} e^{-\tau}. \] - For \( \tau < \): \[ u \ge -\tau, \] so: \[ R_Y(\tau) = \frac{N_}{2} \int_{-\tau}^\infty e^{-2u - \tau} du = e^{-\tau} \frac{N_}{2} \int_{-\tau}^\infty e^{-2u} du = e^{-\tau} \frac{N_}{2} \times \frac{e^{2 \tau}}{2} = \frac{N_}{4} e^{\tau}. \] Combining both: \[ r_h(\tau) = \frac{1}{2} e^{-|\tau|}. \] Therefore: \[ \boxed{ R_Y(\tau) = \frac{N_}{2} \times r_h(\tau) = \frac{N_}{2} \times \frac{1}{2} e^{-|\tau|} = \frac{N_}{4} e^{-|\tau|}. } \] --- **Summary:** - Gaussian PDF: \( f_X(x) = \frac{1}{2 \sqrt{2\pi}} \exp \left( -\frac{(x - 2)^2}{8} \right) \). - Probability \( P( < X < 4) \approx .6827 \). - Median \( X \) value: \( x=2 \). - Autocorrelation of band-limited noise: \( R_X(\tau) = N_ B \cdot \mathrm{sinc}(2 B \tau) \). - At \( B=100\, \mathrm{Hz} \), \( R_X() = 100 \), \( R_X(1/(2B))= \). - Autocorrelation of output of an exponential LTI system driven by white noise: \( R_Y(\tau)= \frac{N_}{4} e^{-|\tau|} \).

# Step-by-Step Explanation of Each Part --- ## 1) Gaussian PDF (mean \( m=2 \), variance \( \sigma^2=4 \)) Given: - Mean \( m=2 \) - Variance \( \sigma^2=4 \) (so \( \sigma = \sqrt{4} = 2 \)) Probability density function: \[ f_X(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp\left( -\frac{(x - m)^2}{2\sigma^2} \right) = \frac{1}{2 \sqrt{2\pi}} \exp \left( -\frac{(x - 2)^2}{8} \right). \] --- ## 2) \( P(0 < X < 4) \) using Standard Normal Table Standardize: \[ Z = \frac{X - m}{\sigma} = \frac{X - 2}{2}. \] Endpoints: \[ \frac{0 - 2}{2} = -1, \quad \frac{4 - 2}{2} = 1. \] Compute probability: \[ P(0 < X < 4) = P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 2\Phi(1) - 1. \] Using \( \Phi(1) \approx 0.84134 \): \[ P(0 < X < 4) \approx 2 \times 0.84134 - 1 = 0.68268 \approx 0.6827. \] *Note:* This matches the common "within one standard deviation" probability (~68.27%). --- ## 3) Value \( x \) for which \( P(X < x) = 0.5 \) For a Gaussian distribution, the median equals the mean: \[ x = m = 2. \] --- ## 4) Autocorrelation of Band-Limited White Noise with Rectangular PSD Given PSD: \[ S_X(f) = \frac{N_0}{2} \quad \text{for } |f| \leq B, \quad \text{and } 0 \text{ otherwise}. \] Autocorrelation: \[ R_X(\tau) = \int_{-\infty}^{\infty} S_X(f) e^{j2\pi f \tau} df = \frac{N_0}{2} \int_{-B}^{B} e^{j2\pi f \tau} df. \] Evaluate integral: \[ \int_{-B}^{B} e^{j2\pi f \tau} df = \frac{e^{j2\pi B \tau} - e^{-j2\pi B \tau}}{j 2\pi \tau} = \frac{\sin(2\pi B \tau)}{\pi \tau}. \] Thus: \[ R_X(\tau) = \frac{N_0}{2} \cdot \frac{\sin(2\pi B \tau)}{\pi \tau} = N_0 B \cdot \mathrm{sinc}(2B\tau), \] where \( \mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \). **Note:** Some sources write the result as \( \frac{N_0}{2} \frac{\sin(2\pi B \tau)}{2 \pi B \tau} \), but the above derivation is correct for \( S_X(f) = \frac{N_0}{2} \) on \( |f| \leq B \). --- ## 5) Numerical: \( R_X(0) \) and \( R_X(1/(2B)) \) with \( B=1000 \, \mathrm{Hz}, N_0=1 \) Using: \[ R_X(\tau) = N_0 B \cdot \mathrm{sinc}(2B\tau). \] - At \( \tau=0 \): \[ R_X(0) = N_0 B \cdot \mathrm{sinc}(0) = N_0 B \cdot 1 = 1 \times 1000 = \boxed{1000}. \] - At \( \tau = \frac{1}{2B} \): \[ 2\pi B \tau = 2\pi B \cdot \frac{1}{2B} = \pi \quad \Rightarrow \quad \sin(\pi) = 0, \] so \[ R_X\left(\frac{1}{2B}\right) = 0. \] **Results:** \[ R_X(0) = 1000, \quad R_X\left(\frac{1}{2B}\right) = 0. \] --- ## 6) Output Autocorrelation of an LTI System with \( h(t)= e^{-t} u(t) \) driven by white noise \( S_X(\omega)= \frac{N_0}{2} \) Given \( h(t) = e^{-t} u(t) \) (causal exponential), the autocorrelation of the output: \[ R_Y(\tau) = \frac{N_0}{2} \int_{-\infty}^{\infty} h(u) h(u + \tau) du. \] Since \( h(t) = 0 \) for \( t < 0 \), the integral reduces to: \[ R_Y(\tau) = \frac{N_0}{2} \int_0^{\infty} e^{-u} e^{-(u + \tau)} \mathbf{1}_{u + \tau \ge 0} du. \] Evaluate separately: - For \( \tau \ge 0 \): \[ R_Y(\tau) = \frac{N_0}{2} \int_0^{\infty} e^{-2u - \tau} du = e^{-\tau} \frac{N_0}{2} \int_0^{\infty} e^{-2u} du = e^{-\tau} \frac{N_0}{2} \cdot \frac{1}{2} = \frac{N_0}{4} e^{-\tau}. \] - For \( \tau < 0 \): \[ R_Y(\tau) = \frac{N_0}{2} \int_{-\tau}^{\infty} e^{-2u - \tau} du = e^{-\tau} \frac{N_0}{2} \int_{-\tau}^{\infty} e^{-2u} du = e^{-\tau} \frac{N_0}{2} \cdot \frac{e^{2\tau}}{2} = \frac{N_0}{4} e^{\tau}. \] Combining both cases: \[ R_Y(\tau) = \frac{N_0}{4} e^{-| \tau |}. \] --- **Summary:** - Gaussian PDF: \( f_X(x) = \frac{1}{2 \sqrt{2\pi}} \exp \left( -\frac{(x - 2)^2}{8} \right) \). - Probability \( P(0 < X < 4) \approx 0.6827 \). - Median \( X \) value: \( x=2 \). - Autocorrelation of band-limited noise: \( R_X(\tau) = N_0 B \cdot \mathrm{sinc}(2 B \tau) \). - At \( B=1000 \, \mathrm{Hz} \), \( R_X(0) = 1000 \), \( R_X(1/(2B))=0 \). - Autocorrelation of output of an exponential LTI system driven by white noise: \( R_Y(\tau)= \frac{N_0}{4} e^{-| \tau |} \).