# Signal Analysis Problem Given: - **Autocorrelation function:** \( R_s(\tau) = 400 \cdot \mathrm{sinc}(200\tau) \cos(100\pi \tau) \) ## (a) Plot the Power Spectral Density (PSD) of \( s(t) \): \( S(f) \) ### Step 1: Fourier Transform of \( R_s(\tau) \) The PSD \( S(f) \) is the Fourier transform of the autocorrelation function \( R_s(\tau) \): \[ S(f) = \mathcal{F}\{ R_s(\tau) \} \] Given \( R_s(\tau) = 400\, \mathrm{sinc}(200\tau) \cos(100\pi\tau) \): - \( \mathrm{sinc}(a\tau) \) Fourier transforms to \( \frac{1}{|a|} \mathrm{rect}\left( \frac{f}{2a} \right) \) - \( \cos(2\pi f_ \tau) \) in time multiplies to frequency shift in \( f \) domain by \( \pm f_ \) Let’s rewrite: - \( \mathrm{sinc}(200\tau) \) FT \( \rightarrow \frac{1}{200} \mathrm{rect}\left( \frac{f}{400} \right) \) - \( \cos(100\pi\tau) = \cos(2\pi(50)\tau) \) (since \( 100\pi = 2\pi \cdot 50 \)), FT \( \rightarrow \) frequency shift by \( \pm 50 \) Hz Thus, by modulation property: \[ S(f) = 400 \cdot \frac{1}{2} \left[ \frac{1}{200} \mathrm{rect}\left( \frac{f-50}{400} \right) + \frac{1}{200} \mathrm{rect}\left( \frac{f+50}{400} \right) \right] \] \[ = 1 \cdot \mathrm{rect}\left( \frac{f-50}{400} \right) + 1 \cdot \mathrm{rect}\left( \frac{f+50}{400} \right) \] ### Step 2: Description of PSD - Each \( \mathrm{rect}\left( \frac{f-f_}{400} \right) \) is a rectangle of width 400 Hz, centered at \( f_ \). - So, the PSD consists of two rectangles: - Centered at \( +50 \) Hz (from \( -150 \) Hz to \( +250 \) Hz) - Centered at \( -50 \) Hz (from \( -250 \) Hz to \( +150 \) Hz) - Amplitude: 1 ### Step 3: Plot Below is a sketch of the PSD (not actual code, but a description): ```plaintext Amplitude | 1 + _______ _______ | | | | | | | | | | +--------|-------|---------|-------|---------- -250 -150 +150 +250 Frequency (Hz) ``` - Left rectangle: from \( f = -250 \) Hz to \( f = +150 \) Hz, centered at \( -50 \) Hz - Right rectangle: from \( f = -150 \) Hz to \( f = +250 \) Hz, centered at \( +50 \) Hz - Overlap from \( -150 \) Hz to \( +150 \) Hz, where the value is 2 --- ## (b) Find the Total Power of \( s(t) \): \( P_s \) Total power is the area under the PSD: \[ P_s = \int_{-\infty}^{\infty} S(f)\,df \] Since the rectangles overlap, calculate the total area: - Each rectangle: width \( 400 \) Hz, height \( 1 \), area \( 400 \) - Overlap region: \( -150 \) Hz to \( +150 \) Hz, width \( 300 \) Hz, counted twice So, total area: \[ P_s = \text{Area of both rectangles} - \text{Area of overlap} \] \[ = 2 \times 400 - 300 = 800 - 300 = 500 \] **Total Power:** \[ \boxed{500} \] --- ## (c) PSD at Output of Filter \( H(f) \): Filter characteristic: \[ H(f) = \begin{cases} 10 & |f| < 50\,\text{Hz} \\ & \text{elsewhere} \end{cases} \] Output PSD: \[ S_y(f) = |H(f)|^2 \cdot S(f) \] So, for \( |f| < 50\,\text{Hz} \), \( |H(f)|^2 = 100 \), elsewhere . - From part (a), \( S(f) = 2 \) in the overlap region (\( -50 < f < 50 \)), since both rectangles overlap there. ### Output PSD: \[ S_y(f) = \begin{cases} 200 & |f| < 50\,\text{Hz} \\ & \text{elsewhere} \end{cases} \] --- ## (d) Total Output Power \( P_y \): \[ P_y = \int_{-\infty}^{\infty} S_y(f)\,df = \int_{-50}^{50} 200\,df = 200 \times (50 - (-50)) = 200 \times 100 = 20{,}000 \] **Total Output Power:** \[ \boxed{20,\!000} \] --- ## Summary Table | Part | Quantity | Result/Expression | |------|-----------------------|------------------------------------------| | (a) | PSD \( S(f) \) | Two rectangles at \( \pm50 \) Hz, width 400 Hz, height 1 | | (b) | Total Power \( P_s \) | \( 500 \) | | (c) | Output PSD \( S_y(f) \) | \( 200 \) for \( |f| < 50 \) Hz, elsewhere | | (d) | Output Power \( P_y \) | \( 20,\!000 \) | --- ### Visual Summary