Let's break down and solve the problem step-by-step. --- # **Step-by-Step Solution** ## **1. Understanding the Signal** - **d(t)**: Triangular pulse centered at the origin. - **Bandwidth:** \( 2\omega_d \) - **Sampled at Nyquist rate.** --- ## **2. Drawing \( d(t) \) and Sampled Signal \( d_s(t) \)** ### **a. \( d(t) \): Triangular Pulse** - Time domain: Looks like a triangle centered at \( t= \). ### **b. Frequency Domain of \( d(t) \)** - Its spectrum is nonzero up to \( 2\omega_d \). ### **c. Sampling at Nyquist Rate** - **Nyquist Rate:** \( 2 \times (\text{bandwidth}) = 2 \times 2\omega_d = 4\omega_d \) - **Nyquist Frequency:** \( 2\omega_d \) (since highest frequency is \( 2\omega_d \)) ### **d. Sampled Signal \( d_s(t) \)** - In time: \( d(t) \) sampled at intervals \( T = \frac{\pi}{\omega_d} \) - In frequency: Replicas of the spectrum every \( 2\omega_d \) --- ## **3. Filtering the Sampled Signal** - Filtering with ideal lowpass filter \( h_p(t) \) with cutoff \( \omega_c \). --- ## **4. Cases for Filtered Signals** ### **Case 1: \( 2.5\omega_d = \omega_c \)** - **Cutoff is wider than the signal bandwidth** (\( \omega_c > 2\omega_d \)): - The original spectrum (up to \( 2\omega_d \)) passes through. - No aliasing, so original signal can be **perfectly recovered**. - **Sketch**: Filter passes the whole original signal and a bit more (extra frequencies pass, but no extra content). ### **Case 2: \( \omega_d = 2\omega_c \implies \omega_c = .5\omega_d \)** - **Cutoff is much less than the signal bandwidth** (\( \omega_c < 2\omega_d \)): - Only part of the spectrum (\( |\omega| < .5\omega_d \)) passes. - **Most of the original signal is lost.** - **Sketch**: Only a small central portion of the original spectrum passes through. --- ## **5. Comments on Signal Recovery** - **Case 1:** The filter passes the entire spectrum of the original triangular pulse. The original signal can be **recovered perfectly**. - **Case 2:** The filter only passes a small fraction of the signal's spectrum. **Most of the original signal is lost**, and only a heavily lowpass-filtered version (smoothed) remains. --- # **Final Answers** ### **Nyquist Rate and Frequency** - **Nyquist Rate:** \( 4\omega_d \) - **Nyquist Frequency:** \( 2\omega_d \) --- ### **Sketches (Description)** #### **Time Domain** - \( d(t) \): Triangular pulse centered at . - \( d_s(t) \): Series of samples (impulses) at intervals \( T \). #### **Frequency Domain** - \( D(\omega) \): Nonzero for \( |\omega| < 2\omega_d \), zero otherwise. - Sampled signal: Replicas of \( D(\omega) \) at intervals \( 2\omega_d \). #### **Filtered Signals** - **Case 1 (\( \omega_c = 2.5\omega_d \))**: Filter passes the entire original spectrum (and then some). - **Case 2 (\( \omega_c = .5\omega_d \))**: Filter passes only the central part; most original energy lost. --- ### **Signal Recovery** - **Case 1:** **Perfect recovery** of \( d(t) \). - **Case 2:** **Severe loss** of information; cannot recover \( d(t) \). --- ## **Summary Table** | Case | Filter Cutoff (\( \omega_c \)) | Recovery of \( d(t) \) | |------|-------------------------------|------------------------| | 1 | \( 2.5\omega_d \) | Perfect | | 2 | \( .5\omega_d \) | Not possible | --- ## **Illustrative Sketch** ``` Frequency axis: |---------------------|---------------------|---------------------| -2ω_d 2ω_d Original spectrum: [=====] Case 1 filter: [=============] Case 2 filter: [==] [=====] = original signal [=============] = filter in case 1 [==] = filter in case 2 ``` --- **In summary:** - To recover a band-limited signal sampled at the Nyquist rate, the reconstruction filter should cover the entire original bandwidth. - If the filter's bandwidth is too narrow, most of the signal's information is lost.