# Step-by-Step Solution ## **Given** - Input spectrum: \[ |X_{in}(f)| = \frac{1}{\sqrt{1 + (.1f)^4}} \] Where \( f \) in kHz. - **Preserve**: \(\)–\(4\) kHz components. - **Aliasing attenuation required**: at least **75 dB**. - **No anti-aliasing filter**: Find minimum sample rate (round up). - **With a 5th order Butterworth LPF, cutoff \( f_c = 6 \) kHz**: Find reduced sampling rate (round up). --- ## **1. No Anti-Aliasing Filter** ### **A. Where Does the Aliased Spectrum Start?** - **Nyquist Criterion**: To avoid aliasing up to \(f_m\), sample at \(f_s \geq 2f_m\). - **Here**: \(f_m = 4\) kHz. - But, since **attenuation is not infinite**, must ensure spectrum above \(f_s - 4\) kHz is **down by at least 75 dB**. ### **B. Where Must the Spectrum Be Down by 75 dB?** - Aliasing from frequencies above \(f_s - 4\) kHz will fold into \(\)–\(4\) kHz after sampling. - So, require: \[ |X_{in}(f)| \leq 10^{-75/20} = 1.778 \times 10^{-4} \] for \(f \geq f_s - 4\) kHz. ### **C. Solve for Frequency \(f\) Where \( |X_{in}(f)| \leq 1.778 \times 10^{-4} \)** \[ |X_{in}(f)| = \frac{1}{\sqrt{1 + (.1f)^4}} \leq 1.778 \times 10^{-4} \] Invert both sides: \[ \sqrt{1 + (.1f)^4} \geq \frac{1}{1.778 \times 10^{-4}} \approx 5625.4 \] \[ 1 + (.1f)^4 \geq (5625.4)^2 = 31644816 \] \[ (.1f)^4 \geq 31644816 - 1 \approx 31644815 \] \[ .1f \geq (31644815)^{1/4} \] Calculate: - \(\log_{10}(31644815) \approx 7.5\), so \((31644815)^{1/4} \approx 10^{7.5/4} = 10^{1.875} \approx 75\) So, \[ .1f \geq 75 \Rightarrow f \geq 750 \text{ kHz} \] ### **D. Where is This in Relation to Sampling Rate?** Recall, the **first alias** of the original spectrum (starting at \(f_s - 4\) kHz) must be at or above \(750\) kHz: \[ f_s - 4 \geq 750 \implies f_s \geq 754 \text{ kHz} \] **Round up:** \[ \boxed{755\ \text{ kHz}} \] --- ## **2. With a 5th Order Butterworth Lowpass Filter (\(f_c = 6\) kHz)** ### **A. Butterworth Filter Attenuation Formula** \[ A(f) = \frac{1}{\sqrt{1 + \left( \frac{f}{f_c} \right)^{2N}}} \] Set \(N = 5\), \(f_c = 6\) kHz. ### **B. Where Does the Filter Attenuate by at Least 75 dB?** \[ A(f) \leq 1.778 \times 10^{-4} \] \[ \sqrt{1 + (f/6)^{10}} \geq 5625.4 \] \[ 1 + (f/6)^{10} \geq 31644816 \] \[ (f/6)^{10} \geq 31644815 \] \[ f/6 \geq (31644815)^{1/10} \] Take logs: \[ \log_{10}(31644815) \approx 7.5 \] \[ \frac{7.5}{10} = .75 \] \[ 10^{.75} \approx 5.62 \] \[ f \geq 6 \times 5.62 = 33.72\ \text{kHz} \] ### **C. Relate to Sampling Rate** Aliased spectrum starts at \(f_s - 4\) kHz: \[ f_s - 4 \geq 33.72 \implies f_s \geq 37.72\ \text{kHz} \] **Round up:** \[ \boxed{38\ \text{ kHz}} \] --- ## **Final Answers** | Condition | Minimum Sampling Rate (kHz) | |------------------------------------------------|-----------------------------| | No anti-aliasing filter | **755 kHz** | | 5th order Butterworth LPF (\(f_c = 6\) kHz) | **38 kHz** | --- ## **Summary Table** | Scenario | Minimum Sampling Rate (rounded up) | |----------------------------------|:----------------------------------:| | **No filter** | **755 kHz** | | **5th order Butterworth, 6 kHz** | **38 kHz** | --- ## **Explanation Recap** - **No filter**: Must wait until input spectrum itself is below \(-75\) dB, which occurs at 750 kHz, so sample at 755 kHz. - **With filter**: 5th-order Butterworth filter gives sufficient attenuation at 33.7 kHz, so sample at 38 kHz. --- ## **Boxed Final Answers** - **Minimum sample rate without anti-aliasing filter:** \(\boxed{755\ \text{ kHz}}\) - **Minimum sample rate with 5th-order Butterworth LPF (\(f_c=6\) kHz):** \(\boxed{38\ \text{ kHz}}\)