Let's break down and solve the DSP question step by step: --- ### **Given:** - **Sampling rate (\( f_s \))** = 1 MHz = 1000 kHz - **Butterworth lowpass filter** with cutoff frequency (\( f_c \)) = 100 kHz - **Aliasing level at the cutoff frequency** must be **< 5%** - **Find:** 1. **Order (\( n \))** of the Butterworth filter 2. **Aliasing percentage value** --- ## **Step 1: Determine the frequency where aliasing occurs** Aliasing occurs at frequencies above the Nyquist frequency (\( f_s/2 \)), which for \( f_s = 1 \) MHz is: \[ f_N = \frac{f_s}{2} = \frac{1,000,000}{2} = 500,000 \text{ Hz} = 500 \text{ kHz} \] --- ## **Step 2: Butterworth Filter Attenuation Formula** The magnitude response of a Butterworth filter is: \[ |H(f)| = \frac{1}{\sqrt{1 + (f/f_c)^{2n}}} \] where \( n \) is the order, \( f_c \) is the cutoff frequency. --- ## **Step 3: Define the Aliasing Level Requirement** We want the output at the **aliasing frequency** (500 kHz) to be less than 5% of the cutoff (100 kHz) level: \[ \frac{|H(f = 500\text{ kHz})|}{|H(f = 100\text{ kHz})|} < 0.05 \] But at the cutoff frequency (\( f_c \)), the Butterworth response is: \[ |H(f_c)| = \frac{1}{\sqrt{2}} \] So, the ratio at \( f = 500 \) kHz is: \[ \text{Aliasing ratio} = \frac{|H(500\text{ kHz})|}{|H(100\text{ kHz})|} = \frac{|H(500\text{ kHz})|}{1/\sqrt{2}} \] But since the **percentage** is compared to the unattenuated input, it's fine to just set: \[ |H(500\text{ kHz})| < 0.05 \times |H(100\text{ kHz})| \] \[ |H(500\text{ kHz})| < 0.05 \times \frac{1}{\sqrt{2}} \] \[ |H(500\text{ kHz})| < 0.03536 \] --- ## **Step 4: Plug into the Butterworth Response** At \( f = 500 \) kHz: \[ |H(500\text{ kHz})| = \frac{1}{\sqrt{1 + (500/100)^{2n}}} = \frac{1}{\sqrt{1 + 5^{2n}}} \] Set this less than 0.03536: \[ \frac{1}{\sqrt{1 + 5^{2n}}} < 0.03536 \] \[ \sqrt{1 + 5^{2n}} > \frac{1}{0.03536} \] \[ 1 + 5^{2n} > \left(\frac{1}{0.03536}\right)^2 \] \[ 1 + 5^{2n} > 799.8 \] \[ 5^{2n} > 798.8 \] --- ## **Step 5: Solve for \( n \)** Take log on both sides: \[ 2n \log(5) > \log(798.8) \] \[ 2n \times 0.69897 > 2.902 \] \[ 2n > \frac{2.902}{0.69897} \] \[ 2n > 4.15 \] \[ n > 2.075 \] Since filter order must be an integer, **round up**: \[ n = 3 \] --- ## **Step 6: Corresponding Aliasing Percentage Calculation** Check the attenuation at \( n = 3 \): \[ |H(500\text{ kHz})| = \frac{1}{\sqrt{1 + (500/100)^{2 \times 3}}} = \frac{1}{\sqrt{1 + 5^6}} = \frac{1}{\sqrt{1 + 15625}} = \frac{1}{\sqrt{15626}} = \frac{1}{125} = 0.008 \] Now, compare to the response at cutoff (\( 1/\sqrt{2} \)): \[ \text{Aliasing percentage} = \frac{0.008}{1/\sqrt{2}} \times 100\% = 0.008 \times \sqrt{2} \times 100\% \approx 1.13\% \] --- ## **Final Answers** ### **1. Order of the anti-aliasing lowpass filter:** \[ \boxed{n = 3} \] ### **2. Corresponding aliasing percentage value:** \[ \boxed{1.13\%} \] --- ### **Summary Table** | Parameter | Value | |-------------------|----------| | Filter order (\( n \)) | 3 | | Aliasing % | 1.13% | Let me know if you need the steps in any other format!