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# Discrete-Time Low-Pass Filter Using Impulse Invariance ## Given Specifications - Magnitude response at passband: \[ |H(e^{j\omega})| = 1 \quad \text{for } 0 \le \omega \le 0.2\pi \] - Magnitude response at stopband: \[ |H(e^{j\omega})| \le 0.1783 \quad \text{for } 0.3\pi \le \omega \le \pi \] - Sampling interval: \[ T = 1 \] - Prototype: **Butterworth filter** --- ## 1. Convert DT Specifications to CT Specifications **Impulse invariance mapping:** \[ \Omega = \frac{\omega}{T} \] Since \( T=1 \): \[ \Omega = \omega \] ### Passband and Stopband Edges - Passband edge: \[ \Omega_p = 0.2\pi \] - Stopband edge: \[ \Omega_s = 0.3\pi \] - Stopband magnitude: \[ |H|_s = 0.1783 \] --- ## 2. Butterworth Magnitude Equation **Butterworth magnitude response:** \[ |H(j\Omega)|^2 = \frac{1}{1+\left(\frac{\Omega}{\Omega_c}\right)^{2N}} \] --- ## 3. Passband Condition - Passband gain = 1 Thus cutoff frequency approximately: \[ \Omega_c \approx \Omega_p = 0.2\pi \] --- ## 4. Stopband Condition \[ |H(j\Omega_s)| = \frac{1}{\sqrt{1+\left(\frac{\Omega_s}{\Omega_c}\right)^{2N}}} \] ### Substitute Values \[ 0.1783 = \frac{1}{\sqrt{1+\left(\frac{0.3\pi}{0.2\pi}\right)^{2N}}} \] This simplifies to: \[ 0.1783 = \frac{1}{\sqrt{1+(1.5)^{2N}}} \] --- ## 5. Solve for Filter Order (N) ### Square Both Sides \[ 0.0318 = \frac{1}{1+(1.5)^{2N}} \] ### Invert \[ 1+(1.5)^{2N} = \frac{1}{0.0318} \] This leads to: \[ 1+(1.5)^{2N} = 31.45 \] Thus: \[ (1.5)^{2N} = 30.45 \] ### Take Logarithm \[ 2N \log(1.5) = \log(30.45) \] Calculating logs: \[ \log(30.45) \approx 1.483 \] \[ \log(1.5) \approx 0.176 \] Therefore: \[ 2N = \frac{1.483}{0.176} \approx 8.43 \] \[ N \approx 4.215 \] **Minimum integer:** \[ N = 5 \] --- ## 6. Continuous-Time Butterworth Transfer Function For **N = 5**, poles of Butterworth filter: \[ s_k = \Omega_c e^{j\left(\frac{\pi}{2}+\frac{(2k+1)\pi}{2N}\right)} \] where \( k=0,1,2,3,4 \) Cutoff frequency: \[ \Omega_c = 0.2\pi = 0.628 \] Thus, the **analog transfer function** is: \[ H(s) = \frac{\Omega_c^5}{(s-s_1)(s-s_2)(s-s_3)(s-s_4)(s-s_5)} \] --- ## 7. Apply Impulse Invariance Impulse invariance relation: \[ H(z) = \sum \frac{A_k}{1-e^{s_kT}z^{-1}} \] where \( T=1 \). Thus poles map as: \[ z_k = e^{s_k} \] --- ## 8. Final Discrete-Time Transfer Function \[ H(z) = \sum_{k=1}^{5} \frac{A_k}{1-e^{s_k}z^{-1}} \] This gives the **digital IIR filter equivalent of the Butterworth analog filter**. --- ## 9. Frequency Response Relation Impulse invariance relation: \[ H(e^{j\omega}) = \frac{1}{T} \sum_{k=-\infty}^{\infty} H_c\left(j\left(\frac{\omega + 2\pi k}{T}\right)\right) \] Since \( T=1 \): \[ H(e^{j\omega}) = \sum_{k=-\infty}^{\infty} H_c(j(\omega + 2\pi k)) \] --- ## ✅ Final Results - **Filter type:** Butterworth Low-Pass - **Filter order:** \[ N = 5 \] - **Passband frequency:** \[ \Omega_p = 0.2\pi \] - **Stopband frequency:** \[ \Omega_s = 0.3\pi \] - **Cutoff frequency:** \[ \Omega_c \approx 0.628 \] - **Discrete-time transfer function:** \[ H(z) = \sum_{k=1}^{5} \frac{A_k}{1-e^{s_k}z^{-1}} \]