Let's break down and solve the problem step by step, referencing the provided text and general DSP concepts. --- # **Step-by-Step Solution** ## **Given:** A discrete-time lowpass filter is to be designed using the **impulse invariance method** applied to a continuous-time Butterworth filter with squared: \[ |H_c(j\Omega)|^2 \] **Specs for discrete-time system:** \[ \begin{align*} .9125 \leq |H(e^{j\omega})|^2 & \leq 1, \quad \leq |\omega| \leq .2\pi \\ |H(e^{j\omega})|^2 & \leq .1778, \quad .5\pi \leq |\omega| \leq \pi \end{align*} \] --- ## **(a) Sketch the Tolerance Bounds** ### **Passband:** - \( \leq |\omega| \leq .2\pi \) - \( .9125 \leq |H(e^{j\omega})|^2 \leq 1 \) ### **Stopband:** - \( .5\pi \leq |\omega| \leq \pi \) - \( |H(e^{j\omega})|^2 \leq .1778 \) **Sketch:** Draw \( |H(e^{j\omega})|^2 \) vs. \( \omega \) from \( \) to \( \pi \): - A horizontal band between .9125 and 1 for \( \) to \( .2\pi \) (passband) - Below .1778 for \( .5\pi \) to \( \pi \) (stopband) --- ## **(b) Find Order \( N \) and \( T_d \) for Butterworth Filter** ### **Step 1: Passband and Stopband Mapping** For impulse invariance: Discrete-time frequency \( \omega \) relates to continuous-time frequency \( \Omega \) by: \[ \omega = \Omega T \] So: - Passband edge: \( \omega_p = .2\pi \implies \Omega_p = \omega_p / T_d \) - Stopband edge: \( \omega_s = .5\pi \implies \Omega_s = \omega_s / T_d \) ### **Step 2: Butterworth Filter Specs** Squared magnitude response: \[ |H_c(j\Omega)|^2 = \frac{1}{1 + (\Omega/\Omega_c)^{2N}} \] - Passband: \( |H_c(j\Omega_p)|^2 \geq .9125 \) - Stopband: \( |H_c(j\Omega_s)|^2 \leq .1778 \) **Convert the bounds:** - Passband: \( \frac{1}{1 + (\Omega_p/\Omega_c)^{2N}} \geq .9125 \) - Stopband: \( \frac{1}{1 + (\Omega_s/\Omega_c)^{2N}} \leq .1778 \) **Solve for N:** Let’s set up the inequalities: **Passband:** \[ .9125 \leq \frac{1}{1 + (\Omega_p/\Omega_c)^{2N}} \implies 1 + (\Omega_p/\Omega_c)^{2N} \leq \frac{1}{.9125} \implies (\Omega_p/\Omega_c)^{2N} \leq \frac{1}{.9125} - 1 = .0958 \] **Stopband:** \[ \frac{1}{1 + (\Omega_s/\Omega_c)^{2N}} \leq .1778 \implies 1 + (\Omega_s/\Omega_c)^{2N} \geq \frac{1}{.1778} \implies (\Omega_s/\Omega_c)^{2N} \geq \frac{1}{.1778} - 1 = 4.625 \] **Combine:** \[ (\Omega_p/\Omega_c)^{2N} \leq .0958,\quad (\Omega_s/\Omega_c)^{2N} \geq 4.625 \] Take logs: \[ \frac{(\Omega_s/\Omega_p)^{2N}}{\left(\Omega_c/\Omega_c\right)^{2N}} \geq \frac{4.625}{.0958} = 48.29 \] \[ \left(\frac{\Omega_s}{\Omega_p}\right)^{2N} \geq 48.29 \implies 2N \cdot \log_{10}(\Omega_s/\Omega_p) \geq \log_{10}(48.29) \implies N \geq \frac{1}{2} \frac{\log_{10}(48.29)}{\log_{10}(\Omega_s/\Omega_p)} \] **Insert \(\Omega_p = .2\pi/T_d\), \(\Omega_s = .5\pi/T_d\):** \[ \Omega_s/\Omega_p = \frac{.5\pi/T_d}{.2\pi/T_d} = 2.5 \] \[ N \geq \frac{1}{2} \frac{\log_{10}(48.29)}{\log_{10}(2.5)} = \frac{1}{2} \cdot \frac{1.684}{.398} = \frac{1}{2} \cdot 4.235 = 2.12 \] **So, \( N = 3 \) (nearest integer above 2.12).** --- ### **Step 3: Determine \( T_d \)** Choose \( T_d \) small enough to avoid aliasing (so that the analog filter’s stopband attenuation is sufficient at frequencies above \(\Omega_s\)). Typically, \( T_d \) is chosen so that the analog filter is negligible for \( \Omega > \pi/T_d \). --- ## **(c) Show that \( T_{d1} = 1 \) is Same as Example 7.2** - In Example 7.2, \( T = 1 \) is used. - The mapping for impulse invariance is \( \omega = \Omega T \). - If you use \( T_{d1} = 1 \), then the discrete and continuous frequency axes align directly. - So, the design and bounds collapse to those in Example 7.2, confirming the results are the same. --- # **Summary Table** | Parameter | Value / Expression | |------------------|--------------------------------------| | \( \Omega_p \) | \( .2\pi/T_d \) | | \( \Omega_s \) | \( .5\pi/T_d \) | | \( N \) | \( 3 \) | | \( T_d \) | To be chosen (see above) | | \( T_{d1} \) | \( 1 \), matches Example 7.2 | --- ## **Conclusion** - **(a)** Sketch the tolerance bounds as described. - **(b)** Find \( N = 3 \); choose \( T_d \) to avoid aliasing. - **(c)** With \( T_{d1} = 1 \), the design reduces to Example 7.2. If you want a graphical sketch or further algebraic detail, let me know!