Let's solve the problem step by step: --- ### **Given:** \[ H(z) = \frac{1 - 2.1z^{-1} + 2.7z^{-2}}{1 + 0.3126z^{-1} + 0.81z^{-2}} \] --- ## **(a) Decompose \( H(z) \) into Minimum-Phase and Allpass Components** ### **Step 1: Find Zeros and Poles** The numerator and denominator can be written as: - Numerator: \( 1 - 2.1z^{-1} + 2.7z^{-2} \) - Denominator: \( 1 + 0.3126z^{-1} + 0.81z^{-2} \) #### **Zeros:** Solve \( 1 - 2.1z^{-1} + 2.7z^{-2} = 0 \) or equivalently, \( z^2 - 2.1z + 2.7 = 0 \) \[ z = \frac{2.1 \pm \sqrt{2.1^2 - 4 \cdot 2.7}}{2} \] \[ = \frac{2.1 \pm \sqrt{4.41 - 10.8}}{2} = \frac{2.1 \pm \sqrt{-6.39}}{2} \] \[ = \frac{2.1 \pm j\sqrt{6.39}}{2} \] So, zeros are **complex conjugate** and outside the unit circle (since the magnitude of each zero \( |z| = \sqrt{2.7} \approx 1.643 \), which is > 1). #### **Poles:** Solve \( 1 + 0.3126z^{-1} + 0.81z^{-2} = 0 \) or \( z^2 + 0.3126z + 0.81 = 0 \) \[ z = \frac{-0.3126 \pm \sqrt{0.3126^2 - 4\times0.81}}{2} \] \[ = \frac{-0.3126 \pm \sqrt{0.0977 - 3.24}}{2} = \frac{-0.3126 \pm \sqrt{-3.1423}}{2} \] \[ = \frac{-0.3126 \pm j\sqrt{3.1423}}{2} \] Poles are inside the unit circle (since \( |z| = \sqrt{0.81} = 0.9 \), which is < 1). --- ### **Step 2: Minimum-Phase/Allpass Decomposition** - **Minimum-phase system:** All zeros inside the unit circle. - **Allpass system:** The *rest* of the zeros, reflected inside the unit circle, and poles matched to make the system allpass. #### **Let’s denote:** Let zeros be \( z_{01}, z_{02} \) (both outside unit circle). Their reflected versions inside the unit circle are \( 1/z_{01}^*, 1/z_{02}^* \). So, - **Minimum-phase system:** Has the same poles, but zeros inside the unit circle (i.e., at \( 1/z_{01}^*, 1/z_{02}^* \)). - **Allpass system:** Has zeros at \( z_{01}, z_{02} \), poles at \( 1/z_{01}^*, 1/z_{02}^* \). Therefore, \[ H(z) = H_{min}(z) \cdot H_{ap}(z) \] Where - \( H_{min}(z) \): Minimum-phase part (all zeros inside unit circle, same poles) - \( H_{ap}(z) \): Allpass part #### **Find explicit forms:** - **Minimum-phase system:** \[ H_{min}(z) = K \frac{(1 - (1/z_{01}^*)z^{-1})(1 - (1/z_{02}^*)z^{-1})}{1 + 0.3126z^{-1} + 0.81z^{-2}} \] - **Allpass system:** \[ H_{ap}(z) = \frac{(1 - z_{01}z^{-1})(1 - z_{02}z^{-1})}{(1 - (1/z_{01}^*)z^{-1})(1 - (1/z_{02}^*)z^{-1})} \] --- ## **(b) Plot Magnitudes** You can plot the magnitude responses using MATLAB, Python, or any DSP tool. - **Observation:** The magnitude response of the allpass system is always 1 for all frequencies. The minimum-phase component carries all the magnitude information. --- ## **(c) Plot Group Delays** - The group delay of \( H(z) \) is the sum of the group delays of the minimum-phase and allpass components. - The minimum-phase system will generally have smaller group delay than the original system. - The allpass system's group delay is nonzero and can be large near the unit circle zeros/poles. --- ## **Summary Table** | Component | Magnitude | Phase/Delay | |-------------------|-----------|--------------------| | \( H_{min}(z) \) | Varies | Minimum-phase | | \( H_{ap}(z) \) | 1 | Allpass (phase only) | | \( H(z) \) | Varies | Original | --- ## **Final Answers** ### **(a) Answer:** Express \( H(z) \) as: \[ H(z) = H_{min}(z) \cdot H_{ap}(z) \] - Where \( H_{min}(z) \) has zeros at \( 1/z_{01}^*, 1/z_{02}^* \) (inside unit circle), same poles. - And \( H_{ap}(z) = \frac{(1 - z_{01}z^{-1})(1 - z_{02}z^{-1})}{(1 - (1/z_{01}^*)z^{-1})(1 - (1/z_{02}^*)z^{-1})} \) (allpass). --- ### **(b) and (c) Answer:** - The magnitude of \( H(z) \) and its minimum-phase component coincide; the allpass part has magnitude 1. - The group delay of \( H(z) \) is the sum of the group delays of minimum-phase and allpass components; the allpass part contributes all the extra delay/phase "wrapping". --- **If you want code for plotting in Python (with `scipy.signal`), let me know!**

Certainly! Let's approach this problem step by step with detailed calculations and explanations. --- ## **Given System Function** \[ H(z) = \frac{1 - 2.1z^{-1} + 2.7z^{-2}}{1 + 0.3126z^{-1} + 0.81z^{-2}} \] --- ## **Part (a): Decompose \( H(z) \) into Minimum-Phase and Allpass Components** ### **Step 1: Express numerator and denominator as polynomials** Numerator: \[ N(z) = 1 - 2.1z^{-1} + 2.7z^{-2} \] which can be written as: \[ N(z) = z^2 - 2.1z + 2.7 \] Denominator: \[ D(z) = 1 + 0.3126z^{-1} + 0.81z^{-2} \] which can be written as: \[ D(z) = z^2 + 0.3126z + 0.81 \] --- ### **Step 2: Find zeros and poles (roots of numerator and denominator)** --- ### **Zeros of numerator polynomial:** Solve \[ z^2 - 2.1z + 2.7 = 0 \] Discriminant: \[ \Delta_{N} = (-2.1)^2 - 4 \times 1 \times 2.7 = 4.41 - 10.8 = -6.39 \] Since discriminant is negative, zeros are complex conjugates: \[ z_{1,2} = \frac{2.1 \pm j \sqrt{6.39}}{2} \] Calculate: \[ \sqrt{6.39} \approx 2.526 \] Thus: \[ z_{1,2} = \frac{2.1 \pm j 2.526}{2} = 1.05 \pm j 1.263 \] - **Magnitude of zeros:** \[ |z| = \sqrt{(1.05)^2 + (1.263)^2} = \sqrt{1.1025 + 1.595} = \sqrt{2.6975} \approx 1.64 \] - **Angles:** \[ \theta = \arctan\left(\frac{1.263}{1.05}\right) \approx \arctan(1.20) \approx 50.2^\circ \] Since magnitude > 1, zeros are **outside the unit circle**. --- ### **Poles of denominator:** Solve: \[ z^2 + 0.3126z + 0.81 = 0 \] Discriminant: \[ \Delta_{D} = (0.3126)^2 - 4 \times 1 \times 0.81 = 0.0977 - 3.24 = -3.1423 \] Zeros are complex conjugates: \[ z_{3,4} = \frac{-0.3126 \pm j \sqrt{3.1423}}{2} \] Calculate: \[ \sqrt{3.1423} \approx 1.772 \] Thus: \[ z_{3,4} = \frac{-0.3126 \pm j 1.772}{2} = -0.1563 \pm j 0.886 \] - **Magnitude:** \[ |z| = \sqrt{(-0.1563)^2 + 0.886^2} \approx \sqrt{0.0244 + 0.785} \approx \sqrt{0.809} \approx 0.899 \] - **Poles are inside the unit circle** (since magnitude < 1). --- ## **Step 3: Reflection of zeros** To obtain a **minimum-phase** system, zeros outside the unit circle should be reflected **inside** the circle: \[ z_{zero}^{\text{reflected}} = \frac{1}{z_{zero}^*} \] - For zeros \( z_{1,2} = 1.05 \pm j 1.263 \) Calculate their conjugates: \[ z_{1}^* = 1.05 - j 1.263 \] \[ z_{2}^* = 1.05 + j 1.263 \] Reflected zeros: \[ z_{1}^{\text{ref}} = \frac{1}{z_{1}^*} = \frac{1}{1.05 - j 1.263} \] \[ z_{2}^{\text{ref}} = \frac{1}{1.05 + j 1.263} \] Calculate magnitude of \( z_{1}^* \): \[ |z_{1}^*| = 1.64 \] so \[ |z_{1}^{\text{ref}}| = \frac{1}{1.64} \approx 0.61 \] Similarly for \( z_{2}^* \). --- ### **Step 4: Form the minimum-phase and allpass components** - **Minimum-phase numerator zeros:** \[ z_{1}^{\text{ref}} \quad \text{and} \quad z_{2}^{\text{ref}} \] - **Allpass zeros:** \[ z_{1} \quad \text{and} \quad z_{2} \] --- ### **Step 5: Construct the polynomials** - **Minimum-phase numerator polynomial:** \[ N_{min}(z) = (1 - z_{1}^{\text{ref}} z^{-1})(1 - z_{2}^{\text{ref}} z^{-1}) \] - **Allpass numerator polynomial:** \[ N_{ap}(z) = (1 - z_{1} z^{-1})(1 - z_{2} z^{-1}) \] - **Polynomials of the minimum-phase numerator:** Calculate \( z_{1}^{\text{ref}} \): \[ z_{1}^{\text{ref}} = \frac{1}{z_{1}^*} = \frac{1}{1.05 - j 1.263} \] Calculate denominator: \[ |z_{1}^*|^2 = 1.64^2 \approx 2.69 \] \[ z_{1}^{\text{ref}} = \frac{1.05 + j 1.263}{2.69} \approx 0.39 + j 0.47 \] Similarly: \[ z_{2}^{\text{ref}} = 0.39 - j 0.47 \] --- ## **Part (a): Final decomposition** \[ \boxed{ H(z) = H_{min}(z) \times H_{ap}(z) } \] where - \( H_{min}(z) \) has numerator zeros at \( z_{1}^{\text{ref}}, z_{2}^{\text{ref}} \), - \( H_{ap}(z) \) has zeros at \( z_{1}, z_{2} \), - Both share the same denominator (poles). --- ## **Part (b): Magnitude responses** - The magnitude of \( H(z) \) is the same as that of \( H_{min}(z) \times H_{ap}(z) \). - The allpass component has magnitude 1 at all frequencies. - The magnitude response of \( H(z) \) and \( H_{min}(z) \) are identical. --- ## **Part (c): Group delay** - The total group delay is the sum of the group delays of the minimum-phase and allpass parts. - The allpass component introduces phase variation but no magnitude change, often causing peaks in group delay near the zeros/poles. --- ## **Summary** | Component | Zeros | Poles | Zero reflection | Nature | |------------|--------|--------|----------------|---------| | \( H_{min}(z) \) | at \( z_{1}^{\text{ref}}, z_{2}^{\text{ref}} \) | same as original | zeros reflected inside | minimum-phase | | \( H_{ap}(z) \) | at original zeros \( z_{1}, z_{2} \) | same as original | zeros outside | allpass | --- ## **Final note:** The key idea is: - Reflect zeros outside the unit circle inside to get the minimum-phase part. - The ratio of the original zeros to their reflected counterparts forms the allpass component. --- Let me know if you'd like the actual polynomial coefficients or MATLAB/Python code for plotting!