Let's solve the given **difference equation** step by step: \[ 0.8 y(n) + 0.1 y(n-3) - 0.3 y(n-5) - 0.4 y(n-2) - 0.5 y(n-1) = x(n) + x(n+1) \] We'll answer each part in detail. --- ## **a. Impulse Response \( y(n) \) for \( x(n) = \delta(n) \)** ### **Step 1: Initial Conditions** Assume the system is **at rest** (causal, initial outputs are zero): \[ y(n) = 0\ \text{for}\ n < 0 \] \[ x(n) = \delta(n) = \begin{cases} 1 & n = 0 \\ 0 & n \ne 0 \end{cases} \] Also, \[ x(n+1) = \delta(n+1) = \begin{cases} 1 & n = -1 \\ 0 & n \ne -1 \end{cases} \] ### **Step 2: Compute \( y(0) \)** Plug \( n = 0 \): \[ 0.8y(0) + 0.1y(-3) - 0.3y(-5) - 0.4y(-2) - 0.5y(-1) = x(0) + x(1) \] All \( y(\cdot) \) for negative indices are zero, and \( x(0) = 1, x(1) = 0 \): \[ 0.8y(0) = 1 \implies y(0) = 1.25 \] --- ### **Step 3: Compute \( y(1) \)** Plug \( n = 1 \): \[ 0.8y(1) + 0.1y(-2) - 0.3y(-4) - 0.4y(-1) - 0.5y(0) = x(1) + x(2) \] Again, \( y(\cdot) \) for negative indices are zero, \( y(0) = 1.25 \), \( x(1) = 0, x(2) = 0 \): \[ 0.8y(1) - 0.5 \times 1.25 = 0 \implies 0.8y(1) = 0.625 \implies y(1) = 0.78125 \] --- ### **Step 4: Compute \( y(2) \)** Plug \( n = 2 \): \[ 0.8y(2) + 0.1y(-1) - 0.3y(-3) - 0.4y(0) - 0.5y(1) = x(2) + x(3) \] \( y(-1) = y(-3) = 0 \), \( y(0) = 1.25, y(1) = 0.78125 \), \( x(2) = x(3) = 0 \): \[ 0.8y(2) - 0.4 \times 1.25 - 0.5 \times 0.78125 = 0 \] \[ 0.8y(2) - 0.5 - 0.390625 = 0 \implies 0.8y(2) = 0.890625 \implies y(2) = 1.11328125 \] --- ### **Step 5: Compute \( y(3) \)** Plug \( n = 3 \): \[ 0.8y(3) + 0.1y(0) - 0.3y(-2) - 0.4y(1) - 0.5y(2) = x(3) + x(4) \] \( y(-2) = 0 \), \( y(0) = 1.25, y(1) = 0.78125, y(2) = 1.11328125 \): \[ 0.8y(3) + 0.125 - 0.4 \times 0.78125 - 0.5 \times 1.11328125 = 0 \] \[ 0.8y(3) + 0.125 - 0.3125 - 0.556640625 = 0 \] \[ 0.8y(3) - 0.744140625 = 0 \implies 0.8y(3) = 0.744140625 \implies y(3) = 0.93017578125 \] Continue similarly for \( y(4), y(5), \ldots \) if needed. --- ## **b. Step response for \( x(n) = u(n) \) (\( x(n) = 1 \) for \( n \ge 0 \))** You can calculate similarly by plugging \( x(n) = 1 \), \( x(n+1) = 1 \) for all \( n \ge 0 \): At \( n = 0: \) \[ 0.8y(0) = 1 + 1 = 2 \implies y(0) = 2.5 \] At \( n = 1: \) \[ 0.8y(1) - 0.5y(0) = 2 \] \[ 0.8y(1) - 0.5 \times 2.5 = 2 \implies 0.8y(1) = 3.25 \implies y(1) = 4.0625 \] At \( n = 2: \) \[ 0.8y(2) - 0.4y(0) - 0.5y(1) = 2 \] \[ 0.8y(2) - 0.4 \times 2.5 - 0.5 \times 4.0625 = 2 \] \[ 0.8y(2) - 1.0 - 2.03125 = 2 \implies 0.8y(2) = 5.03125 \implies y(2) = 6.2890625 \] Continue as needed. --- ## **c. Transfer Function \( H(z) \)** Take the **Z-transform** of both sides (assume zero initial conditions): Left side: \[ 0.8Y(z) + 0.1z^{-3}Y(z) - 0.3z^{-5}Y(z) - 0.4z^{-2}Y(z) - 0.5z^{-1}Y(z) \] \[ = [0.8 + 0.1z^{-3} - 0.3z^{-5} - 0.4z^{-2} - 0.5z^{-1}] Y(z) \] Right side: \[ X(z) + z X(z) = (1 + z)X(z) \] So: \[ [0.8 + 0.1z^{-3} - 0.3z^{-5} - 0.4z^{-2} - 0.5z^{-1}] Y(z) = (1 + z) X(z) \] \[ H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z}{0.8 + 0.1z^{-3} - 0.3z^{-5} - 0.4z^{-2} - 0.5z^{-1}} \] You may want to write numerator and denominator as **polynomials in \( z^{-1} \)** or **z**: Multiply numerator and denominator by \( z^5 \): \[ H(z) = \frac{z^5 + z^6}{0.8z^5 + 0.1z^2 - 0.3 - 0.4z^3 - 0.5z^4} \] Or, as a function of \( z \): \[ H(z) = \frac{z^5(1 + z)}{0.8 z^5 + 0.1 z^2 - 0.3 - 0.4 z^3 - 0.5 z^4} \] --- ## **d. Poles and Zeros on the z-plane** **Zeros:** Solve numerator \( 1 + z = 0 \implies z = -1 \) and \( z^5 = 0 \implies z = 0 \) (multiplicity 5). **Poles:** Solve denominator polynomial: \[ 0.8 z^5 + 0.1 z^2 - 0.3 - 0.4 z^3 - 0.5 z^4 = 0 \] This is a 5th degree polynomial in \( z \). Roots (poles) can be found numerically (e.g., using Python or MATLAB). ### **Plot** - Plot zeros at \( z = 0 \) (5 times) and \( z = -1 \). - Poles at roots of the denominator. --- ## **e. Stability** A discrete-time system is **stable if all poles** are **inside the unit circle** (\( |z| < 1 \)). - Find the roots of the denominator. - If all \( |z_p| < 1 \), system is stable. --- ## **f. Amplitude Response \( |H(e^{j\omega})| \)** Replace \( z = e^{j\omega} \) in \( H(z) \): \[ |H(e^{j\omega})| = \left| \frac{1 + e^{j\omega}}{0.8 + 0.1e^{-j3\omega} - 0.3e^{-j5\omega} - 0.4e^{-j2\omega} - 0.5e^{-j\omega}} \right| \] You can plot this for \( \omega \in [0, \pi] \). --- ## **g. Phase Response** Similarly, the phase is: \[ \angle H(e^{j\omega}) = \arg\left( \frac{1 + e^{j\omega}}{0.8 + 0.1e^{-j3\omega} - 0.3e^{-j5\omega} - 0.4e^{-j2\omega} - 0.5e^{-j\omega}} \right) \] --- # **Summary Table** | Part | Answer | |------|--------| |**a** | Impulse response: \( y(0) = 1.25, y(1) = 0.78125, y(2) = 1.1133, y(3) = 0.9302, ... \) | |**b** | Step response: \( y(0) = 2.5, y(1) = 4.0625, y(2) = 6.2891, ... \) | |**c** | \( H(z) = \frac{1 + z}{0.8 + 0.1z^{-3} - 0.3z^{-5} - 0.4z^{-2} - 0.5z^{-1}} \) | |**d** | Poles: roots of denominator; Zeros: \( z = 0 \) (x5), \( z = -1 \); plot on z-plane | |**e** | Stable if all poles \( |z| < 1 \) | |**f** | Amplitude response: \( |H(e^{j\omega})| \) as given above | |**g** | Phase response: \( \angle H(e^{j\omega}) \) as given above | --- ## **Plots** ### **1. Pole-Zero Plot** - Zeros: at \( z = 0 \) (order 5), \( z = -1 \). - Poles: roots of denominator. ### **2. Frequency Response** - Plot \( |H(e^{j\omega})| \) vs \( \omega \) - Plot \( \angle H(e^{j\omega}) \) vs \( \omega \) --- ## **Python Code for Plots (example)** ```python import numpy as np import matplotlib.pyplot as plt from scipy import signal # Numerator and denominator coefficients (highest power first) # H(z) = (z^5 + z^6) / (0.8z^5 + 0.1z^2 - 0.3 - 0.4z^3 - 0.5z^4) num = [1, 1, 0, 0, 0, 0, 0] # z^6 + z^5 den = [1, -0.5/0.8, -0.4/0.8, 0.1/0.8, 0, -0.3/0.8] # Put in standard form # Adjust numerator and denominator for correct degree num = [1, 1, 0, 0, 0, 0] den = [0.8, -0.5, -0.4, 0.1, 0, -0.3] # z^5 ... z^0 # Pad numerator with zeros to match denominator's degree num = [0, 1, 1, 0, 0, 0