--- title: "Difference equation with input u(2) = −0.2142" subject: "Digital Control System" chapter: 2 key: "semester_6/digital_control_system:difference-equation-given-inputs-neg" source: https://www.bctengineer.com/q/semester_6/digital_control_system/difference-equation-given-inputs-neg site: BCT Engineering --- # Difference equation with input u(2) = −0.2142 Digital Control System · Chapter 2: The Z-Transform - Asked in: 2071 Magh, 4 marks - Video: [SOLUTION OF DIFFERENCE EQUATION](https://www.youtube.com/watch?v=1KuWLNN2SKw) — Department of EC SIRT Bhopal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/difference-equation-given-inputs-neg - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2 ## Question Consider the difference equation x(k+2)-1.3679x(k+1)+0.3679x(k)=0.3679u(k+1)+0.2642u(k), where x(k) is the output and x(k)=0 for k≤ 0, where u(k) is the input and is given by u(k)=0, k<0; u(0)=1; u(1)=0.2142; u(2)=-0.2142; u(k)=0, k=3,4,5,6,… Determine the output x(k). ## Short answer Recursive solution: x(0)=0, x(1)=0.3679, x(2)=0.8463. For k>=3, x(k)=1, a deadbeat response. ## Answer The output is found by recursion, and the z-transform then shows why it becomes constant. #### Step 1: Initial values $x(k) = 0$ for $k \le 0$, so $x(0) = 0$. Put $k = -1$ in the equation, with $x(-1) = 0$, $u(-1) = 0$, $u(0) = 1$: $$ x(1) - 1.3679x(0) + 0.3679x(-1) = 0.3679u(0) + 0.2642u(-1) \;\Rightarrow\; x(1) = 0.3679 $$ #### Step 2: Pulse transfer function With zero initial conditions (the $x(1)$ and $u(0)$ terms cancel), taking z-transforms gives: $$ G(z) = \frac{X(z)}{U(z)} = \frac{0.3679z + 0.2642}{z^2 - 1.3679z + 0.3679} = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} $$ The input is a finite sequence: $$ U(z) = 1 + 0.2142z^{-1} - 0.2142z^{-2} = \frac{z^2 + 0.2142z - 0.2142}{z^2} $$ #### Step 3: Recursive solution $$ x(k+2) = 1.3679x(k+1) - 0.3679x(k) + 0.3679u(k+1) + 0.2642u(k) $$ | $k$ | Calculation | $x(k+2)$ | |---|---|---| | 0 | $1.3679(0.3679) + 0.3679(0.2142) + 0.2642(1)$ | 0.8463 | | 1 | $1.3679(0.8463) - 0.3679(0.3679) - 0.3679(0.2142) + 0.2642(0.2142)$ | 1.0000 | | 2 | $1.3679(1) - 0.3679(0.8463) - 0.2642(0.2142)$ | 1.0000 | | 3 | $1.3679(1) - 0.3679(1)$ | 1.0000 | #### Step 4: Why the output stays at 1 (z-transform) $$ X(z) = \frac{(0.3679z + 0.2642)(z^2 + 0.2142z - 0.2142)}{z^2(z-1)(z-0.3679)} $$ The input numerator has a root at $z = 0.3679$: $$ (0.3679)^2 + 0.2142(0.3679) - 0.2142 = 0.13535 + 0.07880 - 0.2142 \approx 0 $$ So $z^2 + 0.2142z - 0.2142 \approx (z - 0.3679)(z + 0.5821)$. The plant pole at $0.3679$ is cancelled: $$ X(z) = \frac{(0.3679z + 0.2642)(z + 0.5821)}{z^2(z-1)} $$ Only the pole at $z = 1$ remains, apart from poles at the origin, which give a finite transient. Expanding: $$ X(z) = 0.3679z^{-1} + 0.8463z^{-2} + z^{-3} + z^{-4} + \dots $$ #### Result | $k$ | 0 | 1 | 2 | 3 | 4 | 5 | ... | |---|---|---|---|---|---|---|---| | $x(k)$ | 0 | 0.3679 | 0.8463 | 1 | 1 | 1 | ... | This is a **deadbeat** response: the output reaches its final value 1 in three sampling periods and stays there with no ripple. The chosen input sequence cancels the slow plant mode. **Answer:** $x(0) = 0$, $x(1) = 0.3679$, $x(2) = 0.8463$, and $x(k) = 1$ for $k \ge 3$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the output stay constant? A. Input is a step B. x(0) is 1 C. Input numerator has a root at 0.3679, cancelling the plant pole D. Plant has no poles Answer: C. z^2+0.2142z-0.2142 ~ (z-0.3679)(z+0.5821). (Key point: Cancellation) 2. What is this response called? A. Oscillatory B. Unstable C. Critically damped D. Deadbeat Answer: D. Reaches final value in finite steps with no ripple. (Key point: Response type) 3. x(k) for k >= 3? A. 1 B. 1.4284 C. 0.8463 D. 0 Answer: A. x(2)=0.8463, then 1 forever. (Key point: Values) 4. How many sampling periods to reach 1? A. infinitely many B. 3 C. 1 D. 6 Answer: B. x(3) = 1. 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