--- title: "Third-order difference equation: limit x(k+2)/x(k+1)" subject: "Digital Control System" chapter: 2 key: "semester_6/digital_control_system:third-order-difference-limit-k1" source: https://www.bctengineer.com/q/semester_6/digital_control_system/third-order-difference-limit-k1 site: BCT Engineering --- # Third-order difference equation: limit x(k+2)/x(k+1) Digital Control System · Chapter 2: The Z-Transform - Asked in: 2074 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/third-order-difference-limit-k1 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2 ## Question Solve the following difference equation using z transform method. Also determine the value of x(k+2)/x(k+1) as k approaches infinity: x(k+3)-2.2x(k+2)+1.57x(k+1)-0.36x(k)=u(k), where u(k)=1 for all k≥ 0, and x(0)=x(1)=x(2)=0. ## Short answer Closed form: x(k)=100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k. Limit of x(k+2)/x(k+1) as k→∞ is 1. ## Answer Solve by the **z-transform method**, with $u(k)$ a unit step. #### Step 1: Z-transform (zero initial conditions) With $x(0) = x(1) = x(2) = 0$, all initial-condition terms vanish. The input transform is $U(z) = \dfrac{z}{z-1}$: $$ (z^3 - 2.2z^2 + 1.57z - 0.36)X(z) = \frac{z}{z-1} $$ #### Step 2: Factorise the characteristic polynomial Trial roots: $P(0.5) = 0.125 - 0.55 + 0.785 - 0.36 = 0$, so $z = 0.5$ is a root. Dividing out gives $z^2 - 1.7z + 0.72 = (z-0.8)(z-0.9)$. $$ z^3 - 2.2z^2 + 1.57z - 0.36 = (z-0.5)(z-0.8)(z-0.9) $$ $$ X(z) = \frac{z}{(z-1)(z-0.5)(z-0.8)(z-0.9)} $$ #### Step 3: Partial fractions of $X(z)/z$ $$ \frac{X(z)}{z} = \frac{A}{z-1} + \frac{B}{z-0.5} + \frac{C}{z-0.8} + \frac{D}{z-0.9} $$ $$ \begin{aligned} A &= \frac{1}{(0.5)(0.2)(0.1)} = 100 \\ B &= \frac{1}{(-0.5)(-0.3)(-0.4)} = -16.667 \\ C &= \frac{1}{(-0.2)(0.3)(-0.1)} = 166.667 \\ D &= \frac{1}{(-0.1)(0.4)(0.1)} = -250 \end{aligned} $$ #### Step 4: Closed-form solution $$ x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k, \quad k = 0, 1, 2, \dots $$ The exact fractions are $-\tfrac{50}{3}$ and $\tfrac{500}{3}$. #### Check | $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---|---| | Formula | 0 | 0 | 0 | 1 | 3.2 | 6.47 | 10.57 | | Recursion | 0 | 0 | 0 | 1 | 3.2 | 6.47 | 10.57 | The recursion used is $x(k+3) = 2.2x(k+2) - 1.57x(k+1) + 0.36x(k) + 1$. For example, $x(4) = 2.2(1) + 1 = 3.2$. The final value theorem gives $\lim_{z\to1}(z-1)X(z) = \dfrac{1}{0.5\times0.2\times0.1} = 100$, which equals $A$. #### Step 5: Limiting value of $x(k+2)/x(k+1)$ The poles $0.5$, $0.8$ and $0.9$ are inside the unit circle, so every transient term dies out and $x(k) \to 100$: $$ \lim_{k\to\infty}\frac{x(k+2)}{x(k+1)} = \frac{100 + 0}{100 + 0} = 1 $$ The ratio approaches 1 slowly, because the slowest mode $(0.9)^k$ dominates the transient. **Answer:** $x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k$, and $\lim_{k\to\infty} x(k+2)/x(k+1) = 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With x(0)=x(1)=x(2)=0 and u(k)=1, what is X(z) for the third-order equation? A. z/((z-1)(z^3 - 2.2z^2 + 1.57z - 0.36)) B. 1/((z-1)(z^3 - 2.2z^2 + 1.57z - 0.36)) C. z/(z^3 - 2.2z^2 + 1.57z - 0.36) D. (z-1)/(z^3 - 2.2z^2 + 1.57z - 0.36) Answer: A. Initial-condition terms vanish, so the characteristic polynomial times X(z) equals U(z) = z/(z-1). (Key point: Z-transform with zero initial conditions) 2. What are the roots of z^3 - 2.2z^2 + 1.57z - 0.36? A. 0.5, 0.8, 0.9 B. 0.4, 0.8, 1.0 C. 0.5, 0.7, 1.0 D. 0.3, 0.9, 1.0 Answer: A. z = 0.5 is a root by trial; dividing out leaves z^2 - 1.7z + 0.72 = (z-0.8)(z-0.9). (Key point: Factorising the characteristic polynomial) 3. Why is X(z) divided by z before taking partial fractions? A. So each term becomes A/(z-p), and multiplying back by z gives z/(z-p), which inverts to p^k B. To remove the pole at z = 1 C. To make the initial conditions nonzero D. To convert the transform to the s-domain Answer: A. Terms of the form z/(z-p) are in the standard table and invert to p^k. (Key point: Partial fractions of X(z)/z) 4. Why does x(k+2)/x(k+1) tend to 1 as k tends to infinity? A. Poles 0.5, 0.8, 0.9 are inside the unit circle, so transients die and x(k) tends to 100 B. Because the pole at z = 1 has a coefficient of zero C. Because x(k) grows without bound at a constant rate D. Because the poles 0.5, 0.8, 0.9 lie outside the unit circle Answer: A. Both numerator and denominator approach 100, so the ratio is 1. 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