---
title: "Third-order difference equation: limit x(k+2)/x(k)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:third-order-difference-limit-k"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/third-order-difference-limit-k
site: BCT Engineering
---

# Third-order difference equation: limit x(k+2)/x(k)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2077 Chaitra, 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-k
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Consider the difference equation x(k+3)-2.2x(k+2)+1.57x(k+1)-0.36x(k)=u(k), where u(k)=1 for k≥ 0 and x(0)=x(1)=x(2)=0. Obtain the general solution x(k) in a closed form. Find the limiting value of x(k+2)/x(k) when sequence variable approaches infinity.

## Short answer

Using z-transform, x(k)=100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k. Limit of x(k+2)/x(k) 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 $x(k+2)/x(k)$

All the transient terms have $|z| < 1$, so $(0.5)^k, (0.8)^k, (0.9)^k \to 0$:

$$
\lim_{k\to\infty}\frac{x(k+2)}{x(k)} = \lim_{k\to\infty}\frac{100 - 16.667(0.5)^{k+2} + 166.667(0.8)^{k+2} - 250(0.9)^{k+2}}{100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k} = \frac{100}{100} = 1
$$

The system is stable (poles 0.5, 0.8, 0.9 lie inside the unit circle), so $x(k)$ settles at 100 and successive values become equal.

**Answer:** $x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k$, and $\lim_{k\to\infty} x(k+2)/x(k) = 1$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Roots of z^3 - 2.2z^2 + 1.57z - 0.36?
   A. 1, 0.8, 0.9
   B. 0.4, 0.8, 1
   C. 0.5, 0.8, 0.9
   D. 0.5, 1, 0.8
   Answer: C. P(0.5)=0 then z^2-1.7z+0.72=(z-0.8)(z-0.9). (Key point: Characteristic roots)

2. With a unit step input, X(z) = ?
   A. 1/((z-0.5)(z-0.8)(z-0.9))
   B. z^3/(...)
   C. z/((z-0.5)(z-0.8)(z-0.9))
   D. z/((z-1)(z-0.5)(z-0.8)(z-0.9))
   Answer: D. U(z) = z/(z-1) adds a pole at 1. (Key point: Input transform)

3. Constant term in x(k)?
   A. 100
   B. 1
   C. 16.667
   D. 250
   Answer: A. A = 1/(0.5*0.2*0.1) = 100. (Key point: Steady-state term)

4. lim x(k+2)/x(k)?
   A. 1.5
   B. 1
   C. 100
   D. 0.81
   Answer: B. Transient terms vanish, ratio 100/100 = 1. (Key point: Limit)

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