---
title: "Difference equation with step input, x(1)=0.3679"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:difference-equation-step-0-3679"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/difference-equation-step-0-3679
site: BCT Engineering
---

# Difference equation with step input, x(1)=0.3679

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2075 Baisakh, 8 marks; 2070 Magh, 8 marks (asked 2 times)
- 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-step-0-3679
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Solve the following difference equation by use of Z-transform method: x(k+2)+1.379x(k+1)+0.3679x(k)=0.3679u(k), where u(k) is step input and x(0)=0, x(1)=0.3679.

## Short answer

Solve x(k+2)+1.379x(k+1)+0.3679x(k)=0.3679u(k) with x(0)=0, x(1)=0.3679. Z-transform and partial fractions give x(k)=0.1339+0.1490(-0.3616)^k-0.2829(-1.0174)^k. Unstable due to pole magnitude >1.

## Answer

Given (as printed) $x(k+2) + 1.379x(k+1) + 0.3679x(k) = 0.3679u(k)$, $u(k) = 1$ for $k \ge 0$, $x(0) = 0$, $x(1) = 0.3679$.

#### Step 1: z-transform

$$
[z^2X - z^2x(0) - zx(1)] + 1.379[zX - zx(0)] + 0.3679X = 0.3679\frac{z}{z-1}
$$

$$
\begin{aligned}
(z^2 + 1.379z + 0.3679)X(z) &= 0.3679z + \frac{0.3679z}{z-1} = \frac{0.3679z^2}{z-1} \\
X(z) &= \frac{0.3679z^2}{(z-1)(z^2+1.379z+0.3679)}
\end{aligned}
$$

#### Step 2: roots of the characteristic equation

$$
z = \frac{-1.379 \pm \sqrt{1.379^2 - 4(0.3679)}}{2} = \frac{-1.379 \pm \sqrt{0.4300}}{2} = \frac{-1.379 \pm 0.6558}{2}
$$

$p_1 = -0.3616$, $\quad p_2 = -1.0174$

#### Step 3: partial fractions of $X(z)/z = \frac{0.3679z}{(z-1)(z-p_1)(z-p_2)}$

$$
\begin{aligned}
A &= \frac{0.3679(1)}{(1-p_1)(1-p_2)} = \frac{0.3679}{2.7469} = 0.1339 \\
B &= \frac{0.3679p_1}{(p_1-1)(p_1-p_2)} = \frac{0.3679(-0.3616)}{(-1.3616)(0.6558)} = 0.1490 \\
C &= \frac{0.3679p_2}{(p_2-1)(p_2-p_1)} = \frac{0.3679(-1.0174)}{(-2.0174)(-0.6558)} = -0.2829
\end{aligned}
$$

$$
X(z) = 0.1339\frac{z}{z-1} + 0.1490\frac{z}{z+0.3616} - 0.2829\frac{z}{z+1.0174}
$$

#### Result

$$
x(k) = 0.1339 + 0.1490(-0.3616)^k - 0.2829(-1.0174)^k
$$

#### Check

| k | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $x(k)$ | 0 | 0.3679 | -0.1394 | 0.4248 | -0.1666 | 0.4414 |

These agree with the recursion $x(k+2) = -1.379x(k+1) - 0.3679x(k) + 0.3679$. Since $|p_2| > 1$, the response oscillates with slowly growing amplitude, so the final value theorem cannot be applied.

**Answer:** $x(k) = 0.1339 + 0.1490(-0.3616)^k - 0.2829(-1.0174)^k$.

**Note:** this equation is usually printed in textbooks with $-1.3679$ (poles at $1$ and $0.3679$). With $x(k+2) - 1.3679x(k+1) + 0.3679x(k) = 0.3679u(k)$ and the same data, $X(z) = \frac{0.3679z^2}{(z-1)^2(z-0.3679)}$ and $x(k) = 0.5820k - 0.3388 + 0.3388(0.3679)^k$, i.e. $0, 0.3679, 0.8712, 1.4242, \dots$ (a ramp-like rise).

## Check yourself

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

1. X(z) =
   A. 0.3679 z/((z-1)(z^2+1.379z+0.3679))
   B. 0.3679/(z^2 + 1.379z + 0.3679)
   C. 0.3679 z^2/((z-1)^2 (z-0.3679))
   D. 0.3679 z^2/((z-1)(z^2 + 1.379z + 0.3679))
   Answer: D. As printed. (Key point: X(z))

2. Characteristic roots are about:
   A. -0.3616 and -1.0174
   B. 0.3616 and 1.0174
   C. -1 and -0.3679
   D. 0.5 and 0.5
   Answer: A. Quadratic formula. (Key point: Poles)

3. Final value theorem cannot be applied because:
   A. There is no pole at 1
   B. |p2| > 1, so the response grows
   C. x(0) = 0
   D. The input is a step
   Answer: B. Unstable pole. (Key point: Stability)

4. With -1.3679 the poles are:
   A. -1 and -0.3679
   B. 0 and 1
   C. 1 and 0.3679
   D. 0.5 twice
   Answer: C. Gives ramp-like rise. (Key point: Textbook version)

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