---
title: "x(k) for k = 0 to 4 from X(z)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:xk-first-five-values"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/xk-first-five-values
site: BCT Engineering
---

# x(k) for k = 0 to 4 from X(z)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2072 Asoj, 6 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/xk-first-five-values
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find x(k) for k=0,1,2,3,4 when X(z) is given by X(z)=(10z+5)/((z-1)(z-2)).

## Short answer

By long division or partial fractions: x(0)=0, x(1)=10, x(2)=35, x(3)=85, x(4)=185. Closed form: x(k) = −15 + 12.5(2)^k for k ≥ 1.

## Answer

The first few values are found by **direct (long) division**. A closed form from partial fractions confirms them.

#### Method 1: Long division

$$
X(z) = \frac{10z + 5}{(z-1)(z-2)} = \frac{10z + 5}{z^2 - 3z + 2} = \frac{10z^{-1} + 5z^{-2}}{1 - 3z^{-1} + 2z^{-2}}
$$

Divide $10z^{-1} + 5z^{-2}$ by $1 - 3z^{-1} + 2z^{-2}$:

| Step | Quotient term | Remainder after subtraction |
|---|---|---|
| 1 | $10z^{-1}$ | $35z^{-2} - 20z^{-3}$ |
| 2 | $35z^{-2}$ | $85z^{-3} - 70z^{-4}$ |
| 3 | $85z^{-3}$ | $185z^{-4} - 170z^{-5}$ |
| 4 | $185z^{-4}$ | $385z^{-5} - 370z^{-6}$ |

$$
X(z) = 0 + 10z^{-1} + 35z^{-2} + 85z^{-3} + 185z^{-4} + \dots
$$

Equivalently, the recursion $x(k) = 3x(k-1) - 2x(k-2)$ holds for $k \ge 3$: for example $x(3) = 3(35) - 2(10) = 85$ and $x(4) = 3(85) - 2(35) = 185$.

#### Method 2: Closed form (check)

$$
\frac{X(z)}{z} = \frac{10z+5}{z(z-1)(z-2)} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2}
$$

$$
A = \frac{5}{(-1)(-2)} = 2.5, \quad B = \frac{15}{(1)(-1)} = -15, \quad C = \frac{25}{(2)(1)} = 12.5
$$

$$
X(z) = 2.5 - \frac{15z}{z-1} + \frac{12.5z}{z-2}
$$

$$
x(k) = 2.5\,\delta(k) - 15 + 12.5(2)^k
$$

- $x(0) = 2.5 - 15 + 12.5 = 0$
- $x(1) = -15 + 25 = 10$
- $x(2) = -15 + 50 = 35$
- $x(3) = -15 + 100 = 85$
- $x(4) = -15 + 200 = 185$

#### Result

| $k$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $x(k)$ | 0 | 10 | 35 | 85 | 185 |

The pole at $z = 2$ lies outside the unit circle, so $x(k)$ grows without bound.

**Answer:** $x(0) = 0$, $x(1) = 10$, $x(2) = 35$, $x(3) = 85$, $x(4) = 185$. In general $x(k) = -15 + 12.5(2)^k$ for $k \ge 1$.

## Check yourself

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

1. To find the first values by division, X(z) is first written as:
   A. (10z^{-1} + 5z^{-2})/(1 - 3z^{-1} + 2z^{-2})
   B. (10z + 5)/(1 - 3z + 2z^2)
   C. (10 + 5z^{-1})/(1 - 3z^{-1} + 2z^{-2})
   D. (10z^{-1} + 5z^{-2})/(1 + 3z^{-1} + 2z^{-2})
   Answer: A. Dividing numerator and denominator by z^2 gives powers of z^{-1}; the coefficient of z^{-k} is x(k). (Key point: Long division in z^{-1})

2. What are x(0), x(1), x(2)?
   A. 0, 10, 35
   B. 10, 35, 85
   C. 0, 10, 30
   D. 5, 10, 35
   Answer: A. The series is 10z^{-1} + 35z^{-2} + 85z^{-3} + 185z^{-4}, with no z^0 term. (Key point: Values)

3. For k >= 3, x(k) equals:
   A. 3x(k-1) - 2x(k-2)
   B. 3x(k-1) + 2x(k-2)
   C. -3x(k-1) + 2x(k-2)
   D. 2x(k-1) - 3x(k-2)
   Answer: A. The denominator 1 - 3z^{-1} + 2z^{-2} gives this; e.g. x(3) = 3(35) - 2(10) = 85. (Key point: Recursion)

4. Why does x(k) grow without bound?
   A. The pole at z = 2 is outside the unit circle
   B. The pole at z = 1 is double
   C. The numerator has a zero at z = 0
   D. The poles are complex
   Answer: A. The term 12.5(2)^k dominates: x(k) = -15 + 12.5(2)^k for k >= 1. (Key point: Closed form behaviour)