---
title: "Initial value theorem; find x(0), x(1)"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:initial-value-theorem"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/initial-value-theorem
site: BCT Engineering
---

# Initial value theorem; find x(0), x(1)

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2073 Bhadra, 1+4 marks
- Video: [Initial and Final Value Theorem - Z Transform - Signals and Systems](https://www.youtube.com/watch?v=dYzftpt6ZLc) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/initial-value-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

State initial value theorem for z transform. Find the initial value x(0) and x(1) for the function X(z) = (1 - e^T) z⁻¹ / ((1 - z⁻¹)(1 - e^{-T} z⁻¹)).

## Short answer

Initial value theorem: x(0)=lim_{z→∞} X(z); x(1)=lim_{z→∞} z[X(z)-x(0)]. For X(z)=(1-e^T)z^{-1}/((1-z^{-1})(1-e^{-T}z^{-1})), x(0)=0, x(1)=1-e^T.

## Answer

#### Initial value theorem

If X(z) = Z[x(k)] and the limit exists, then

x(0) = lim_{z→∞} X(z).

Further, x(1) = lim_{z→∞} z [X(z) − x(0)].

(Reason: X(z) = x(0) + x(1)z⁻¹ + x(2)z⁻² + …, and every term except x(0) tends to 0 as z → ∞.)

#### Finding x(0) and x(1)

X(z) = (1 − eᵀ) z⁻¹ / [ (1 − z⁻¹)(1 − e⁻ᵀ z⁻¹) ]

**x(0):** as z → ∞, z⁻¹ → 0, so the numerator → 0 and the denominator → 1.

x(0) = 0

**x(1):** multiply X(z) by z (x(0) = 0):

```text
z X(z) = (1 − eᵀ) / [ (1 − z⁻¹)(1 − e⁻ᵀ z⁻¹) ]
lim (z → ∞) = (1 − eᵀ) / (1 · 1)
```

x(1) = 1 − eᵀ

**Check.** Expand in powers of z⁻¹: X(z) = (1 − eᵀ) z⁻¹ [1 + (1 + e⁻ᵀ) z⁻¹ + …]. The z⁻¹ coefficient is 1 − eᵀ, confirming x(1). (For T = 0.9 the series gives x(1) = −1.4596 = 1 − e^{0.9}.)

**Answer:** x(0) = 0, x(1) = 1 − eᵀ.

## Check yourself

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

1. The initial value theorem says
   A. x(0) = lim as z -> infinity of X(z)
   B. x(0) = lim as z -> 0 of X(z)
   C. x(0) = lim as z -> 1 of (z-1)X(z)
   D. x(0) = X(1)
   Answer: A. Only x(0) survives when z^-k terms vanish. (Key point: Statement)

2. Why does the theorem work?
   A. The terms all cancel at infinity
   B. X(z) = x(0) + x(1)z^-1 + ..., and all terms except x(0) tend to 0 as z -> infinity
   C. z^-k grows as z -> infinity
   D. X(z) is constant
   Answer: B. Negative powers vanish. (Key point: Why)

3. For the given X(z), x(0) is
   A. 1 - e^T
   B. 1
   C. 0 since the numerator has z^-1 and the denominator tends to 1
   D. infinity
   Answer: C. Numerator -> 0. (Key point: x(0))

4. x(1) is found from
   A. lim X(z) = 1 - e^T
   B. lim z X(z) = e^-T
   C. lim (z-1) X(z)
   D. lim z X(z) = 1 - e^T
   Answer: D. Multiplying by z removes the z^-1 factor. (Key point: x(1))

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