---
title: "z-transform of sin αk"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:z-transform-sin-alpha-k"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-sin-alpha-k
site: BCT Engineering
---

# z-transform of sin αk

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2068 Magh, 2.5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-sin-alpha-k
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find the z-transform of sin αk.

## Short answer

Applying Euler's formula and Z[a^k]=z/(z-a) gives Z[sin αk] = z sin α / (z^2 - 2z cos α + 1), for |z| > 1.

## Answer

**Idea.** Write sin αk using Euler's formula and use Z[aᵏ] = z/(z − a).

sin αk = (e^{jαk} − e^{−jαk}) / (2j)

```text
Z[sin αk] = (1/2j) [ z/(z − e^{jα}) − z/(z − e^{−jα}) ]
          = (1/2j) · z (e^{jα} − e^{−jα}) / [ z² − z(e^{jα} + e^{−jα}) + 1 ]
          = (1/2j) · z (2j sin α) / (z² − 2z cos α + 1)
```

**Answer:** Z[sin αk] = z sin α / (z² − 2z cos α + 1), for |z| > 1.

## Check yourself

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

1. sin(alpha k) is written as
   A. (e^(j alpha k) - e^(-j alpha k))/(2j)
   B. (e^(j alpha k) + e^(-j alpha k))/2
   C. (e^(j alpha k) - e^(-j alpha k))/2
   D. (e^(j alpha k) + e^(-j alpha k))/(2j)
   Answer: A. That is Euler's formula for sine. (Key point: Euler form)

2. Each exponential is transformed using
   A. Z[b^k] = 1/(z - b)
   B. Z[b^k] = z/(z - b)
   C. Z[b^k] = z/(z + b)
   D. Z[b^k] = b/(z - 1)
   Answer: B. With b = e^(+-j alpha). (Key point: Standard pair)

3. The denominator becomes z^2 - 2z cos alpha + 1 because
   A. e^(j alpha) + e^(-j alpha) = 2 sin alpha
   B. e^(j alpha) e^(-j alpha) = 0
   C. (z - e^(j alpha))(z - e^(-j alpha)) = z^2 - 2z cos alpha + 1
   D. the roots are real
   Answer: C. Sum of the poles is 2 cos alpha and their product is 1. (Key point: Denominator)

4. Z[sin alpha k] =
   A. z cos alpha/(z^2 - 2z cos alpha + 1), |z| > 1
   B. z sin alpha/(z^2 - 2z sin alpha + 1), |z| > 1
   C. z sin alpha/(z^2 - 2z cos alpha + 1), |z| < 1
   D. z sin alpha/(z^2 - 2z cos alpha + 1), |z| > 1
   Answer: D. The 2j in the numerator cancels the 1/2j, leaving sin alpha. (Key point: Result and ROC)

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