---
title: "z-transform of cosh t sinh t"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:z-transform-cosh-sinh"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-cosh-sinh
site: BCT Engineering
---

# z-transform of cosh t sinh t

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2071 Bhadra
- Video: [Z Transforms of 𝑪𝒐𝒔𝒉 𝒏𝜽 & 𝑺𝒊𝒏𝒉 𝒏𝜽 || 18mat31 || Dr Prashant Patil](https://www.youtube.com/watch?v=YsqIFJMwpok) — Dr Prashant Patil
- 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-cosh-sinh
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find the z-transform of x(z) = cosh t sinh t.

## Short answer

Since cosh t sinh t = ½ sinh 2t, its Z-transform is half of Z[sinh 2t], giving z sinh 2T / [2(z^2 - 2z cosh 2T + 1)]. For T=1, that is z sinh 2 / [2(z^2 - 2z cosh 2 + 1)], ROC |z|>e^2.

## Answer

The z-transform of a sampled signal x(k) is X(z) = Σ_{k=0}^{∞} x(k) z⁻ᵏ. Here the signal is sampled at t = kT; take T = 1 (for a general T, replace 2 by 2T).

**Step 1: simplify the product.**

cosh t sinh t = ½ sinh 2t = ¼ (e²ᵗ − e⁻²ᵗ)

**Step 2: use Z[aᵏ] = z/(z − a).** With t = k and T = 1:

```text
Z[cosh t sinh t] = ¼ [ z/(z − e²) − z/(z − e⁻²) ]
                 = ¼ · z (e² − e⁻²) / (z² − z(e² + e⁻²) + 1)
                 = ½ · z sinh 2 / (z² − 2z cosh 2 + 1)
```

**Answer:** X(z) = z sinh 2 / [2 (z² − 2z cosh 2 + 1)], for |z| > e².

For general sampling period T: X(z) = z sinh 2T / [2 (z² − 2z cosh 2T + 1)].

Numerically, sinh 2 = 3.6269 and cosh 2 = 3.7622, so X(z) = 1.8134 z/(z² − 7.5244 z + 1). (Checked numerically at z = 20: both sides give 0.144778.)

## Check yourself

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

1. cosh t sinh t equals
   A. (1/2) sinh 2t = (1/4)(e^2t - e^-2t)
   B. cosh 2t
   C. (1/2) cosh 2t
   D. (1/4)(e^2t + e^-2t)
   Answer: A. Using sinh 2t = 2 sinh t cosh t. (Key point: Simplify product)

2. Which pair converts the exponentials?
   A. Z[a^k] = 1/(z - a)
   B. Z[a^k] = z/(z - a) with a = e^2 and e^-2
   C. Z[a^k] = z/(z + a)
   D. Z[a^k] = a z/(z - 1)
   Answer: B. e^(2k) = (e^2)^k, so each exponential is geometric. (Key point: Standard pair)

3. The result is
   A. z cosh 2/[2(z^2 - 2z sinh 2 + 1)]
   B. z sinh 2/(z^2 + 2z cosh 2 + 1)
   C. z sinh 2/[2(z^2 - 2z cosh 2 + 1)]
   D. z sinh 2/[2(z^2 - 2z cosh 2 - 1)]
   Answer: C. Combining (1/4)[z/(z-e^2) - z/(z-e^-2)] gives this. (Key point: Final form)

4. For sampling period T, how does the result change?
   A. Multiply the whole result by T
   B. Replace z by zT
   C. Nothing changes
   D. Replace 2 by 2T in sinh and cosh
   Answer: D. t = kT gives exponentials e^(+-2T). (Key point: General T)

## Related in Applied Mathematics

- [z-transform of cosh(at) sin(bt)](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Az-transform-cosh-at-sin-bt) — Applied Mathematics, chapter 2, asked 2×