---
title: "Z-transform of cos ωt"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-cos-wt"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-cos-wt
site: BCT Engineering
---

# Z-transform of cos ωt

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2071 Magh, 4 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/z-transform-cos-wt
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the z-transform of x(t)={cos ωt, t≥ 0; 0, t<0}

## Short answer

The z-transform of sampled cosine is X(z)= z(z-cosωT)/(z^2 - 2z cosωT + 1).

## Answer

The z-transform of the sampled cosine is

$$
X(z) = \frac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \frac{z(z - \cos\omega T)}{z^2 - 2z\cos\omega T + 1}
$$

The derivation follows.

#### Step 1: Sample the signal

With sampling period $T$, $x(kT) = \cos\omega kT$ for $k \ge 0$. Write it with Euler's formula:

$$
\cos\omega kT = \frac{1}{2}\left(e^{j\omega kT} + e^{-j\omega kT}\right)
$$

#### Step 2: Use the exponential pair

$$
\mathcal{Z}\left[e^{-akT}\right] = \sum_{k=0}^{\infty}e^{-akT}z^{-k} = \frac{1}{1 - e^{-aT}z^{-1}}, \quad |z| > |e^{-aT}|
$$

With $a = \mp j\omega$:

$$
X(z) = \frac{1}{2}\left[\frac{1}{1 - e^{j\omega T}z^{-1}} + \frac{1}{1 - e^{-j\omega T}z^{-1}}\right]
$$

#### Step 3: Combine the fractions

$$
\begin{aligned}
X(z) &= \frac{1}{2}\cdot\frac{(1 - e^{-j\omega T}z^{-1}) + (1 - e^{j\omega T}z^{-1})}{1 - (e^{j\omega T} + e^{-j\omega T})z^{-1} + z^{-2}} \\
&= \frac{1}{2}\cdot\frac{2 - 2z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \\
&= \frac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}}
\end{aligned}
$$

Here $e^{j\omega T} + e^{-j\omega T} = 2\cos\omega T$ was used.

Multiplying numerator and denominator by $z^2$:

$$
X(z) = \frac{z(z - \cos\omega T)}{z^2 - 2z\cos\omega T + 1}, \quad |z| > 1
$$

#### Remarks

- The poles are at $z = e^{\pm j\omega T}$, on the unit circle. This is expected for an undamped oscillation.
- Check: $x(0) = \lim_{z\to\infty}X(z) = 1 = \cos 0$.
- In the same way, $\mathcal{Z}[\sin\omega kT] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$.

**Answer:** $X(z) = \dfrac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \dfrac{z(z-\cos\omega T)}{z^2 - 2z\cos\omega T + 1}$.

## Check yourself

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

1. How is cos(wkT) expressed to find its z-transform?
   A. (1/2)(e^{jwkT} + e^{-jwkT})
   B. (1/2j)(e^{jwkT} - e^{-jwkT})
   C. e^{jwkT} + e^{-jwkT}
   D. (1/2)(e^{jwkT} - e^{-jwkT})
   Answer: A. Cosine is half the sum of the complex exponentials; the 1/(2j) form with a minus sign is for sine. (Key point: Euler's formula)

2. What is the z-transform of cos(wkT)?
   A. z(z - cos wT)/(z^2 - 2z cos wT + 1)
   B. z sin wT/(z^2 - 2z cos wT + 1)
   C. z(z + cos wT)/(z^2 - 2z cos wT + 1)
   D. z(z - cos wT)/(z^2 - 2z cos wT - 1)
   Answer: A. Combining the two geometric-series fractions and using e^{jwT}+e^{-jwT} = 2cos wT gives this result. (Key point: Resulting transform)

3. Where are the poles of the z-transform of cos(wkT)?
   A. z = e^{±jwT}, on the unit circle
   B. z = ±1
   C. Inside the unit circle at radius e^{-wT}
   D. At z = 0
   Answer: A. The denominator z^2 - 2z cos wT + 1 has roots e^{±jwT}, matching an undamped oscillation. (Key point: Pole locations)

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