---
title: "DTFT of cos ω₀n"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-cosine"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-cosine
site: BCT Engineering
---

# DTFT of cos ω₀n

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2081 Asoj, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-cosine
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find Fourier Transform of DT signal cos ω₀n

## Short answer

The DTFT of cos ω₀n is an impulse train: X(e^jω) = π ∑_l [δ(ω−ω₀−2πl) + δ(ω+ω₀−2πl)], periodic with 2π. Within one period [−π,π) there are two impulses of area π at ±ω₀.

## Answer

The signal $x[n] = \cos\omega_0 n$ is not absolutely summable, so its DTFT contains impulses.

#### Step 1: Euler's formula
$$
\cos\omega_0 n = \frac{1}{2}e^{j\omega_0 n} + \frac{1}{2}e^{-j\omega_0 n}
$$

#### Step 2: DTFT of a complex exponential
$$
e^{j\omega_0 n} \xleftrightarrow{DTFT} 2\pi\sum_{l=-\infty}^{\infty}\delta(\omega - \omega_0 - 2\pi l)
$$
**Check:** over one period the inverse DTFT is $\frac{1}{2\pi}\int_{2\pi}2\pi\delta(\omega - \omega_0)e^{j\omega n}d\omega = e^{j\omega_0 n}$. The impulses repeat every $2\pi$ because every DTFT is $2\pi$-periodic.

#### Step 3: Linearity
$$
\boxed{X(e^{j\omega}) = \pi\sum_{l=-\infty}^{\infty}\left[\delta(\omega - \omega_0 - 2\pi l) + \delta(\omega + \omega_0 - 2\pi l)\right]}
$$
In the range $-\pi \le \omega < \pi$ this is two impulses of area $\pi$ at $\omega = \pm\omega_0$.

```text
 X(e^jw)
     pi          pi
      ^           ^
      |           |
 --+--+-----+-----+--+--- w
  -pi -w0   0    w0  pi
 (pattern repeats every 2pi)
```

## Check yourself

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

1. Why does the DTFT of cos(ω0 n) contain impulses?
   A. Because it is not absolutely summable
   B. Because it is causal
   C. Because it is complex
   D. Because ω0 is irrational
   Answer: A. cos(ω0 n) does not decay, so the DTFT sum does not converge normally; it exists only in the impulse sense. (Key point: Why impulses)

2. The DTFT of e^(jω0 n) is:
   A. δ(ω - ω0) in one period only
   B. 2π Σ δ(ω - ω0 - 2πl)
   C. π Σ δ(ω - ω0 - 2πl)
   D. 1/(1 - e^(-j(ω - ω0)))
   Answer: B. The inverse DTFT over one period of 2π δ(ω - ω0) gives e^(jω0 n); impulses repeat every 2π. (Key point: Complex exponential pair)

3. In -π <= ω < π, the DTFT of cos(ω0 n) is:
   A. Two impulses of area π at ±ω0
   B. Two impulses of area 2π at ±ω0
   C. One impulse of area π at ω0
   D. Two impulses of area 1/2 at ±ω0
   Answer: A. Each of the two exponentials has weight 1/2, giving 2π(1/2) = π at ω0 and -ω0. (Key point: Final result)

4. Why do the impulses repeat every 2π?
   A. Because cos is periodic in n
   B. Because every DTFT is periodic with period 2π
   C. Because of Euler's formula
   D. Because ω0 is small
   Answer: B. The DTFT of any discrete signal is 2π-periodic. (Key point: Periodicity)

## Close variants set in other exams

- [Fourier transform of e^(−at)cos ω₀t](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-damped-cosine) — Signals and Systems, chapter 3, asked 4×
- [DTFT of aⁿu[n]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-geometric-sequence) — Signals and Systems, chapter 3, asked 4×
- [Fourier transform of e^(−2|t−1|)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-shifted-two-sided-exponential) — Signals and Systems, chapter 3, asked 4×

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