---
title: "Short note: Frequency shifting of DTFT"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:short-note-dtft-frequency-shifting"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/short-note-dtft-frequency-shifting
site: BCT Engineering
---

# Short note: Frequency shifting of DTFT

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2076 Baisakh, 3 marks
- Video: [Frequency shifting property of DTFT ll Example and solution](https://www.youtube.com/watch?v=hAHE3FnBSpQ) — easy study plus
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/short-note-dtft-frequency-shifting
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Write a short note on the frequency shifting property of DTFT.

## Short answer

Multiplying a sequence by e^{jω0 n} shifts its DTFT: e^{jω0 n}x[n] ↔ X(e^{j(ω-ω0)}). Proof: Σ e^{jω0 n}x[n]e^{-jωn} = Σ x[n]e^{-j(ω-ω0)n} = X(e^{j(ω-ω0)}). The spectrum moves rigidly; modulation by cos(ω0 n) yields two copies at ±ω0.

## Answer

**Frequency shifting property:** Multiplying a sequence by a complex exponential $e^{j\omega_0 n}$ shifts its DTFT by $\omega_0$:

$$
x[n] \leftrightarrow X(e^{j\omega}) \;\Rightarrow\; e^{j\omega_0 n}x[n] \leftrightarrow X(e^{j(\omega-\omega_0)})
$$

**Proof:**

$$
\begin{aligned}
\sum_{n=-\infty}^{\infty} e^{j\omega_0 n}x[n]\,e^{-j\omega n} &= \sum_{n=-\infty}^{\infty} x[n]\,e^{-j(\omega-\omega_0)n} \\
&= X(e^{j(\omega-\omega_0)})
\end{aligned}
$$

**Points to note:**

- The whole spectrum moves right by $\omega_0$; the shape does not change. The shifted spectrum is still periodic with period $2\pi$.
- It is the dual of the time-shifting property ($x[n-n_0] \leftrightarrow e^{-j\omega n_0}X(e^{j\omega})$).
- **Modulation:** Since $\cos\omega_0 n = \frac{1}{2}(e^{j\omega_0 n} + e^{-j\omega_0 n})$,
  $x[n]\cos\omega_0 n \leftrightarrow \frac{1}{2}\left[X(e^{j(\omega-\omega_0)}) + X(e^{j(\omega+\omega_0)})\right]$.
- Special case $\omega_0 = \pi$: $(-1)^n x[n] \leftrightarrow X(e^{j(\omega-\pi)})$, which turns a low-pass spectrum into a high-pass one.

**Example:** $a^n u[n] \leftrightarrow \frac{1}{1-ae^{-j\omega}}$, so $e^{j\pi n/4}a^n u[n] \leftrightarrow \frac{1}{1-ae^{-j(\omega-\pi/4)}}$; the peak moves from $\omega = 0$ to $\omega = \pi/4$.

## Check yourself

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

1. If x[n] <-> X(e^{jω}), what is the DTFT of e^{jω0 n} x[n]?
   A. X(e^{j(ω-ω0)})
   B. X(e^{j(ω+ω0)})
   C. e^{jω0} X(e^{jω})
   D. X(e^{jω0 ω})
   Answer: A. Multiplying by e^{jω0 n} shifts the spectrum right by ω0. (Key point: Statement)

2. What is the DTFT of x[n] cos(ω0 n)?
   A. X(e^{j(ω-ω0)}) only
   B. (1/2)[X(e^{j(ω-ω0)}) + X(e^{j(ω+ω0)})]
   C. (1/2)[X(e^{j(ω-ω0)}) - X(e^{j(ω+ω0)})]
   D. cos(ω0) X(e^{jω})
   Answer: B. cos is the sum of two complex exponentials, each with half amplitude, giving two shifted copies. (Key point: Modulation)

3. What does multiplying a sequence by (-1)^n do to a low-pass spectrum?
   A. Reverses it in time
   B. Leaves it unchanged
   C. Shifts it by π, making it high-pass
   D. Halves its bandwidth
   Answer: C. (-1)^n = e^{jπn}, so the spectrum shifts by π and the low-frequency content moves to ω = π. (Key point: Special case ω0 = π)

## Related in Signals and Systems

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- [Time shifting property of CTFT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-time-shifting) — Signals and Systems, chapter 3, asked 5×

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