---
title: "DTFT conjugation and frequency shifting"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-conjugate-frequency-shift"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-conjugate-frequency-shift
site: BCT Engineering
---

# DTFT conjugation and frequency shifting

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2082 Chaitra, 4 marks
- Video: [4.12 Time shifting property of DTFT | frequency shifting property of DTFT | properties with proof](https://www.youtube.com/watch?v=yw0pQ9ap9OA) — JOTHI ECE VIDEOS
- 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-conjugate-frequency-shift
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

State and prove conjugate and frequency shifting properties of discrete time Fourier transform.

## Short answer

Conjugation: x*[n] ↔ X*(e^{−jω}) (sum conjugate). Frequency shifting: e^{jω₀n}x[n] ↔ X(e^{j(ω−ω₀)}) (exponential factor). Both properties follow directly from the DTFT definition.

## Answer

Let $x[n] \xleftrightarrow{DTFT} X(e^{j\omega}) = \sum_n x[n]e^{-j\omega n}$.

#### Conjugation property
**Statement:**
$$
x^*[n] \xleftrightarrow{DTFT} X^*(e^{-j\omega})
$$
**Proof:**
$$
\sum_{n}x^*[n]e^{-j\omega n} = \left[\sum_{n}x[n]e^{j\omega n}\right]^* = \left[\sum_{n}x[n]e^{-j(-\omega)n}\right]^* = X^*(e^{-j\omega})
$$
**Consequence:** if $x[n]$ is real, $X(e^{j\omega}) = X^*(e^{-j\omega})$. The magnitude is then even and the phase is odd.

#### Frequency shifting property
**Statement:**
$$
e^{j\omega_0 n}x[n] \xleftrightarrow{DTFT} X(e^{j(\omega - \omega_0)})
$$
**Proof:**
$$
\sum_{n}e^{j\omega_0 n}x[n]e^{-j\omega n} = \sum_{n}x[n]e^{-j(\omega - \omega_0)n} = X(e^{j(\omega - \omega_0)})
$$
Multiplying by $e^{j\omega_0 n}$ shifts the whole periodic spectrum by $\omega_0$. **Example:** $(-1)^n x[n] = e^{j\pi n}x[n] \leftrightarrow X(e^{j(\omega - \pi)})$, which turns a low-pass spectrum into a high-pass one.

## Check yourself

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

1. The DTFT of x*[n] is:
   A. X*(e^(jω))
   B. X*(e^(-jω))
   C. X(e^(-jω))
   D. -X*(e^(jω))
   Answer: B. Conjugating the sum turns e^(-jωn) into e^(jωn) = e^(-j(-ω)n), giving X*(e^(-jω)). (Key point: Conjugation property)

2. For a real x[n], the DTFT magnitude and phase are:
   A. Magnitude odd, phase even
   B. Magnitude even, phase odd
   C. Both even
   D. Both odd
   Answer: B. X(e^(jω)) = X*(e^(-jω)) means magnitude is even and phase is odd. (Key point: Real signal consequence)

3. The DTFT of e^(jω0 n) x[n] is:
   A. X(e^(j(ω + ω0)))
   B. X(e^(j(ω - ω0)))
   C. e^(-jωω0) X(e^(jω))
   D. X(e^(jω0))
   Answer: B. The exponent combines to e^(-j(ω - ω0)n), so the spectrum shifts by ω0. (Key point: Frequency shift)

4. Multiplying x[n] by (-1)^n does what to a low-pass spectrum?
   A. Leaves it unchanged
   B. Shifts it by π, making it high-pass
   C. Reverses its phase only
   D. Halves its amplitude
   Answer: B. (-1)^n = e^(jπn), which shifts the spectrum by π so low frequencies move to the highest frequency. (Key point: (-1)^n modulation)

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