---
title: "Frequency shifting property of DTFS"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:dtfs-frequency-shifting"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-frequency-shifting
site: BCT Engineering
---

# Frequency shifting property of DTFS

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2081 Chaitra, 3 marks
- Video: [Frequency Shift | Properties of DTFS | EnggClasses](https://www.youtube.com/watch?v=50O1EuLXo4Q) — EnggClasses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-frequency-shifting
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

State and prove frequency shifting property of discrete time Fourier series.

## Short answer

Multiplying by e^{j M (2π/N) n} shifts coefficients: e^{j M ω0 n} x[n] ↔ a_{k-M}. Proof: factor e^{-j(k-M)ω0 n} in the DTFS sum.

## Answer

#### Statement
If $x[n]$ is periodic with period $N$ and $x[n] \overset{FS}{\longleftrightarrow} a_k$, then for an integer $M$:

$$
e^{jM(2\pi/N)n}\,x[n] \overset{FS}{\longleftrightarrow} a_{k-M}
$$

Multiplying by a complex exponential at the $M$th harmonic shifts the coefficient sequence by $M$.

#### Proof
$y[n] = e^{jM(2\pi/N)n}x[n]$ is also periodic with period $N$ (both factors are). Its coefficients are

$$
\begin{aligned}
b_k &= \frac{1}{N}\sum_{n=\langle N\rangle}e^{jM(2\pi/N)n}x[n]\,e^{-jk(2\pi/N)n} \\
&= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-j(k-M)(2\pi/N)n} \\
&= a_{k-M}
\end{aligned}
$$

**Example:** for $N = 4$, multiplying by $e^{j\pi n/2}$ ($M = 1$) moves $a_0$ to position $k = 1$, $a_1$ to $k = 2$, and so on (cyclically, since $a_k$ has period 4).

## Check yourself

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

1. What is the DTFS of e^(jM(2pi/N)n) x[n]?
   A. a_(k+M)
   B. a_(k-M)
   C. a_k e^(jM)
   D. M a_k
   Answer: B. The sequence shifts by M places. (Key point: Statement)

2. How does the exponential disappear in the proof?
   A. It combines with e^(-jk(2pi/N)n) to give e^(-j(k-M)(2pi/N)n)
   B. It integrates to zero
   C. It equals 1
   D. It is conjugated
   Answer: A. Combining exponents makes the sum the analysis equation at index k - M. (Key point: Proof step)

3. Why is the shift cyclic in DT?
   A. Because a_k has period N
   B. Because x[n] is real
   C. Because M is real
   D. Because N is odd
   Answer: A. a_(k+N) = a_k, so shifting wraps around. (Key point: Cyclic shift)

## Close variants set in other exams

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

## Related in Signals and Systems

- [DTFS frequency shifting and convolution](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-shifting-convolution) — Signals and Systems, chapter 2
- [Frequency shifting property of CTFT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-frequency-shifting) — Signals and Systems, chapter 3, asked 3×
- [Derive the discrete-time Fourier series](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-derivation) — Signals and Systems, chapter 2, asked 15×
- [Time-scaling property of DTFS](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtfs-time-scaling) — Signals and Systems, chapter 2, asked 6×
- [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×
- [DTFT conjugation and frequency shifting](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-conjugate-frequency-shift) — Signals and Systems, chapter 3

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