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

# DTFT linearity and time shifting

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2081 Chaitra, 6 marks
- Video: [Properties of Discrete Time Fourier Transform | Time Shifting Property | Signals and Systems](https://www.youtube.com/watch?v=_gR-mI3EUTs) — Ekeeda
- 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-linearity-time-shift
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

State and prove the linearity and time shifting properties of DTFT

## Short answer

Linearity: a x₁[n]+b x₂[n] ↔ a X₁(e^{jω})+b X₂(e^{jω}) by linearity of summation. Time shifting: x[n−n₀] ↔ e^{−jω n₀} X(e^{jω}), adding linear phase −ω n₀ without changing magnitude spectrum.

## Answer

Let $x_1[n] \leftrightarrow X_1(e^{j\omega})$ and $x_2[n] \leftrightarrow X_2(e^{j\omega})$, where $X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$.

#### Linearity
**Statement:** for constants $a$ and $b$,
$$
a\,x_1[n] + b\,x_2[n] \xleftrightarrow{DTFT} a\,X_1(e^{j\omega}) + b\,X_2(e^{j\omega})
$$
**Proof:**
$$
\begin{aligned}
\sum_{n=-\infty}^{\infty}\left(a\,x_1[n] + b\,x_2[n]\right)e^{-j\omega n} &= a\sum_{n}x_1[n]e^{-j\omega n} + b\sum_{n}x_2[n]e^{-j\omega n} \\
&= a\,X_1(e^{j\omega}) + b\,X_2(e^{j\omega})
\end{aligned}
$$
The step is valid because summation is a linear operation.

**Example:** $x[n] = \delta[n] + 2(0.5)^n u[n]$ has
$$
X(e^{j\omega}) = 1 + \frac{2}{1 - 0.5e^{-j\omega}}
$$

#### Time shifting
**Statement:** for an integer delay $n_0$,
$$
x[n - n_0] \xleftrightarrow{DTFT} e^{-j\omega n_0}X(e^{j\omega})
$$
**Proof:** let $y[n] = x[n - n_0]$:
$$
Y(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n - n_0]e^{-j\omega n}
$$
Put $m = n - n_0$. The limits stay $-\infty$ to $\infty$:
$$
\begin{aligned}
Y(e^{j\omega}) &= \sum_{m=-\infty}^{\infty}x[m]e^{-j\omega(m + n_0)} \\
&= e^{-j\omega n_0}\sum_{m=-\infty}^{\infty}x[m]e^{-j\omega m} = e^{-j\omega n_0}X(e^{j\omega})
\end{aligned}
$$

**Interpretation:**
- $|Y(e^{j\omega})| = |X(e^{j\omega})|$, so a delay does not change the magnitude spectrum.
- $\angle Y(e^{j\omega}) = \angle X(e^{j\omega}) - \omega n_0$. A delay adds a phase that is linear in $\omega$. This is why a linear-phase filter delays all frequency components by the same amount and causes no phase distortion.

**Example:** $\delta[n - 3] \leftrightarrow e^{-j3\omega}$, and $(0.5)^{n-1}u[n-1] \leftrightarrow \frac{e^{-j\omega}}{1 - 0.5e^{-j\omega}}$.

## Check yourself

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

1. Why is the DTFT linear?
   A. Summation is a linear operation, so a and b come out of the sums
   B. Because e^(-jωn) is real
   C. Because the DTFT is periodic
   D. Because x[n] must be causal
   Answer: A. The sum of (a x1 + b x2)e^(-jωn) splits into a times one sum plus b times the other. (Key point: Why linearity holds)

2. The DTFT of x[n - n0] is:
   A. e^(jωn0) X(e^(jω))
   B. e^(-jωn0) X(e^(jω))
   C. X(e^(j(ω - n0)))
   D. n0 X(e^(jω))
   Answer: B. With m = n - n0, e^(-jω(m+n0)) factors as e^(-jωn0) e^(-jωm). (Key point: Time-shift result)

3. What does a delay of n0 samples do to the spectrum?
   A. Changes magnitude by n0
   B. Leaves magnitude unchanged and subtracts ωn0 from the phase
   C. Shifts the spectrum by n0 in frequency
   D. Makes the spectrum real
   Answer: B. |e^(-jωn0)| = 1 so magnitude is unchanged, while the phase gets a term linear in ω. (Key point: Effect of delay)

4. What is the DTFT of δ[n - 3]?
   A. e^(j3ω)
   B. 3
   C. e^(-j3ω)
   D. 1/(1 - e^(-j3ω))
   Answer: C. δ[n] has DTFT 1; the 3-sample delay multiplies it by e^(-j3ω). (Key point: Application)

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