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

# DTFT convolution property

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2082 Chaitra, 4 marks; 2075 Baisakh, 6 marks; 2083 Bhadra (new course), 4 marks (asked 3 times)
- Video: [Properties of Discrete Time Fourier Transform | Convolution Property | Signals and Systems](https://www.youtube.com/watch?v=IBl-oUYg760) — 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-convolution-property
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

State and prove the convolution property of discrete time Fourier transform (convolution in time domain results in multiplication in frequency domain).

## Short answer

For DTFT: x[n]*h[n] ↔ X(e^{jω})H(e^{jω}). Proof: insert convolution sum into analysis equation, exchange summation order, let m=n-k to get e^{-jωk}H(e^{jω}), then sum over k yields X(e^{jω})H(e^{jω}).

## Answer

#### Statement
If $x[n] \xleftrightarrow{DTFT} X(e^{j\omega})$ and $h[n] \xleftrightarrow{DTFT} H(e^{j\omega})$, then
$$
y[n] = x[n]*h[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k] \;\xleftrightarrow{DTFT}\; Y(e^{j\omega}) = X(e^{j\omega})H(e^{j\omega})
$$
Convolution in time becomes multiplication in frequency.

#### Proof
Using $X(e^{j\omega}) = \sum_n x[n]e^{-j\omega n}$:
$$
Y(e^{j\omega}) = \sum_{n=-\infty}^{\infty}\left[\sum_{k=-\infty}^{\infty}x[k]h[n-k]\right]e^{-j\omega n}
$$
Interchange the order of summation:
$$
Y(e^{j\omega}) = \sum_{k=-\infty}^{\infty}x[k]\sum_{n=-\infty}^{\infty}h[n-k]e^{-j\omega n}
$$
Put $m = n - k$ in the inner sum (so $n = m + k$):
$$
\sum_{m=-\infty}^{\infty}h[m]e^{-j\omega(m+k)} = e^{-j\omega k}H(e^{j\omega})
$$
Hence
$$
Y(e^{j\omega}) = H(e^{j\omega})\sum_{k=-\infty}^{\infty}x[k]e^{-j\omega k} = X(e^{j\omega})H(e^{j\omega})
$$

#### Significance and example
- $H(e^{j\omega})$ is the frequency response of the DT LTI system. The output spectrum is the input spectrum shaped by $H$.
- **Example:** with $x[n] = a^n u[n]$ and $h[n] = b^n u[n]$ ($|a|, |b| < 1$):
$$
Y(e^{j\omega}) = \frac{1}{(1-ae^{-j\omega})(1-be^{-j\omega})}
$$
Partial fractions give $y[n] = \frac{a^{n+1} - b^{n+1}}{a - b}u[n]$ for $a \ne b$, which matches the direct convolution sum.

## Check yourself

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

1. If y[n] = x[n] * h[n], then its DTFT is:
   A. X(e^{jω}) * H(e^{jω})
   B. X(e^{jω}) H(e^{jω})
   C. X(e^{jω}) + H(e^{jω})
   D. X(e^{jω}) / H(e^{jω})
   Answer: B. Convolution in time becomes multiplication in frequency. (Key point: Statement)

2. In the proof, the inner sum Σ h[n-k]e^{-jωn} with m = n - k becomes:
   A. H(e^{jω})
   B. e^{-jωk} H(e^{jω})
   C. e^{jωk} H(e^{jω})
   D. h[k]
   Answer: B. e^{-jωn} = e^{-jω(m+k)}, so the factor e^{-jωk} comes out and the rest is H(e^{jω}). (Key point: Proof substitution)

3. After taking out H(e^{jω}), what remains is Σ x[k] e^{-jωk}. This is:
   A. H(e^{jω})
   B. X(e^{jω})
   C. Y(e^{jω})
   D. x[0]
   Answer: B. By definition it is the DTFT of x[n], giving Y = X H. (Key point: Last step of proof)

4. For x[n] = a^n u[n] and h[n] = b^n u[n], Y(e^{jω}) is:
   A. 1/(1 - (a+b)e^{-jω})
   B. 1/((1 - a e^{-jω})(1 - b e^{-jω}))
   C. 1/(1 - ab e^{-jω})
   D. (1 - a e^{-jω})(1 - b e^{-jω})
   Answer: B. Each factor 1/(1 - a e^{-jω}) is the DTFT of a^n u[n], and they multiply. (Key point: Example)

## Close variants set in other exams

- [CTFT convolution property](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-convolution-property) — Signals and Systems, chapter 3, asked 12×

## Related in Signals and Systems

- [DTFT properties; convolution property](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-properties-convolution) — Signals and Systems, chapter 3, asked 2×
- [CTFT convolution; DTFT of 2ⁿu[n]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actft-convolution-dtft-2n) — Signals and Systems, chapter 3
- [Fourier series vs transform; DTFT of aⁿ](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Afs-vs-ft-dtft-geometric) — Signals and Systems, chapter 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×
- [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
- [Convolution property of CTFS](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Actfs-convolution-property) — Signals and Systems, chapter 2

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