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

# Convolution property of DTFS

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2078 Chaitra, 2+3 marks
- Video: [Convolution | Properties of DTFS | EnggClasses](https://www.youtube.com/watch?v=OUj_d14LbI8) — 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-convolution-property
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

State and prove convolution property of discrete time periodic signals.

## Short answer

For periodic signals x[n] and y[n] with period N and FS coefficients a_k and b_k, their periodic convolution z[n] = Σ_r x[r] y[n−r] has coefficients c_k = N a_k b_k. The proof substitutes the convolution sum into the analysis equation and uses orthogonality of exponentials.

## Answer

#### Statement
Let $x[n]$ and $y[n]$ be periodic with the same period $N$, with $x[n] \overset{FS}{\longleftrightarrow} a_k$ and $y[n] \overset{FS}{\longleftrightarrow} b_k$. Their **periodic convolution**

$$
z[n] = \sum_{r=\langle N\rangle}x[r]\,y[n - r]
$$

is periodic with period $N$, and

$$
z[n] \overset{FS}{\longleftrightarrow} c_k = N a_k b_k
$$

Convolution in time becomes multiplication of coefficients.

#### Proof
With $\omega_0 = 2\pi/N$:

$$
\begin{aligned}
c_k &= \frac{1}{N}\sum_{n=\langle N\rangle}z[n]e^{-jk\omega_0 n} = \frac{1}{N}\sum_{n=\langle N\rangle}\sum_{r=\langle N\rangle}x[r]\,y[n-r]\,e^{-jk\omega_0 n} \\
&= \frac{1}{N}\sum_{r=\langle N\rangle}x[r]\,e^{-jk\omega_0 r}\sum_{n=\langle N\rangle}y[n-r]\,e^{-jk\omega_0(n-r)}
\end{aligned}
$$

Put $m = n - r$; since $y$ is periodic, $m$ still runs over one full period, so the inner sum is $\sum_{m=\langle N\rangle}y[m]e^{-jk\omega_0 m} = N b_k$. Then

$$
c_k = \frac{1}{N}\sum_{r=\langle N\rangle}x[r]e^{-jk\omega_0 r}\cdot N b_k = N\,a_k\,b_k
$$

Hence $\sum_{r=\langle N\rangle}x[r]y[n-r] \overset{FS}{\longleftrightarrow} N a_k b_k$. (The dual property: multiplication $x[n]y[n]$ gives the periodic convolution of coefficients $\sum_{l=\langle N\rangle}a_l b_{k-l}$.)

## Check yourself

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

1. If z[n] is the periodic convolution of x[n] and y[n] (period N), its DTFS coefficients are:
   A. a_k b_k
   B. N a_k b_k
   C. a_k + b_k
   D. (1/N) a_k b_k
   Answer: B. Periodic convolution in time becomes N a_k b_k in the coefficients. (Key point: Statement of property)

2. In the proof, why can the inner sum over m = n - r be taken over any one period?
   A. Because x[n] is zero outside one period
   B. Because y[m] e^{-jk w0 m} is periodic with period N
   C. Because r is always zero
   D. Because the exponent vanishes
   Answer: B. Since y and the exponential are both periodic with N, the summand is periodic and any N consecutive values give the same sum, N b_k. (Key point: Change of variable in proof)

3. Where does the factor N in c_k = N a_k b_k come from?
   A. From the 1/N in the definition of a_k multiplied by the inner sum equal to N b_k
   B. From the period of x[n] being N
   C. From the number of harmonics
   D. From Parseval's relation
   Answer: A. The inner sum equals N b_k, which cancels the leading 1/N, leaving N a_k b_k. (Key point: Origin of the factor N)

4. The dual of this property says that x[n]y[n] corresponds to:
   A. N a_k b_k
   B. a_k b_k
   C. Periodic convolution of the coefficients, sum of a_l b_{k-l}
   D. Sum of a_k + b_k
   Answer: C. Multiplication in time becomes periodic convolution of coefficients. (Key point: Dual property)

## Related in Signals and Systems

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