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# Discrete Time Circular Convolution and the DTFS

Module by: Justin Romberg. E-mail the author

Summary: This module describes the circular convolution algorithm and an alternative algorithm

Note: You are viewing an old version of this document. The latest version is available here.

## Introduction

This module relates circular convolution of periodic signals in one domain to multiplication in the other domain.

You should be familiar with Discrete-Time Convolution, which tells us that given two discrete-time signals xn x n , the system's input, and hn h n , the system's response, we define the output of the system as

yn=xn*hn= k =xkhnk y n x n h n k x k h n k
(1)
When we are given two DFTs (finite-length sequences usually of length NN), we cannot just multiply them together as we do in the above convolution formula, often referred to as linear convolution. Because the DFTs are periodic, they have nonzero values for nN n N and thus the multiplication of these two DFTs will be nonzero for nN n N . We need to define a new type of convolution operation that will result in our convolved signal being zero outside of the range n=01N1 n 0 1 N 1 . This idea led to the development of circular convolution, also called cyclic or periodic convolution.

## Signal Circular Convolution

Given a signal fn f n with Fourier coefficients c k c k and a signal gn g n with Fourier coefficients d k d k , we can define a new signal, vn v n , where vn=fngn v n f n g n We find that the Fourier Series representation of vn v n , a k a k , is such that a k = c k d k a k c k d k . fngn f n g n is the circular convolution of two periodic signals and is equivalent to the convolution over one interval, i.e. fngn= n =0N η =0Nfηgnη f n g n n 0 N η 0 N f η g n η .

### Note:

Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients.
This is proved as follows
a k =1N n =0Nvne(j ω 0 kn)=1N2 n =0N η =0Nfηgnηe(ω j 0 kn)=1N η =0Nfη(1N n =0Ngnηe(j ω 0 kn))= ν ,ν=nη:1N η =0Nfη(1N ν =ηNηgνe(j ω 0 (ν+η)))=1N η =0Nfη(1N ν =ηNηgνe(j ω 0 kν))e(j ω 0 kη)=1N η =0Nfηdke(j ω 0 kη)= d k (1N η =0Nfηe(j ω 0 kη))= c k d k a k 1 N n 0 N v n j ω 0 k n 1 N 2 n 0 N η 0 N f η g n η ω j 0 k n 1 N η 0 N f η 1 N n 0 N g n η j ω 0 k n ν ν n η 1 N η 0 N f η 1 N ν η N η g ν j ω 0 ν η 1 N η 0 N f η 1 N ν η N η g ν j ω 0 k ν j ω 0 k η 1 N η 0 N f η d k j ω 0 k η d k 1 N η 0 N f η j ω 0 k η c k d k
(2)

### Circular Convolution Formula

What happens when we multiply two DFT's together, where Yk Y k is the DFT of yn y n ?

Yk=FkHk Y k F k H k
(3)
when 0kN1 0 k N 1

Using the DFT synthesis formula for yn y n

yn=1N k =0N1FkHkej2πNkn y n 1 N k 0 N 1 F k H k j 2 N k n
(4)

And then applying the analysis formula Fk= m =0N1fme(j)2πNkn F k m 0 N 1 f m j 2 N k n

yn=1N k =0N1 m =0N1fme(j)2πNknHkej2πNkn= m =0N1fm(1N k =0N1Hkej2πNk(nm)) y n 1 N k 0 N 1 m 0 N 1 f m j 2 N k n H k j 2 N k n m 0 N 1 f m 1 N k 0 N 1 H k j 2 N k n m
(5)
where we can reduce the second summation found in the above equation into h ( ( n m ) ) N =1N k =0N1Hkej2πNk(nm) h ( ( n m ) ) N 1 N k 0 N 1 H k j 2 N k n m yn= m =0N1fmh ( ( n m ) ) N y n m 0 N 1 f m h ( ( n m ) ) N which equals circular convolution! When we have 0nN1 0 n N 1 in the above, then we get:
ynfnhn y n f n h n
(6)

#### Note:

The notation represents cyclic convolution "mod N".

#### Alternative Convolution Formula

##### Alternative Circular Convolution Algorithm
• Step 1: Calculate the DFT of fn f n which yields Fk F k and calculate the DFT of hn h n which yields Hk H k .
• Step 2: Pointwise multiply Yk=FkHk Y k F k H k
• Step 3: Inverse DFT Yk Y k which yields yn y n

Seems like a roundabout way of doing things, but it turns out that there are extremely fast ways to calculate the DFT of a sequence.

To circularily convolve 2 2 N N-point sequences: yn= m =0N1fmh ( ( n m ) ) N y n m 0 N 1 f m h ( ( n m ) ) N For each n n : N N multiples, N1 N 1 additions

N N points implies N2 N 2 multiplications, N(N1) N N 1 additions implies ON2 O N 2 complexity.

### Steps for Circular Convolution

We can picture periodic sequences as having discrete points on a circle as the domain

Shifting by mm, fn+m f n m , corresponds to rotating the cylinder mm notches ACW (counter clockwise). For m=-2 m -2 , we get a shift equal to that in the following illustration:

To cyclic shift we follow these steps:

1) Write fn f n on a cylinder, ACW

2) To cyclic shift by mm, spin cylinder m spots ACW fnf (( n + m )) N f n f (( n + m )) N

#### Notes on circular shifting

f (( n + N )) N =fn f (( n + N )) N f n Spinning NN spots is the same as spinning all the way around, or not spinning at all.

f (( n + N )) N =f (( n - ( N - m ) )) N f (( n + N )) N f (( n - ( N - m ) )) N Shifting ACW mm is equivalent to shifting CW Nm N m

f (( - n )) N f (( - n )) N The above expression, simply writes the values of fn f n clockwise.

#### Example 1: Convolve (n = 4)

• h (((m()() N h ( ( m ) ) N

Multiply fm f m and sum sum to yield: y0=3 y 0 3

• h ((1((m()() N h ( ( 1 m ) ) N

Multiply fm f m and sum sum to yield: y1=5 y 1 5

• h ((2((m()() N h ( ( 2 m ) ) N

Multiply fm f m and sum sum to yield: y2=3 y 2 3

• h ((3((m()() N h ( ( 3 m ) ) N

Multiply fm f m and sum sum to yield: y3=1 y 3 1

### Exercise

Take a look at a square pulse with a period of T.

For this signal c k ={1N  if  k=012sinπ2kπ2k  otherwise   c k 1 N k 0 1 2 2 k 2 k

Take a look at a triangle pulse train with a period of T.

This signal is created by circularly convolving the square pulse with itself. The Fourier coefficients for this signal are a k = c k 2=14sin2π2kπ2k2 a k c k 2 1 4 2 k 2 2 k 2

#### Exercise 1

Find the Fourier coefficients of the signal that is created when the square pulse and the triangle pulse are convolved.

##### Solution

a k = undefined k = 0 1 8 s i n 3 [ π 2 k ] [ π 2 k ] 3 otherwise a k = undefined k = 0 1 8 s i n 3 [ π 2 k ] [ π 2 k ] 3 otherwise

## Circular Shifts and the DFT

### Theorem 1: Circular Shifts and DFT

If fn DFT Fk f n DFT F k then f (( n - m )) N DFT e(i2πNkm)Fk f (( n - m )) N DFT 2 N k m F k (i.e. circular shift in time domain = phase shift in DFT)

#### Proof

fn=1Nk=0N1Fkei2πNkn f n 1 N k 0 N 1 F k 2 N k n
(7)
so phase shifting the DFT
fn=1Nk=0N1Fke(i2πNkn)ei2πNkn=1Nk=0N1Fkei2πNk(nm)=f (( n - m )) N f n 1 N k 0 N 1 F k 2 N k n 2 N k n 1 N k 0 N 1 F k 2 N k n m f (( n - m )) N
(8)

## Conclusion

Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients in the frequency domain.

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