# 7.4 Discrete time circular convolution and the dtfs

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This module describes the circular convolution algorithm and an alternative algorithm

## 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 $x(n)$ , the system's input, and $h(n)$ , the system's response, we define the output of the system as

$y(n)=(x(n), h(n))=\sum_{k=()}$ x k h n k
When we are given two DFTs (finite-length sequences usually oflength $N$ ), we cannot just multiply them together as we do in the above convolutionformula, often referred to as linear convolution . Because the DFTs are periodic, they have nonzero values for $n\ge N$ and thus the multiplication of these two DFTs will be nonzero for $n\ge N$ . We need to define a new type of convolution operation that will result in our convolved signal being zerooutside of the range $n=\{0, 1, \dots , N-1\}$ . This idea led to the development of circular convolution , also called cyclic or periodic convolution.

## Signal circular convolution

Given a signal $f(n)$ with Fourier coefficients ${c}_{k}$ and a signal $g(n)$ with Fourier coefficients ${d}_{k}$ , we can define a new signal, $v(n)$ , where $v(n)=⊛(f(n), g(n))$ We find that the Fourier Series representation of $v(n)$ , ${a}_{k}$ , is such that ${a}_{k}={c}_{k}{d}_{k}$ . $⊛(f(n), g(n))$ is the circular convolution of two periodic signals and is equivalent to the convolution over one interval, i.e. $⊛(f(n), g(n))=\sum_{n=0}^{N} \sum_{\eta =0}^{N} f(\eta )g(n-\eta )$ .

Circular convolution in the time domain is equivalent to multiplication of the Fourier coefficients.
This is proved as follows
${a}_{k}=\frac{1}{N}\sum_{n=0}^{N} v(n)e^{-(j{\omega }_{0}kn)}=\frac{1}{N^{2}}\sum_{n=0}^{N} \sum_{\eta =0}^{N} f(\eta )g(n-\eta )e^{-(\omega {j}_{0}kn)}=\frac{1}{N}\sum_{\eta =0}^{N} f(\eta )\frac{1}{N}\sum_{n=0}^{N} g(n-\eta )e^{-(j{\omega }_{0}kn)}=\forall \nu , \nu =n-\eta \colon \frac{1}{N}\sum_{\eta =0}^{N} f(\eta )\frac{1}{N}\sum_{\nu =-\eta }^{N-\eta } g(\nu )e^{-(j{\omega }_{0}(\nu +\eta ))}=\frac{1}{N}\sum_{\eta =0}^{N} f(\eta )\frac{1}{N}\sum_{\nu =-\eta }^{N-\eta } g(\nu )e^{-(j{\omega }_{0}k\nu )}e^{-(j{\omega }_{0}k\eta )}=\frac{1}{N}\sum_{\eta =0}^{N} f(\eta )d_{k}e^{-(j{\omega }_{0}k\eta )}={d}_{k}\frac{1}{N}\sum_{\eta =0}^{N} f(\eta )e^{-(j{\omega }_{0}k\eta )}={c}_{k}{d}_{k}$

## Circular convolution formula

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

$Y(k)=F(k)H(k)$
when $0\le k\le N-1$

Using the DFT synthesis formula for $y(n)$

$y(n)=\frac{1}{N}\sum_{k=0}^{N-1} F(k)H(k)e^{(j\frac{2\pi }{N}kn)}$

And then applying the analysis formula $F(k)=\sum_{m=0}^{N-1} f(m)e^{(-j\frac{2\pi }{N}kn)}$

$y(n)=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} f(m)e^{(-j\frac{2\pi }{N}kn)}H(k)e^{(j\frac{2\pi }{N}kn)}=\sum_{m=0}^{N-1} f(m)\frac{1}{N}\sum_{k=0}^{N-1} H(k)e^{(j\frac{2\pi }{N}k(n-m))}$
where we can reduce the second summation found in the above equation into $h({\left(\left(n-m\right)\right)()}_{N})=\frac{1}{N}\sum_{k=0}^{N-1} H(k)e^{(j\frac{2\pi }{N}k(n-m))}$ $y(n)=\sum_{m=0}^{N-1} f(m)h({\left(\left(n-m\right)\right)()}_{N})$ which equals circular convolution! When we have $0\le n\le N-1$ in the above, then we get:
$y(n)\equiv ⊛(f(n), h(n))$
The notation $⊛$ represents cyclic convolution "mod N".

## Alternative circular convolution algorithm

• Step 1: Calculate the DFT of $f(n)$ which yields $F(k)$ and calculate the DFT of $h(n)$ which yields $H(k)$ .
• Step 2: Pointwise multiply $Y(k)=F(k)H(k)$
• Step 3: Inverse DFT $Y(k)$ which yields $y(n)$

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 $2$ $N$ -point sequences: $y(n)=\sum_{m=0}^{N-1} f(m)h({\left(\left(n-m\right)\right)()}_{N})$ For each $n$ : $N$ multiples, $N-1$ additions

$N$ points implies $N^{2}$ multiplications, $N(N-1)$ additions implies $O(N^{2})$ complexity.

## Steps for circular convolution

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

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

To cyclic shift we follow these steps:

1) Write $f(n)$ on a cylinder, ACW

2) To cyclic shift by $m$ , spin cylinder m spots ACW $\to (f(n), f({\left(\left(n+m\right)\right)}_{N}))$

## Notes on circular shifting

$f({\left(\left(n+N\right)\right)}_{N})=f(n)$ Spinning $N$ spots is the same as spinning all the way around, or not spinning at all.

$f({\left(\left(n+N\right)\right)}_{N})=f({\left(\left(n-\left(N-m\right)\right)\right)}_{N})$ Shifting ACW $m$ is equivalent to shifting CW $N-m$

$f({\left(\left(-n\right)\right)}_{N})$ The above expression, simply writes the values of $f(n)$ clockwise.

## Convolve (n = 4)

• $h({\left((\left(, -, m, \right), \right))}_{N})$

Multiply $f(m)$ and $\mathrm{sum}$ to yield: $y(0)=3$

• $h({\left((\left(, 1, -, m, \right), \right))}_{N})$

Multiply $f(m)$ and $\mathrm{sum}$ to yield: $y(1)=5$

• $h({\left((\left(, 2, -, m, \right), \right))}_{N})$

Multiply $f(m)$ and $\mathrm{sum}$ to yield: $y(2)=3$

• $h({\left((\left(, 3, -, m, \right), \right))}_{N})$

Multiply $f(m)$ and $\mathrm{sum}$ to yield: $y(3)=1$

## Exercise

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

For this signal ${c}_{k}=\begin{cases}\frac{1}{N} & \text{if k=0}\\ \frac{1}{2}\frac{\sin (\frac{\pi }{2}k)}{\frac{\pi }{2}k} & \text{otherwise}\end{cases}$

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 ${a}_{k}={c}_{k}^{2}=\frac{1}{4}\frac{\sin (\frac{\pi }{2}k)^{2}}{(\frac{\pi }{2}k)^{2}}$

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

${a}_{k}=\left\{\begin{array}{cc}\text{undefined}\hfill & k=0\hfill \\ \frac{1}{8}\frac{si{n}^{3}\left[\frac{\pi }{2}k\right]}{{\left[\frac{\pi }{2}k\right]}^{3}}\hfill & \text{otherwise}\hfill \end{array}\right)$

## Circular shifts and dft

If $f(n)\stackrel{\text{DFT}}{↔}F(k)$ then $f({\left(\left(n-m\right)\right)}_{N})\stackrel{\text{DFT}}{↔}e^{-(i\frac{2\pi }{N}km)}F(k)$ ( i.e. circular shift in time domain = phase shift in DFT)

$f(n)=\frac{1}{N}\sum_{k=0}^{N-1} F(k)e^{i\frac{2\pi }{N}kn}$
so phase shifting the DFT
$f(n)=\frac{1}{N}\sum_{k=0}^{N-1} F(k)e^{-(i\frac{2\pi }{N}kn)}e^{i\frac{2\pi }{N}kn}=\frac{1}{N}\sum_{k=0}^{N-1} F(k)e^{i\frac{2\pi }{N}k(n-m)}=f({\left(\left(n-m\right)\right)}_{N})$

## Conclusion

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

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