7.4 Discrete time circular convolution and the dtfs

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

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 formula

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

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

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

Alternative convolution formula

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: $y(n)=\sum_{m=0}^{N-1} f(m)h({\left(\right(n-m\left)\right)()}_N)$ 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 $$\to (f(n), f({((n+m))}_N))$$

Notes on circular shifting

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

$f({((n+N))}_N)=f({((n-(N-m)))}_N)$ Shifting ACW $m$ is equivalent to shifting CW $N-m$

$f({((-n))}_N)$ The above expression, simply writes the values of $f(n)$ clockwise.

Convolve (n = 4)

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

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

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

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

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

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

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

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

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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 $a_k=c_k^2=\frac{1}{4}\frac{\sin (\frac{\pi }{2}k)^2}{(\frac{\pi }{2}k)^2}$

Circular shifts and the dft

Circular shifts and dft

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

Circular convolution demonstration

Conclusion

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