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6.5: Discrete Time Circular Convolution and the DTFS

Last updated

Jun 4, 2024
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- 6.4: Properties of the DTFS
- 7: Discrete -Time Fourier Transform (DTFT)

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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 (Section 4.3), 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

$\begin{align} y[n] &=x[n] * h[n] \nonumber \\ &=\sum_{k=-\infty}^{\infty} x[k] h[n-k] \end{align} \nonumber$

When we are given two DFTs (finite-length sequences usually of length $N$ ), 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 $n≥N$ and thus the multiplication of these two DFTs will be nonzero for $n≥N$ . We need to define a new type of convolution operation that will result in our convolved signal being zero outside of the range $n={0,1,…,N−1}$ . This idea led to the development of circular convolution, also called cyclic or periodic convolution.

Signal Circular Convolution

Given a signal $f[n]$ with Fourier coefficients $c_k$ and a signal $g[n]$ with Fourier coefficients $d_k$ , we can define a new signal, $v[n]$ , where $v[n]=f[n] \circledast g[n]$ . We find that the Fourier Series representation of $v[n]$ , $a_k$ , is such that $a_k=c_kd_k$ . $f[n] \circledast g[n]$ is the circular convolution (Section 7.5) of two periodic signals and is equivalent to the convolution over one interval, i.e. $\displaystyle{f[n] \circledast g[n] = \sum_{n=0}^{N} \sum_{\eta=0}^{N} f[\eta] g[n-\eta]}$ .

Note

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

This is proved as follows

$\begin{align} a_{k} &=\frac{1}{N} \sum_{n=0}^{N} v[n] e^{-\left(j \omega_{0} k n\right)} \nonumber \\ &=\frac{1}{N^{2}} \sum_{n=0}^{N} \sum_{n=0}^{\eta} f[\eta] g[n-\eta] e^{-\left(j \omega_{0} k n\right)} \nonumber \\ &=\frac{1}{N} \sum_{\eta=0}^{N} f[\eta]\left(\frac{1}{N} \sum_{n=0}^{N} g[n-\eta] e^{-\left(j \omega_{0} k n\right)}\right) \nonumber \\ &=\left(\frac{1}{N} \sum_{\eta=0}^{N} f[\eta]\left(\frac{1}{N} \sum_{\nu=-\eta}^{N-\eta} g[\nu] e^{-\left(j \omega_{0}(\nu+\eta)\right)}\right)\right) \quad , \quad \nu = n - \eta \nonumber \\ &=\frac{1}{N} \sum_{\eta=0}^{N} f[\eta]\left(\frac{1}{N} \sum_{\nu=-\eta}^{N-\eta} g[\nu] e^{-\left(j \omega_{0} k \nu\right)}\right) e^{-\left(j \omega_{0} k \eta\right)} \nonumber \\ &=\frac{1}{N} \sum_{\eta=0}^{N} f[\eta] d_{k} e^{-\left(j \omega_{0} k \eta\right)} \nonumber \\ &=d_{k}\left(\frac{1}{N} \sum_{\eta=0}^{N} f[\eta] e^{-\left(j \omega_{0} k \eta\right)}\right) \nonumber \\ &=c_{k} d_{k} \end{align} \nonumber$

Circular Convolution Formula

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

$Y[k] = F[k]H[k] \nonumber$

when $0≤k≤N−1$

Using the DFT synthesis formula for $y[n]$

$y[n]=\frac{1}{N} \sum_{k=0}^{N-1} F[k] H[k] e^{j \frac{2 \pi}{N} k n} \nonumber$

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

$\begin{align} y[n] &=\frac{1}{N} \sum_{k=0}^{N-1} \sum_{m=0}^{N-1} f[m] e^{(-j) \frac{2 \pi}{N} k n} H[k] e^{j \frac{2 \pi}{N} k n} \nonumber \\ &=\sum_{m=0}^{N-1} f[m]\left(\frac{1}{N} \sum_{k=0}^{N-1} H[k] e^{j \frac{2 \pi}{N} k(n-m)}\right) \end{align} \nonumber$

where we can reduce the second summation found in the above equation into $h\left[((n-m))_{N}\right]=\frac{1}{N} \sum_{k=0}^{N-1} H[k] e^{j \frac{2 \pi}{N} k(n-m)} y[n]=\sum_{m=0}^{N-1} f[m] h\left[((n-m))_{N}\right]$ which equals circular convolution! When we have $0≤n≤N−1$ in the above, then we get:

$y[n] \equiv f[n] \circledast h[n] \nonumber$

Note

The notation $\circledast$ represents cyclic convolution "mod N".

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 circularly convolve 2 $N$ -point sequences: $y[n]=\sum_{m=0}^{N-1} f[m] h\left[((n-m))_{N}\right]$ . 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 (Section 6.1) 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

$f[n] \rightarrow f\left[((n+m))_{N}\right] \nonumber$

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 mm is equivalent to shifting CW $N−m$

$f[((−n))_N]$ The above expression, simply writes the values of $f[n]$ clockwise.

Example $\PageIndex{1}$

Convolve (n = 4)

Figure $\PageIndex{8}$ : Two discrete-time signals to be convolved.

$h[((−(m()()_N]$

Multiply $f[m]$ and sum to yield: $y[0]=3$

$h[((1(−(m()()_N]$

Multiply $f[m]$ and sum to yield: $y[1]=5$

$h[((2(−(m()()_N]$

Multiply $f[m]$ and sum to yield: $y[2]=3$

$h[((3(−(m()()_N]$

Multiply $f[m]$ and sum to yield: $y[3]=1$

Exercise

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

For this signal $c_{k}=\left\{\begin{array}{l} \frac{1}{N} \text { if } k=0 \\ \frac{1}{2} \frac{\sin \left(\frac{\pi}{2} k\right)}{\frac{\pi}{2} k} \text { otherwise } \end{array}\right.$

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 $a_{k}=c_{k}^{2}=\frac{1}{4} \frac{\sin ^{2}}{\left(\frac{x}{2} k\right)}$

Exercise $\PageIndex{1}$

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

Answer: $a_{k}=\left\{\begin{array}{ll} \text { undefined } & k=0 \\ \frac{1}{8} \frac{\sin ^{3}\left[\frac{\pi}{2} k\right]}{\left[\frac{\pi}{2} k\right]^{3}} & \text { otherwise } \end{array}\right.$

Circular Shifts and the DFT

Theorem $\PageIndex{1}$ : Circular Shifts and DFT

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

Proof

$f[n]=\frac{1}{N} \sum_{k=0}^{N-1} F[k] e^{j \frac{2 \pi}{N} k n} \nonumber$

so phase shifting the DFT

$\begin{align} f[n] &=\frac{1}{N} \sum_{k=0}^{N-1} F[k] e^{-\left(j \frac{2 \pi}{N} k n\right)} e^{j \frac{2 \pi}{N} k n} \nonumber \\ &=\frac{1}{N} \sum_{k=0}^{N-1} F[k] e^{j \frac{2 \pi}{N} k(n-m)} \nonumber \\ &=f\left[((n-m))_{N}\right] \end{align}$

Circular Convolution Demonstration

circularshiftsDemo — Figure $\PageIndex{13}$ : Interact (when online) with a Mathematica CDF demonstrating Circular Shifts.

Conclusion

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