# 1.10 Filtering with the dft

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The ideas of using the DFT to filter a signal and recover a signal from a noisy transmission are addressed based on the ideas of the DFT and convolution.

## Introduction

$y(n)=(x(n), h(n))=\sum$ x k h n k
$Y()=X()H()$
Assume that $H()$ is specified.

How can we implement $X()H()$ in a computer?

Discretize (sample) $X()$ and $H()$ . In order to do this, we should take the DFTs of $x(n)$ and $h(n)$ to get $X(k)$ and $X(k)$ . Then we will compute $\stackrel{~}{y}(n)=\mathrm{IDFT}(X(k)H(k))$ Does $\stackrel{~}{y}(n)=y(n)$ ?

Recall that the DFT treats $N$ -point sequences as if they are periodically extended ( ):

## Compute idft of y[k]

$\stackrel{~}{y}(n)=\frac{1}{N}\sum_{k=0}^{N-1} Y(k)e^{i\times 2\pi \frac{k}{N}n}=\frac{1}{N}\sum_{k=0}^{N-1} X(k)H(k)e^{i\times 2\pi \frac{k}{N}n}=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} x(m)e^{-(i\times 2\pi \frac{k}{N}m)}H(k)e^{i\times 2\pi \frac{k}{N}n}=\sum_{m=0}^{N-1} x(m)\frac{1}{N}\sum_{k=0}^{N-1} H(k)e^{i\times 2\pi \frac{k}{N}(n-m)}=\sum_{m=0}^{N-1} x(m)h({\left(\left(nm\right)\right)}_{N})$
And the IDFT periodically extends $h(n)$ : $\stackrel{~}{h}(n-m)=h({\left(\left(nm\right)\right)}_{N})$ This computes as shown in :

$\stackrel{~}{y}(n)=\sum_{m=0}^{N-1} x(m)h({\left(\left(nm\right)\right)}_{N})$
is called circular convolution and is denoted by .

## Dft pair

Note that in general:

## Regular vs. circular convolution

To begin with, we are given the following two length-3 signals: $x(n)=\{1, 2, 3\}$ $h(n)=\{1, 0, 2\}$ We can zero-pad these signals so that we have the following discrete sequences: $x(n)=\{, 0, 1, 2, 3, 0, \}$ $h(n)=\{, 0, 1, 0, 2, 0, \}$ where $x(0)=1$ and $h(0)=1$ .

• Regular Convolution:
$y(n)=\sum_{m=0}^{2} x(m)h(n-m)$
Using the above convolution formula (refer to the link if you need a review of convolution ), we can calculate the resulting value for $y(0)$ to $y(4)$ . Recall that because we have two length-3 signals, our convolved signal will be length-5.
• $n=0$ $\{, 0, 0, 0, 1, 2, 3, 0, \}$ $\{, 0, 2, 0, 1, 0, 0, 0, \}$
$y(0)=1\times 1+2\times 0+3\times 0=1$
• $n=1$ $\{, 0, 0, 1, 2, 3, 0, \}$ $\{, 0, 2, 0, 1, 0, 0, \}$
$y(1)=1\times 0+2\times 1+3\times 0=2$
• $n=2$ $\{, 0, 1, 2, 3, 0, \}$ $\{, 0, 2, 0, 1, 0, \}$
$y(2)=1\times 2+2\times 0+3\times 1=5$
• $n=3$
$y(3)=4$
• $n=4$
$y(4)=6$

• Circular Convolution:
$\stackrel{~}{y}(n)=\sum_{m=0}^{2} x(m)h({\left(\left(nm\right)\right)}_{N})$
And now with circular convolution our $h(n)$ changes and becomes a periodically extended signal:
$h({\left(\left(n\right)\right)}_{N})=\{, 1, 0, 2, 1, 0, 2, 1, 0, 2, \}$
• $n=0$ $\{, 0, 0, 0, 1, 2, 3, 0, \}$ $\{, 1, 2, 0, 1, 2, 0, 1, \}$
$\stackrel{~}{y}(0)=1\times 1+2\times 2+3\times 0=5$
• $n=1$ $\{, 0, 0, 0, 1, 2, 3, 0, \}$ $\{, 0, 1, 2, 0, 1, 2, 0, \}$
$\stackrel{~}{y}(1)=1\times 1+2\times 1+3\times 2=8$
• $n=2$
$\stackrel{~}{y}(2)=5$
• $n=3$
$\stackrel{~}{y}(3)=5$
• $n=4$
$\stackrel{~}{y}(4)=8$

illustrates the relationship between circular convolution and regularconvolution using the previous two figures:

## Regular convolution from periodic convolution

• "Zero-pad" $x(n)$ and $h(n)$ to avoid the overlap (wrap-around) effect. We will zero-pad the two signals to a length-5 signal (5being the duration of the regular convolution result): $x(n)=\{1, 2, 3, 0, 0\}$ $h(n)=\{1, 0, 2, 0, 0\}$
• Now take the DFTs of the zero-padded signals:
$\stackrel{~}{y}(n)=\frac{1}{N}\sum_{k=0}^{4} X(k)H(k)e^{i\times 2\pi \frac{k}{5}n}=\sum_{m=0}^{4} x(m)h({\left(\left(nm\right)\right)}_{5})$
Now we can plot this result ( ):

We can compute the regular convolution result of a convolution of an $M$ -point signal $x(n)$ with an $N$ -point signal $h(n)$ by padding each signal with zeros to obtain two $M+N-1$ length sequences and computing the circular convolution (or equivalently computing the IDFT of $H(k)X(k)$ , the product of the DFTs of the zero-padded signals) ( ).

## Dsp system

• Sample finite duration continuous-time input $x(t)$ to get $x(n)$ where $n=\{0, , M-1\}$ .
• Zero-pad $x(n)$ and $h(n)$ to length $M+N-1$ .
• Compute DFTs $X(k)$ and $H(k)$
• Compute IDFTs of $X(k)H(k)$ $y(n)=\stackrel{~}{y}(n)$ where $n=\{0, , M+N-1\}$ .
• Reconstruct $y(t)$

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