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Filtering with the DFT

Module by: Robert Nowak. E-mail the author

Summary: 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

Figure 1
Figure 1 (sec11_fig1.png)

y⁢n=x⁢n*h⁢n=∑ k =−∞∞x⁢k⁢h⁢n−k y n x n h n k x k h n k
(1)
Y⁢ω=X⁢ω⁢H⁢ω Y ω X ω H ω
(2)
Assume that H⁢ω H ω is specified.

Exercise 1

How can we implement X⁢ω⁢H⁢ω X ω H ω in a computer?

Solution

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

Recall that the DFT treats NN-point sequences as if they are periodically extended (Figure 2):

Figure 2
Figure 2 (sec11_fig2.png)

Compute IDFT of Y[k]

y ~ ⁢n=1N⁢∑ k =0N−1Y⁢k⁢ei⁢2⁢π⁢kN⁢n=1N⁢∑ k =0N−1X⁢k⁢H⁢k⁢ei⁢2⁢π⁢kN⁢n=1N⁢∑ k =0N−1∑ m =0N−1x⁢m⁢e−(i⁢2⁢π⁢kN⁢m)⁢H⁢k⁢ei⁢2⁢π⁢kN⁢n=∑ m =0N−1x⁢m⁢(1N⁢∑ k =0N−1H⁢k⁢ei⁢2⁢π⁢kN⁢(n−m))=∑ m =0N−1x⁢m⁢h⁢ (( n − m )) N y ~ n 1 N k N 1 0 Y k 2 k N n 1 N k N 1 0 X k H k 2 k N n 1 N k N 1 0 m N 1 0 x m 2 k N m H k 2 k N n m N 1 0 x m 1 N k N 1 0 H k 2 k N n m m N 1 0 x m h (( n − m )) N
(3)
And the IDFT periodically extends h⁢n h n : h ~ ⁢n−m=h⁢ (( n − m )) N h ~ n m h (( n − m )) N This computes as shown in Figure 3:

Figure 3
Figure 3 (sec11_fig3.png)

y ~ ⁢n=∑ m =0N−1x⁢m⁢h⁢ (( n − m )) N y ~ n m N 1 0 x m h (( n − m )) N
(4)
is called circular convolution and is denoted by Figure 4.

Figure 4: The above symbol for the circular convolution is for an NN-periodic extension.
Figure 4 (sec11_fig4.png)

DFT Pair

Figure 5
Figure 5 (sec11_fig5.png)

Note that in general:

Figure 6
Figure 6 (sec11_fig6.png)

Example 1: Regular vs. Circular Convolution

To begin with, we are given the following two length-3 signals: x⁢n=123 x n 1 2 3 h⁢n=102 h n 1 0 2 We can zero-pad these signals so that we have the following discrete sequences: x⁢n=…01230… x n … 0 1 2 3 0 … h⁢n=…01020… h n … 0 1 0 2 0 … where x⁢0=1 x 0 1 and h⁢0=1 h 0 1 .

  • Regular Convolution:
    y⁢n=∑ m =02x⁢m⁢h⁢n−m y n m 2 0 x m h n m
    (5)
    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 y 0 to y⁢4 y 4 . Recall that because we have two length-3 signals, our convolved signal will be length-5.
    • n=0 n 0 …0001230… … 0 0 0 1 2 3 0 … …0201000… … 0 2 0 1 0 0 0 …
      y⁢0=1×1+2×0+3×0=1 y 0 1 1 2 0 3 0 1
      (6)
    • n=1 n 1 …001230… … 0 0 1 2 3 0 … …020100… … 0 2 0 1 0 0 …
      y⁢1=1×0+2×1+3×0=2 y 1 1 0 2 1 3 0 2
      (7)
    • n=2 n 2 …01230… … 0 1 2 3 0 … …02010… … 0 2 0 1 0 …
      y⁢2=1×2+2×0+3×1=5 y 2 1 2 2 0 3 1 5
      (8)
    • n=3 n 3
      y⁢3=4 y 3 4
      (9)
    • n=4 n 4
      y⁢4=6 y 4 6
      (10)

Figure 7: Result is finite duration, not periodic!
Regular Convolution Result
Regular Convolution Result (sec11_fig7.png)

  • Circular Convolution:
    y ~ ⁢n=∑ m =02x⁢m⁢h⁢ (( n − m )) N y ~ n m 2 0 x m h (( n − m )) N
    (11)
    And now with circular convolution our h⁢n h n changes and becomes a periodically extended signal:
    h⁢ (( n )) N =…102102102… h (( n )) N … 1 0 2 1 0 2 1 0 2 …
    (12)
    • n=0 n 0 …0001230… … 0 0 0 1 2 3 0 … …1201201… … 1 2 0 1 2 0 1 …
      y ~ ⁢0=1×1+2×2+3×0=5 y ~ 0 1 1 2 2 3 0 5
      (13)
    • n=1 n 1 …0001230… … 0 0 0 1 2 3 0 … …0120120… … 0 1 2 0 1 2 0 …
      y ~ ⁢1=1×1+2×1+3×2=8 y ~ 1 1 1 2 1 3 2 8
      (14)
    • n=2 n 2
      y ~ ⁢2=5 y ~ 2 5
      (15)
    • n=3 n 3
      y ~ ⁢3=5 y ~ 3 5
      (16)
    • n=4 n 4
      y ~ ⁢4=8 y ~ 4 8
      (17)

Figure 8: Result is 3-periodic.
Circular Convolution Result
Circular Convolution Result (sec11_fig8.png)

Figure 9 illustrates the relationship between circular convolution and regular convolution using the previous two figures:

Figure 9: The left plot (the circular convolution results) has a "wrap-around" effect due to periodic extension.
Circular Convolution from Regular
Circular Convolution from Regular (sec11_fig9.png)

Regular Convolution from Periodic Convolution

  1. "Zero-pad" x⁢n x n and h⁢n h n to avoid the overlap (wrap-around) effect. We will zero-pad the two signals to a length-5 signal (5 being the duration of the regular convolution result): x⁢n=12300 x n 1 2 3 0 0 h⁢n=10200 h n 1 0 2 0 0
  2. Now take the DFTs of the zero-padded signals:
    y ~ ⁢n=1N⁢∑ k =04X⁢k⁢H⁢k⁢ei⁢2⁢π⁢k5⁢n=∑ m =04x⁢m⁢h⁢ (( n − m )) 5 y ~ n 1 N k 4 0 X k H k 2 k 5 n m 4 0 x m h (( n − m )) 5
    (18)
Now we can plot this result (Figure 10):

Figure 10: The sequence from 0 to 4 (the underlined part of the sequence) is the regular convolution result. From this illustration we can see that it is 5-periodic!
Figure 10 (no_image.png)

General Result:

We can compute the regular convolution result of a convolution of an MM-point signal x⁢n x n with an NN-point signal h⁢n h n by padding each signal with zeros to obtain two M+N−1 M N 1 length sequences and computing the circular convolution (or equivalently computing the IDFT of H⁢k⁢X⁢k H k X k , the product of the DFTs of the zero-padded signals) (Figure 11).

Figure 11: Note that the lower two images are simply the top images that have been zero-padded.
Figure 11 (sec11_fig11.png)

DSP System

Figure 12: The system has a length NN impulse response, h⁢n h n
Figure 12 (no_image.png)

  1. Sample finite duration continuous-time input x⁢t x t to get x⁢n x n where n=0…M−1 n 0 … M 1 .
  2. Zero-pad x⁢n x n and h⁢n h n to length M+N−1 M N 1 .
  3. Compute DFTs X⁢k X k and H⁢k H k
  4. Compute IDFTs of X⁢k⁢H⁢k X k H k y⁢n= y ~ ⁢n y n y ~ n where n=0…M+N−1 n 0 … M N 1 .
  5. Reconstruct y⁢t y t

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