Digital Signal Processing (EE313):
DFT as a filtering tool
Krishnan C.M.C
Assistant Professor, E&E,
NITK Surathkal
2
Significance of Linear Convolution
Filter
𝑥(𝑛)
ℎ(𝑛)
𝑦(𝑛)
Linear Filtering
𝑦 𝑛 = 𝑥 𝑛 ∗ℎ 𝑛
𝑦 𝑛 = ෍ 𝑥 𝑚 ℎ(𝑛 − 𝑚)
𝑚
DTFT
𝑥(𝑛)
𝑋(𝜔)
𝐻(𝜔)
I DTFT
×
EE313, Dept. of E & E, NITK Surathkal
𝑌(𝜔)
𝑦(𝑛)
3
Linear Convolution with DFT
DTFT
𝑋(𝜔)
𝑥(𝑛)
𝐻(𝜔)
I DTFT
×
𝑦(𝑛) = ෍ 𝑥 𝑚 ℎ(𝑛 − 𝑚)
𝑌(𝜔)
𝑚
DFT
𝑥(𝑛)
𝑋(𝑘)
𝐻(𝑘)
I DFT
×
𝑌(𝑘)
EE313, Dept. of E & E, NITK Surathkal
=
෍
𝑚=0→𝑁−1
𝑥 𝑚 ℎ 𝑛−𝑚
𝑁
4
Linear Convolution
Linear convolution of two sequencesℎ 𝑛 = 1↑ , −1,2 (impulse response),
𝑥 𝑛 = 0↑ , 1,2,3,4,5 (ramp input) demonstrated graphically
𝑥(𝑚)
0
1
2
𝟑
4
5
𝑦 𝑛 = ෍ 𝑥 𝑚 ℎ(𝑛 − 𝑚)
(𝑛 = 0)
𝑚
ℎ(−𝑚) 2 −1
(𝑛 = 1)
ℎ(1 − 𝑚) 2
1
𝑦(0) = 0 × 1 = 𝟎
−1
𝑦 1 = 0 × −1 + 1 × 1 = 𝟏
1
(𝑛 = 6)
2
ℎ(6 − 𝑚)
−1 1
𝑦 6 = 4 × 2 + 5 × −1 = 𝟑
2
𝑦 7 = 5 × 2 = 𝟏𝟎
(𝑛 = 7)
ℎ(7 − 𝑚)
𝑥(𝑛)
𝑦(𝑛)
0
1
2
𝟑
4
5
0
1
1
𝟑
5
7
EE313, Dept. of E & E, NITK Surathkal
−1 1
3
10
Total length (6+3-1=8 samples)
Tail (3-1=2 samples)
5
Linear convolution using circular convolution
ℎ 𝑛 = 1↑ , −1,2
𝑥 𝑛 = 0↑ , 1,2,3,4,5
𝑋(𝑘)
6 − point
×
ℎ(𝑛) DFT
𝐻(𝑘)
𝑥(𝑛)
𝑥(𝑛)
0
1
2
𝟑
4
5
𝑦(𝑛)
0
1
1
𝟑
5
7
6 − point
I DFT
𝑌(𝑘)
3
10
𝑦1 (𝑛)
0 1 1 𝟑 5 7 3 10
6 samples
0 1 1 𝟑 5 7 3 10
3 11 1 𝟑 5 7
EE313, Dept. of E & E, NITK Surathkal
0 1 1 𝟑 5 7
6
Linear convolution using circular convolution
ℎ 𝑛 = 1↑ , −1,2
𝑥 𝑛 = 0↑ , 1,2,3,4,5
𝑥(𝑛)
0
1
2
𝟑
4
5
𝑦(𝑛)
0
1
1
𝟑
5
7
𝟑 𝟏𝟏 1
𝟑
5
7
3
10
(8-point)
𝑦𝑎𝑙𝑖𝑎𝑠 (𝑛)
Points to note:
(6-point)
Option-1 • For a circular convolution (DFT multiplication) to mimic linear
convolution (DTFT multiplication)
 Increase the N of the N-point DFT to 𝑁 = 𝐿 + 𝑀 − 1 (here 6 + 3 − 1 = 8)
 By giving “space” (i.e., 𝑁 ≥ 𝐿 + 𝑀 − 1) you avoid time domain alias.
Option-2 • If you want to keep 𝑁 = 𝐿, the output will have Time domain alias.
 First M-1 samples of output 𝑦 𝑛 will be wrong
 The contribution of the last M-1 samples of input 𝑥 𝑛 will be forgotten.
EE313, Dept. of E & E, NITK Surathkal