CS654: Digital Image Analysis Lecture 14: Properties of DFT

Recap of Lecture 13 Introduction to DFT 1D and 2D DFT - Unitary Separability of DFT Computational complexity Improvement in computational complexity

Outline of Lecture 14 Properties of DFT Translation Periodicity Conjugate symmetry Distributivity Scaling Average value Convolution

Numerical example 𝐴= 1 2 1 1 1 1 1 −𝑗 −1 𝑗 1 −1 1 −1 1 𝑗 −1 −𝑗 A 4×4 DFT transformation matrix can be written as 𝐴= 1 2 1 1 1 1 1 −𝑗 −1 𝑗 1 −1 1 −1 1 𝑗 −1 −𝑗 𝑢= 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 𝑽=𝑨𝑼𝑨 𝑣= 1 −1 1 −1 0 0 0 0 0 0 0 0 0 0 0 0

Translation property of DFT Let, the input image 𝑢 𝑚,𝑛 is translated to a location 𝑢(𝑚 − 𝑚 0 ,𝑛− 𝑛 0 ) 𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁 𝑣 𝑡 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘(𝑚− 𝑚 0 )+𝑙(𝑛− 𝑛 0 )) 𝑁 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑙𝑛) 𝑁 exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁 𝑣 𝑡 𝑘,𝑙 =𝑣(𝑘,𝑙)exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁

Translation property 𝑣 𝑡 𝑘,𝑙 =𝑣(𝑘,𝑙)exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁 DFT of translated image 𝑣 𝑘− 𝑘 0 ,𝑙− 𝑙 0 =𝑢(𝑚,𝑛)exp⁡ 𝑗2𝜋( 𝑘 0 𝑚+ 𝑙 0 𝑛) 𝑁 Inverse DFT 𝑢(𝑚,𝑛)exp⁡ 𝑗2𝜋 𝑘 0 𝑚+ 𝑙 0 𝑛 𝑁 ↔𝑣(𝑘− 𝑘 0 ,𝑙− 𝑙 0 ) 𝑢(𝑚− 𝑚 0 ,𝑛− 𝑛 0 )↔𝑣(𝑘,𝑙)exp⁡ 𝑗2𝜋(𝑘 𝑚 0 + 𝑛 0 𝑙) 𝑁

Periodicity 𝑣 𝑘,𝑙 =𝑣 𝑚+𝑁,𝑛 =𝑣 𝑚,𝑛+𝑁 =𝑣(𝑚+𝑛,𝑛+𝑁) ∀𝑘,𝑙 𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁 𝑣 𝑘+𝑁,𝑙+𝑁 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋 𝑁 (𝑘+𝑁)𝑚+ 𝑙+𝑁 𝑛 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁 𝑒𝑥𝑝 −𝑗2𝜋(𝑥+𝑦)

Conjugate symmetry 𝑣 𝑁 2 ±𝑘, 𝑁 2 ±𝑙 = 𝑣 ∗ 𝑁 2 ∓𝑘, 𝑁 2 ∓𝑙 0≤𝑘,𝑙≤ 𝑁 2 −1 When 𝑢(𝑚,𝑛) is real 𝑣 𝑁 2 ±𝑘, 𝑁 2 ±𝑙 = 𝑣 ∗ 𝑁 2 ∓𝑘, 𝑁 2 ∓𝑙 0≤𝑘,𝑙≤ 𝑁 2 −1 𝑘=0 𝑁 𝑁 2 1-D Example 𝑁 2 , 𝑁 2 2-D Example

Conjugate symmetry 𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁 𝑘= 𝑁 2 ±𝑘,𝑙= 𝑁 2 ±𝑙 𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋 𝑁 2 +𝑘 𝑚+ 𝑁 2 +𝑙 𝑛 𝑁

Distributivity Scaling DFT of sum of two signals is equal to the sum of their individual summations 𝐷𝐹𝑇 𝑢 1 𝑚,𝑛 + 𝑢 2 (𝑚,𝑛) =𝐷𝐹𝑇 𝑢 1 (𝑚,𝑛) +𝐷𝐹𝑇 𝑢 2 (𝑚,𝑛) Scaling 𝑢(𝑎𝑚,𝑏𝑛)↔ 1 |𝑎𝑏| 𝑣 𝑘 𝑎 , 𝑙 𝑏 𝑎,𝑏 are scaling parameters

Average value 𝑢 𝑚,𝑛 = 1 𝑁 2 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢(𝑚,𝑛) 𝑢 𝑚,𝑛 = 1 𝑁 2 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢(𝑚,𝑛) Average value of image 𝑣 𝑘,𝑙 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 exp⁡ −𝑗2𝜋(𝑘𝑚+𝑛𝑙) 𝑁 For 𝑘=0,𝑙=0 𝑣 0,0 = 1 𝑁 𝑚=0 𝑁−1 𝑛=0 𝑁−1 𝑢 𝑚,𝑛 =𝑁 𝑢 𝑚,𝑛 DC Component of an image

Rotation 𝑢 𝑟,𝜃 ↔𝑣(𝜔,𝜙) 𝑢 𝑟,𝜃+ 𝜃 0 ↔𝑣(𝜔,𝜙+ 𝜃 0 ) 𝑚=𝑟𝑐𝑜𝑠θ 𝑛=𝑟𝑠𝑖𝑛θ Polar coordinate in source domain Instead of working in the Cartesian coordinate, we are working in the polar coordinate 𝑘=𝜔𝑐𝑜𝑠𝜙 𝑛=𝜔𝑠𝑖𝑛𝜙 Polar coordinate in target domain If we have 𝑢 𝑟,𝜃 ↔𝑣(𝜔,𝜙) 𝑢 𝑟,𝜃+ 𝜃 0 ↔𝑣(𝜔,𝜙+ 𝜃 0 ) Then,

Convolution Let there be two images of different size 𝑢 2 𝑚,𝑛 = (0,0) (𝑁−1) ≠𝟎 (𝑀−1) 𝑢 1 (𝑚,𝑛) ℎ(𝑚,𝑛) 𝑚 𝑛 𝒉,𝒎 =𝟎 𝑢 2 𝑚,𝑛 = 𝑚 ′ =0 𝑁−1 𝑛 ′ =0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑢 1 𝑚 ′ , 𝑛 ′ ; 0≤𝑚,𝑛≤𝑁−1 ℎ 𝑚,𝑛 𝑐 =ℎ(𝑚 𝑚𝑜𝑑 𝑁,𝑛 𝑚𝑜𝑑 𝑁)

Circular Symmetry 𝒉 𝒎− 𝒎 ′ ,𝒏− 𝒏 ′ 𝒄 𝑢 2 𝑚,𝑛 = (0,0) (𝑁−1) ≠𝟎 (𝑀−1) 𝑢 1 (𝑚,𝑛) ℎ(𝑚,𝑛) 𝑚 𝑛 𝒉,𝒎 =𝟎 𝒉 𝒎− 𝒎 ′ ,𝒏− 𝒏 ′ 𝒄 𝑚′ 𝑛′ 𝑢 2 𝑚,𝑛 = 𝑚 ′ =0 𝑁−1 𝑛 ′ =0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑢 1 𝑚 ′ , 𝑛 ′ ; 0≤𝑚,𝑛≤𝑁−1 𝑢 1 (𝑚′,𝑛′) Computational complexity??

2D DFT for h in frequency domain 𝐷𝐹𝑇 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 = 𝑚=0 𝑁−1 𝑛=0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑊 𝑁 𝑚𝑘+𝑛𝑙 = 𝑊 𝑁 𝑚 ′ 𝑘+ 𝑛 ′ 𝑙 𝑚=0 𝑁−1 𝑛=0 𝑁−1 ℎ 𝑚− 𝑚 ′ ,𝑛− 𝑛 ′ 𝑐 𝑊 𝑁 (𝑚− 𝑚 ′ )𝑘+(𝑛− 𝑛 ′ )𝑙 Let, 𝑝=𝑚−𝑚′ and 𝑞=𝑛−𝑛′ = 𝑊 𝑁 𝑚 ′ 𝑘+ 𝑛 ′ 𝑙 𝑝=−𝑚′ 𝑁−1−𝑚′ 𝑞=−𝑛′ 𝑁−1−𝑛′ ℎ 𝑝,𝑞 𝑐 𝑊 𝑁 𝑝𝑘+𝑞𝑙 = 𝑊 𝑁 𝑚 ′ 𝑘+ 𝑛 ′ 𝑙 𝑝=0 𝑁−1 𝑞=0 𝑁−1 ℎ 𝑝,𝑞 𝑐 𝑊 𝑁 𝑝𝑘+𝑞𝑙

DFT of Convolution Function 𝐷𝐹𝑇 𝑢 2 𝑚,𝑛 𝑁 =𝐷𝐹𝑇 ℎ 𝑚,𝑛 𝑁 𝐷𝐹𝑇{ 𝑢 1 (𝑚,𝑛)} DFT of a two dimensional circular convolution of two arrays is the product of their DFTs

Thank you Next Lecture: Hadamard Transform