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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 DFT transformation matrix can be written as
𝐴=12 [1 1 1 11 − 𝑗 −1 𝑗1 −1 1 −11 𝑗 −1 − 𝑗
] 𝑢=[0 0 1 00 0 1 00 0 1 00 0 1 0
]𝑣=[ 1 −1 1 −1
0 0 0 00 0 0 00 0 0 0
]𝑽=𝑨𝑼𝑨
Translation property of DFT
• Let, the input image is translated to a location
𝑣𝑡 (𝑘 , 𝑙 )=1𝑁 ∑𝑚=0
𝑁−1
∑𝑛= 0
𝑁− 1
𝑢 (𝑚 ,𝑛) exp [− 𝑗 2𝜋 (𝑘(𝑚−𝑚0)+ 𝑙(𝑛−𝑛0))𝑁 ]
¿ 1𝑁 ∑𝑚=0
𝑁− 1
∑𝑛=0
𝑁 −1
𝑢 (𝑚 ,𝑛) exp [− 𝑗2𝜋 (𝑘𝑚+𝑙𝑛)𝑁 ]exp [ 𝑗 2𝜋 (𝑘𝑚0+𝑛0 𝑙)
𝑁 ]𝑣𝑡 (𝑘 , 𝑙 )=𝑣 (𝑘 ,𝑙)exp [ 𝑗 2𝜋 (𝑘𝑚0+𝑛0𝑙)
𝑁 ]
Translation property
𝑣𝑡 (𝑘 , 𝑙 )=𝑣 (𝑘 ,𝑙)exp [ 𝑗 2𝜋 (𝑘𝑚0+𝑛0𝑙)𝑁 ]DFT of translated image
Inverse DFT 𝑣 (𝑘−𝑘0 , 𝑙− 𝑙0 )=𝑢 (𝑚 ,𝑛)exp [ 𝑗 2𝜋 (𝑘0𝑚+𝑙0𝑛)𝑁 ]
𝑢(𝑚 ,𝑛)exp [ 𝑗2𝜋 (𝑘0𝑚+ 𝑙0𝑛)𝑁 ]↔𝑣 (𝑘−𝑘0 ,𝑙− 𝑙0)
𝑢(𝑚−𝑚0 ,𝑛−𝑛0)↔𝑣 (𝑘 , 𝑙)exp [ 𝑗 2𝜋 (𝑘𝑚0+𝑛0𝑙)𝑁 ]
Periodicity
𝑣 (𝑘 , 𝑙 )=𝑣 (𝑚+𝑁 ,𝑛)=𝑣 (𝑚 ,𝑛+𝑁 )=𝑣 (𝑚+𝑛 ,𝑛+𝑁 )
𝑣 (𝑘 , 𝑙 )= 1𝑁 ∑𝑚=0
𝑁− 1
∑𝑛=0
𝑁 −1
𝑢 (𝑚 ,𝑛 ) exp [ − 𝑗2𝜋 (𝑘𝑚+𝑛𝑙)𝑁 ]
𝑣 (𝑘+𝑁 , 𝑙+𝑁 )= 1𝑁 ∑𝑚=0
𝑁 −1
∑𝑛=0
𝑁−1
𝑢 (𝑚 ,𝑛) exp [− 𝑗 2𝜋𝑁 {(𝑘+𝑁 )𝑚+(𝑙+𝑁 )𝑛}]¿ 1𝑁 ∑𝑚=0
𝑁− 1
∑𝑛=0
𝑁 −1
𝑢 (𝑚 ,𝑛) exp [− 𝑗2𝜋 (𝑘𝑚+𝑛𝑙)𝑁 ]𝑒𝑥𝑝 [− 𝑗2𝜋 (𝑥+𝑦 )]
∀𝑘 ,𝑙
Conjugate symmetry
𝑣 (𝑁2 ±𝑘 , 𝑁2 ± 𝑙)=𝑣∗(𝑁2 ∓𝑘 ,𝑁2 ∓𝑙) 0≤𝑘 , 𝑙≤𝑁2−1
𝑘=0 𝑁𝑁2
1-D Example
(𝑁2 ,𝑁2 )
2-D Example
When is real
Conjugate symmetry
𝑣 (𝑘 , 𝑙 )= 1𝑁 ∑𝑚=0
𝑁− 1
∑𝑛=0
𝑁 −1
𝑢 (𝑚 ,𝑛 ) exp [ − 𝑗2𝜋 (𝑘𝑚+𝑛𝑙)𝑁 ]𝑘=𝑁2±𝑘 , 𝑙=
𝑁2± 𝑙
𝑣 (𝑘 , 𝑙 )= 1𝑁 ∑𝑚=0
𝑁− 1
∑𝑛=0
𝑁 −1
𝑢 (𝑚 ,𝑛 ) exp [ − 𝑗2𝜋 ((𝑁2 +𝑘)𝑚+(𝑁2 +𝑙 )𝑛)𝑁 ]
Distributivity
• DFT of sum of two signals is equal to the sum of their individual summations
𝐷𝐹𝑇 {𝑢1 (𝑚 ,𝑛)+𝑢2(𝑚 ,𝑛) }=𝐷𝐹𝑇 {𝑢1(𝑚 ,𝑛)}+𝐷𝐹𝑇 {𝑢2(𝑚 ,𝑛) }
Scaling𝑢(𝑎𝑚 ,𝑏𝑛)↔
1
¿ 𝑎𝑏∨¿𝑣 (𝑘𝑎 , 𝑙𝑏 )¿
are scaling parameters
Average value
𝑢 (𝑚 ,𝑛)= 1𝑁 2 ∑
𝑚=0
𝑁 −1
∑𝑛=0
𝑁−1
𝑢(𝑚 ,𝑛)Average value of image
𝑣 (𝑘 , 𝑙 )= 1𝑁 ∑𝑚=0
𝑁− 1
∑𝑛=0
𝑁 −1
𝑢 (𝑚 ,𝑛 ) exp [ − 𝑗2𝜋 (𝑘𝑚+𝑛𝑙)𝑁 ]
For
𝑣 (0,0 )= 1𝑁 ∑𝑚=0
𝑁−1
∑𝑛=0
𝑁− 1
𝑢 (𝑚 ,𝑛 )=𝑁𝑢 (𝑚 ,𝑛)
DC Component of an image
Rotation
𝑚=𝑟𝑐𝑜𝑠θ
𝑛=𝑟𝑠𝑖𝑛θPolar coordinate in source domain
𝑘=𝜔𝑐𝑜𝑠𝜙
𝑛=𝜔𝑠𝑖𝑛𝜙
Polar coordinate in target domain
Instead of working in the Cartesian coordinate, we are working in the polar coordinate
𝑢 (𝑟 ,𝜃 )↔𝑣 (𝜔 ,𝜙)If we have
𝑢 (𝑟 , 𝜃+𝜃0 )↔𝑣 (𝜔 ,𝜙+𝜃0)Then,
Convolution
• Let there be two images of different size
(0,0) (𝑁−1)
(𝑁−1)
≠𝟎(𝑀−1)
(𝑀−1)
𝑢1(𝑚 ,𝑛)
h (𝑚 ,𝑛)
𝑚
𝑛
(𝒉 ,𝒎 )=𝟎
h (𝑚 ,𝑛)𝑐=h(𝑚𝑚𝑜𝑑𝑁 ,𝑛𝑚𝑜𝑑𝑁 )
𝑚 ′
𝑛 ′
Circular Symmetry
(0,0) (𝑁−1)
(𝑁−1)
≠𝟎(𝑀−1)
(𝑀−1)
𝑢1(𝑚 ,𝑛)
h (𝑚 ,𝑛)𝑚
𝑛
(𝒉 ,𝒎 )=𝟎
𝑢1(𝑚 ′ ,𝑛 ′)
𝒉 (𝒎−𝒎′ ,𝒏−𝒏′ )𝒄
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′ )𝑐𝑊 𝑁𝑚𝑘+𝑛𝑙
¿𝑊 𝑁𝑚′ 𝑘+𝑛′ 𝑙∑
𝑚= 0
𝑁−1
∑𝑛=0
𝑁−1
h (𝑚−𝑚′ ,𝑛−𝑛′ )𝑐𝑊 𝑁(𝑚−𝑚 ′)𝑘+(𝑛−𝑛′ )𝑙
Let, and
¿𝑊 𝑁𝑚′ 𝑘+𝑛′ 𝑙 ∑
𝑝=−𝑚 ′
𝑁−1−𝑚 ′
∑𝑞=−𝑛 ′
𝑁−1−𝑛 ′
h (𝑝 ,𝑞 )𝑐𝑊 𝑁𝑝𝑘+𝑞𝑙
¿𝑊 𝑁𝑚′ 𝑘+𝑛′ 𝑙∑
𝑝=0
𝑁−1
∑𝑞=0
𝑁−1
h (𝑝 ,𝑞)𝑐𝑊 𝑁𝑝𝑘+𝑞𝑙
DFT of Convolution Function
𝐷𝐹𝑇 {𝑢2 (𝑚 ,𝑛) }𝑁=𝐷𝐹𝑇 {h (𝑚 ,𝑛) }𝑁𝐷𝐹𝑇 {𝑢1(𝑚 ,𝑛)}
DFT of a two dimensional circular convolution of two arrays is the product of their DFTs
Thank youNext Lecture: Hadamard Transform