Frequency Decomposition The base frequency or the fundamental - - PowerPoint PPT Presentation

The base frequency or the fundamental frequency is the lowest frequency. All multiples of the fundamental frequency are known as harmonics. A given signal can be constructed back from its frequency decomposition by a weighted addition of the fundamental frequency and all the harmonic frequencies

Frequency Decomposition

IIT Bombay Slide 21 GNR607 Lecture 18-19 B. Krishna Mohan

Click to edit Master text styles Second level

• Third level
• Fourth level
• Fifth level

Image reconstruction by weighted summation of sinusiodal functions Input signal Sinusoid 1 Sinusoid 2 Sinusoid 3 Sinusoid 4 Reconstructed signals IIT Bombay Slide 22 GNR607 Lecture 18-19 B. Krishna Mohan

Different forms of Fourier Transform

• Continuous Fourier Transform
• Fourier Series

IIT Bombay Slide 23 GNR607 Lecture 18-19 B. Krishna Mohan

2

1 ( ) ( ) 2

j ux

F u f x e dx

π

π

+∞ − −∞

=

∫

( ) cos(2 ) sin(2 ) 1 ( )cos(2 ) 2 1 ( )sin(2 ) 2

n n n n n

f x a a nx b nx where a f x nx dx b f x nx dx

π π π π

π π π π π π

+∞ =−∞ − −

= + + = =

∑ ∫ ∫

Continuous Fourier Transform

• In the continuous domain, the basis functions of the Fourier

transform are the complex exponentials e-j2πux

• These functions extend from -∞ to + ∞
• These are continuous functions, and exist everywhere

Please recall that e-j2πux = cos(2πux) – j.sin(2πux) (according to Euler’s identity)

IIT Bombay Slide 24 GNR607 Lecture 18-19 B. Krishna Mohan

Real and Imaginary Parts of Fourier Transform

Real part Imaginary Part

IIT Bombay Slide 25 GNR607 Lecture 18-19 B. Krishna Mohan

2

1 ( ) ( ) 2

j ux

F u f x e dx

π

π

+∞ − −∞

=

∫

1 1 ( ) ( )cos(2 ) ( )sin(2 ) 2 2 F u f x ux dx j f x ux dx π π π π

+∞ +∞ −∞ −∞

= −

∫ ∫

Fourier Series

• Fourier series is computed for periodic functions,

where the coefficients of the sinusoids are computed over one period of the function.

• It is assumed that the function has finite energy

within one period so that the integral can be computed

IIT Bombay Slide 26 GNR607 Lecture 18-19 B. Krishna Mohan

1-D FT

f(x) |F(u)| u

IIT Bombay Slide 27 GNR607 Lecture 18-19 B. Krishna Mohan

u x

• T/2

T/2 1 [ ]

/2 2 /2 2 /2 2 /2 2 /2 2 ( /2) 2 2

1 ( ) ( ) ( ) 1 1 ( ) 1. ( ) , { / 2, / 2} 2 1 1 ( ) 2 2 2 1 cos sin cos sin 2 sin s

T j ux T j ux T j ux T j uT j u T j uT j uT

F u f x e dx T e e dx x T T T T j u e e e e T j u j u j uT uT j uT uT j uT j uT uT uT

π π π π π π π

π π π π π π π π π π π

− − − − − − − − −

= = = ∈ − −     = − = −     − −   = + − + = =

∫ ∫

inc( ) uT

π

1-D FT

u

IIT Bombay Slide 27a GNR607 Lecture 18-19 B. Krishna Mohan

Input Image

IIT Bombay Slide 28 GNR607 Lecture 18-19 B. Krishna Mohan

Fourier Transform Magnitude

IIT Bombay Slide 29 GNR607 Lecture 18-19 B. Krishna Mohan

Points to be noted  The basis functions are continuous and extend all over the input domain  An abrupt cut-off in one domain leads to infinite extent

• f the transformed function, with the waves or ripples

decreasing in magnitude from zero-frequency  The width of the rectangle in one domain is inversely proportional to the spacing of the ripples in the other domain

Observations

IIT Bombay Slide 30 GNR607 Lecture 18-19 B. Krishna Mohan

Discrete Fourier Transform

• In the discrete domain, the Discrete Fourier Transform (DFT)

is defined as

• u is the frequency variable. weights e-j2πux
• F(u) one term of Fourier Transform
• The input sequence finite length N
• u=0 zero-frequency term, and u=N-1 highest frequency

IIT Bombay Slide 31 GNR607 Lecture 18-19 B. Krishna Mohan

1 2 /

1 ( ) ( )

N j un N n

F u f n e N

π − − =

=

∑

Inverse Fourier Transform

• Inverse Fourier transform is given by
• function f(x) has contributions from ALL

frequencies F(u).

• If F(u) = 0 for u ≥T, then the function f(x) is said to

be band-limited.

IIT Bombay Slide 32 GNR607 Lecture 18-19 B. Krishna Mohan

2

( ) ( )

j ux

f x F u e du

π ∞ −∞

= ∫

Inverse Discrete Fourier Transform

• Inverse discrete Fourier transform (IDFT) is given

by

• Each element in the input domain is a weighted

combination of ALL elements of the frequency domain

IIT Bombay Slide 33 GNR607 Lecture 18-19 B. Krishna Mohan

1 2 /

( ) ( )

N j un N n

f n F u e

π − + =

=∑

Real and Imaginary Parts of Fourier Transform

IIT Bombay Slide 34 GNR607 Lecture 18-19 B. Krishna Mohan

2

1 ( ) ( ) 2

j ux

F u f x e dx

π

π

+∞ − −∞

=

∫

1 ( ) ( )cos(2 ) 2 1 ( )sin(2 ) 2 F u f x ux dx j f x ux dx π π π π

+∞ −∞ +∞ −∞

= −

∫ ∫

Magnitude and Phase of Fourier Transform

• Given the real and imaginary parts of the Fourier

transform of f(x), we can write

• F(u) = FR(u) + j FI(u)
• Fourier magnitude = [FR(u)2 + FI(u)2]0.5
• Fourier phase = Arctan[FI(u)/FR(u)]

IIT Bombay Slide 35 GNR607 Lecture 18-19 B. Krishna Mohan

Fourier Magnitude and Phase

• Fourier magnitude denotes strength of Fourier

components

• Fourier phase controls the relative positioning of

features

• If parts of the image are positionally interchanged,

the Fourier magnitude remains unchanged, while the phase will be different

• Fourier phase is more important for recon-structing

the input image from the transformed version This is an important property to be noted

IIT Bombay Slide 36 GNR607 Lecture 18-19 B. Krishna Mohan

Two-dimensional Fourier Transform

• In case of two-dimensional data (e.g., images), the

2-D Fourier transform is defined by

• The basis functions in the 2-D Fourier transform

are defined by e-j2π(ux+vy)

• For each pair of (u,v), we have a basis image

generated from different values of (x,y).

IIT Bombay Slide 37 GNR607 Lecture 18-19 B. Krishna Mohan

2 ( )

1 ( , ) ( , ) 2

j ux vy

F u v f x y e dxdy

π

π

∞ − + −∞

=

∫

Basis images or basis matrices are

two-dimensional (2-D) versions of basis vectors The process of transforming an image from the spatial domain into another domain, or mathematical space, amounts to projecting the image onto the basis images The transform coefficient is obtained by taking the inner product of the image with the basis image Frequency transforms can be applied to the entire image or smaller blocks Basis Images IIT Bombay Slide 37a GNR607 Lecture 18-19 B. Krishna Mohan

2-D Discrete Fourier Transform

• The 2-D DFT is defined by
• The inverse 2-D DFT is defined by

IIT Bombay Slide 38 GNR607 Lecture 18-19 B. Krishna Mohan Basis image

1 1 2

1 ( , ) ( , )

um vn M N j M N m n

F u v f m n e MN

π   − − − +  ÷   = =

=

∑∑

1 1 2

( , ) ( , )

um vn M N j M N m n

f m n F u v e

π   − − + +  ÷   = =

= ∑∑

Image Transform

• Input Image

…

Basis Images

IIT Bombay Slide 38a GNR607 Lecture 18-19 B. Krishna Mohan

Unit of Spatial Frequency

• For time domain functions, frequency is

expressed in cycles/second

• For images, frequency is expressed in cycles

per unit distance

• How much is the change in intensity over 1 mm

distance in image or n pixels in the image?

IIT Bombay Slide 39 GNR607 Lecture 18-19 B. Krishna Mohan

Example

Given f(n) = [3,2,2,1], corresponding to the brightness values of one row of a digital image. Find F (u) in both rectangular form, and in exponential form

IIT Bombay Slide 40 GNR607 Lecture 18-19 B. Krishna Mohan

Example

IIT Bombay Slide 41 GNR607 Lecture 18-19 B. Krishna Mohan

2 1.1/ 4 2 2.1/4 2 3.1/4

1 (0) [3 2 2 1] 2 4 1 (1) [3 2 2 1. ] 4 1 1 [3 2 2 ] [1 ] 4 4

j j j

F F e e e j j j

π π π − − −

= + + + = = + + + = + − + = −

Example Contd.

IIT Bombay Slide 42 Therefore F(u) = [2 ¼ (1-j) ½ ¼ (1+j) ] GNR607 Lecture 18-19 B. Krishna Mohan

2 3.1/ 4 2 3.2/ 4 2 3.3/ 4

1 1 (2) [3 2 2 1] 4 2 1 (3) [3 2 2 1. ] 4 1 1 [3 2 2 ] [1 ] 4 4

j j j

F F e e e j j j

π π π − − −

= − + − = = + + + = + − − = +

IIT Bombay Slide 43 GNR607 Lecture 18-19 B. Krishna Mohan

Magnitude-Phase Form

F(0)= 2 = 2 + j0  Mag=sqrt(22 + 02)=2; Phase=tan- 1(0/2)=0 F(1) = ¼ (1-j) = ¼ - j ¼  Mag= ¼ sqrt(12 + (-1)2)=0.35; Phase = tan-1(-(1/4) / (1/4)) = tan-1(-1) =

• π/4

F(2) = ½ = ½ + j0  Mag = sqrt(( ½ )2 + 02) = ½ Phase = tan-1( 0 / (1/2) ) = 0 F(3) = ¼ (1+j) = ¼ - j ¼  Mag= ¼ sqrt(12 + (-1)2)=0.35; Phase = tan-1((1/4) / (1/4)) = tan-1(1) = π/4

Fourier Transform Calculation

Given f(n) = [ 3 2 2 1 ] F(u) = [2 ¼ (1-j) ½ ¼ (1+j) ] In phase magnitude form, M(u) = [2 0.35 ½ 0.35 ] Φ(u) = [0 −π/4 0 π/4 ] Calculate the above for f(n) = [ 0 0 4 4 4 4 0 0] Plot f(n), F(u), M(u) and Φ(u) graphically IIT Bombay Slide 44 GNR607 Lecture 18-19 B. Krishna Mohan