Filtering in the Frequency Domain · Dr. Praveen Sankaran Department of ECE NIT Calicut January 11,...

Preliminary ConceptsFourier Transform of Sampled Functions

Filtering in the Frequency Domain

Dr. Praveen Sankaran

Department of ECE

NIT Calicut

January 11, 2013

Dr. Praveen Sankaran DIP Winter 2013


Outline

1 Preliminary Concepts

2 Fourier Transform of Sampled FunctionsContinuous FunctionsDiscrete Fourier Transform (DFT)



Signal

A measurable phenomenon that changes over time or throughoutspace.

Case in hand for this class → an image.

Signals undulate → variation in pixel values.

Frequency-domain representation

An exact description of a signal in terms of its undulations.

Any Real Signal has a Frequency-Domain Representation.



Fourier Series

Any periodic function can be expressed as the sum of sines and/orcosines of di�erent frequencies, each multiplied by a di�erentcoe�cient.

The sinusoids are called �basis functions�.

The multipliers are called �Fourier coe�cients�.



Fourier Series

f (t) =∞

∑n=−∞

cnej 2πnT t (1)

T = period

cn =1

T

∫ T/2

−T/2f (t)e−j

2πnT tdt (2)

n = 0,±1, · · ·

Note →e jθ = cosθ + jsinθ



Non-periodic Functions?

Any Real Signal has a Frequency-Domain Representation. Sowhat about non-periodic functions?

Fourier Transform

ℑ{f (t)}= F (µ) =∫

∞

−∞

f (t)e−j2πµtdt (3)

µ = frequency.

f (t) =∫

∞

−∞

F (µ)e j2πµtdµ (4)

Fourier transform pair.

Note → integral of the square of f (t) =�nite.



Fourier Transform Example



Impulses

A unit impulse of a continuous variable t located at t = 0, isde�ned as,

δ (t) =

{∞ if t = 0

0 if t 6= 0(5)

and, ∫∞

−∞

δ (t)dt = 1 (6)

Unit Discrete Impulse

δ [x ] =

{1 if x = 0

0 if x 6= 0(7)



Sifting Property

Sifting Property ∫∞

−∞

f (t)δ (t)dt = f (0) (8)

if f (t) is continuous at t = 0

∫∞

−∞

f (t)δ (t− t0)dt = f (t0) (9)

Discrete Form∞

∑x=−∞

f [x ]δ [x ] = f [x ] (10)

∞

∑x=−∞

f [x ]δ [x− x0] = f [x0] (11)



Impulse Train

s (t) =∞

∑n=−∞

δ (t−n∆T )



Convolution

Convolution

f (t)?h (t) =∫

∞

−∞

f (τ)h (t− τ) (12)

Convolution Theorem

ℑ(f (t)?h (t)) = H (µ)F (µ) (13)

ℑ(f (t)h (t)) = H (µ)?F (µ) (14)



Continuous FunctionsDiscrete Fourier Transform (DFT)

Outline






Sampling

1 Sampling theorem?

2 Aliasing?




Fourier Transform of an Impulse

F (µ) =∫

∞

−∞f (t)e−j2πµtdt

f (t) =∫

∞

−∞F (µ)e j2πµtdµ

Note that the di�erence is in the sign of the exponential. Thusif,

f (t)→ F (µ), thenF (t)→ f (−µ)

ℑ(δ (t− t0)) = e−j2πµt0

Then, e−j2πtt0 (notice the variable replacement), will have atransform, δ (−µ− t0).

If we put −t0 = a, transform of e j2πat is δ (a−µ) or δ (µ−a).

So we have, ℑ

(ej2πnt∆T

)= δ

(µ− n

∆T

)Dr. Praveen Sankaran DIP Winter 2013



Fourier Series of an Impulse Train

Let f̃ [t] represent the product of f (t) and an impulse train.(sampled function).

F̃ (µ) = ℑ

{f̃ [t]

}= F (µ)?S (µ) (15)

This is obtained from equation 14, convolution theorem.So now we need to �nd the transform of an impulse train,

S (µ) = ℑ(s (t)) = ℑ(∑

∞n=−∞ δ (t−n∆T )

),

which is periodic.

We �rst �nd the Fourier series expansion for the impulse trainas,

s (t) =∞

∑n=−∞

cnej 2πnt

∆T

cn = 1

∆T

∫ ∆T2

−∆T2

s (t)e−j2πnt∆T dt = 1

∆T

s (t) = 1

∆T

∞

∑n=−∞

ej 2πnt

∆T




Fourier Transform of Sampled Functions

So now, S (µ) = ℑ

(1

∆T

∞

∑n=−∞

ej2πnt∆T

)

S (µ) = 1∆T

ℑ

(∞

∑n=−∞

ej2πnt∆T

), the term inside the function

summation we recognize as being that of impulse!

So now,

S (µ) =1

∆T

∞

∑n=−∞

δ

(µ− n

∆T

)(16)

Remember - F̃ (µ) = F (µ)?S (µ), leading to,

F̃ (µ) =1

∆T

∞

∑n=−∞

F(

µ− n

∆T

)(17)

Note → notice that F (µ) is a continuous function. And sinceF̃ (µ) consists of copies of F (µ), this is also continuous.




Fourier Transform - Illustration




Reconstruction Problem: Illustration




Reconstruction Problem

1 Generate two sinusoids with frequencies F0 = 18Hz and

F1 =−78Hz . Sample both at Fs = 1Hz . Generate stem plots

for each, with a di�erent color for each. Plot the originalcontinuous signals over the sampled one, again each colorcoded. What do you observe? Determine the sampling rate atwhich you can avoid a problem here? Repeat the procedure atthis new sampling frequency. What do you observe?

2 You have done such a process in your DSP lab.

1 You may try this out in Matlab.




Illustration




Reconstruction Function?

f (t) = ℑ−1 {F (µ)}

= ℑ−1{H (µ) F̃ (µ)

}= h (t)? f̃ [t]




Outline






Continuous Transform Revisited

F̃ (µ) =∫

∞

−∞

f̃ [t]e−j2πµtdt (18)

=∫

∞

−∞

∞

∑n=−∞

f (t)δ (t−n∆T )e−j2πµtdt (19)

=∞

∑n=−∞

∫∞

−∞

f (t)δ (t−n∆T )e−j2πµtdt (20)

=∞

∑n=−∞

f [n]e−j2πµn∆T (21)

Remember → F̃ (µ) = 1∆T

∞

∑n=−∞

F(

µ− n

∆T

).

f [n] is discrete, F̃ (µ) is continuous.




Illustration - FT

We have F̃ (µ) is continuous, repeating itself with period 1∆T

.

Consider one period → µ = 0 to 1∆T

.




Getting the DFT

What we want

K equally spaced samples of F̃ (µ) between µ = 0 andµ = 1

∆T.

What we will do -

Take samples at following frequencies -

µ =k

K∆Tk = 0,1, · · ·K −1 (22)

Substituting above µ to F̃ (µ) =∞

∑n=−∞

f [n]e−j2πµt ,

F [k] =K−1

∑n=0

f [n]e−j2πnk/K k = 0,1, · · ·K −1 (23)




Getting the DFT

What we want


∆T.

What we will do -


µ =k

K∆Tk = 0,1, · · ·K −1 (22)


∑n=−∞

f [n]e−j2πµt ,

F [k] =K−1

∑n=0

f [n]e−j2πnk/K k = 0,1, · · ·K −1 (23)




Getting the DFT

What we want


∆T.

What we will do -


µ =k

K∆Tk = 0,1, · · ·K −1 (22)


∑n=−∞

f [n]e−j2πµt ,

F [k] =K−1

∑n=0

f [n]e−j2πnk/K k = 0,1, · · ·K −1 (23)




DFT Pair

DFT

F [k] =K−1

∑n=0

f [n]e−j2πnk/K k = 0,1, · · ·K −1

Inverse DFT

f [n] =1

K

K−1

∑k=0

F [k]ej2πkn/K n = 0,1, · · ·K −1 (24)




Convolution

f [n]?h [n] =K−1

∑k=0

f [m]h [n−m] n = 0,1, · · ·K −1 (25)

Circular convolution.




Sample Calculation

In terms of variable x (f [x ]). x = 0, 1, 2, 3.

Calculate,

F [0] , F [1] , F [2] , F [3]f [0] , f [1] , f [2] , f [3] from the above.




Summary

Fourier series.

Fourier transforms.

Sampling, sampling theorem.

Fourier transform of sampled functions.

DFT, IDFT, Circular convolution.




Assignment?

1 Perform histogram equalization of image provided. Save theinput image histogram and the output image histogram as.pgm images.

2 Repeat histogram equalization on the processed output ofprevious question. Save the histogram as .pgm. Repeat thisprocess a few times.

1 Do you notice any di�erence over successive histograms?

3 Using images provided, perform Histogram Matching.

Save histograms of both input image and the reference image.Save histogram of processed image.(All in .pgm format).




References

Dr. Richard Alan Peters II:http://archive.org/details/Lectures_on_Image_Processing


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Filtering in the Frequency Domain · Dr. Praveen Sankaran Department of ECE NIT Calicut January 11,...

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