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Semester -3

EE 3101

Introduction to system theory

Lecture : 9

DISTRIBUTION THEORY AND FOURIER TRANSFORM OF SOME SPECIAL FUNCTIONS

Compiled by:

Sakshi Mishra: BE/1239/09

Abhishek Koul: BE/1242/09

Prateek Tripathi: BE/1237/09

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1. GATE FUNCTIONA unit gate function is defined as a gate pulse of unit height and unit width,centered at the origin

0 |t| > T

Rect(t/2T) = {

1/2 |t| = T

1 |t| < T

Here ,the function exists as a pulse for a duration of 2T seconds. The inverse of factor of variable t

indicates the width of the pulse. In the given figure,the width is 2T.

2. SINC FUNCTION

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The sinc function sinc(x), also called the "sampling function," is a function that arises frequently in signalprocessing and the theory of Fourier transform. It is also known asfiltering or interpolating function. The full

name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc." There are two

definitions in common use. The one adopted in this work defines

a. sinc(x) is an even function of x.

b. Sinc (x) = 0 when sin(x) = 0 except at x = 0.

c. Using LHospital rule

Lim sin(x) / x = Lim cos(x) = 1

x0 x0

d. This has the normalization

-+

sinc(x)dx =

The half-infinite integral of sinc(x) can be derived using contour integration. In the above figure, consider the

path = 1 + 12 + 2 + 21 Now write z = Rei. On an arc, dz = iRe

iand on the x-axis, dz = e

idR. Write

-+

sinc(x)dx = I (ei

/z) dz

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where I denotes the imaginary part. Now define

The same result is arrived at using the method of complex residues by noting

I = 0 +1/2

(2i Res f(z) )

= i (z 0) (eiz

/ z) | z = 0

= i eiz

| z =0

= i

Since the integrand is symmetric , we have

0+

sinc(x)dx = /2

This is an improper integral.

We thus have the following result:

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But the expression on the right, considered as a function of k, is the sign function. We have here one of the

simplest examples of an integral representation of a function; the need for such a representation often arises in

applied mathematics. The integral on the left is known as Dirichlets discontinuity factor.

3. TRANSFORM OF GATE AND SINC FUNCTIONSa. TRANSFORM OF GATE FUNCTION

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b. TRANSFORM OF SINC FUNCTION

f(x) = sinc(x)

Then

F(j) =sinc(t).e

jtdt

=(sin(at)/at). e

jtdt

=( sin(at).cos(t)dt )/at

I =(1

/2a)

(sin(t + at).dt )/t +

II (1/2a)

(sin(t - at).dt )/t

=(/2a) rect(

t/2a)

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c. BANDWIDTH OF RECT(t/2T)

The spectrum F() peaks at = 0 and decays at higher frequencies. Therefore rect(t/2T) is a

lowpass signal with most of the signal energy concentrated in lower frequency components. But ,

because the spectrum exists for t = . However , much of the spectrum is concentrated within the

first lobe. Therefore , a rough estimate of the bandwidth is 2T

d. WINDOWED SINC FILTER

Insignal processing, a sinc filter is an idealizedfilterthat removes all frequency components

above a given bandwidth, leaves the low frequencies alone, and haslinear phase. The filter's

impulse responseis asinc functionin the time domain, and itsfrequency responseis arectangular

function.

http://en.wikipedia.org/wiki/Signal_processinghttp://en.wikipedia.org/wiki/Signal_processinghttp://en.wikipedia.org/wiki/Signal_processinghttp://en.wikipedia.org/wiki/Filter_%28signal_processing%29http://en.wikipedia.org/wiki/Filter_%28signal_processing%29http://en.wikipedia.org/wiki/Filter_%28signal_processing%29http://en.wikipedia.org/wiki/Linear_phasehttp://en.wikipedia.org/wiki/Linear_phasehttp://en.wikipedia.org/wiki/Linear_phasehttp://en.wikipedia.org/wiki/Impulse_responsehttp://en.wikipedia.org/wiki/Impulse_responsehttp://en.wikipedia.org/wiki/Sinc_functionhttp://en.wikipedia.org/wiki/Sinc_functionhttp://en.wikipedia.org/wiki/Sinc_functionhttp://en.wikipedia.org/wiki/Frequency_responsehttp://en.wikipedia.org/wiki/Frequency_responsehttp://en.wikipedia.org/wiki/Frequency_responsehttp://en.wikipedia.org/wiki/Rectangular_functionhttp://en.wikipedia.org/wiki/Rectangular_functionhttp://en.wikipedia.org/wiki/Rectangular_functionhttp://en.wikipedia.org/wiki/Rectangular_functionhttp://en.wikipedia.org/wiki/Rectangular_functionhttp://en.wikipedia.org/wiki/Rectangular_functionhttp://en.wikipedia.org/wiki/Frequency_responsehttp://en.wikipedia.org/wiki/Sinc_functionhttp://en.wikipedia.org/wiki/Impulse_responsehttp://en.wikipedia.org/wiki/Linear_phasehttp://en.wikipedia.org/wiki/Filter_%28signal_processing%29http://en.wikipedia.org/wiki/Signal_processing

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It is an "ideal"low-pass filterin the frequency sense, perfectly passing low frequencies, perfectly

cutting all the frequencies.

Real-time filters can only approximate this ideal, since an ideal sinc filter (aka rectangular filter)

acausal and produces infinite delay.

Windowed-sinc filters are used to separate one band of frequencies from another. They are verystable, produce few surprises, and can be pushed to incredible performance levels. These

exceptional frequency domain characteristics are obtained at the expense of poor performance in

the time domain, including excessive ripple and overshoot in the step response. When carried out

Convolving an input signal with this filter kernel provides aperfectlow-passfilter. The problem is, the sinc function continues to both negative and positiveinfinity without dropping to zero amplitude. While this infinite length is not

a problem for mathematics

To get around this problem, we will make two modifications to the sinc

function in (b), resulting in the waveform shown in (c). First, it is truncatedtoM%1 points, symmetrically chosen around the main lobe, whereMis an

even number. All samples outside theseM%1 points are set to zero, or simplyignored. Second, the entire sequence is shifted to the right so that it runs from

0 toM. This allows the filter kernel to be represented using onlypositiveindexes. While many programming languages allow negative indexes, they area nuisance to use. The sole effect of thisM/2 shift in the filter kernel is toshift the output signal by the same amount.

Since the modified filter kernel is only an approximation to the ideal filter

kernel, it will not have an ideal frequency response. To find the frequencyresponse that is obtained, the Fourier transform can be taken of the signal in

(c), resulting in the curve in (d). It's a mess! There is excessive ripple in thepassband and poor attenuation in the stopband (recall the Gibbs effectdiscussed in Chapter 11). These problems result from the abrupt discontinuity

at the ends of the truncated sinc function. Increasing the length of the filterkernel does not reduce these problems; the discontinuity is significant no matter

how longMis made.

Fortunately, there is a simple method of improving this situation. Figure (e)shows a smoothly tapered curve called a Blackman window. Multiplying the

truncated-sinc, (c), by the Blackman window, (e), results in the windowedsinc

filter kernel shown in (f). The idea is to reduce the abruptness of thetruncated ends and thereby improve the frequency response. Figure (g) shows

this improvement. The passband is now flat, and the stopband attenuation isso good it cannot be seen in this graph.

Several different windows are available, most of them named after their

original developers in the 1950s. Only two are worth using, the Hammingwindow and the Blackman window These are given by:

http://en.wikipedia.org/wiki/Low-pass_filterhttp://en.wikipedia.org/wiki/Low-pass_filterhttp://en.wikipedia.org/wiki/Low-pass_filterhttp://en.wikipedia.org/wiki/Low-pass_filter

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Figure 16-2a shows the shape of these two windows forM 50 (i.e., 51 total points in the curves).Which of these two windows should you use? It's a trade-off between parameters. As shown in

Fig. 16-2b, the Hamming window has about a 20% faster roll-offthan the Blackman. However, (c)shows that the Blackman has a better stopband attenuation. To be exact,the stopband attenuation

for the Blackman is -74dB (-0.02%), while theHamming is only -53dB (-0.2%). Although it

cannot be seen in these graphs,the Blackman has apassband ripple of only about 0.02%, while theHamming is typically 0.2%. In general, the Blackman should be your first choice; aslow roll-off is easier to handle than poor stopband attenuation.There are other windows youmight hear about, although they fall short of theBlackman and Hamming. The Bartlett window is

a triangle, using straight lines for the taper. The Hanning window, also called the raised cosine

window, is given by: w[i] 0.5& 0.5 cos(2Bi/M) . These two windows have about the same roll-off speed as the Hamming, but worse stopband attenuation (Bartlett: -25dB or 5.6%, Hanning -

44dB or 0.63%). You might also hear on a rectangular window. This is the same as no window,just a truncation of the tails (such as in Fig. 16-1c). While the roll-off is -2.5 times faster than the

Blackman, the stopband attenuation is only -21dB (8.9%).

FIGURE 16-1

Derivation of the windowed-sinc filter kernel. The frequency response of the ideal low-pass filter is shown

in (a), with the corresponding filter kernel in (b), a sinc function. Since the sinc is infinitely long, it must be

truncated to be used in a computer, as shown in (c). However, this truncation results in undesirable changes

in the frequency response, (d). The solution is to multiply the truncated-sinc with a smooth window, (e),

resulting in the windowed-sinc filter kernel, (f). The frequency response of the windowed-sinc, (g), is smooth

and well behaved. These figures are not to scale.

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FIGURE 16-2

Characteristics of the Blackman and Hamming

windows. The shapes of these two windows areshown in (a), and given by Eqs. 16-1 and 16-2. As

shown in (b), the Hamming window results in about

20% faster roll-off than the Blackman window.

However, the Blackman window has better stopbandattenuation (Blackman: 0.02%, Hamming:

0.2%), and a lower passband ripple (Blackman:

0.02% Hamming: 0.2%).

4. DISTRIBUTION THEORYDistribution states the relation between following improper integrals

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a. Lim -(sin(t)/) = (t)

t

b. -cos(t) d = ()

c. 0sin(t) () = 1/t

d. 0sin(t) dt =

1/

Lim -

(sin(t)/) = (t)t

In the above expression, the number of oscillations per unit length of the sinc function approaches

infinity. Nevertheless, the expression always oscillates inside an envelope of 1/(t), and

approaches zero for any nonzero value ofx. This complicates the informal picture of ( t) as being

zero for allx except at the pointx = 0 and illustrates the problem of thinking of the delta function

as a function rather than as a distribution.

-cos(t) d = ()

Fourier transform

dtett tj )()]([F 10

t

tje

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]1[)(1 Ft

detj

1

2

1

de tj

2

1

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5. FOURIER TRANSFORM OF FUNCTIONS USING DISTRIBUTION THEORYa. Fourier transform of constant

b. Fourier transform of exponential wave ejt

c. Fourier transform of sinusoidal functions

Using the exponential wave fourier transform and eulers theorem,we get

dettj

2

1)(

dtjt )sin(cos

2

1

td

jtd sin

2cos

2

1

0

cos1

td

dAeAjFtj

][)( F

dteA

tj )(

2

12

)(2 A

)(2]1[ F

)(2][0

0 tjeF

)()(cos000

Ft

)()(sin 000 jjtF

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d. Fourier transform of unit step function

F(u(t)) = 0

e-jt

dt

= 0

cos(t)dt - 0

sin(t)dt

= () +1/j

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e. Fourier transform of periodic function

We can express a periodic function as

The FT of a periodic function consists of a sequence of equidistant impulses located at the harmonic

frequencies of the function.

Tectf

n

tjn

n

2,)(

00

n

tjnnectfjF

0)]([)( FF

n

tjnn ec ][

0F

n

n nc )(2 0

n

n nc )(2 0

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6. FOURIER TRANSFORM OF NON PERIODIC FUNCTION

a. f(t)=unit step function[u(t)]

F(jw)= () -jwt(dt)

or,F(jw)= ()

-jwt

(dt)

=

-jwt(dt) [using u(t)=1]

= () ()

Using distribution theory, ()

=1/w & () ()

=(w)-j/w

=(w)+1/jw

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b. f(t)=signum function[sgn(t)]

Sgn(t)=-1 ,t0

F(jw)= ()

-jwt(dt)

or,F(jw)= ()

-jwt(dt)+ ()

-jwt(dt)

=

-jwt

(dt)+

-jwt

(dt) [using the definition of the function]

=

jwt

(dt) +

-jwt

(dt)

= (

-jwt

-ejwt

)(dt)

= ()()

=-2j () ()

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Using distribution theory, ()

=1/w

=-2j/w

=2/wj

Since,signum function is odd, fourier transform is completely imaginary.

c. TRUNCATED SINUSOID

f(t)=cos(wct).u(t)

F(jw)= ()

-jwt

(dt)

or,F(jw)= ()

cos(wct)e

-jwt(dt)

=1/2 ()

{e

jw1t+e

-jw1t}e

-jwt(dt)

=1/2 ()

[ejt(w1-w)+e-jt(w+w1)](dt)

=1/2{ ()

e-jt(w-w1)(dt)+ ()

e-jt(w+w1)}

=1/2[(w wc)+ (w +wc)]+1/2{1/j(w-wc)+1/j(w+wc)}

=/2[(w wc)+ (w +wc)]+jw/(wc2-w2)