Convolution FT 0

8/2/2019 Convolution FT 0

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Convolution and FourierTransform

Dr. Hariharan Muthusamy,

School of Mechatronic Engineering,

Universiti Malaysia Perlis.


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Convolution

Consider a linear system whose behaviour is specified by the impulse

response h(n). To find the output signal be y(n), resolve the signal x(n) into

a weighted sum of impulses, and the linearity and time shift property of the

LTI system.

The above expression gives the response y(n) of the LTI system as a function

of the input signal x(n) and impulse response h(n) is called the convolutionsum.

Steps involved in finding out the convolution sum

1. Folding: Fold the signal h(k) about the origin, ie at k=0;

2. Shifting: Shift h(-k) to the right by no,if no is positive or shift h(-k) to theleft by no, if no is negative to obtain h(no k).

3. Multiplication: Multiply x(k) by h(no k) to obtain the product sequence

yo(k) = x(k) h(no k)

k

k)x(k)h(ny(n)


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4. Summation: Sum all the values of

the product sequence yo(k) toobtain the value of the output at

time n = no

Find the convolution of two finite

duration sequences

otherwise

1n1

0

1x(n)

otherwise

1n1

0

1h(n)


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2. Find the convolution of the two signals x(n) = u(n) and h(n) =

anu(n), |a|=0.


6/29

Fourier Transform When the repetition period T becomes infinity, the waveform f(t) become

non-periodic, the separation between two adjacent harmonics will be zero.

Assuming f(t) is initially periodic, we have

dtef(t)Tc)F(jn

Let

,ecTTf(t)

dtef(t)

T

1c

where

,ecf(t)

T/2

T/2

tjn

no

tjn

n

T/2

T/2

tjn

n

tjn

n

o

o

o

o

As T ->, the fundamental

frequency becomes infinitelysmall, hence o -> d

dtef(t))F(j

de)F(j

2

1f(t)

e)F(jn2

1

ecTT

1f(t)

tj

tj

o

tjn

o

tjn

n

o

o


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Properties of Fourier Transform

Linearity

FT obeys superposition and homogeneity principles.

According to this property FT of the sum of two signals is

equal to sum of FT of individual signals.F[f1(t)]= F1(j) F[f2(t)]= F2(j)

F[a1f1(t)+b1f2(t)] = a1 F1(j) + b1F2(j)

Where a1 and b1 are constants.

Time Shifting Time Reversal

F[f(t)]= F(j) F[f(t)]= F(j)

F[f(t-t0) = e-jt0 F(j) F[f(-t)]= F(-j)


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Frequency Shifting

According the this property multiplication of the signal ejot with f(t)shifts the frequency spectrum by o

F[f(t)]= F(j)

F[f(t) ejot ]= F[j(- o)]

Time Scaling:

F[f(t)]= F(j) F[f(at)]= 1/|a| F(j /a), when a>1 f(at) is a compressed

version of the signal f(t) by a factor a.When a < 1, f(at) is an expanded version of the signal x(t) by a factor

a

Differentiation in time

Differentiating a signal results in a multiplication of the Fourier

transform by j, F[f(t)]= F(j) then F[d/dt(f(t))]= j F(j)


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Differentiation in Frequency

Multiplication of signal x(t) with time t results in differentiation of

frequency spectrum F(j)

F[f(t)]= F(j) then F[t f(t)]= j d/d F(j)

Time Integration

Integration of a signal results in a division of Fourier transform by j.

However to account for the dc or average value that can result from

integration, we must add the term F(0)()

F[f(t)] = F(j), then

)F(j)(jdt

f(t)d

Fn

n

n

F(0)(F(j(j

1f(()F

t


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Conjugation

F[f(t)]= X(j), F[f*(t)] = F*(-j)

Fourier transform of complex and real functions

f(t) =fR(t)+jfI(t)Fourier transform of f(t) is given by

F[f(t)]=

Auto-correlation

F[Rff()] = F[f(t) * f*(-t)] = ff(j) = |F(j|2

Duality

F[f(t)]= F(j)

F[f(t)]= 2f(-j)

dtef(t))F(jtj


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Convolution

The convolution of two signals results in the multiplication of their Fourier

transforms in the frequency domain.

F[f(t)*h(t)]= F(j) H(j)

The output of a system can be obtained from the convolution of input

signal and the system impulse response

d))h(tf(y(t)


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Fourier transform of Single Gate Function

otherwise

T/2tT/2for

0

1,f(t)

2

TTsinc

2T

2

Tsin

T.

ej

1

dt.e1

dtef(t))F(j

T/2

T/2

tj

tj

T/2

T/2

tj

The amplitude spectrum is shown in Fig. At = 0, sin

c(T/2) = T, therefore F(j) at = 0 is equal to T.

At T/2 = n, sin c(T/2) = 0,


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Fourier Transform of Rectangular Pulse

otherwise

Tt0for

0

1,f(t)

2

T

2

Tsin

eT

2j

ee2e

eej

e

ej

1

dt.e1

dtef(t))F(j

T/2j-

T/2jT/2jT/2j

T/2jT/2jT/2j

T

0

tj

tj

T

0

tj

2

TsincTe

T/2j


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Fourier transform of a rectangular pulse 2 seconds

long with a magnitude of 10 volts

csine20

sine20

2j

eee20

eej

e10

j

1e

10

j

e10

dt0.e1

dtef(t))F(j

j

j

jjj-

jjj

j2-

tj

tj

T/2

T/2

tj


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Exponential Pulse

0tfor

0tfor

0

,ef(t)

-at

atan

a

1

ja

1

dte

0tfor0f(t)since,dtee0

dtef(t)dtef(t)

dtef(t))F(j

1

22

0

t)j(a

0

tjat

0

tj

0

tj

tj


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Fourier transform for the double exponential pulse

0tfor

0tfor

e

,ef(t)

at

-at

22

0

)(

ajF

a2

ja

1

ja

1

dtedte

dteedtee

dtef(t)dtef(t)

dtef(t))F(j

0

t)j(a

-

t)j(a

0

tjat

0

tjat

0

tj

0

tj

tj


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Fourier transform of triangular pulse

f(t) = A (1+ 2t/T), -T/2


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4

Tsi n2

T

4A

2

Tcos1

T

4A

dt

si ni

2Tsi n

2

T

T

4A

2Tsi n

2A

dt2tcostT

2Adtcoso2A

dtetdtetT2AdtedteAF j

obt ai nwei nt egral s,t hirdandfirstt hei n-t oChangi ng

2

2

2

T/ 2

0

T/ 2

0

T/ 2

0

T/ 2

0

j

T/ 2

0

j

T/ 2

0

T/ 2

0

j j


19/29

Impulse Function (Unit Impulse)

The impulse function, which has an infinite amplitude and is infinitely narrow. This isdefined as (t) = 0 for all values except at t = 0

The Fourier transform or the impulse function (t) is obtained as

The frequency spectrum of the impulse function

(t) has a constant amplitude and extends

over positive and negative frequencies.

The inverse Fourier transform of the

unit impulse is given by

1dte(t))F(j tj

)(2

1.

Therefore2

1f(t)

t2

1

d)(2

ede)(

2

1f(t)

tjtj

Hence, the Fourier transform of the constant

function is an impulse at the origin with an area

equal to 2


20/29

Signum Function

The signum function denoted by sgn(t) is defined by

0tif

0tif

0tif

1

0

1

f(t)

j

2

j

1

j

1

j

e

j

e

dte(1)dte1)(

dtesgn(t))F(j

0

tj0

tj

0

0

tjtj

tj


21/29

Unit Step Function

The unit step function is obtained by suddenly closing a switch of a DC

circuit. For easier analysis, the waveform of the unit step function splitinto two component waveforms.

The first waveform is similar to the signum function with half amplitude.

Therefore, the Fourier transform function is given by

F1(j) = (2/j) = 1/j

The second waveform is related to the unit impulse function and hence its

Fourier transform is given by

F2(j) = [2 ()] = ()

Therefore, the Fourier transform of the step function becomes

F(j) = F1(j) + F2(j)

=1/j + ()


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Sinusoidal Functions

The Fourier transforms of the sinusoidal functions cos ot and sin ot areobtained as given below.

F[cos ot] =

Using the transform pair

F[cos ot] = [( o) + (+o)]

Similarly

F[sin ot] = [( + o) + ( - o)]

2

eeF

tjtj oo

getwe),(2e otj o


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FOURIER TRANSFORM OF POWER AND ENERGY

SIGNALS The average power of a signal x(t) over a single period (t1, t1+T)

is given by

Where x(t) is a complex periodic signal.

A signal f(t) is called a power signal, if the average power

expressed by

If x(t) is bounded, P

is finite. Every bounded and periodic signal is

a power signal. But it is true that a power signal is not

necessarily a bounded and periodic signal.

dtx(t)

T

1P

2Tt

t

av

1

i

T

T

2

tx x(t)

2T

1LtP


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Energy Signal

A signal x(t) is called an Energy signal, if its total energy over the interval

(- , ) is finite, that is

For a digital signal, the energy is defined by

As n ->, the energy of period signals becomes infinite, whereas the energy

of aperiodic pulse-signals have a finite value.

dtx(t)LtE

T

T

2

Tx

2

n

x(n)E


25/29

Determine the signal energy and signal power for (a). f(t) = e-3|t|, (b).

f(t) = e-3t

31e62

dte2

dtedte

dteE

06t

0

6t

0

6t

0

6t

2

|t|3

The signal power P

= 0,

since Eis finite. Hence,

the signal f(t) is an energysignal

TT

T

T

T

e6

2

6T

T

6t-

3t-

e6

1

dte

dteE

As T -> , ET approaches infinity.

Its average power is

Hence, e-3t is neither an energy signal nor a power

signal.

T

eeLtE

TLtP

TT

TT

T 82

166


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Compute the signal energy for x(t) = e-4t u(t)

Find the Fourier transform of the signal f(t) shown in Fig.

f(t) = (A/T) t, for 0 < t < T, A , for T < t < 2T

Tj2T/2jT/2jT/2j2

Tj2Tj

2

2T

T

tjT

0

2

tjtj

2T

T

tj

T

0

tj-

eA

jeeeT

A

eA

j1e

T

A

j

eA

)j(

e

j

et

T

A

dteAdtetT

AE

Tj2T/2j

Tj2T/2j

Tj2T/2j

2

e2Tsince

jA

eA

j2

Tsince

Aj

e

A

j2

T

sineT

2A

j


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Obtain the Fourier transform of the trapezoidal pulse is shown in Fig.

f(t) = At/(tp ta) + Atp/(tp ta), for tp < t < -ta

= A, for ta < t < ta

= Atp/(tp ta) - At/(tp ta), for tp < t < -ta

Therefore,

p

a

p

a

a

p

a

p

a

p

t

t ap

tjt

t ap

tj

p

t

t ap

tj

t

t ap

tjp

t

t ap

tj

dt

tt

Atedt

tt

etAdt

tt

Ate

dttt

eAtdt

tt

Ate)F(j


28/29

j

eeA

j

ee

tt

At

ee1

ee1

j

eet

tt

A

jeA

je

je

ttAt

e

j

tee

j

te

tt

A

dteAdtedtett

At

dtetdtettt

A

aaaa

ppaa

aa

a

a

p

a

a

p

p

a

a

p

a

a

a

p

p

a

a

p

p

a

tjtjtjtj

ap

p

tjtj

2

tjtj

2

tjtj

a

ap

t

t

tjt

t

tjt

t

tj

ap

p

t

t

2

tjtjt

t

2

tjtj

ap

t

t

tj

t

t

t

t

tjtj

ap

p

t

t

t

t

tjtj

ap


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pa2

pa2

ap

p

ap

p

a

aa

ap

p

p2a2a

a

ap

tcostcos2

tcostcos2

tt

t

tt

t1tsin

2A

t2sinA

t2sin1

tt

At

t2cos1

t2cos1

tsin2t

tt

A

Convolution FT 0

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