24 2 Properties Fourier Trnsform

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Properties of the

Fourier Transform

24.2Introduction

In this Section we shall learn about some useful properties of the Fourier transform which enableus to calculate easily further transforms of functions and also in applications such as electroniccommunication theory.

Prerequisites

Before starting this Section you should . . .

be aware of the basic definitions of theFourier transform and inverse Fouriertransform

Learning Outcomes

On completion you should be able to . . .

state and use the linearity property and thetime and frequency shift properties of Fouriertransforms

state various other properties of the Fouriertransform

14 HELM (2005):Workbook 24: Fourier Transforms
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1. Linearity properties of the Fourier transform(i) Iff(t), g(t) are functions with transforms F(), G() respectively, then

F{f(t) +g(t)} =F() +G()

i.e. if we add 2 functions then the Fourier transform of the resulting function is simply the sum ofthe individual Fourier transforms.

(ii) Ifk is any constant,

F{kf(t)} =kF()

i.e. if we multiply a function by any constant then we must multiply the Fourier transform by thesame constant. These properties follow from the definition of the Fourier transform and from theproperties of integrals.

Examples

1.

F{2etu(t) + 3e2tu(t)} = F{2etu(t)} + F{3e2tu(t)}

= 2F{etu(t)} + 3F{e2tu(t)}

= 2

1 +i+

3

2 +i

2.

If f(t) =

4 3 t 30 otherwise

then f(t) = 4p3(t)

so F() = 4P3() = 8

sin 3

using the standard result for F {pa(t)}.

TaskskIf f(t) =

6 2 t 20 otherwise write down F().

Your solution

Answer

We have f(t) = 6p2(t) so F() =12

sin2.

HELM (2005):Section 24.2: Properties of the Fourier Transform

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2. Shift properties of the Fourier transformThere are two basic shift properties of the Fourier transform:

(i) Time shift property: F{f(t t0)} =eit0F()

(ii) Frequency shift property F{ei0t

f(t)} = F( 0).Here t0, 0 are constants.

In words, shifting (or translating) a function in one domain corresponds to a multiplication by acomplex exponential function in the other domain.

We omit the proofs of these properties which follow from the definition of the Fourier transform.

Example 2

Use the time-shifting property to find the Fourier transform of the function

g(t) =

1 3 t 50 otherwise

t

g(t)

1

3 5

Figure 4

Solution

g(t) is a pulse of width 2 and can be obtained by shifting the symmetrical rectangular pulse

p1(t) =

1 1 t 10 otherwise

by 4 units to the right.

Hence by putting t0= 4 in the time shift theorem

G() = F{g(t)} = e4i2

sin .



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TaskskVerify the result of Example 2 by direct integration.

Your solution

Answer

G() = 53

1eitdt =

eiti

53

= e5i e3i

i =e4

iei ei

i

=e4i2

sin

,

as obtained using the time-shift property.

TaskskUse the frequency shift property to obtain the Fourier transform of themodulated wave

g(t) =f(t)cos 0t

where f(t) is an arbitrary signal whose Fourier transform is F().

First rewrite g(t) in terms of complex exponentials:

Your solution

Answer

g(t) =f(t)

ei0t +ei0t

2

=

1

2f(t)ei0t +

1

2f(t)ei0t


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Now use the linearity property and the frequency shift property on each term to obtain G():

Your solution

AnswerWe have, by linearity:

F{g(t)} =12F{f(t)ei0t} +12 F{f(t)e

i0t}

and by the frequency shift property:

G() =1

2F( 0) +

1

2F(+0).

F()

112

0 0

G()

3.Inversion of the Fourier transformFormal inversion of the Fourier transform, i.e. finding f(t) for a given F(), is sometimes possibleusing the inversion integral (4). However, in elementary cases, we can use a Table of standard Fourier

transforms together, if necessary, with the appropriate properties of the Fourier transform.The following Examples and Tasks involve such inversion.



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Example 3

Find the inverse Fourier transform of F() = 20sin5

5 .

Solution

The appearance of the sine function implies that f(t) is a symmetric rectangular pulse.

We know the standard form F{pa(t)} = 2asin a

a or F1{2a

sin a

a } =pa(t).

Putting a= 5 F1{10sin5

5 } =p5(t). Thus, by the linearity property

f(t) = F1{20sin5

5

} = 2p5(t)

f(t)

2

5 5 t

Figure 4

Example 4

Find the inverse Fourier transform ofG() = 20sin5

5 exp (3i).

Solution

The occurrence of the complex exponential factor in the Fourier transform suggests the time-shift

property with the time shift t0= +3 (i.e. a right shift).From Example 3

F1{20sin5

5 } = 2p5(t) so g(t) = F

1{20sin5

5 e3i} = 2p5(t 3)

2

t

g(t)

2 8

Figure 5


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TaskskFind the inverse Fourier transform of

H() = 6sin2

e4i.

Firstly ignore the exponential factor and find the inverse Fourier transform of the remaining terms:

Your solution

Answer

We use the result: F1{2asin a

a } =pa(t)

Putting a= 2 gives F1{2sin2

} =p2(t) F

1{6sin2

} = 3p2(t)

Now take account of the exponential factor:

Your solution

AnswerUsing the time-shift theorem for t0= 4

h(t) = F1{6sin2

e4i} = 3p2(t 4)

t

h(t)

3

2 6



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Example 5Find the inverse Fourier transform of

K() = 2

1 + 2( 1)i

Solution

The presence of the term ( 1) instead of suggests the frequency shift property.

Hence, we consider first

K() = 2

1 + 2i.

The relevant standard form is

F{etu(t)} = 1

+i or F1{

1

+i} =etu(t).

Hence, writing K() = 112

+ik(t) =e

1

2tu(t).

Then, by the frequency shift property with 0 = 1 k(t) = F1{

2

1 + 2( 1)i} =e

1

2teitu(t).

Here k(t) is a complex time-domain signal.

TaskskFind the inverse Fourier transforms of

(a) L() = 2sin {3( 2)}

( 2) (b) M() =

ei

1 +i

Your solution


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Answer(a) Using the frequency shift property with 0= 2

l(t) = F1{L()} =p3(t)ei2t

(b) Using the time shift property with t0 = 1

m(t) = e(t+1)u(t+ 1)

t

m(t)

1

4. Further properties of the Fourier transform

We state these properties without proof. As usual F() denotes the Fourier transform off(t).

(a) Time differentiation property:

F{f(t)} =iF()

(Differentiating a function is said to amplify the higher frequency components because ofthe additional multiplying factor .)

(b) Frequency differentiation property:

F{tf(t)} =idF

d or F{(it)f(t)} =

dF

d

Note the symmetry between properties (a) and (b).

(c) Duality property:

If F{f(t)} =F() then F{F(t)} = 2f().

Informally, the duality property states that we can, apart from the 2 factor, interchange the timeand frequency domains provided we put rather than in the second term, this corresponding toa reflection in the vertical axis. Iff(t) is even this latter is irrelevant.

For example, we know that if f(t) =p1(t) =

1 1< t


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Graphically:

t

F

P1()

P1(t) 2p1()

2

1 1

F

p1(t)

t 1 1

1

Figure 6

TaskskRecalling the Fourier transform pair

f(t) =

e2t t >0e2t t


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Answer

We have F{f(t)} F {e2|t|} = 4

4 +2. F {

1

4e2|t|} =

1

4 +2 (by linearity)

F{ 1

4 +t2

} = 21

4

e2|| =

2

e2|| =G() (by duality).

t

F

F

t

f(t) F()

g(t) G()

1

14

2

1

(b) Use the modulation property based on the frequency shift property:

Your solution

AnswerWe have h(t) =g(t)cos2t. F {g(t)cos 0t} =

12

(G( 0) +G(+ 0)),

so with 0= 2 F {h(t)} =

4 e2|2| +e2|+2|

=H()

H()

2 2



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Exercises

1. Using the superposition and time delay theorems and the known result for the transform of therectangular pulse p(t), obtain the Fourier transforms of each of the signals shown.

12 0 1 2

xa(t)(a)

t12 0 1 2

xb (t)(b)

t

12 0 1 2

xc(t)

(c)

t31 2

xd(t)

(d)

t

11

1 1

2 2

1

2. Obtain the Fourier transform of the signal

f(t) =etu(t) +e2tu(t)

where u(t) denotes the unit step function.

3. Use the time-shift property to obtain the Fourier transform of

f(t) =

1 1 t 3

0 otherwise

Verify your result using the definition of the Fourier transform.

4. Find the inverse Fourier transforms of

(a) F() = 20sin(5)

5 e3i

(b) F() = 8

sin 3 ei

(c) F() = ei

1 i

5. Iff(t) is a signal with transform F() obtain the Fourier transform off(t)cos(0t)cos(0t).


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Answer

1. Xa() = 4

sin(

2)cos(

3

2 )

Xb() = 4i sin( 2 ) sin(32 )

Xc() = 2

[sin(2) + sin()]

Xd() = 2

sin(

3

2) + sin(

2

e3i/2

2. F() = 3 + 2i

2 2 + 3i (using the superposition property)

3. F() = 2sin

e2i

4. (a) f(t) =

2 2< t

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24 2 Properties Fourier Trnsform