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Chapter 5: Chapter 5:
Fourier Fourier TransformTransform
FOURIER TRANSFORM:FOURIER TRANSFORM:
2
Definition of the Fourier transformsDefinition of the Fourier transformsRelationship between Laplace Transforms and
Fourier TransformsFourier transforms in the limitProperties of the Fourier TransformsCircuit applications using Fourier TransformsParseval’s theoremEnergy calculation in magnitude spectrum
Definition of Fourier Definition of Fourier TransformsTransforms
3
dtetf
tfFF
tj
)(
)()(
Fourier Transforms:
Inverse Fourier Transforms:
4
dteF
FFtf
tj
)(2
1
)()( 1
Example 1:Obtain the Fourier Transform for thefunction below:
0
1
f ( t)
t
ate
5
Solution:
Given function is:
00
0)(
t
tetf
at
6
Fourier Transforms:
7
ja
eja
dtedtee
dtetfF
tja
tjatjat
tj
1
1
)(
)()(
0
)(
0
)(
0
FOURIER TRANSFORM:FOURIER TRANSFORM:
8
Definition of the Fourier transformsRelationship between Laplace Transforms and Relationship between Laplace Transforms and
Fourier TransformsFourier TransformsFourier transforms in the limitProperties of the Fourier TransformsCircuit applications using Fourier TransformsParseval’s theoremEnergy calculation in magnitude spectrum
Relationship between Fourier Transforms and Laplace Transforms
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There are 3 rules apply to the use of Laplace transforms to find Fourier Transforms of such functions.
Rule 1:If f(t)=0 for t<=0-
Replace s=jω
jstfLtfF )()(
10
Example:
11
0cos
00)(
tte
ttf
oat
Replace s=jω
12
22
22
)(
)()(
o
jso
aj
aj
as
astfF
Rule 2: Inverse negative function
13
jstfLtfF )()(
Example:
14
0cos
00)(
tte
ttf
oat
0cos
00)(
tte
ttf
oat
Negative
Fourier Transforms
15
22
22
)(
)(
)()(
o
jso
js
aj
aj
as
as
tfLtfF
Rule 3:Add the positive and negative function
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0)()(
0)()(
ttftf
ttftf
)()()( tftftf
Thus,
17
jsjs tfLtfL
tfFtfFtfF
)()(
)()()(
Example 1:
18
at
at
etf
etf
)(
)(
as
tfL
astfL
1)(
1)(
Fourier transforms:
19
22
2
11
11)(
a
a
ajaj
asastfF
jsjs
Example 2:
Obtain the Fourier Transforms for the function below:
0sin
00)(
tte
ttf
oat
20
Solution:
21
22
22
)(
)(
)()(
o
o
jso
o
js
aj
as
sFF
Example 3:
22
0
00)(
tte
ttf
at
Solution:
23
2
2
)(
1
)(
1
)()(
aj
as
tfLF
js
js
Example 4:
24
0
0)(
tte
ttetf
at
at
Solution:
25
at
at
tetf
tetf
)(
)(
2
2
1)(
1)(
astfL
astfL
26
222
22
22
4
)(
1
)(
1
)(
1
)(
1)(
a
aj
jaaj
asastfF
jsjs
FOURIER TRANSFORM:FOURIER TRANSFORM:
27
Definition of the Fourier transformsRelationship between Laplace Transforms and
Fourier TransformsFourier transforms in the limitFourier transforms in the limitProperties of the Fourier TransformsCircuit applications using Fourier TransformsParseval’s theoremEnergy calculation in magnitude spectrum
Fourier Transforms in the limitFourier transform for signum function
(sgn(t))
1 .0
0
-1 .0
t
s g n ( t)
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)()()sgn( tutut 0)()(lim)sgn(0
tuetuet tt
1 .0
0
-1 .0
t
f ( t)
)(tue t
)( tue t
29
30
22
2
11
11)(
j
jj
sstfF
jsjs
assume ε→0,
31
j
tF2
)sgn(
Fourier Transforms for step function:
)sgn(2
1
2
1)( ttu
j
tFFtuF
1
)sgn(2
1
2
1)(
32
Fourier Transforms for cosine function
tjetf 0)(
)(2 00 tjeF
33
22
2cos
00
00
0
tjtj
tjtj
ee
eet
34
Thus,
00
00
0
222
12
1cos 00
tjtj eFeFtF
35
FOURIER TRANSFORM:FOURIER TRANSFORM:
36
Definition of the Fourier transformsRelationship between Laplace Transforms and
Fourier TransformsFourier transforms in the limitProperties of the Fourier TransformsProperties of the Fourier TransformsCircuit applications using Fourier TransformsParseval’s theoremEnergy calculation in magnitude spectrum
Properties of Fourier Transforms
Multiplication by a constant
)()( FtfF
KFtKfF )(37
Addition and subtraction
33
22
11
)(
)(
)(
FtfF
FtfF
FtfF
)()()(
)()()(
321
321
FFF
tftftfF
38
Differentiation
Fjdt
tfdF
Fjdt
tdfF
nn
n
)(
)(
39
Integration
)0()(
)()(
Fj
FtgF
dxxftgt
40
Scaling
01
)(
aa
Fa
atfF
41
Time shift
)()( 0 FeatfF tj
42
Frequency shift
)()( 00 FtfeF tj
43
Modulation
00
0
2
1
2
1
)()cos(
FF
tftF
44
Convolution in time domain
)()()()( 2121 FFtftf
45
Convolution in frequency domain:
)()(2
1)()( 2121
FFtftf
46
Example 1:
Determine the inverse Fourier Transforms for the function below:
86)(
410)(
2
jj
jF
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Solution:
48
24
)2)(4(
410
86)(
410)(
2
s
B
s
A
ss
s
ss
ssF
LAPLACELAPLACETRANSFORMSTRANSFORMS
A and B value: 818 BA
2
8
4
18
jjF
49
)()818()( 24 tueetf tt
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