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Chapter 4 The Fourier Transform
EE 207 Dr. Adil Balghonaim
0
0
sinc 2
nA nXT
0
0
| | sinc 2n
A nXT
0
0
0
180
nsinc > 0
2
nsinc < 0
2
n
o
0 0Frequency incremen (t )n
0
0
2The spacing between spectrum lines is T
0( ) jn t
p n
n
Xx t e
0
00
1 ( ) jn t
T
n x t e dtT
X
Let xp(t) be a periodical wave, then expanding the periodical function
Rewriting xp(t) and Xn
00
0
( )) (n jn t
pn
x nt eX
00
( ) nX nf X
0
00
0( )jn
n
n te nX
0
0
0( )2
jn to x t e dt
0
0
01 ( )2
jn tx t e dt
f
f
f
1T
1
1T
2
1T
1T
1T
period 0 T
0T 0 0 n
0 Continuous Variable n
0 0( )n 0
00
0( )( ) jn tn
pn
Xx t e n
0T
0
0
00
0
1( ) ( )2
jn tnX n x t e dX t
( ) x t Aperiodic Signal
( ) p
x t Periodic Signal
( ) p
x t Periodic Signal
0
00 0( ) ( ) ( ) jn
n
t
px t X e nn
0T
( ) p
x t Periodic Power Signal( ) x t Aperiodic Energy Signal
( ) ( ) j tx t X e d
1( ) ( )
2j tX x t e dt
Fourier Transform Pairs
0
00 0( ) ( ) ( ) jn
n
t
px t X e nn
0
0
00
0
1( ) ( )2
jn tnX n x t e dX t
0 0 0n 0( ) n d
( ) ( ) j tx t X e d
1( ) ( )2
j tX x t e dt
Fourier Transform Pairs
Sufficient conditions for the existence of the Fourier transform are
On any finite interval, a. ( ) is bounded; b. ( ) has a finite number of maxima
f tf t
( Dirichlet conditions )
1.
and minima; and c. ( ) has a finite number of discontinuities.
f (t) is absolutely integrable; that is, ( )
Note that these are conditions and not condi
f t
f t dt
2.
sufficient necessary tions
you can have a function that is not absolutely integrable however it has Fourier Transform like cos( ) (will be shown later)tNote
( ) ( ) j tx t X e d
1( ) ( )2
j tX x t e dt
Fourier Transform Pairs
absolutely integrable; that is, ( ) f t dt
Examples of functions that is not absolutely integrable
e , cos( ), sin( ), ( )t t t u t
cos( ), sin( ), ( ) has Fourier Transform (will be shown later)t t u t
e does not have Fourier Transform however e ( ) does havet tu t
0)()( atuetx at
( )
0( ) ( )at j t a j tX j e u t e dt e dt
Finding the Fourier Transform
1a
( )
0
1
( )a j te
a j
1
( )a j
0)()( atuetx at 1( )
( )X j
a j
0)()( atuetx at 1( )
( )X j
a j
Example Find the Fourier Transform for the following function
aa( ) ( ) j tX x t e dt
1
1
j te dt
1
1
j tej
2sinc( )
j je ej
22
j je ej
2 sin( )
sin( )2
(1) ( 1)j je ej
a ( ) 2sinc( )X
2
a( )X
22
a ( ) 2sinc( )X
aa
( )| ( )|( ) jXX e
2
0
a| ( )|X
22
0
( )f
2 2
2
a( )X
22
Example
0
b( )x t
t1 1
1
1bb( ) ( ) j tX x t e dt
0 1
1 0
(1) ( 1) j t te dt e dt
2sinc
2j
bb
( )| ( )|( ) jXX e
b2sin| |
2) c(X
0
b( )x t
t1 1
1
1
2b ( ) sinc
2X j
bb
( )| ( )|( ) jXX e
0
1
b| ( )|X
b22 sin|
2) c| (X
2 2
0
b( )x t
t1 1
1
1
2b ( ) sinc
2X f j
2
f0
( )f
2
2
2b ( ) sinc
2X j
bb
( )| ( )|( ) jXX e
oAlways 0 it add no angle (0 )
It was shown previously
aa( ) ( ) j tX x t e dt
1
1
j te dt
1
1
j tej
2sinc( )
j je ej
22
j je ej
2 sin( )
sin( )2
(1) ( 1)j je ej
The Fourier Transform for the following function
( ) ( ) j tX x t e dt
2
2
j te dt
2
2
j tej
4sinc(2 )
2 2j je ej
2 24
2 2
j je ej
4 sin(2 )
2
sin(2 )42
(2) ( 2)j je ej
2sinc( )
4sinc(2 )
(2 )sinc( )A T T
trect function defintion
T
2 2T T
u t u t
Example Find the Fourier Transform for the delta function x(t) = d(t)
( ) ( ) j tX x t e dt
( ) j tt e dt
0
( )j t
te t
(1) ( )t dt
1
( )t 1
1 1 1 11 1 1 1[ ( ) ( ) ] [ ( ) ( ) ] j tF a x t a x t a x t a x t e dt
1 1 2 2( ) + ( )j t j ta x t e dt a x t e dt
1 21 2( ) + ( )j t j ta x t e dt a x t e dt
1 21 2( ) + ( )a X a X
1 21 1 1 1 1 2[ ( ) ( ) ] ( ) + ( )F a x t a x t a X a X 1-Linearity
Proof
Properties of the Fourier Transform
( ) ( ) j tX x t e dt
2 0 2 4
4 2 0 2
(1) (2) + (2) (1)j t j t j t j te dt e dt e dt e dt
Direct Method
( )X 1( )X 2( )X
(1)(4)sinc(4 ) (2)(2)sinc(2 ) +
( )sinc( )A T T
4sinc(4 ) 4sinc(2 )+
Using Fourier Transform Properties
Let ( ) ( )x t X Then1( ) ta
aax X
Proof
Let 0 > a ( ) ( ) j tF x t x t e ta da
Change of variable 't ta
''
'( ) ( )a
jt
tdea tt
aF x x
'
' '1 ( )j
at
x e tdta
'
' '1 ( )j t
ax e dt ta
2-Time-Scaling (compressing or expanding)
'
' '1( ) ( )j
at
F x t x e ta t da
Let ' a
'' '1( ) ( ) jF x t x e dt ta
a
'( )X
'
1( ) ( )F x t Xaa
1aa
X
Now Let 0 < a
( ) ( | | ) j tF x t x ta e dta
Change of variable ' | |t ta
|' |
''
( ) ( )| |
aj
t
dF x t x eta
ta
| |a at t
| |
'' '1 ( )
| |
ja
t
t te da
x
'
| |ddtat
'
'
= =
tt
tt
| |aX
1( )| || |
F x t Xaaa
Since | | a a 0 < a 1( )| || |
F x t Xaaa
) 2sinc( )(aX
2(2sinc(2 ))
what is the fourier transform of
Let
11
21 2 ( ) aX X
since ( )1 2
a
tx t x
2 2 )(aX
4sinc(2 )
00 Let ( ) Then ( ) ( ) ( )e tjx t x t tX X
Proof
( ) ( ) j tY y t e dt
0Let ( ) ( )y t x t t
0( ) j tx t e dtt
' 0Change of variable t tt 0' ( )'
( ) ( ) j t tY tx e dt
0
''( ) ( ) t jj ttY x e e dt
0''( )t tjje x e dtt
( )X
0
0( ) ( )e tjt tx X
3-Time-Shifting
Example Find the Fourier Transform of the pulse function
0
( )x t
t2
1
Solution
From previous Example
a ( ) 2sin (2 )X c
Since ( ) ( )1ax t x t 1( )
2sinc(2 ) = e j
2sinc(2 ) = e j
0 ( ) ( )e a
tjX X
4-Time Transformation
0
0 1Let ( ) Then ( ) | |
( ) e aj t
x t x ata
ta
X X
Proof
0 0 Let > 0 ( ) ( ) j ta F x at xt ttat e d
' 0Change of variable t tat
0
0
'
''( ) ( )
tj
a
t
F x dat xt eatt
0
'
t at t
'
dt adt
'
'
= =
tt
tt
0 '
' '
( )
tj
a jtae x e d
at t
0
0
' '
' '0
'
'
( ) ( ) ( )
jj a ja a
tttt
eF x at x e x ea a
t dtt t dt
0
'
' '( )j
a ja
tt
e x et ta
d
Xa
0
tj
ae X aa
0
0Similarly < 0 ( ) | |
tjaeF x at X aa
a t
0
01( ) | |
e atj
x tataa
X
Find the Fourier Transform for the Rect function g(t)
From Fourier Transform Pairs (Table 5.2)
sinc2T
T
sinc2
= 3rect 0.5 2t
0
01( ) | |
e atj
x tataa
X
(2)0.5
1( )=3 sinc
0.5 0.5e
jG
4=6sinc0.5
e j
If ( ) ( ) ( ) 2 ) (Xx t tX x then
2W
2W
2W
t
2 sinc(2 ) W Wt
Find the of 2 sinc(2 ) W WtF.T
5-Duality ازدواجية
2W
2W
2W
t
12 sinc(2 ) ( ) ( 4 )2
W Wt X t W
2
sinc2
( ) = X
2
Step 1 from Known transform from the F.T Table
Step 2
t
2
0
1
t
( ) = rectt
x t
2
2W0
1
rect4W
2W
rect4W
Even Function
( ) 2 ( )tX x
1 ( ) ( )2
2 x x
2 21 1( ) ( ) ( ( ) )x t x t X X
Multiplication in Frequency Convolution in Time
Proof
1 21 2( ) ( ) = ( ) ( ) x t x t x x t d
221( ) ( )
2j tx t X e d
( )
221( ) ( )
2 j tx t X e d
2
1 ( )2
j t jX e e d
6- The convolution Theorem
1 21 2( ) ( ) = ( ) ( ) x t x t x x t d
Now substitute x2(t-l) ( as the inverse Fourier Transform) in the convolution integral
1 21 2
1( ) ( ) = ( ) ( )
2
j t j
x t x x X e e d dt
221( ) ( )
2j t jx t X e e d
Exchanging the order of integration , we have
2 11 21( ) ( ) = ( ) ( )
2j
j tx t x t X x e e dd
1
( )X
2 11 21( ) ( ) = ( ) ( )
2 j tx t x t X X e d
Inverse Fourier Transform
2 21 1( ) ( ) ( ( ) )x t x t X X
1 21 2
1( ) ( ) = ( ) ( )
2
j t j
x t x x X e e d dt
2 21 11( ) ( ) ) ( )
2( Xx t x t X
Proof
Similar to the convolution theorem , left as an exercise
The multiplication Theorem
1 2 1 21Applying (the multiplication Theorem ( ) ( ) ( ) )
2 x t t Xx X
Applying the multiplication Theorem
rect rectt t
Find the Fourier Transform of following
Solution
Since rect
t
t
2
2 0
rect
t
t
2
2 0
trit
t 0
convolution
Time
sinc( )f sinc( )f 2 2sinc ( )f Frequencymultiplication
System Analysis with Fourier Transform
= ( ) ( ) x h t d
( )x t
( )h ty(t) = ( ( ) )x t h t
( )X ( )H ( )Y ( ) ( )X H
( ) ) ) ( (Y X H y(t) = ( ( ) )x t h t
convolution in time
multiplication in Frequency
impulse response
convolution in timemultiplication in Frequency
Proof
( ) ( ) j tY y t e dt
0
Let ( ) ( )e tjy t x t
0( )e j t j tx t e dt
' 0Change of variable
0( )( ) j tx t e dt
'
( ) ( ) j tY x t e dt
'( )X 0( )X
0
0 Let ( ) Then ( ) ( ) ( e ) tjx t x tX X
6- Frequency Shifting
0
0 Let ( ) Then ( ) ( ) ( e ) tjx t x tX X
0
0Similarly ( ) e ) (tjx t X 0
0( ) e ( )tjx t X
0 0
e e2
j t j t
0 01 12 2
( ) = .j t j t
e eX FT
0
0 ( ) e ( ) tjx t X Since
Example Find the Fourier Transform for 0( ) cos(2 )x t A t
0( ) cos(2 )x t t 0 01 12 2
j t j te e
0 0[ ]1 12 2
[ ]= . .j t j te eFT FT
( ) 1t Since 1 2 ( ) duality
0
0[ ]. (1) 2 ( )j teFT
0 01 1
2 ( ) 2 ( )2 2
( ) = X
0 00cos(2 ) ( ) ( )t
Find the Fourier Transform of the function
0( ) rect cos( )2t
x t t
0co s ( ) t
t0
r e c t 2t
t11
0
X
0( ) r e c t co s ( )
2t
x t t
t11 0
0( ) r e c t co s ( )
2t
x t t
t12
12
0
0( ) r e c t c o s ( t )
2t
x t
0 0t t
r e c t
2 2
j jt e e
0 01 1( ) r e c t r e c t
2 22 2
j t j tt tx t e e
0
0( )e ( )j t
x t X
rect 2sinc(2 )2t
Since and
0 00( ) r e c t c o s ( )
2 s (inc(2 )) ( )s inc(2 )t
x t t
Therefore
Method 1
Method 2
0( ) r e c t co s ( )
2t
x t t
t12
12
0
0( ) r e c t c o s ( t )
2t
x t
1 1 221( ) ( ) ( ) ( )
2x t Xt Xx
Since
rect 2sinc(2 )2t
0 00cos(2 ) ( ) ( )t
0( ) r e c t c o s ( t )
2t
x t
0rect cos( t)2
( )= . .t
X FT FT
0 01 2sinc(2 ) ( ) ( )
2( )= X
0 0( )s inc(2 ) s inc(2( ))( )X
7-Differentiation
Let ( ) ( ) ( ) ( )( ) d
dtx t x t j XX
in general ( ) ( ) ( )n
nn
d
dtx t j X
( ) ( ) j tdx t dx t
dt dtF e dt
Proof
( ) ( ) j tdx t dx t
dt dtF e dt
Using integration by parts
j tu e ) ( )(dx tdt
dv dt dx t j tdu j e
( )v x t
b b
b
aa a
udv uv vdu
( ) ( ) + ( )
j tj tdx t
dtF x t e j x t e dt
( ) ( ) ( )nn
nd
dtx t j X
Proof
( ) ( ) ( )d
dtx t j X
( ) ( ) + ( )
j tj tdx t
dtF x t e j x t e dt
( ) x t dt
Since x(t) is absolutely integrable
( ) ( ) ( ) ( ) + ( ) j j j tx e x e j f x t e dt
( ) 0 ( ) 0
xx
( )
( )( ) ( ) 0j jx e x e
( )
( ) j tdx t
dtF j x t e dt
( )j X
( ) ( )dx t
dtj X ( ) ( ) ( )n
n
nd x t
dt
j X
1( ) ( ) (0) ( )t
x d X Xj
Example Find the Fourier Transform of the unit step function u(t)
( ) = ( ) t
u t d 0
(1)1 () )1 (j
( ) 1 t
1 ( ) ( )u tj
7- Integration
? 0
Proof
[ ( )] ( ) j tF.T t t e dt
If ( ) ( ) t X ?
0
( )j t
t
e t dt
(1) ( )t dt
1
( ) 1 t
(DC) 1 2 ( )
2 ( ) k k
From duality
2 ( )
2j
1
0
1 22
0
1Applying the multiplication Theorem ( ) ( ) ( )( ) 2
and ( ) we ha ve
( ) ( )
x t x t X X
G G
0 0( )( ) ( )G G
( ) ( ) j tF f t e dt
Since 1 ) 2 (
Then
0
( )k
Find the Transfer Function for the following RC circuit
C
R
( )x t
( )y t
( ) ( ) ( ) dy tRC y t x t t
dt
C
R
( )t
( )h t
( ) ( ) ( )dh tRC h t t
dt
1( ) ( )t
R Ch t e u tR C
we can find h(t) by solving differential equation as follows
Method 1
C
R
( )x t
( )y t
( ) ( ) ( )dy tRC y t x t
dt
( )FT ( ) FT ( ) dy tRC y t x t
dt
( ) ( ) ( ) ( ) RC j Y Y X
( ) 1 ( ) ( ) j RC Y X
( )( )
( ) Y
HX
1( ) 1j RC
We will find h(t) using Fourier Transform Method rather than solving differential equation as follows
Method 2
C
R
( )x t
( )y t
( ) ( ) ( )dy tRC y t x t
dt
( )( )
( ) Y
HX
1( ) 1j RC
(1/ )( ) (1/ )
RCj RC
1( ) ( ) > 0 tx t e u tj
1( ) ( )t
R Ch t e u tR C
From Table 4-2
Method 3
i(t)
RRV ( )t
RV ( ) ( )t Ri t
I(f)
RRV ( )f
RV ( ) ( )RI
i(t)
L LV ( )t
L
( )V ( )
di tt L
dt
I(f)
L LV ( )f
LV ( ) ( ) ( )L j I
i(t)
C CV ( )t
( )( ) CdV t
i t Cdt
I(f)
C CV ( )f
C( ) ( )V ( )I C j
C1
V ( ) ( )( )
f I fj C
( ) ( )j L I
In this method we are going to transform the circuit to the Fourier domain . However we first see the FT on Basic elements
Method 3
( )i t
( )v t
( )I f
( )V f
( )Z f
Resistor
( ) Inductor
1Capacitor
R
Z f j L
j C
C
R
( )x t
( )y t
Method 3
R
( )X
( )Y
1j C
Fourier Transform
1
( ) ( )1
j CY f X
Rj C
1( )
1X
j RC
( ) 1( )
( ) 1
YH
X j RC
1( ) ( )t
R Ch t e u tR C
C
R
( )x t
( )y t
R
( )X
( )Y
1j C
Fourier Transform
( ) 1( )
( ) 1
YH
X j RC
2
1| ( ) |
1 ( )H
RC
( )H | ( ) |H
( )H 1tan ( )RC
2
1| ( ) |
1 ( )H
RC
( )H 1tan ( )RC
1( )
1H
j RC
C
R
( )x t
( )y t
Find y(t) if the input x(t) is
( ) ( )tx t A e u t
Method 1 ( convolution method)
Using the time domain ( convolution method , Chapter 3)
y( ) = ( ) ) (t x t h t
1( ) ( )t
R Ch t e u tR C
Example
C
R
( )x t
( )y t
( ) ( )tx t A e u t
/ ( )(1/ )
A RCj RC j
1 1( ) 1/1AY jRC jRC
Using partial fraction expansion (will be shown later)
From Table 5-2 /( ) ( )1
t RC tAy t e e u tRC
( ) AXj
1( ) ( )t
R Ch t e u tR C
(1/ )( )
(1/ ) ( )RCH
RC j
( ) ) ( ( )Y X H
Method 2 Fourier Transform
Sine Y(w) is not on the Fourier Transform Table 5-2
Example
Find y(t)
Method 1 ( convolution method) y( ) = ( ) ) (t x t h t
( ) ) ( ( )Y X H
Method 2 Fourier Transform
( )F
Therefor , to obtain the energy spectral density function of the signal t :f fE
2 2
2 | ( ) | | ( ) |t ( ) | ( ) | fold about =0
2
F Ff F F
2| ( ) |
2
F 2
fold about =0
| ( ) | F
We find the energy contained in some band of frequencies of particular interest by
finding the area under the energy spectral density curve over that band of frequencies
power spctral density (PSD)
