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8/16/2019 Chapter 3 Part 2 Fourier Transform Properties
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FOURIER TRANSFORMPROPERTIES
2
Linearity(superposit
ion)Time shift Time sa!e
"ua!ity
Fre#uenyshift
(Mo$u!ation)
%on&o!ution
"i'erentiation
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LINEARIT SUPERPOSITION
3
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E+amp!e: Find the Fourier Transform for the signalg(t)
)2
()6
(5.0
)()()(
:
21
t rect t rect
t g t g t g
Solution
+=
+=
)(sin2)3(sin3
)()()(
:
21
ω ω
ω ω ω
cc
GGG
Transform Fourier In
+=
+=
v(t)
A
-τ/2 τ/2 t
=
2sin)(
ωτ τ ω c AV
= τ
t Arect t v )(
5
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Time delay of a signal causes a linear phaseshift in the spectrum of the signal.
If the signal is delayed by , the phase
spectrum will be shifted by . The amplitude remains the same
TIME S,IFT
0)()(
)()(
0
t jeV t t v
V t v
ω ω
ω
−⇔−
⇔
0t
0t ω −
6
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E+amp!e: Find the Fourier Transform for the signal
So!ution*2/
2
|5.0|2
4
123 ω
ω
jt ee −−−
+⇔
5.023
−− t e
7
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E+amp!e: Find the Fourier Transform for the signalof Figure (a).
Solution:2/
2sinc)( ωτ ωτ
τ ω je AV −
=
Similar magnitude with
signal but a
phase shift of
τ
t Arect
2/τ ω −
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⇔a
ω V
av(at)
1a: a real constant
⇒ =
τ
t At v rect)(
=2
sinc)V( ωτ τ ω A
==τ
t At vt v
2rect)2()(if 1
=
=⇒4
sinc222
1)(Therefore 1
ωτ τ ω ω
AV V
• If a!, the signal v(t) will be ompresse$- but this willproduce e+pansion in fre#ueny $omain.
• "r else If a#!, the signal v(t) will be e+pen$e$- but this willproduce ompression in fre#ueny $omain.
TIME S%ALE9
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E+amp!e of TimeSa!e
10
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)(2)(
)()(
ω π
ω
−⇔⇔
vt V
V t v
−⇔
⇔
τ A
t A
Aτ
t A
ω π
τ τ
ωτ τ
rect22
sinc
2sincrect
E+amp!e: Find the Fourier Transform for
=2
sinc)( τ τ t At g
"UALIT 11
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FRE/UEN% S,IFT(MO"ULATION)
[ ]
[ ])()(21sin)(
)()(2
1cos)(
000
000
ω ω ω ω ω
ω ω ω ω ω
+−−=
++−=
V V j
t t v
V V t t v
)()( ω V t v =
)()(0
0 ω ω ω −=−
V et v t j
13
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E+amp!eFind the Fourier Transform for the signal g(t)$f(t)cos(2%t)
where f(t)$2rect(t).
=2
sinc2)( ω
ω F So!ution*
[ ]
+
+ −
=
++
−=
++−=
2
20sinc2
20sinc
2
20sinc2
2
20sinc2
2
1
)20()20(2
1
)(
ω ω
ω ω
ω ω ω F F G
14
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( ) ( )
( ) ( )
( ) ( ) ( ) ( )ω ω
ω
ω
2121
22
11
V V t vt v
V t v
V t v
⇔∗
⇔
⇔
%ON0OLUTION15
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&'ample: ien two signals as below:
)()(1 t uet f at −= a > 0
)()(2 t uet f bt −= b > 0
So!ution*
+
−
+−
=
++=
+
+
=
=
ω ω
ω ω
ω ω
ω ω ω
jb jaab
jb ja
jb ja
F F F
111
))((
1
11
)()()( 21
[ ])()(1
)( t uet ueab
t f bt at −− −−
=
)()()( 21 t f t f t f ∗=
16
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( )ω V v(t) ⇔
"IFFERENTIATION17
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E+amp!e*
)50rect()50rect()( .t A.t Adt t dg −++=
)1()(2)1()(
2
2
−+−+= t At At A
dt
t g d δ δ δ
18
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So!ution* 2n$ or$er $i'erentia!
[ ]
=
=
−=−=−
+−=
−+−+=
−
2sinc
2sin
4)(
2sin4)1(cos2)(
2)()(
)1()(2)1()(
22
2
22
2
2
2
ω ω
ω ω
ω ω ω ω
ω ω
δ δ δ
ω ω
A A
G
A AG
Ae A AeG j
t At At A F dt
t g d F
j j
20
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( )[ ] ( )
( )
( )ω πδ
ω δ π
π
ω ω δ
π
ω δ ω
2 1
2
1
2
1
2
11
⇔
⇔
∫ == ∞
∞−
− d e F t j
IN0ERSE FOURIERTRANSFORM
( )t v( )ω V ( )[ ] ( )∫ = ∞
∞−
−ω ω
π
ω ω d eV V F t j
2
11
21
