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Fourier Transform
Content
Introduction
Fourier

Integral
Fourier Transform
Properties of Fourier Transform
Convolution
Parseval’s Theorem
Continuous-Time
Fourier Transform
Introduction
The Topic
Periodic

Discrete
Time

Fourier
Fourier
Series
Series

Discrete
Discrete
Fourier
Fourier
Transform
Transform

Aperiodic

Continuous
Time

Continuous
Continuous
Fourier
Fourier
Transform
Transform

Fourier
Fourier
Transform
Transform
Review of Fourier Series
 Deal

with continuous-time periodic signals.
 Discrete frequency spectra.
A Periodic Signal
A Periodic Signal

f(t)
t
T

2T

3T
Two Forms for Fourier Series
Sinusoidal
a0 ∞
2πnt ∞
2πnt
f (t ) = + ∑ an cos
+ ∑ bn sin
Form
2 n =1
T
T
n =1
2 T /2
a0 = ∫
f (t )dt
−T / 2
T

Complex
Form:

f (t ) =

∞

∑ cn e

n = −∞

jnω0t

2 T /2
an = ∫
f (t ) cos nω0tdt
T −T / 2
2 T /2
bn = ∫
f (t ) sin nω0tdt
T −T / 2

1
cn =
T

∫

T /2

−T / 2

f (t )e − jnω0t dt
How to Deal with Aperiodic Signal?
A Periodic Signal
A Periodic Signal

f(t)
t
T

If T→∞, what happens?
Continuous-Time
Fourier Transform
Fourier Integral
Fourier Integral
fT (t ) =

∞

∑c e

n = −∞

n

jnω0t

1
cn =
T

∫

T /2

−T / 2

fT (t )e − jnω0t dt

∞

 1 T /2

=∑ ∫
fT (τ)e − jnω0 τ dτ e jnω0t
−T / 2

n = −∞  T
1 ∞  T /2
=
fT (τ)e − jnω0 τ dτ ω0 e jnω0t
∑


2π n = −∞  ∫−T / 2
1 ∞  T /2
=
fT (τ)e − jnω0 τ dτ e jnω0t ∆ω
∑


2π n = −∞  ∫−T / 2
1 ∞ ∞
=
fT (τ)e − jωτ dτ e jωt dω



2π ∫−∞  ∫−∞

ω0 =

2π
T

1 ω0
=
T 2π

Let ∆ω = ω0 =

2π
T

T → ∞ ⇒ dω = ∆ω ≈ 0
Fourier Integral
1 ∞ ∞
− jωτ
 e jωt dω
f (t ) =
∫−∞ ∫−∞ f (τ)e dτ

2π 
F(jω )

1 ∞
jω t
f (t ) =
∫−∞ F ( jω)e dω
2π
∞

F ( jω) = ∫ f (t )e
−∞

− jω t

dt

Synthesis
Analysis
Fourier Series vs. Fourier Integral
Fourier
Series:

f (t ) =

cn e jnω0t
∑

Period Function

n = −∞

1
cn =
T

Fourier
Integral:

∞

∫

T /2

−T / 2

Discrete Spectra

fT (t )e − jnω0t dt

1 ∞
f (t ) =
F ( jω)e jωt dω
2π ∫−∞
∞

F ( jω) = ∫ f (t )e − jωt dt
−∞

Non-Period
Function

Continuous Spectra
Continuous-Time
Fourier Transform
Fourier Transform
Fourier Transform Pair
Inverse Fourier Transform:

1 ∞
f (t ) =
F ( jω)e jωt dω
2π ∫−∞

Synthesis

Fourier Transform:
∞

F ( jω) = ∫ f (t )e
−∞

− jωt

dt

Analysis
Existence of the Fourier Transform

Sufficient Condition:
f(t) is absolutely integrable, i.e.,

∫

∞

−∞

| f (t ) |dt < ∞
Continuous Spectra
∞

F ( jω) = ∫ f (t )e − jωt dt
−∞

F ( jω) = FR ( jω) + jFI ( jω)

=| F ( jω) | e
Magnitude

jφ ( ω )
Phase

FI(jω)

|
ω)
j
|F(
φ(ω)

FR(jω)
Example
1

-1

f(t)
t

1

1 − jωt
F ( jω) = ∫ f (t )e dt = ∫ e dt =
e
−∞
−1
− jω
j − jω
2 sin ω
jω
= (e − e ) =
ω
ω
∞

− jω t

1

1

− jωt

−1
Example

F(ω
F(ω) )

33
22
11
00

-1

1

-1
f(t)

-10
-10

-5
-5

00

55

33
22

t

|F(ω
|F(ω)|)|

-1

10
10

11 1
00
-10
-10
1
44

− jω t

arg[F(ω
arg[F(ω)])]

1 − jωt
F ( jω) = ∫ f (t )e dt = ∫ e dt =
e
−∞
−1
− jω
22
j − jω
2 sin ω
jω
00
= (e − e ) = -10 -5-5 00 55 10
-10
10
ω
ω
∞

-5
-5

− jωt

00

55

10
10

1

−1
Example
f(t)

e−αt
t
∞

F ( jω) = ∫ f (t )e

− jω t

−∞
∞

=∫ e
0

− ( α + jω ) t

∞

dt = ∫ e −αt e − jωt dt
0

1
dt =
α + jω
Example
f(t)

1
1
|F(jω)|
|F(jω)|

=2
αα =2

0.5
0.5

0
0
2
2

∞

−∞
∞

=∫ e
0

arg[F(jω)]
arg[F(jω)]

F ( jω) = ∫ f (t )e

-10
-10

− jω t

0
0

-2
-2

− ( α + jω ) t

e−αt

-5
-5

0
0

5
5

t 10
10

∞

dt = ∫ e −αt e − jωt dt
0

1
dt =
α + jω
-10
-10

-5
-5

0
0

5
5

10
10
Continuous-Time
Fourier Transform
Properties of
Fourier Transform
Notation
F [ f (t )] = F ( jω)

F [ F ( jω)] = f (t )
-1

f (t ) ←
→ F ( jω)
F
Linearity
a1 f1 (t ) + a2 f 2 (t ) ←
→ a1 F1 ( jω) + a2 F2 ( jω)
F

orrk !!
Wo k
H om e W
!!Home
Time Scaling
1  ω
f (at ) ←
→
F j 
|a|  a
F

orrk !!
Wo k
H om e W
!!Home
Time Reversal
f ( −t ) ←
→ F ( − jω)
F

Pf) F [ f (−t )] = ∞ f (−t )e − jωt dt = t =∞ f (−t )e − jωt dt
∫−∞
∫t =−∞
=∫

− t =∞

−t = −∞

= −∫

=∫
f (t )e jωt d ( −t )
f (t )e d ( −t )
−t = −∞

t = −∞

t =∞

∞

− t =∞

j ωt

f (t )e dt = ∫
j ωt

t =∞

t = −∞

f (t )e jωt dt

= ∫ f (t )e jωt dt = F (− jω)
−∞
Time Shifting
f (t − t0 ) ←
→ F ( jω) e
F

− jωt 0

Pf) F [ f (t − t )] = ∞ f (t − t )e − jωt dt = t =∞ f (t − t )e − jωt dt
0
0
0
∫−∞
∫t =−∞
=∫

t +t0 =∞

=e

− j ωt 0

=e

− j ωt 0

t + t 0 = −∞

f (t )e − jω(t +t0 ) d (t + t0 )

∫

t =∞

∫

∞

t = −∞

−∞

f (t )e − jωt dt

− jω t
f (t )e − jωt dt = F ( jω)e 0
Frequency Shifting (Modulation)
f (t )e
Pf)

jω0t

¬  F [ j (ω − ω0 ) ]
→

F [ f (t )e

F

jω 0 t

∞

] = ∫ f (t )e jω0t e − jωt dt
−∞
∞

= ∫ f (t )e − j ( ω−ω0 )t dt
−∞

= F [ j (ω − ω0 )]
Symmetry Property
F [ F ( jt )] = 2πf (−ω)
Proof

∞

2πf (t ) = ∫ F ( jω)e jωt dω
−∞

∞

2πf (−t ) = ∫ F ( jω)e − jωt dω
−∞

Interchange symbols ω and t
∞

2πf (−ω) = ∫ F ( jt )e − jωt dt = F [ F ( jt )]
−∞
Fourier Transform for
Real Functions
If f(t) is a real function, and F(jω) = FR(jω) + jFI(jω)
F(−jω) = F*(jω)

∞

F ( jω) = ∫ f (t )e

− jωt

−∞
∞

dt

F * ( jω) = ∫ f (t )e dt = F (− jω)
−∞

jωt
Fourier Transform for
Real Functions
If f(t) is a real function, and F(jω) = FR(jω) + jFI(jω)
F(−jω) = F*(jω)
FR(jω) is even, and FI(jω) is odd.
F R jω ) = F R F (− jω ) = − F (jω )
(−
(jω ) I
I
Magnitude spectrum |F(jω)| is even, and
phase spectrum φ(ω) is odd.
Fourier Transform for
Real Functions
If f(t) is real and even
F(jω) is real
Pf)
Even

If f(t) is real and odd

√

f (t ) = f (−t )

F(jω) is pure imaginary
Pf)
Odd

F ( jω) = F (− jω)

Real

F (− jω) = F * ( jω)
F ( jω) = F * ( jω)

√

f (t ) = − f (−t )
F ( jω) = − F (− jω)

Real

F (− jω) = F * ( jω)
F ( jω) = − F * ( jω)
Example:
F [ f (t )] = F ( jω)
Sol)

F [ f (t ) cos ω0t ] = ?

1
f (t ) cos ω0t = f (t )(e jω0t + e − jω0t )
2
1
1
jω 0 t
F [ f (t ) cos ω0t ] = F [ f (t )e ] + F [ f (t )e − jω0t ]
2
2
1
1
= F [ j (ω − ω0 )] + F [ j (ω + ω0 )]
2
2
Example:
1

−d/2

f(t)=wd(t)cosω0t

wd(t)
t

d/2

−d/2

d/2

t

2  ωd 
Wd ( jω) = F [ wd (t )] = ∫ e dt = sin 

−d / 2
ω  2 
d
d
sin (ω − ω0 ) sin (ω + ω0 )
2
2
=
+
F ( jω) = F [ wd (t ) cos ω0t ]
ω − ω0
ω + ω0
d /2

− jωt
1.5
1.5
d=2
d=2
ω0=5ππ
ω =5

1
1

0

F(jω)
F(jω)

Example:

0.5
0.5
0
0

-0.5
-0.5 -60
-60

1

−d/2

-40
-40

-20
-20

0
0

20
20

40
40

ω
ω

60
60

f(t)=wd(t)cosω0t

wd(t)
t

d/2

−d/2

d/2

t

2  ωd 
Wd ( jω) = F [ wd (t )] = ∫ e dt = sin 

−d / 2
ω  2 
d
d
sin (ω − ω0 ) sin (ω + ω0 )
2
2
=
+
F ( jω) = F [ wd (t ) cos ω0t ]
ω − ω0
ω + ω0
d /2

− jωt
1

Example:
sin at
f (t ) =
πt
Sol)

wd(t)
t

−d/2

d/2

F ( jω) = ?

2  ωd 
Answer is
Wd ( jω) = sin 

just
ω  2 
opposite to
as expected
 2  td  
F [Wd ( jt )] = F  sin    = 2πwd (−ω)
 2 
t
0 ω <| a |
 sin at 
F [ f (t )] = F 
 = w2 a (−ω) = 1 ω >| a |
 πt 

Fourier Transform of f’(t)
f (t ) ←
→ F ( jω) and lim f (t ) = 0
F

t → ±∞

f ' (t ) ←F jωF ( jω)
→
Pf) F [ f ' (t )] = ∞ f ' (t )e − jωt dt
∫−∞
= f (t )e

− j ωt ∞
−∞

= jωF ( jω)

∞

+ jω∫ f (t )e − jωt dt
−∞
Fourier Transform of f (t)
(n)

f (t ) ←
→ F ( jω) and lim f (t ) = 0
F

t → ±∞

f ( n ) (t ) ←F ( jω) n F ( jω)
→

orrk !!
Wo k
H om e W
!!Home
Fourier Transform of f (t)
(n)

f (t ) ←
→ F ( jω) and lim f (t ) = 0
F

t → ±∞

f ( n ) (t ) ←F ( jω) n F ( jω)
→

orrk !!
Wo k
H om e W
!!Home
Fourier Transform of Integral
f (t ) ←
→ F ( jω) and
F

∫

∞

−∞

f (t )dt = F ( 0 ) = 0

 t f ( x)dx  = 1 F ( jω)
F ∫
 −∞
 jω


Let φ(t ) =

∫

t

−∞

f ( x)dx

lim φ(t ) = 0
t →∞

F [φ' (t )] = F [ f (t )] = F ( jω) = jωΦ ( jω)
1
Φ ( jω) =
F ( jω)
jω
The Derivative of Fourier Transform
dF ( jω)
F [− jtf (t )] ←
→
dω
F

Pf)

∞

F ( jω) = ∫ f (t )e − jωt dt
−∞

∞
dF ( jω) d ∞
∂ − j ωt
− j ωt
=
∫−∞ f (t )e dt = ∫−∞ f (t ) ∂ω e dt
dω
dω
∞

= ∫ [− jtf (t )]e − jωt dt = F [− jtf (t )]
−∞
You!!
nk You
!!Tha nk
Tha

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fourier transforms

  • 2. Content Introduction Fourier Integral Fourier Transform Properties of Fourier Transform Convolution Parseval’s Theorem
  • 5. Review of Fourier Series  Deal with continuous-time periodic signals.  Discrete frequency spectra. A Periodic Signal A Periodic Signal f(t) t T 2T 3T
  • 6. Two Forms for Fourier Series Sinusoidal a0 ∞ 2πnt ∞ 2πnt f (t ) = + ∑ an cos + ∑ bn sin Form 2 n =1 T T n =1 2 T /2 a0 = ∫ f (t )dt −T / 2 T Complex Form: f (t ) = ∞ ∑ cn e n = −∞ jnω0t 2 T /2 an = ∫ f (t ) cos nω0tdt T −T / 2 2 T /2 bn = ∫ f (t ) sin nω0tdt T −T / 2 1 cn = T ∫ T /2 −T / 2 f (t )e − jnω0t dt
  • 7. How to Deal with Aperiodic Signal? A Periodic Signal A Periodic Signal f(t) t T If T→∞, what happens?
  • 9. Fourier Integral fT (t ) = ∞ ∑c e n = −∞ n jnω0t 1 cn = T ∫ T /2 −T / 2 fT (t )e − jnω0t dt ∞  1 T /2  =∑ ∫ fT (τ)e − jnω0 τ dτ e jnω0t −T / 2  n = −∞  T 1 ∞  T /2 = fT (τ)e − jnω0 τ dτ ω0 e jnω0t ∑   2π n = −∞  ∫−T / 2 1 ∞  T /2 = fT (τ)e − jnω0 τ dτ e jnω0t ∆ω ∑   2π n = −∞  ∫−T / 2 1 ∞ ∞ = fT (τ)e − jωτ dτ e jωt dω    2π ∫−∞  ∫−∞ ω0 = 2π T 1 ω0 = T 2π Let ∆ω = ω0 = 2π T T → ∞ ⇒ dω = ∆ω ≈ 0
  • 10. Fourier Integral 1 ∞ ∞ − jωτ  e jωt dω f (t ) = ∫−∞ ∫−∞ f (τ)e dτ  2π  F(jω ) 1 ∞ jω t f (t ) = ∫−∞ F ( jω)e dω 2π ∞ F ( jω) = ∫ f (t )e −∞ − jω t dt Synthesis Analysis
  • 11. Fourier Series vs. Fourier Integral Fourier Series: f (t ) = cn e jnω0t ∑ Period Function n = −∞ 1 cn = T Fourier Integral: ∞ ∫ T /2 −T / 2 Discrete Spectra fT (t )e − jnω0t dt 1 ∞ f (t ) = F ( jω)e jωt dω 2π ∫−∞ ∞ F ( jω) = ∫ f (t )e − jωt dt −∞ Non-Period Function Continuous Spectra
  • 13. Fourier Transform Pair Inverse Fourier Transform: 1 ∞ f (t ) = F ( jω)e jωt dω 2π ∫−∞ Synthesis Fourier Transform: ∞ F ( jω) = ∫ f (t )e −∞ − jωt dt Analysis
  • 14. Existence of the Fourier Transform Sufficient Condition: f(t) is absolutely integrable, i.e., ∫ ∞ −∞ | f (t ) |dt < ∞
  • 15. Continuous Spectra ∞ F ( jω) = ∫ f (t )e − jωt dt −∞ F ( jω) = FR ( jω) + jFI ( jω) =| F ( jω) | e Magnitude jφ ( ω ) Phase FI(jω) | ω) j |F( φ(ω) FR(jω)
  • 16. Example 1 -1 f(t) t 1 1 − jωt F ( jω) = ∫ f (t )e dt = ∫ e dt = e −∞ −1 − jω j − jω 2 sin ω jω = (e − e ) = ω ω ∞ − jω t 1 1 − jωt −1
  • 17. Example F(ω F(ω) ) 33 22 11 00 -1 1 -1 f(t) -10 -10 -5 -5 00 55 33 22 t |F(ω |F(ω)|)| -1 10 10 11 1 00 -10 -10 1 44 − jω t arg[F(ω arg[F(ω)])] 1 − jωt F ( jω) = ∫ f (t )e dt = ∫ e dt = e −∞ −1 − jω 22 j − jω 2 sin ω jω 00 = (e − e ) = -10 -5-5 00 55 10 -10 10 ω ω ∞ -5 -5 − jωt 00 55 10 10 1 −1
  • 18. Example f(t) e−αt t ∞ F ( jω) = ∫ f (t )e − jω t −∞ ∞ =∫ e 0 − ( α + jω ) t ∞ dt = ∫ e −αt e − jωt dt 0 1 dt = α + jω
  • 19. Example f(t) 1 1 |F(jω)| |F(jω)| =2 αα =2 0.5 0.5 0 0 2 2 ∞ −∞ ∞ =∫ e 0 arg[F(jω)] arg[F(jω)] F ( jω) = ∫ f (t )e -10 -10 − jω t 0 0 -2 -2 − ( α + jω ) t e−αt -5 -5 0 0 5 5 t 10 10 ∞ dt = ∫ e −αt e − jωt dt 0 1 dt = α + jω -10 -10 -5 -5 0 0 5 5 10 10
  • 21. Notation F [ f (t )] = F ( jω) F [ F ( jω)] = f (t ) -1 f (t ) ← → F ( jω) F
  • 22. Linearity a1 f1 (t ) + a2 f 2 (t ) ← → a1 F1 ( jω) + a2 F2 ( jω) F orrk !! Wo k H om e W !!Home
  • 23. Time Scaling 1  ω f (at ) ← → F j  |a|  a F orrk !! Wo k H om e W !!Home
  • 24. Time Reversal f ( −t ) ← → F ( − jω) F Pf) F [ f (−t )] = ∞ f (−t )e − jωt dt = t =∞ f (−t )e − jωt dt ∫−∞ ∫t =−∞ =∫ − t =∞ −t = −∞ = −∫ =∫ f (t )e jωt d ( −t ) f (t )e d ( −t ) −t = −∞ t = −∞ t =∞ ∞ − t =∞ j ωt f (t )e dt = ∫ j ωt t =∞ t = −∞ f (t )e jωt dt = ∫ f (t )e jωt dt = F (− jω) −∞
  • 25. Time Shifting f (t − t0 ) ← → F ( jω) e F − jωt 0 Pf) F [ f (t − t )] = ∞ f (t − t )e − jωt dt = t =∞ f (t − t )e − jωt dt 0 0 0 ∫−∞ ∫t =−∞ =∫ t +t0 =∞ =e − j ωt 0 =e − j ωt 0 t + t 0 = −∞ f (t )e − jω(t +t0 ) d (t + t0 ) ∫ t =∞ ∫ ∞ t = −∞ −∞ f (t )e − jωt dt − jω t f (t )e − jωt dt = F ( jω)e 0
  • 26. Frequency Shifting (Modulation) f (t )e Pf) jω0t ¬  F [ j (ω − ω0 ) ] → F [ f (t )e F jω 0 t ∞ ] = ∫ f (t )e jω0t e − jωt dt −∞ ∞ = ∫ f (t )e − j ( ω−ω0 )t dt −∞ = F [ j (ω − ω0 )]
  • 27. Symmetry Property F [ F ( jt )] = 2πf (−ω) Proof ∞ 2πf (t ) = ∫ F ( jω)e jωt dω −∞ ∞ 2πf (−t ) = ∫ F ( jω)e − jωt dω −∞ Interchange symbols ω and t ∞ 2πf (−ω) = ∫ F ( jt )e − jωt dt = F [ F ( jt )] −∞
  • 28. Fourier Transform for Real Functions If f(t) is a real function, and F(jω) = FR(jω) + jFI(jω) F(−jω) = F*(jω) ∞ F ( jω) = ∫ f (t )e − jωt −∞ ∞ dt F * ( jω) = ∫ f (t )e dt = F (− jω) −∞ jωt
  • 29. Fourier Transform for Real Functions If f(t) is a real function, and F(jω) = FR(jω) + jFI(jω) F(−jω) = F*(jω) FR(jω) is even, and FI(jω) is odd. F R jω ) = F R F (− jω ) = − F (jω ) (− (jω ) I I Magnitude spectrum |F(jω)| is even, and phase spectrum φ(ω) is odd.
  • 30. Fourier Transform for Real Functions If f(t) is real and even F(jω) is real Pf) Even If f(t) is real and odd √ f (t ) = f (−t ) F(jω) is pure imaginary Pf) Odd F ( jω) = F (− jω) Real F (− jω) = F * ( jω) F ( jω) = F * ( jω) √ f (t ) = − f (−t ) F ( jω) = − F (− jω) Real F (− jω) = F * ( jω) F ( jω) = − F * ( jω)
  • 31. Example: F [ f (t )] = F ( jω) Sol) F [ f (t ) cos ω0t ] = ? 1 f (t ) cos ω0t = f (t )(e jω0t + e − jω0t ) 2 1 1 jω 0 t F [ f (t ) cos ω0t ] = F [ f (t )e ] + F [ f (t )e − jω0t ] 2 2 1 1 = F [ j (ω − ω0 )] + F [ j (ω + ω0 )] 2 2
  • 32. Example: 1 −d/2 f(t)=wd(t)cosω0t wd(t) t d/2 −d/2 d/2 t 2  ωd  Wd ( jω) = F [ wd (t )] = ∫ e dt = sin   −d / 2 ω  2  d d sin (ω − ω0 ) sin (ω + ω0 ) 2 2 = + F ( jω) = F [ wd (t ) cos ω0t ] ω − ω0 ω + ω0 d /2 − jωt
  • 33. 1.5 1.5 d=2 d=2 ω0=5ππ ω =5 1 1 0 F(jω) F(jω) Example: 0.5 0.5 0 0 -0.5 -0.5 -60 -60 1 −d/2 -40 -40 -20 -20 0 0 20 20 40 40 ω ω 60 60 f(t)=wd(t)cosω0t wd(t) t d/2 −d/2 d/2 t 2  ωd  Wd ( jω) = F [ wd (t )] = ∫ e dt = sin   −d / 2 ω  2  d d sin (ω − ω0 ) sin (ω + ω0 ) 2 2 = + F ( jω) = F [ wd (t ) cos ω0t ] ω − ω0 ω + ω0 d /2 − jωt
  • 34. 1 Example: sin at f (t ) = πt Sol) wd(t) t −d/2 d/2 F ( jω) = ? 2  ωd  Answer is Wd ( jω) = sin   just ω  2  opposite to as expected  2  td   F [Wd ( jt )] = F  sin    = 2πwd (−ω)  2  t 0 ω <| a |  sin at  F [ f (t )] = F   = w2 a (−ω) = 1 ω >| a |  πt  
  • 35. Fourier Transform of f’(t) f (t ) ← → F ( jω) and lim f (t ) = 0 F t → ±∞ f ' (t ) ←F jωF ( jω) → Pf) F [ f ' (t )] = ∞ f ' (t )e − jωt dt ∫−∞ = f (t )e − j ωt ∞ −∞ = jωF ( jω) ∞ + jω∫ f (t )e − jωt dt −∞
  • 36. Fourier Transform of f (t) (n) f (t ) ← → F ( jω) and lim f (t ) = 0 F t → ±∞ f ( n ) (t ) ←F ( jω) n F ( jω) → orrk !! Wo k H om e W !!Home
  • 37. Fourier Transform of f (t) (n) f (t ) ← → F ( jω) and lim f (t ) = 0 F t → ±∞ f ( n ) (t ) ←F ( jω) n F ( jω) → orrk !! Wo k H om e W !!Home
  • 38. Fourier Transform of Integral f (t ) ← → F ( jω) and F ∫ ∞ −∞ f (t )dt = F ( 0 ) = 0  t f ( x)dx  = 1 F ( jω) F ∫  −∞  jω   Let φ(t ) = ∫ t −∞ f ( x)dx lim φ(t ) = 0 t →∞ F [φ' (t )] = F [ f (t )] = F ( jω) = jωΦ ( jω) 1 Φ ( jω) = F ( jω) jω
  • 39. The Derivative of Fourier Transform dF ( jω) F [− jtf (t )] ← → dω F Pf) ∞ F ( jω) = ∫ f (t )e − jωt dt −∞ ∞ dF ( jω) d ∞ ∂ − j ωt − j ωt = ∫−∞ f (t )e dt = ∫−∞ f (t ) ∂ω e dt dω dω ∞ = ∫ [− jtf (t )]e − jωt dt = F [− jtf (t )] −∞