The document provides an overview of Fourier transforms. It begins by introducing Fourier series which deals with continuous-time periodic signals and results in discrete frequency spectra. It then discusses how the Fourier integral and continuous Fourier transform can deal with aperiodic signals by providing continuous spectra. The continuous Fourier transform represents a function as an integral of its frequencies, while the inverse transform uses this representation to recover the original function. The properties of the Fourier transform discussed include linearity, time scaling, time reversal, time shifting, and frequency shifting. Real functions have special properties where the Fourier transform is always real or pure imaginary. Examples are provided to illustrate how to calculate the Fourier transform of simple functions.

1. Fourier Transform

2. Content
Introduction
Fourier
Integral
Fourier Transform
Properties of Fourier Transform
Convolution
Parseval’s Theorem

3. Continuous-Time
Fourier Transform
Introduction

4. 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

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?

8. Continuous-Time
Fourier Transform
Fourier Integral

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

12. Continuous-Time
Fourier Transform
Fourier Transform

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

20. Continuous-Time
Fourier Transform
Properties of
Fourier Transform

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 )]
−∞

40. You!!
nk You
!!Tha nk
Tha