Signal Processing Chapter 3

1. BELT4093
SIGNALS &
NETWORKS
Chapter 3
Fourier Transform
by Dr. Siti Nur Sakinah Jamaludin, 2019
Acknowledgement to Nurul Wahidah Arshad, FKEE, UMP

2. 1
2
3
4
Definition & Properties of FT
FT using Derivative Technique
Inverse FT
Applications
Fourier
Transform
(FT)

3. At the end of Chapter 3, student should be able
to:
• Determine the Fourier Transforms using its
table and properties.
• Determine the Fourier Transforms using
derivative technique.
• Transform the signal from frequency domain
into time domain using inverse Fourier
Transform.
• Discuss Fourier Transform application in
amplitude modulation.
Learning
Outcomes
CHAPTER 3
3

4. 3.1 Definition & Properties 4
• Same as Fourier series but for aperiodic functions.
Ex: unit step u(t), exp function eat
• The trick is to extend the period T in Fourier series:
• When T increased the only a single function remains
• So, the function will no longer be periodic
Fourier
Transform
Fourier Series is about …
approximating functions on [-L, L].
How do we approximate function on [-∞, ∞]???

5. Definition of Fourier Transform 5
• The Fourier transform is an integral transformation of f(t) from time domain to
frequency domain
• The inverse Fourier transform:

6. Practice Problem 18.2





1
1
)
(t
f
1
0
,
0
1
,





t
t
   



















j
j
j
j
j
e
j
e
j
j
e
j
e
dt
e
dt
e
e
t
f
F
j
j
t
j
t
j
t
j
t
j
t
j
)
1
(cos
2
2
cos
2
1
1
1
1
)
(
)
(
1
0
0
1
1
0
0
1












































7. Practice Problem 18.3




0
)
(
at
e
t
f






t
t
0
,
0
,









j
a
j
a
j
a
e
dt
e
dt
e
dt
e
e
e
t
f
F
t
j
a
t
j
a
t
j
t
j
at
t
j




































1
0
1
)
0
(
)
(
)
(
0
)
(
0
)
(
0
0

8. Fourier transform pairs

9. • Linearity
• Time scaling
• Time shifting
• Frequency shifting
• Time differentiation
• Time integration
• Reversal
• Duality
Properties of Fourier Transform
Useful in finding the transforms of complicated functions
from the transform of simple functions:

10. Fourier transform of a linear
combination of functions is the same
as the linear combination of the
transforms of the individual functions
Properties of
Fourier
Transform -
Linearity

11. – Time expansion |a| > 1 will causes
frequency compression
– Time compression |a| < 1 will causes
frequency expansion
Properties of
Fourier
Transform –
Time Scaling
Example 1
𝑓 𝑡 = 𝑠𝑔𝑛 𝑡 ↔ 𝐹 𝜔 =
2
𝑗𝜔
Find 𝐺 𝜔 if 𝑔 𝑡 = 𝑓(4𝑡)
𝐺 𝜔 = ℱ 𝑓(4𝑡 ] =
1
|4|
𝐹
𝜔
4
=
1
|4|
.
2
𝑗(𝜔
4)
=
2
𝑗𝜔

12. – Delay in time domain causes phase shift
in frequency domain
– Recall in phasors, complex number z can
be express in exponential form as:
Properties of
Fourier
Transform -
Time shifting 
 j
re
r
z 


Example 2
𝑓 𝑡 = 𝑠𝑔𝑛 𝑡 ↔ 𝐹 𝜔 =
2
𝑗𝜔
Find 𝐺 𝜔 if 𝑔 𝑡 = 𝑓(𝑡 − 2)
𝐺 𝜔 = ℱ 𝑓(𝑡 − 2 ] = 𝑒−𝑗𝜔2𝐹 𝜔 =
2𝑒−𝑗𝜔2
𝑗𝜔

13. A frequency shift in frequency domain adds
a phase shift in time function
Properties of
Fourier
Transform -
Frequency
shifting





 )
(
)
(
)
( 0
0 




 f
d
f
* Special case: sifting property of the impulse function
Example 3
𝑓 𝑡 = 𝑠𝑔𝑛 𝑡 ↔ 𝐹 𝜔 =
2
𝑗𝜔
Find 𝐺 𝜔 if 𝑔 𝑡 = 𝑓(𝑡)𝑒𝑗5𝑡
𝐺 𝜔 = ℱ 𝑓(𝑡 𝑒𝑗5𝑡
] = 𝐹 𝜔 − 5 =
2
𝑗(𝜔 − 5)
=
2
𝑗𝜔 − 𝑗5

14. – Fourier transform of the derivative f(t) is
obtained by multiplying the transform of f(t)
by jω
– The repeated derivative will give:
Properties of
Fourier
Transform -
Time
differentiation Example 4
𝑓 𝑡 = 𝑠𝑔𝑛 𝑡 ↔ 𝐹 𝜔 =
2
𝑗𝜔
Find 𝐺 𝜔 if 𝑔 𝑡 = 𝑓2(𝑡)
𝐺 𝜔 = ℱ 𝑓2(𝑡 ] = (𝑗𝜔)2𝐹 𝜔 = −𝜔 ∙
2
𝑗

15. Reversing the f(t) in time domain will result in
reversing the F(ω) in frequency domain
Properties of
Fourier
Transform -
Reversal

16. Properties of Fourier Transform
Summarized table of the Fourier transform properties

17. Differentiate the signal until
you get impulse response
3.2 Derivative
Methode
Remember from chapter 1?
• Differentiate Unit Ramp
become Unit Step
• Derivative of the Unit Step
function is Impulse Function

18. Practice
Problem 18.5














8
2
4
2
4
2
8
2
)
(
t
t
t
t
t
f
4
3
,
3
2
,
2
3
,
3
4
,












t
t
t
t










2
2
2
2
)
(
' t
f
4
3
,
3
2
,
2
3
,
3
4
,












t
t
t
t
0
f ’(t)
t
4
3
2
1
–4 –3 –2 –1
2
-2
First Derivative:

19. Practice
Problem 18.5
0
f “(t)
t
4
3
2
1
–4 –3 –2 –1
2 2 2 2
–4
–4
)
4
(
2
)
3
(
4
)
2
(
2
)
2
(
2
)
3
(
4
)
4
(
2
)
(
'
'












t
t
t
t
t
t
t
f






Left-hand side:
    )
(
)
(
)
(
'
' 2
2



 F
F
j
t
f
F 


Right-hand side:
𝐹
2𝛿(𝑡 + 4) − 4𝛿(𝑡 + 3)
+2𝛿(𝑡 + 2) + 2𝛿(𝑡 − 2)
−4𝛿(𝑡 − 3) + 2𝛿(𝑡 − 4)
= 2𝑒𝑗4𝜔
− 4𝑒𝑗3𝜔
+ 2𝑒𝑗2𝜔
+2𝑒−𝑗2𝜔 − 4𝑒−𝑗3𝜔 + 2𝑒−𝑗4𝜔
Second derivative:
Transform using table:

20. Practice
Problem 18.5
Balancing both sides:
Group the transform:







 4
3
2
2
3
4
2
2
4
2
2
4
2
)
( j
j
j
j
j
j
e
e
e
e
e
e
F 









     
2
2
2
2
2
3
3
4
4
2
cos
4
3
cos
8
4
cos
4
2
cos
4
3
cos
8
4
cos
4
2
4
2
)
(

































 j
j
j
j
j
j
e
e
e
e
e
e
F

21. Problem
18.5






1
1
)
(
t
t
t
h
1
0
,
0
1
,





t
t
0
h’(t)
t
1
–1
1
-2
-1




1
1
)
(
' t
h
1
0
,
0
1
,





t
t )
(
2 t


First Derivative:

22. Problem
18.5
)
1
(
)
(
'
2
)
1
(
)
(
'
' 



 t
t
t
t
h 


0
h’’(t)
t
1
–1
1
-2
-1
 Left-hand side:
    )
(
)
(
)
(
'
' 2
2



 F
F
j
t
h
F 


 Right-hand side:
 
)
(sin
2
2
sin
2
2
)
1
(
)
(
'
2
)
1
(






















j
j
j
e
j
e
t
t
t
F
j
j
Second derivative:
Transform using table:
• Balancing both sides:
2
2
)
sin
(
2
)
(
)
(sin
2
)
(














j
F
j
F

23. • Given:
• Find Fourier transform for time-scaling property first:
• Then, the Fourier transform for δ(t) and δ’(t):
• So, the Fourier transform should be:
Problem
18.15c
)
2
(
'
)
3
(
)
( t
t
t
f 
 















2
.
2
1
3
.
3
1
)
(


 F
F
F
δ(t) before
transform
δ’(t) before
transform




j
j
t
F
t
F



)
1
(
)]
(
'
[
1
)]
(
[

 j
F
2
1
3
1
)
( 


24. • Refer to the table of Fourier
transform properties and
pairs
• Are used for transforming
back from frequency domain
(ω) to the time domain (t).
3.3 Inverse
Fourier
Transform

25. Inverse Fourier
Transform
• When fraction is involved, there are 3
possible form of F()
– Simple poles:
– Repeated poles:
– Complex poles:
n
n
n p
s
k
p
s
k
p
s
k
p
s
p
s
p
s
s
N











 2
2
1
1
2
1 )
(
)
)(
(
)
(
n
n
n
p
s
k
p
s
k
p
s
k
p
s
s
N
)
(
)
(
)
(
)
(
)
(
2
2
1
1









b
as
s
k
s
k
b
as
s
s
N





 2
2
1
2
)
(

26. Practice
Problem
18.6a
• Given:
• To avoid complex calculation using the jω, let:
• Using fraction method, we get:
)
2
)(
4
)(
1
(
)
2
3
(
6
)
(





j
j
j
j
H






j
s 
3
)
2
)(
1
(
0
0
6
5
0
)
2
)(
3
(
0
30
2
0
0
)
1
)(
3
(
6
)
4
)(
1
(
)
2
)(
1
(
)
2
)(
4
(
)
2
3
(
6
2
4
1
)
2
)(
4
)(
1
(
)
2
3
(
6






































C
C
B
B
A
A
s
s
C
s
s
B
s
s
A
s
s
C
s
B
s
A
s
s
s
s
,
2
,
4
,
1






s
s
s
for
for
for

27. Practice
Problem
18.6a
• So:
• Using Fourier transform pairs table:
  )
(
3
5
2
)
(
3
)
(
5
)
(
2
)
(
2
4
2
4
t
u
e
e
e
t
u
e
t
u
e
t
u
e
t
h
t
t
t
t
t
t













28. • Given:
• To avoid complex calculation using the jω, let:
• For the first half of the transform we get:
• For the second half of the transform we get:
• So for both halves, the inverse Fourier transform is:
Practice
Problem
18.6b
16
)
1
(
)
1
(
2
1
)
(
)
( 2












j
j
j
Y

j
s 
)
(
1
)
(
1
t
u
j
F 











)
(
4
cos
2
4
)
1
(
)
1
(
2
16
)
1
(
)
1
(
2
2
2
1
2
1
t
tu
e
j
j
F
j
j
F t



































)
(
)
4
cos
2
1
(
)
(
4
cos
2
)
(
)
(
t
u
t
e
t
tu
e
t
u
t
y
t
t







29. Problem
18.26a
• Given:
• Using the time shift property, we get:
• Inverse transform of F(ω):
• Property statement make the inverse transform becomes:











 



 

j
e
j
e
G j
j
1
1
1
)
( 2
2
  )
2
(
)
(
1
1 1
2
1






















t
f
F
e
F
j
e
F a
j
j




t in the inverse
transform of F(ω)
must be t-2
  )
(
1
1
)
( 1
1
t
u
e
j
F
F
F t














  )
2
(
)
(
)
( )
2
(
1


 


t
u
e
t
g
G
F t


30. 3.4 APPLICATIONS OF FOURIER
TRANSFORM
– Transmission through space is only efficient for high frequency above 20kHz
– Information signals (speech and music) contained low frequency 50Hz to 20kHz
– Is a process whereby the amplitude of the carrier is controlled by the modulating
signal.
30
Amplitude Modulation (AM)

31. 31
http://www.technologyuk.net/telecommunications/telecom-principles/amplitude-modulation.shtml
https://en.m.wikipedia.org/wiki/File:Illustration_of_Amplitude_Modulation.png
A simplified AM radio transmitter-receiver system is shown
below.
(Carrier signal)

32. Amplitude
Modulation
– Information signals:
– Carrier signal:
– Value of ωc:
– AM signal:
t
V
t
m m
m 
cos
)
( 
t
V
t
c c
c 
cos
)
( 
m
c 
 
t
t
m
V
t
f c
c 
cos
)]
(
1
[
)
( 

– The Fourier transform of the AM signal:
– Where M(ω) is the Fourier transform of m(t)
– The lower sideband (LSB) is the amplitude in frequency ωc – ωm
– The upper sideband (USB) is the amplitude in frequency ωc + ωm
 
)
(
)
(
2
)]
(
)
(
[
]
cos
)
(
[
]
cos
[
)
(
c
c
c
c
c
c
c
c
c
c
M
M
V
V
t
t
m
V
F
t
V
F
F

























33. Information signal
Carrier signal
AM signal
• In real world, m(t) is non-sinusoidal &
band limited signal means it have
frequency spectrum range limited
• To avoid interference, carriers for AM
radio stations are spaced 10kHz apart
Amplitude
Modulation

34. Practice
Problem
18.11
If a 2 MHz carrier is modulated by a 4 KHz
information signal, determine the frequencies of
the three components of the AM signal that results
• Given:
– The carrier:
– The modulating signal:
• So the 3 components are:
– The lower side band:
– The carrier:
– The upper side band:
MHz
fc 2

kHz
fm 4

MHz
004
.
2
2004000
4000
2000000 


MHz
996
.
1
1996000
4000
2000000 


MHz
fc 2


35. Problem
18.58
Given, AM signal:
Determine the carrier frequency, LSB & USB.
• We get:
• The upper side band:
• The lower side band:
t
t
t
f 4
10
cos
)
200
cos
4
1
(
10
)
( 

 

kHz
f
V
c
c
c
c
5
2
10
10
4










Hz
f
V
m
m
m
m
100
2
200
4









kHz
1
.
5
100
5000 

kHz
9
.
4
100
5000 


36. 36
Figure below shows a demodulation system using Amplitude Modulation of a
receiver. The received modulated signal spectrum is:
Determine and sketch the spectrum of the:
(i) Demodulated signal, y(t)
(ii) Message signal, m(t)
Example 5

37. 𝐢) 𝐃𝐞𝐦𝐨𝐝𝐮𝐥𝐚𝐭𝐞𝐝 𝐬𝐢𝐠𝐧𝐚𝐥
𝑥 𝜔 = 2𝜋 𝛿 𝜔 + 160 + 𝛿 𝜔 − 160 −
𝜋
2
[𝛿 𝜔 + 190 + 𝛿 𝜔 − 190 ]
−
𝜋
2
𝛿 𝜔 + 130 + 𝛿 𝜔 − 130
∴ 𝑥 𝑡 = 2 cos 160𝑡 −
1
2
𝑐𝑜𝑠190𝑡 −
1
2
cos 130𝑡
37
𝑦 𝑡 = 𝑥 𝑡 × 𝑐(𝑡)
= (2 cos 160𝑡)(cos 160𝑡) − (
1
2
cos 190𝑡)(cos 160𝑡) − (
1
2
cos 130𝑡)(cos 160𝑡)
= [cos 320𝑡 + cos 0 𝑡] −
1
4
cos 350𝑡 −
1
4
cos 30𝑡 −
1
4
cos 290𝑡 −
1
4
cos 30𝑡
= 1 −
1
2
cos 30𝑡 −
1
4
cos 290𝑡 + cos 320𝑡 −
1
4
cos 350𝑡

38. 38
ii) Message signal, m(t) is the demodulated signal that has passed through a
low pass filter with gain = 2 and allowable frequency ||≤100;
∴ 𝑚 𝑡 = 2 − cos 30𝑡

39. Sampling – For digital system nowadays
– Can be done by using train of pulses or impulses
• The sampling interval is Ts
• The sampling rate (frequency) is fs
= 1/Ts
• The sampled signal g(t) is:
• The Fourier transform is:
Continuous (analog)
signal to be sampled
Train of impulses
Sampled (digital) signal












n
s
s
n
s
s nT
t
nT
g
nT
t
t
g
t
g )
(
)
(
)
(
)
(
)
( 













n
s
s
n
T
jn
s
s n
G
T
e
nT
g
G s
)
(
1
)
(
)
( 

 

40. Sampling
• In order to ensure optimum recovery of the original signal, the sampling
frequency must be at least twice the highest frequency in the modulating signal
• Where W is the highest frequency (band-limited) of the modulating signal
W
f
T
s
s
2
1


Nyquist rate:
W
fs 2

Nyquist interval:
W
Ts
2
1


41. Practice
Problem
18.12
• Given:
– Audio signal is band-limited:
• Maximum sampling interval:
kHz
5
.
12
s
T
T
kHz
kHz
f
W
f
s
s
s
s

40
25000
1
25000
1
25
)
5
.
12
(
2
2








42. Problem
18.65
• Given:
– Highest frequency of a voice signal:
• Nyquist rate of the sampler:
kHz
4
.
3
kHz
fs
8
.
6
6800
)
3400
(
2




43. Problem
18.67
• Given:
– Signal:
• We get:
– The Nyquist rate:
– The Nyquist interval:


200
200
sin
)
(
t
t
g 
Hz
W
f 100
2
200
200








Hz
W
fs 200
)
100
(
2
2 


ms
f
T
s
s 5
200
1
1




44. Thank you
44
Conclusion of Fourier Transform
Can you?
• Determine the Fourier Transforms using its
table and properties.
• Determine the Fourier Transforms using
derivative technique.
• Transform the signal from frequency domain
into time domain using inverse Fourier
Transform.
• Discuss Fourier Transform application in
amplitude modulation.