FOURIER TRANSFORMS
JELMAAN FOURIER:
• Definition of the Fourier transforms
• Relationship between Laplace Transforms
and Fourier Transforms
• Fourier transforms in the limit
• Properties of the Fourier Transforms
• Circuit applications using Fourier
Transforms
• Parseval’s theorem
• Energy calculation in magnitude spectrum
CIRCUIT APPLICATION USING
FOURIER TRANSFORMS
• Circuit element in frequency domain:
RR
L  jL
1
C
jC
Example 1:
• Obtain vo(t) if vi(t)=2e-3tu(t)
Solution:
• Fourier Transforms for vi
2
Vi ( ) 
3  j
Transfer function:
Vo ( )
1 / j
H ( ) 

Vi ( ) 2  1 / j
1

1  j 2
Thus,
Vo ( )  Vi ( ) H ( )
2

(3  j )(1  j 2 )
1

(3  j )(0.5  j )
From partial fraction:
 0.4
0.4
Vo ( ) 

3  j 0.5  j
• Inverse Fourier Transforms:
vo (t )  0.4(e
0.5t
3t
 e )u (t )
Example 2:
• Determine vo(t) if vi(t)=2sgn(t)=-2+4u(t)
Solution:
4
vi  2 sgn( t )  Vi ( ) 
j
4
H ( ) 
4  j
Vo ( )  H ( )Vi ( )
16

j  ( 4  j )
A
B


j 4  j 
4
4
Vo ( ) 

j 4  j
4 t
vo (t )  2 sgn( t )  4e u (t )
JELMAAN FOURIER:
• Definition of the Fourier transforms
• Relationship between Laplace Transforms
and Fourier Transforms
• Fourier transforms in the limit
• Properties of the Fourier Transforms
• Circuit applications using Fourier
Transforms
• Parseval’s theorem
• Energy calculation in magnitude spectrum
PARSEVAL’S THEOREM
Energy absorbed by a function f(t)

W1   f (t )dt

2
Parseval’s theorem stated that energy
also can be calculate using,




2
1


f (t )dt 
F

d


2 
2
• Parseval’s theorem also can be
written as:

2

1
2


f
(
t
)
dt

F

d

 

 0
PARSEVAL’S THEOREM
DEMONSTRATION
• If a function,
f (t )  e
a t
• Integral left-hand side:



e
2a t
0

dt   e dt   e
2 at

0
2 at 0
e

2a
 2 at
 2 at 
e

2
a

1
1 1



2a 2a a
0
dt
• Integral right-hand side:
1



0
2
4a
d
2
2 2
(a   )

1  
1
1  

 tan

2 
2
2
 2a    a a
a 0
4a
2
2

 1
 0 
 0  0 

2a
 a
JELMAAN FOURIER:
• Definition of the Fourier transforms
• Relationship between Laplace Transforms
and Fourier Transforms
• Fourier transforms in the limit
• Properties of the Fourier Transforms
• Circuit applications using Fourier
Transforms
• Parseval’s theorem
• Energy calculation in magnitude spectrum
ENERGY CALCULATION IN
MAGNITUDE SPECTRUM
• Magnitude of the Fourier Transforms
squared is an energy density (J/Hz)
1



0

F 2f  2 df  2  F 2f  df
2
0
2
• Energy in the frequency band
from ω1 and ω2:
W1 
1
2

 
F   d
2
1
1

2
1


2
1
F   d 
2
2
2
 F  
1
2
d
Example 1:
• The current in a 40Ω resistor is:
2t
i  20 e u(t ) A
• What is the percentage of the total
energy dissipated in the resistor can
be associated with the frequency band
0 ≤ ω ≤ 2√3 rad/s?
Solution:
• Total energy dissipated in the resistor:

W40  40  400e dt
 4t
0

 4t
e
 16000
4
 4000 J
0
Check the answer with
parseval’s theorem:
• Fourier Transform of the
current:
20
F ( ) 
2  j
• Magnitude of the current:
F ( ) 
20
4
2
40

400
W40 
d

2

0

4


16000  1

1


tan
  2
2 0 
8000   

   4000 J
 2
• Energy associated with the frequency
band:
W40 
40


2 3
0
400
d

2
4
2 3

16000  1
1 


tan
  2
20 

8000    8000

J
 
 3
3
• Percentage of the total energy
associated:
8000 / 3

100  66.67%
4000
Example 2:
• Calculate the percentage of output
energy to input energy for the filter
below:
5t
vi  15e u (t )V
10k

vi


10F
vo

• Energy at the input filter:

Wi   (15e
5t 2
0

10t
) dt
e
 225
 22.5 J
 10 0
• Fourier Transforms for the output
voltage:
Vo ( )  Vi ( ) H ( )
15
Vi ( ) 
5  j
1 / RC
10
H ( ) 

1 / RC  j 10  j
• Thus,
150
Vo ( ) 
(5  j )(10  j )
22500
2
Vo ( ) 
2
2
( 25   )(100   )
• Energy at the output filter:
Wo 
1



0
22500
d

2
2
(25   )(100   )

1   300
300 
 

2
2
0
0
  25  
100   
300  1    1   

      15 J

  5  2  10  2 
• Thus the percentage:
15

(100 )  66.67%
22.5