The Fourier Transform
Derivation
Assume that we have a generalized, time-limited pulse centered at t = 0 as shown below.
f(t)
1.2
A
1
0.8
0.6
0.4
0.2
0
-30
-2T
-25
-20
-15
-T
-10
-5
00
5
10
15
T
20
25
30
2T
The Fourier Transform of this pulse can be developed by starting with a periodic version of this pulse
where the original pulse now repeats every T seconds.
fT(t)
1.2
A
1
0.8
0.6
0.4
0.2
0
-30
-2T
-25
-20
-15
-T
-10
-5
0
5
10
15
T
20
25
30
2T
Note:
lim
f T (t )  f (t )
T  
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 2
fT(t) is periodic with period T so we can express it by its exponential Fourier series as
fT (t ) 

F
n  
n
jn 0 t
*
where
T
1 2
Fn   fT (t ) *   jn 0 t dt
T T
2
and
 0  2 T
Now let’s make a small change in notation
1. n = n*0
2. F(n) = T*Fn
We now have
1 
fT (t )   F ( n ) * 
T n  
T
j n t
Fn 
and
2
f
T
T
(t ) *   j n t dt
2
The sum can be rewritten as
fT (t ) 
0
2

 F (
n  
) *
j n t
n
) *
j n t
n
or
fT (t ) 
1
2

 F (
n  

Taking the limit as T
lim
fT (t )  f (t ) 
T
 
0
1
2
 
lim
  F ( n ) * 
T
   n  
But 0 = 2/T so for large T let 0
f (t ) 
1
2
 
lim   F ( n ) * 
T    n  
or since T
J. N. Denenberg
j n t
j n t

0 

 and the limit becomes

 

 implies that 
0 and the sum, in the limit, becomes an integral
February 6, 2016
Fourier Transform
f (t ) 
1
2

 F ( ) * 
Page 3
j t
d and
F ( ) 


f
T
(t ) * 
 j t
dt

This pair of equations defines the Fourier Transform
1. F() is the Fourier Transform of f(t)
2. f(t) is the inverse Fourier Transform of F()
3. F() is also called the Spectral Density of f(t) as it describes how the energy of the original
pulse is distributed as a function of frequency (in radians per second)
I use a backwards upper case script “F” to denote taking the Fourier Transform of a function and the
same symbol with a “-1” superscript to denote taking the inverse Fourier Transform.
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 4
Example 1
Take the Fourier Transform of the single-sided exponential
f(t)=U(t)*exp(-at)
2
1.5
1
0.5
0
-2
-1
0
1
2
3
4
5
-0.5
-1

F ( )   U (t ) *  at  jt dt


F ( )     at  jt dt
0

F ( )     ( a  j )t dt
0
F ( ) 

1
*  ( a  j )t |
0
a  j
F ( ) 
1
a  j
Note that the Fourier Transform is complex. It has a magnitude and a phase. The magnitude is found by
multiplying it by its complex conjugate and taking the square root.
F ( ) 
1
1
*
a  j a  j 
F ( ) 
1
a 2
2
2
F ( ) 
2
1
a 2
2
J. N. Denenberg
This is the magnitude
February 6, 2016
Fourier Transform
Page 5
Now find the phase. First, find the real and imaginary parts.
F ( ) 
1
a  j
F ( ) 
1
a  j
*
a  j a  j
F ( ) 
a  j
a
j
 2
 2
2
2
2
a 
a 
a 2
Therefore the real part is
ReF ( ) 
a
a 2
2
and the imaginary part is
ImF ( ) 

a 2
2
The phase is then given by
 ImF ( )
 
  tan 1  

a
 ReF ( )
  tan 1 
Note: The ArcTan function of your calculator can lie! Its answers always fall
between ±90 (±π/2) and the real answer can be in one of the other two
quadrants. You should draw a picture to adjust your result as required.
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 6
Singularity Functions
We run into special functions when taking the Fourier Transform of functions that have infinite energy.
The first of these special functions is the Delta Function
(t )  lim G (t )
  
Where G(t) is any function from the set of all functions having the properties

1.
 G (t )dt  1

2.
For all t  0
lim G (t )  0
  
Sifting Property of the Delta Function
Integrating the product of the Delta Function with a “well-behaved” function results in “sampling” the
“well-behaved” function at the time that the Delta Function goes to infinity. Or
f ( t 0 ) if
b
 f (t ) * (t  t
0
a t 0 b
)dt 
a
0
eleswhere
Proof
Use Integration by parts
b
b
 U (t )dV (t )  U (t )V (t )  V (t )dU (t )
b
a
a
a
Let U(t) = f(t) and dV(t) = (t-t0)dt
b

b
f (t ) * (t  t 0 )dt  f (t )U (t  t 0 ) a   f ' (t ) *U (t )dt
b
a
a
Case 1: a < t0 < b
b
 f (t ) * (t  t
b
0
)dt  f (b)  0   f ' (t ) * U (t )dt
a
t0
b
 f (t ) * (t  t
0
)dt  f (b)  f (t ) t
0
)dt  f (b)  f (b)  f (t 0 )
0
)dt  f (t 0 )
b
0
a
b
 f (t ) * (t  t
a
b
 f (t ) * (t  t
Q.E.D
a
Case 2: t0 < a or t0 > b
b

b
f (t ) * (t  t 0 )dt  0  0   0dt  0 Q.E.D
a
J. N. Denenberg
a
February 6, 2016
Fourier Transform
Page 7
Example 2
Take the Fourier Transform of a constant
f(t)=A
2
1.5
A
1
0.5
0
-2

F ( ) 
 A
j t
-1
dt
0
1
2
-0.5

Here the integral can’t be directly computed, we have to approach it as a limiting case. Let’s replace the
-1equals the constant as its parameter approaches zero, the
constant with a parameterized function that
double-sided exponential function:
f (t )  A
a t
Now the Transform becomes:
Fa ( ) 

 A
a t

 jt dt 
0


0
 at jt
 at jt
 A  dt   A  dt
Let u = - in the first integral
0


0
Fa ( )   A at  j ( u )t dt   A at  jt dt
From our first example this is:
Fa ( ) 
A
A
2 Aa

 2
a  j a  j a   2
Now we need to take the limit as a
J. N. Denenberg
0 to get F()
February 6, 2016
3
Fourier Transform
Page 8
F ( )  lim Fa ( )
a
 0
2 Aa
F ( )  lim 2

2
a
 0 a  
0 if   0
 if   0
so this is a -function that goes to  at  = 0 if its integral is a constant.

I
 2A a
2

a
d
2
Let a*x = 

a
adx
2
a
1

x

I  2A 
2



1
dx
1  x2

I  2A 
1

I  2 A * tan x

    
I  2 A *     
 2  2 
I  2A
Therefore
F    2A *  
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 9
Exercises:
1:
Find the Fourier Transforms for each of the two pulses
2
f 1(t)
1.5
t
1
0.5
0
-2
-1.5
-0.5
t
-1
0
0.5
t
1
1.5
2t
-0.5
-1
3
2.5
t
2
1.5
1
0.5
-2
-1.5
-1
t
-0.5
0
0
t
0.5
1
1.5
2t
-0.5
-1
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 10
2:
Find the transfer function for the simple RC low-pass filter
3:
Determine the Fourier Transform of the RC low-pass filter output due to each of the pulses in
part 1
4:
Find the limit of each of the results in part 3 as t
J. N. Denenberg
0
February 6, 2016
Fourier Transform
Page 11
Properties of the Fourier Transform
Symmetry Property
If
f(t)
F()
Then F(t)
2 f(-)
Proof:
1
f (t ) 
2

 F ( ) * 
j t
d
 j t
d

Therefore

2 * f  t  
 F  

Let u =  and v = t

2 * f  v  
 F u 
 juv
du

Now let  = v and t = u
2 * f    

 F t 
 j t
dt

Therefore
F(t)
2 f(-)
And if f(t) is an even function
F(t)
2 f()
Linearity Property
If
f1(t)
F1()
And
f2(t)
F2()
Then [a*f1(t) + b*f1(t)]
[a*F1() + b*F2()]
Proof:
Results due to the linearity of integration
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 12
Scaling Property
If
f(t)
F()
Then for a real
1  
F 
a a
f(a*t)
Proof:
 f a * t  

 f a * t 
 j t
dt

case 1: a > 0 Let x = a*t
 f a * t  


f x 
j

a
x

1
dx
a


j x
1
 f a * t    f  x  a dx
a 
or
 f a * t  
1  
F 
a a
case 2: a < 0 Again let x = a*t
 f a * t  


f x 
j

a

x
1
dx
a
(Note the limits are now backwards)


j x
1
 f a * t     f  x  a dx
a 
or
 f a * t   
1  
F 
a a
Therefore including both cases
f(a*t)
1  
F 
a a
Q. E. D.
Note: The compression of a function in the time domain results in an expansion in the frequency
domain and vice versa.
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 13
Frequency Shifting
If
f t   F  
f t  * 
Then
j 0 t
 F    0 
Proof:

F   
 f t  * 
 jt
dt

F    0  

 f t  * 
 j   0 t
dt

F    0  


  f t  *  * 
j
0t
 j t
dt

or

F    0    f t  * 
j 0 t

Q. E. D.
Note: The Modulation Theorem (very important in communications)
Remember Euler’s Identities
cosx  
 jx    jx
2
and
sin x  
 jx    jx
2j
therefore
f t  cosx  
f t  *  jx  f t  *   jx
2
or
f t cosx 

F    0   F    0 
2
similarly
f t sin x  
f t  *  jx  f t  *   jx
2j
or
f t sin x  
 j
J. N. Denenberg
F    0   F    0 
2
February 6, 2016
Fourier Transform
Page 14
Time Shifting
If
f t   F  
Then
f t  t 0   F  0  *   jt0
Proof:
1
f t  
2
f t  t0  
f t  t0  

 F   * 
j t
d

1
2
1
2

 F   * 
j  t  t 0 
d


 F   * 
 j t 0
* 
j t
d

f t  t0   F   *   j t 0
Q. E. D.
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 15
Time Differentiation and Integration
If
f t   F  
Then
d
 f t    j F  
dt
t
And
 f  d 

1
F  
j
Proof:
First for differentiation (part 1)
f t  
1
2

 F   * 
j t
d


d
 f t   d  1
dt
dt  2
d
 f t   1
dt
2
d
 f t   1
dt
2
d
 f t   1
dt
2

 F   * 

d 

j t


d 
 F   * dt  d
j t


 F   * j * 
j t
d


  j F  * 
j t
d

or
d
 f t    j F  
dt
J. N. Denenberg
Q. E. D. for part 1
February 6, 2016
Fourier Transform
Page 16
Now for integration (part 2)
f t  
t

1
2

 F   * 
j t
d

f  d 

 1
  2
t

 F   * 
j 


d d

Interchanging the order of integration
t
1
 f  d  2
t
1
 f  d  2
t
 f  d 

1
2
 t j 
 F   *  d  d



 1
 F   *  j 
j t


 1

  j F   * 

 d

j t
d

or
t
 f  d 

1
F  
j
J. N. Denenberg
Q. E. D. for part 2
February 6, 2016
Fourier Transform
Page 17
Frequency Differentiation
If
f t   F  
 jt n f t  
Then
dn
F  
dt n
Proof:
F   

 f t  * 
 j t
dt



dn
dn 


F


f t  *   jt dt 
n
n 
dt
dt  




dn
d n  j t
F     f t  * n 
dt
dt n
dt


dn
n
F     f t  *  jt    jt dt
n
dt




dn
n
F      jt  * f t  *   jt dt
dt n

or
dn
 jt  f t   n F  
dt
n
J. N. Denenberg
Q. E. D.
February 6, 2016
Fourier Transform
Page 18
The Convolution Theorem
Definition:
f1 t   f 2 t  
the convolution of two functions f1 t  and


f1   * f 2 t   d 

f2 t  is defined as:

 f   * f t   d
2
1

Time Convolution
If
f1 t   F1  
And
f 2 t   F2  
Then
 f1 t   f 2 t   F1   * F2  
Proof:
F   

 f t  * 
 j t
dt

Therefore
 f1 t   f 2 t  



 j t
t  
 f1 t   f 2 t  


  
 

  f1   * f 2 t   d  dt
  

 

f1    f 2 t    *   jt dt  d
t  

Let u = t -  in the inner integral
 f1 t   f 2 t  


t  
 f1 t   f 2 t  


t  
 

f1    f 2 u  *   j u  du  d
u  

 

f1    j   f 2 u  *   ju du  d
u  

Since the inner integral is no longer a function of , it can be brought out as a constant and this leaves
 f1 t   f 2 t  

 f  
1
t  
 j
d *

 f u  * 
2
 ju
du
u  
or
 f1 t   f 2 t    f1 *  f 2 u  Q. E. D
J. N. Denenberg
February 6, 2016
Fourier Transform
Page 19
Frequency Convolution
If
f1 t   F1  
And
f 2 t   F2  
Then
f1 t  * f 2 t  
1
F1    F2  
2
Proof: Same method as for time convolution
J. N. Denenberg
February 6, 2016