Fourier transform

1. 1
Fourier Transform
Z. Aliyazicioglu
Electrical & Computer Engineering Dept.
Cal Poly Pomona
ECE 307
Fourier Transform
The Fourier transform (FT) is the extension of the Fourier series to
nonperiodic signals. The Fourier transform of a signal exist if
satisfies the following condition.
The Fourier transform
The inverse Fourier transform (IFT) of X(ω) is x(t)and given by
2
( )
x t dt
∞
−∞
< ∞
∫
( ) ( ) j t
X x t e dt
ω
ω
∞
−
−∞
= ∫
1
( ) ( )
2
j t
x t X e d
ω
ω ω
π
∞
−∞
= ∫

2. 2
Fourier Transform
Also, The Fourier transform can be defined in terms of frequency of
Hertz as
and corresponding inverse Fourier transform is
2
( ) ( ) j ft
X f x t e dt
π
∞
−
−∞
= ∫
2
( ) ( ) j ft
x t X f e df
π
∞
−∞
= ∫
Fourier Transform
Determine the Fourier transform of a rectangular pulse
shown in the following figure
Example:
-a/2 a/2
h
t
x(t)
/ 2
2 2
/ 2
( )
sin( )
2 2
sin( )
2
2
sinc
2
a a a
j j
j t
a
h
X he dt e e
j
a
h a
ha
a
a
ha
ω ω
ω
ω
ω
ω
ω
ω
ω
ω
π
−
−
−
 
= = −
 
−  
= =
 
=  
 
∫

3. 3
Fourier Transform
Example: To find in frequency domain,
( )
/ 2 2 2
2 2 2
/ 2
( )
2
sin( )
sin( )
sinc
a fa fa
j j
j ft
a
h
X f he dt e e
j f
h fa
fa ha
f fa
ha fa
π π
π
π
π
π
π π
−
−
−
 
= = −
 
−  
= =
=
∫
>> h=1;
>> a=1;
>> f=-3.5:0.01:3.5;
>> w=2*pi*f;
>> x=h*a*sinc(w*a/(2*pi));
>> plot (w,x)
>> title ('X(omega)')
>> xlabel('omega');
>>
1,
1
( ) 2sinc
2
h
a
X
ω
ω
π
=
=
 
=  
 
Fourier Transform
1,
2
2
( ) 2sinc
2
h
a
X
ω
ω
π
=
=
 
=  
 
>> h=1;
>> a=1;
>> f=-3.5:0.01:3.5;
>> w=2*pi*f;
>> x=abs(h*a*sinc(w*a/(2*pi)));
>> subplot (2,1,1)
>> plot (w,x)
>> title ('|X(omega)|')
>> xlabel('omega')
>> xp=phase(h*a*sinc(w*a/(2*pi)));
>> subplot (2,1,2)
>> plot (w,xp)
>> title ('phase X(omega)')
>> xlabel('omega')

4. 4
Fourier Transform
Determine the Fourier transform of the Delta function δ(t)
Example
0
( ) ( ) 1
j t j
X t e dt e
ω ω
ω δ
∞
− −
−∞
= = =
∫
1
X(ω)
ω
Fourier Transform
Properties of the Fourier Transform
We summarize several important properties of the Fourier Transform as follows.
1. Linearity (Superposition)
1 1
( ) ( )
x t X ω
⇔ 2 2
( ) ( )
x t X ω
⇔
1 1 2 2 1 1 2 2
( ) ( ) ( ) ( )
a x t a x t a X a X
ω ω
+ ⇔ +
Then,
If and
Proof:
[ ]
1 1 2 2 1 1 2 2
1 1 2 2
( ) ( ) ( ) ( )
( ) ( )
j t j t j t
a x t a x t e dt a x t e dt a x t e dt
a X a X
ω ω ω
ω ω
∞ ∞ ∞
− − −
−∞ −∞ −∞
+ = +
= +
∫ ∫ ∫

5. 5
Fourier Transform
Properties of the Fourier Transform
2. Time Shifting
Then,
If
Proof:
( ) ( )
x t X ω
⇔
0
0
( ) ( ) j t
x t t X e ω
ω −
− ⇔
0
t t
τ = − 0
t t
τ
= + dt dτ
=
0
0
0
( )
0
( ) ( )
( )
( )
j t
j t
j t j
j t
x t t e dt x e d
e x e d
e X
ω τ
ω
ω ωτ
ω
τ τ
τ τ
ω
∞ ∞
− +
−
−∞ −∞
∞
− −
−∞
−
− =
=
=
∫ ∫
∫
Let then and
Fourier Transform
Let
0
( ) ( )
y t x t t
= −
0 0
0
( )
( ( ) )
( ) ( ) ( )
( )
j t j t
j X
j X t
Y X e X e e
X e
ω ω
ω
ω ω
ω ω ω
ω
− −
∠
∠ −
= =
=
0
( ( ) )
( )
( ) ( ) j X t
j Y
Y e X e ω ω
ω
ω ω ∠ −
∠
=
Therefore, the amplitude spectrum of the time shifted signal is the
same as the amplitude spectrum of the original signal, and the phase
spectrum of the time-shifted signal is the sum of the phase spectrum of
the original signal and a linear phase term.

6. 6
Fourier Transform
Example: Determine the Fourier transform of the following time
shifted rectangular pulse.
0 a
h
t
x(t)
2
( ) sinc
2
a
j
a
X ha e
ω
ω
ω
π
−
 
=  
 
>> h=1;
>> a=1;
>> f=-3.5:0.01:3.5;
>> w=2*pi*f;
>> x=abs(h*a*sinc(w*a/(2*pi)).*exp(-
j*w*1/2));
>> subplot (2,1,1)
>> plot (w,x)
>> title ('|X(omega)|')
>> xlabel('omega')
>> xp=phase(h*a*sinc(w*a/(2*pi)).*exp(-
j*w.*1/2));
>> subplot (2,1,2)
>> plot (w,xp)
>> xlabel('omega')
>> title ('phaseX(omega)')
Fourier Transform
3. Time Scaling
then
Proof:
If , a>0 then
( ) ( )
x t X ω
⇔
If
1
( ) ( )
x at X
a a
ω
⇔
at
τ = /
t a
τ
= (1/ )
dt a dτ
=
Let then and
1
( ) ( )
1
( )
j
j t a
x at e dt x e d
a
X
a a
ω
τ
ω
τ τ
ω
∞ ∞
−
−
−∞ −∞
=
=
∫ ∫
If , a<0 then
1
( ) ( )
1 1
( ) ( )
j
j t a
j
a
x at e dt x e d
a
x e d X
a a a
ω
τ
ω
ω
τ
τ τ
ω
τ τ
∞ ∞
−
−
−∞ −∞
∞
−
−∞
=
= =
∫ ∫
∫

7. 7
Fourier Transform
Example. if , then find the Fourier transform of the
following signals
1
( 2 ) ( )
2 2
x t X
ω
−
− ⇔
( /5) 5 (5 )
x t X ω
⇔
2
1
( 5( 2)) ( )
5 5
j
x t X e ω
ω −
−
− − ⇔
( ) ( )
x t X ω
⇔
a.
b.
c.
Example: Find the Fourier transform of the following signal.
1 1
( ) ( ) ( ) sinc
2
x t t X
ω
ω
π
 
= ∏ ⇔ =  
 
2 2 1
1 1
( ) (5 ) ( ) ( ) sinc
5 5 5 10
x t t X X
ω ω
ω
π
 
= ∏ ⇔ = =  
 
3 3 1
( ) ( /5) ( ) 5 (5 ) 5sinc
0.4
x t t X X
ω
ω ω
π
 
= ∏ ⇔ = =  
 
a.
b.
c.
Fourier Transform
4. Duality (Symmetry)
If then
( ) ( )
x t X ω
⇔
( ) 2 ( )
X t x
π ω
⇔ − ( ) ( )
X t x f
⇔ −
or
Proof: Since t and ω are arbitrary variables in the inverse Fourier
transform
1
( ) ( )
2
j t
x t X e d
ω
ω ω
π
∞
−∞
= ∫
we can replace ω with t and t with - ω to get
1
( ) ( )
2
j t
x X t e dt
ω
ω
π
∞
−
−∞
− = ∫
{ }
( ) ( ) 2 ( )
j t
X t X t e dt x
F ω
π ω
∞
−
−∞
= = −
∫
Therefore,

8. 8
Fourier Transform
Similarly, if we can replace f with t and t with -f in the inverse
Fourier transform
2
( ) ( ) j ft
x t X f e df
π
∞
−∞
= ∫
2
( ) ( ) j ft
x f X t e df
π
∞
−
−∞
− = ∫
{ }
( ) ( )
X t x f
F = −
to get
Therefore,
then
Let
Fourier Transform
Example: Applying symmetry property,
( ) ( ) ( ) 1
x t t X
δ ω
= ⇔ =
( ) 1 ( ) 2 ( ) 2 ( )
x t X ω πδ ω πδ ω
= ⇔ = − = ( is even function)
( )
δ ω
( ) 1 ( ) ( ) ( )
x t X f f f
δ δ
= ⇔ = − =
or
Example:
( ) ( ) sinc
2
t a
x t rect X a
a
ω
ω
π
   
= ⇔ =
   
   
( ) sinc ( ) 2 2
2
ta
x t a X rect rect
a a
ω ω
ω π π
π
−
     
= ⇔ = =
     
     
2
a
c
π
= 2
a c
π
=
( )
1
( ) sinc ( ) 2
2 2
x t a ct X rect rect
c c c
ω ω
ω π
π π
   
= ⇔ = =
   
   

9. 9
Fourier Transform
Time Reversal
If then
( ) ( )
x t X ω
⇔
( ) ( )
x t X ω
− ⇔ −
Proof: Let . Then and
t τ
− = t τ
= − dt dτ
= −
( )
( ) ( ) ( )
j t j
x t e dt x e d X
ω ω τ
τ τ ω
∞ ∞
− − −
−∞ −∞
− = − = −
∫ ∫
Fourier Transform
Frequency Shifting
If then
( ) ( )
x t X ω
⇔
( ) ( )
c
j t
c
x t e X
ω
ω ω
−
⇔ −
( )
( ) ( ) ( )
c c
j t j t
j t
c
x t e e dt x t e dt X
ω ω ω
ω
ω ω
∞ ∞
− −
−
−∞ −∞
= = −
∫ ∫
Proof:

10. 10
Fourier Transform
Determine the Fourier transform of and
Example: cos ct
ω sin ct
ω
[ ]
1 1
( ) cos ( ) ( ) ( )
2 2
c c
j t j t
c c c
x t t e e X
ω ω
ω ω π δ ω ω δ ω ω
−
= = + ⇔ = − + +
[ ]
1 1 1
( ) cos ( ) ( ) ( )
2 2 2
c c
j t j t
c c c
x t t e e X f f f f f
ω ω
ω δ δ
−
= = + ⇔ = − + +
or
f
1/2
fc
-fc
The phase spectrum is zero everywhere.
X(f)
Fourier Transform
[ ]
1 1
( ) sin ( ) ( ) ( )
2 2
c c
j t j t
c c c
x t t e e X j
j j
ω ω
ω ω π δ ω ω δ ω ω
−
= = − ⇔ = − − − +
[ ]
1 1
( ) sin ( ) ( ) ( )
2 2 2
c c
j t j t
c c c
j
x t t e e X f f f f f
j j
ω ω
ω δ δ
− −
= = − ⇔ = − − +
f
π/2
-fc
fc
-π/2
f
1/2
fc
-fc
|X(f)|
θ(f)

11. 11
Fourier Transform
7. Modulation
If then
Proof:
( ) ( )
x t X ω
⇔
[ ]
1
( )cos( ) ( ) ( )
2
c c c
x t t X X
ω ω ω ω ω
⇔ − + +
[ ]
( ) ( )
1
( )cos( ) ( )
2
1
( ) ( )
2
1
( ) ( )
2
c c
c c
j t j t
j t j t
c
j t j t
c c
x t t e dt x t e e e dt
x t e dt x t e dt
X X
ω ω
ω ω
ω ω ω ω
ω
ω ω ω ω
∞ ∞
− −
−∞ −∞
∞ ∞
− − − +
−∞ −∞
 
= +
 
 
= +
 
 
= − + +
∫ ∫
∫ ∫
Fourier Transform
8. Time Differentiation:
If then
Proof:
( ) ( )
x t X ω
⇔
( )
( )
dx t
j X
dt
ω ω
⇔
( )
( ) ( )
n
n
n
d x t
j X
dt
ω ω
⇔
General case
Taking the derivative of the inverse Fourier transform
1
( ) ( )
2
j t
x t X e d
ω
ω ω
π
∞
−∞
= ∫
( ) 1
( )
2
j t
dx t
j X e d
dt
ω
ω ω ω
π
∞
−∞
= ∫
( )
( )
dx t
j X
dt
ω ω
⇔
we obtain
Therefore

12. 12
Fourier Transform
9. Time Differentiation:
If then
Proof:
( ) ( )
x t X ω
⇔ General case
Taking derivative of Fourier Transform
( )
( )
dX j
tx t j
d
ω
ω
⇔
( )
( )
n
n n
n
d X
t x t j
d
ω
ω
⇔
( ) ( ) j t
X x t e dt
ω
ω
∞
−
−∞
= ∫
( )
( ) ( ) j t
dX
jt x t e dt
d
ω
ω
ω
∞
−
−∞
= −
∫
with respect to ω, we obtain
( )
( )
dX j
tx t j
d
ω
ω
⇔
Therefore
Fourier Transform
10 Conjugate
If then
Proof:
( ) ( )
x t X ω
⇔
If x(t) is real so that
* *
( ) ( )
x t X ω
⇔ −
*
* ( ) *
( ) ( ) ( )
j t j t
x t e dt x t e dt X
ω ω
ω
∞ ∞
− − −
−∞ −∞
 
= = −
 
 
∫ ∫
*
( ) ( )
x t x t
= *
( ) ( )
X X
ω ω
= −

13. 13
Fourier Transform
11. Convolution
Proof:
Interchanging the order of integration, we obtain
If , , and
( ) ( )
x t X ω
⇔ ( ) ( )
h t H ω
⇔ ( ) ( )
y t Y ω
⇔
( ) ( ) * ( ) ( ) ( )
y t h t x t h x t d
τ τ τ
∞
−∞
= = −
∫
( ) ( ) ( )
Y H X
ω ω ω
=
( ) ( ) ( ) j t
Y h x t d e dt
ω
ω τ τ τ
∞ ∞
−
−∞ −∞
 
= −
 
 
∫ ∫
( ) ( ) ( ) j t
Y h x t e dt d
ω
ω τ τ τ
∞ ∞
−
−∞ −∞
 
= −
 
 
∫ ∫
( ) ( ) ( ) ( ) ( )
( ) ( )
j j
Y h X e d X h e d
X H
ωτ ωτ
ω τ ω τ ω τ τ
ω ω
∞ ∞
− −
−∞ −∞
= =
=
∫ ∫
Fourier Transform
12. Multiplication
or
If , and
1 2 1 2 1 2
1 1
( ) ( ) ( ) * ( ) ( ) ( )
2 2
x t x t X X X v X v dv
ω ω ω
π π
∞
−∞
⇔ = −
∫
1 1
( ) ( )
x t X ω
⇔ 2 2
( ) ( )
x t X ω
⇔
1 2 1 2 1 2
( ) ( ) ( ) * ( ) ( ) ( )
x t x t X f X f X v X f v dv
∞
−∞
⇔ = −
∫

14. 14
Fourier Transform
13. Parseval’s Theorem
Proof
If , then total normalized(based on one ohms
resistor) energy E of and x(t) is given by
1 1
( ) ( )
x t X ω
⇔
2 2 2
1
( ) ( ) ( )
2
E x t dt X d X f df
ω ω
π
∞ ∞ ∞
−∞ −∞ −∞
= = =
∫ ∫ ∫
2 * *
1
( ) ( ) ( ) ( ) ( )
2
j t
x t dt x t x t dt x t X e d dt
ω
ω ω
π
∞ ∞ ∞ ∞
−
−∞ −∞ −∞ −∞
 
= =  
 
∫ ∫ ∫ ∫
Interchanging the order of integration, we obtain
Fourier Transform
2 *
*
2
1
( ) ( ) ( )
2
1
( ) ( )
2
1
( )
2
j t
x t dt X x t e dt d
X X d
X d
ω
ω ω
π
ω ω ω
π
ω ω
π
∞ ∞ ∞
−
−∞ −∞ −∞
∞
−∞
∞
−∞
 
=  
 
=
=
∫ ∫ ∫
∫
∫
Proof (cont)