Fourier Transforms of
Special Functions
主講者:虞台文
Content
Introduction
More on Impulse Function
Fourier Transform Related to Impulse Function
Fourier Transform of Some Special Functions
Fourier Transform vs. Fourier Series
Introduction
Sufficient condition for the existence of a
Fourier transform
| f (t ) |dt
That is, f(t) is absolutely integrable.
However, the above condition is not the
necessary one.
Some Unabsolutely Integrable Functions
Functions: cos t, sin t,…
Unit Step Function: u(t).
Sinusoidal
Generalized
–
–
Functions:
Impulse Function (t); and
Impulse Train.
Fourier Transforms of
Special Functions
More on
Impulse Function
Dirac Delta Function
0 t 0
(t )
t 0
and
(t )dt 1
Also called unit impulse function.
0
t
Generalized Function
The value of delta function can also be defined
in the sense of generalized function:
(t )(t )dt (0)
(t): Test Function
We shall never talk about the value of (t).
Instead, we talk about the values of integrals
involving (t).
Properties of Unit Impulse Function
(t t0 )(t )dt (t0 )
Pf)
Write t as t + t0
(t t0 )(t )dt (t )(t t0 )dt
(t0 )
Properties of Unit Impulse Function
1
(at )(t )dt | a | (0)
Pf) Write t as t/a
Consider a>0
(at )(t )dt
1
t
(t ) dt
a
a
1
(0)
|a|
Consider a<0
(at )(t )dt
1
t
(t ) dt
a
a
1
(0)
|a|
Properties of Unit Impulse Function
f (t )(t ) f (0)(t )
Pf)
[ f (t )(t )](t )dt (t )[ f (t )(t )]dt
f (0)(0)
f (0) (t )(t )dt
[ f (0)(t )](t )dt
Properties of Unit Impulse Function
f (t )(t ) f (0)(t )
Pf)
1
(at )
(t )
|a|
1
1
(at )(t )dt | a | (0) | a | (t )(t )dt
1
(t )(t )dt
| a |
Properties of Unit Impulse Function
f (t )(t ) f (0)(t )
1
(at )
(t )
|a|
t(t ) 0
(t ) (t )
Generalized Derivatives
The derivative f’(t) of an arbitrary
generalized function f(t) is defined by:
f ' (t )(t )dt f (t )' (t )dt
Show that this definition is consistent to the ordinary
definition for the first derivative of a continuous function.
f ' (t )(t )dt f (t )(t ) f (t )' (t )dt
=0
Derivatives of the -Function
' (t )(t )dt (t )' (t )dt ' (0)
d(t )
' (t )
,
dt
d(t )
' (0)
dt t 0
(t )(t )dt (1) (0)
( n)
n
d
(t )
(n)
(t )
,
n
dt
n
( n)
n
d
(t )
(n)
(0)
dt n t 0
Product Rule
[ f (t )(t )]' f ' (t )(t ) f (t )' (t )
Pf)
[ f (t )(t )]' (t )dt [ f (t )(t )]' (t )dt (t )[ f (t )' (t )]dt
(t ){[ f (t )(t )]' f ' (t )(t )}dt
(t )[ f (t )(t )]' dt (t )[ f (t )' (t )]dt
' (t )[ f (t )(t )]dt (t )[ f (t )' (t )]dt
[' (t ) f (t ) (t ) f ' (t )](t )dt
Product Rule
f (t )' (t ) f (0)' (t ) f ' (0)(t )
Pf)
f (t )' (t ) [ f (t )(t )]' f (t )' (t )
[ f (0)(t )]'
f (0)' (t )
f ' (0)(t )
Unit Step Function u(t)
Define
u(t )(t )dt (t )dt
0
u(t)
0
t
1 t 0
u (t )
0 t 0
Derivative of the Unit Step Function
Show
that u ' (t ) (t )
u' (t )(t )dt u(t )' (t )dt
' (t )dt
0
[() (0)] (0)
(t )(t )dt
Derivative of the Unit Step Function
(t)
u(t)
Derivative
0
t
0
t
Fourier Transforms of
Special Functions
Fourier Transform
Related to
Impulse Function
Fourier Transform for (t)
(t )
1
F
F [(t )] (t )e
jt
jt
e
dt
t 0
1
F(j)
(t)
F
0
t
1
0
Fourier Transform for (t)
Show that
1 jt
(t )
e d
2
1 jt
1 jt
e d
1e d
(t ) F [1]
2
2
1
1 j t
e d converges to
The integration
2
in the sense of generalized function.
(t )
Fourier Transform for (t)
1
Show that (t ) cos td
0
1
1 jt
(cos t j sin t )d
(t )
e d
2
2
1
j
cos td
sin td
2
2
1
Converges to (t) in the sense of
cos td generalized function.
0
Two Identities for (t)
1 jxy
( y )
e dx
2
1
( y ) cos xydx
0
These two ordinary integrations themselves are meaningless.
They converge to (t) in the sense of generalized function.
Shifted Impulse Function
(t t0 ) e
F
jt 0
Use the fact F [ f (t t0 )] F ( j)e
jt 0
(t t0)
|F(j)|
F
0
t0
t
1
0
Fourier Transforms of
Special Functions
Fourier Transform of a
Some Special Functions
Fourier Transform of a Constant
f (t ) A
F ( j) A2()
F
F ( j) F [ A] Ae jt d
1 j ( ) t
2A e
dt
2
2A()
Fourier Transform of a Constant
f (t ) A
F ( j) A2()
F
F(j)
0
A2()
F
A
t
0
Fourier Transform of Exponential Wave
f (t ) e
F [ f (t )e
j0t
j0t
F ( j) 2( 0 )
F
] F [ j ( 0 )]
F [1] 2()
F [e
j0t
] 2( 0 )
Fourier Transforms of Sinusoidal Functions
cos 0t ( 0 ) ( 0 )
F
sin 0t j( 0 ) j( 0 )
F
F(j)
(+0) (0)
f(t)=cos0t
t
F
0
0
0
Fourier Transform of Unit Step Function
Let F [u (t )] F ( j)
F [u (t )] F ( j)
u (t ) u (t ) 1 (except for t 0)
F [u (t ) u (t )] F [1]
F [u (t )] F [u (t )] 2()
F ( j) F ( j) 2()
F(j)=?
Can you guess it?
Fourier Transform of Unit Step Function
Guess F ( j) k() B()
k
F ( j) F ( j) k() k() B() B()
2k() B() B()
0
B() must be odd
F ( j) F ( j) 2()
Fourier Transform of Unit Step Function
k
Guess F ( j) k() B()
u ' (t ) (t )
F [u (t )] F ( j)
1
B()
j
F [u ' (t )] F [(t )] 1
F [u ' (t )] jF ( j)
j[() B()]
j() jB()
0
Fourier Transform of Unit Step Function
Guess F ( j) k() B()
1
u (t )
()
j
F
k
1
B()
j
Fourier Transform of Unit Step Function
|F(j)|
f(t)
F
1
t
0
1
u (t )
()
j
F
()
0
Fourier Transforms of
Special Functions
Fourier Transform vs.
Fourier Series
Find the FT of a Periodic Function
Sufficient condition --- existence of FT
| f (t ) |dt
Any periodic function does not satisfy this
condition.
How to find its FT (in the sense of general
function)?
Find the FT of a Periodic Function
We can express a periodic function f(t) as:
f (t )
c e
n
n
jn0t
,
2
0
T
F ( j) F [ f (t )] F cn e jn0t cnF [e jn0t ]
n
n
c 2( n )
n
n
0
2 c n ( n 0 )
n
Find the FT of a Periodic Function
We can express a periodic function f(t) as:
f (t )
c e
n
jn0t
n
,
2
0
T
F ( j) 2 cn ( n0 )
n
The FT of a periodic function consists of a sequence of
equidistant impulses located at the harmonic frequencies
of the function.
Example:
Impulse Train
3T 2T T
T (t )
0
T
2T
3T
t
(t nT )
n
Find the FT of the
impulse train.
Example:
Impulse Train
3T 2T T
0
T
2T
3T
t
1
jn0t
Find
the
FT
of
the
T (t ) (t nT ) T (t ) e
impulse T
train.
n
n
cn
2
F [T (t )]
( n0 )
Example:
T n
0
Impulse Train
3T 2T T
0
T
2T
3T
t
1
jn0t
Find
the
FT
of
the
T (t ) (t nT ) T (t ) e
impulse T
train.
n
n
cn
2
F [T (t )]
( n0 )
Example:
T n
0
Impulse Train
3T 2T T
0
T
2T
3T
0
20 30
t
F
2/T
30 20 0
0
Find Fourier Series Using
Fourier Transform
f(t)
t
T/2
f (t )
c e
n
jn0t
n
1
cn
T
Fo ( j) f o (t )e jt
T / 2
T /2
T / 2
f (t )e
jn0t
1
cn Fo ( jn0 )
T
T /2
T/2
f (t )e jt
T/2
fo(t)
t
T/2
Sampling the Fourier Transform of fo(t) with period
2/T,
we can find
the Fourier
Series of f (t).
Find
Fourier
Series
Using
Fourier Transform
f(t)
t
T/2
f (t )
c e
n
jn0t
n
1
cn
T
Fo ( j) f o (t )e jt
T / 2
T /2
T / 2
f (t )e
jn0t
1
cn Fo ( jn0 )
T
T /2
T/2
f (t )e jt
T/2
fo(t)
t
T/2
Example:
The Fourier Series of a Rectangular Wave
f(t)
1
1
d
f (t )
0
jn0t
c
e
n
n
fo(t)
t
t
0
Fo ( j)
d /2
e jt dt
d / 2
2 d
1
sin
cn Fo ( jn0 )
2
T
2
1
n0 d
n0 d
sin
sin
Tn0 2 n 2
F ( j) 2 cn ( n0 )
Example:
n
The Fourier Transform of a Rectangular Wave
f(t)
1
d
f (t )
t
0
jn0t
c
e
n
n
F [f(t)]=?
2 n0 d
F ( j) sin
( n0 )
2
n n
1
cn Fo ( jn0 )
T
2
1
n0 d
n0 d
sin
sin
Tn0 2 n 2
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