M. J. Roberts - 2/18/07
Web Appendix H - Derivations of the
Properties of the Continuous-Time
Fourier Transform
H.1 Numerical Computation of the CTFT
In cases in which the signal to be transformed is not readily describable by a
mathematical function or the Fourier-transform integral cannot be done analytically, we
can sometimes find an approximation to the CTFT numerically using the DFT which was
()
first introduced in Chapter 8. The CTFT of a signal x t is
( ) x (t ) e
X f =
j 2 ft
.
dt
If we apply this to a signal that is causal we get
( )
()
X f = x t e j 2 ft dt .
0
We can write this integral in the form
( )
( n+1)Ts
n=0
nTs
X f =
()
x t e j 2 ft dt .
()
(
)
If Ts is small enough, the variation of x t in the time interval nTs t < n + 1 Ts is small
and the CTFT can be approximated by
( )
( ) X f x nTs
n=0
( n+1)Ts
e j 2 ft dt .
nTs
or
( )
( )
X f x nTs
n=0
e
j 2 fnTs
(
)
j 2 f n+1 Ts
e
j2 f
or
j 2 fTs
1 e
X f j2 f
( )
x ( nT ) e
s
j 2 fnTs
= Ts e
n=0
j fTs
( ) x ( nT ) e
sinc Ts f
s
n=0
H-1
j 2 fnTs
M. J. Roberts - 2/18/07
(Figure H-1).
x(t)
Ts
t
n=N F
n=0
Figure H-1 A signal and multiple intervals on which the CTFT integral can be evaluated
()
If x t is an energy signal then beyond some finite time its size must become negligible
and we can replace the infinite range of n in the summation with a finite range 0 n < N F
yielding
( )
X f Ts e
j fTs
N F 1
( ) x ( nT ) e
sinc Ts f
j 2 fnTs
s
.
n=0
Now if we compute the CTFT only at integer multiples of f s / N F = f F , which is the
frequency-domain resolution of this approximation to the CTFT,
( )
X kf F Ts e
j kf F Ts
N F 1
(
sinc Ts kf F
) x ( nT ) e
j 2 knf F Ts
s
n=0
or
( )
X kf F Ts e
j k / N F
N F 1
(
sinc k / N F
) x ( nT ) e
j 2 kn/ N F
s
.
n=0
( )
The summation in this equation is the DFT of x nTs . Therefore
( )
X kf F Ts e
j k / N F
(
)
( ( ))
sinc k / N F DFT x nTs
()
where the notation DFT means “discrete Fourier transform of”. For k << N F ,
( ( )) .
( )
X kf F Ts DFT x nTS
(H.1)
So if the signal to be transformed is a causal energy signal and we sample it over a time
containing practically all of its energy and if the samples are close enough together that
the signal does not change appreciably between samples, the approximation in (H.1)
becomes accurate for k << N F .
H-2
M. J. Roberts - 2/18/07
H.2 Linearity
()
()
()
Let and be constants and let z t = x t + y t . Then
( ) Z f =
()
()
()
()
( )
( )
x t + y t e j 2 ft dt = x t e j 2 ft dt + y t e j 2 ft dt = X f + y f
and the linearity property is
()
()
( )
( )
( )
( )
F
x t + y t X f + y f
or
()
()
F
x t + y t X j + y j .
H.3 Time Shifting and Frequency Shifting
() (
)
()
Let t0 be any real constant and let z t = x t t0 . Then the CTFT of z t is
( )
Z f =
()
x (t t ) e
j 2 ft
z t e dt =
j 2 ft
0
dt
Make the change of variable = t t0 and d = dt . Then
( ) x ( ) e
Z f =
(
j 2 f +t0
)
d = e
x ( ) e
j 2 ft0
j 2 f d = e
j 2 ft0
( )
X f
and the time-shifting property is
(
)
( )
j 2 ft0
(
)
( )
j t0
F
x t t0 X f e
or
F
x t t0 X j e
( )
.
(
)
()
Let f0 be any real constant and let Z f = X f f0 . Then z t is
() ( )
z t =
Z f e j 2 ft df =
X( f f )e
0
j 2 ft
df
Make the change of variable = f f0 and d = df . Then
() ( )
z t =
X e
(
)
j 2 + f0 t
d = e
j 2 f0 t
X ( ) e
H-3
j 2 t
d = e
j 2 f0 t
()
x t
M. J. Roberts - 2/18/07
and the time-shifting property is
()
x t e
or
()
x t e
+ j 2 f0 t
+ j 0 t
(
F
X f f0
((
F
X j 0
)
))
.
H.4 Time Scaling and Frequency Scaling
() ( )
Let a be any real constant other than zero and let z t = x at . Then the CTFT of
()
z t is
( )
Z f =
()
x ( at ) e
j 2 ft
z t e dt =
j 2 ft
dt
Make the change of variable = at and d = adt . Then, if a > 0 ,
( ) ()
Z f =
x e j 2 f / a
and if a < 0 ,
( ) x ( ) e
Z f =
j 2 f / a
1 f
d 1
j 2 f / a = x e ( ) d = X a a a
a
()
1
1 f
d
j 2 f / a = x e ( ) d = X .
a a a
a
()
( ) (
) (
)
Therefore, in either case, Z f = 1 / a X f / a and the time-scaling property is
(
( )
) (
)
( )
(
) (
F
F
x at 1 / a X f / a or x at 1 / a X j / a
)
In the time-scaling property, let b = 1 / a . Then
( )
( )
F
x t / b b X bf
( )
(
F
or x t / b b X jb
)
and the frequency scaling property is
(1 / b ) x (t / b) X ( bf )
F
(
)( )
(
)
F
or 1 / b x t / b X jb .
H.5 Transform of a Conjugate
( )
The inverse CTFT of X f is,
H-4
M. J. Roberts - 2/18/07
( ( )) ( ) X ( f ) e
=x t =
F -1 X f
j 2 ft
df .
()
The complex conjugate of x t is
*
x t = X f e j 2 ft df = X * f e j 2 ft df .
*
()
( )
( )
Make the change of variable = f and d = df we get
()
( )
x t = X e
*
*
j 2 t
d =
X ( ) e
j 2 t
d .
*
( )
F 1 X* f The conjugation property is
()
( )
F
x * t X* f
()
(
H.6 Multiplication - Convolution Duality
()
()
Let the convolution of x t and y t be
( ) ( ) ( ) x ( ) y (t ) d .
z t = x t y t =
()
The CTFT of z t is
( ) z (t ) e
Z f =
j 2 ft
dt
or
( )
Z f =
j 2 ft
x y t d e dt .
() (
)
Reversing the order of integration,
( )
Z f =
j 2 ft
x
dt d .
y t e
()
)
F
or x * t X * j .
(
)
(
)
F y t Then, using the time-shifting property,
H-5
M. J. Roberts - 2/18/07
( ) x ( ) e
Z f =
( )
j 2 f Y f d
( )
Since Y f is not a function of ,
( )
( ) x ( ) e
Z f =Y f
j 2 f d ,
()
F x ( )
( ) ( )
and finally Z f = X f Y f . The time-domain convolution property is
() ()
( ) ( )
F
x t y t X f Y f
( )
() ()
( ) ( )
F
or x t y t X j Y j
( )
Let the convolution of X f and Y f be
( )
( ) ( )
Z f = X f Y f .
()
Then z t is
( ) Z( f ) e
z t =
j 2 ft
X( f ) Y( f )e
df =
j 2 ft
df
or
()
z t =
j 2 ft
X Y f d e df .
() (
)
Reversing the order of integration,
()
z t =
j 2 ft
X
df d .
Y f e
()
(
)
()
e j 2 t y t
Then, using the frequency-shifting property,
( ) ( ) X ( ) e
z t =y t
j 2 t
() ()
d = x t y t
()
=x t
and the time-domain multiplication property is
() ()
( ) ( )
F
x t y t X f Y f
() ()
F
or x t y t H-6
( ) ( )
1
X j Y j .
2
M. J. Roberts - 2/18/07
The extra 1 / 2 is there because of the change of variable in converting f to .
H.7 Time Differentiation
()
The time-domain function x t can be expressed as
( ) X( f )e
x t =
j 2 ft
df .
Differentiating both sides with respect to time,
( ( ))
( )
( )
(
( ))
( ( ))
( )
d
d
j 2 ft
j 2 ft
1
x t =
X f e df = j2 f X f e df = F j2 f X f
dt dt
Therefore the time-differentiation property is
( ( ))
( )
d
F
x t j2 f X f
dt
or
d
F
x t j X j .
dt
H.8 Transforms of Periodic Signals
()
If a time signal x t is periodic it can be represented exactly by a complex CTFS
(by letting TF = T0 ).
property),
Therefore, for periodic signals (using the frequency-shifting
() x t =
k = X k e
( )
j 2 kf F t
( ) X k ( f kf ) ,
X f =
F
k = F
or
( ) X k e
x t =
(
)
j k F t
k = ( )
F
X j = 2
X k ( k ) .
k = F
H.9 Parseval’s Theorem
The product of two energy signals in the time domain corresponds to the
convolution of their transforms in the frequency domain.
H-7
M. J. Roberts - 2/18/07
() ()
( ) ( ) X ( ) Y ( f ) d
F x t y t = X f Y f =
.
(H.2)
From the definition of the CTFT,
( ) ( ) x (t ) y (t ) e
F x t y t =
j 2 ft
dt .
(H.3)
Combining (H.2) and (H.3),
x (t ) y (t ) e
j 2 ft
X ( ) Y ( f ) d .
dt =
This relation holds for any value of f . Setting f = 0 ,
() ()
x t y t dt =
X ( ) Y ( ) d .
This is known as the generalized form of Parseval’s theorem. For the special case in
()
()
which y t = x * t , this becomes
() ()
x t x * t dt =
()
x t
2
dt =
( ) ( )
( ) ( )
X X d =
()
2
X X * d =
X ( )
2
d
( )
F
Y * f . So, ultimately, we have the
where we have used yy* = y and y* t equality of total signal energy in the time and frequency domains
x (t )
2
dt =
X( f )
2
df .
H.10Integral Definition of an Impulse
The definition of the Fourier transform pair
( )
( ( )) x ( t ) e
X f =F x t =
j 2 ft
()
dt , x t = F
can be used to prove a handy result. Since
H-8
( X ( f )) = X ( f ) e
-1
+ j 2 ft
df
M. J. Roberts - 2/18/07
( ) X( f )e
x t =
+ j 2 ft
(H.4)
df
and (making the change of variable t in the definition of the forward CTFT)
( ) x ( ) e
X f =
j 2 f d
(H.5)
we can combine (H.4) and (H.5), to get
x t = x e j 2 f d e+ j 2 ft df
()
()
or
( ) x ( ) e
x t =
(
j 2 f t )
d df .
(H.6)
Rearranging to do the f integration first in (H.6),
j 2 f (t ) x t = x e
df d .
()
()
(H.7)
In words, this integral says that if we take any arbitrary signal x(t) replace t by , and
j 2 f t ( )
multiply that by a function
e
df and then integrate over all we get x(t) back.
We could re-write (H.7) in the form
( ) x ( ) g (t ) d
x t =
where
( ) e
g t =
j 2 ft
df .
Recall the sampling property of the impulse
g (t ) (t t ) dt = g (t ) .
0
0
The only way the equality
( ) x ( ) g (t ) d
x t =
H-9
M. J. Roberts - 2/18/07
(
) (
)
can be satisfied is if g t = t . That is, if
e
( )
j 2 f t (
)
df = t .
This is another valid definition of a unit impulse. A more common form of this result is
e
j 2 xy
()
dy = x
.
H.11Duality
()
( )
()
If the CTFT of x t is X f , what is the CTFT of X t ?
( ) ()
X f =
x t e j 2 ft dt =
x ( ) e
j 2 f d
Therefore, replacing f by t in the right side,
( ) x ( ) e
X t =
j 2 t
d .
The CTFT is
j 2 t
F X t = F x e
d = x e j 2 t d e j 2 ft dt
( ( ))
()
()
or
( ( )) x ( ) e
F X t =
(
)
j 2 + f t
dtd .
Using the integral definition of an impulse,
e
j 2 xy
()
dy = x ,
we can write
e
(
j 2 t + f
)
(
)
dt = + f .
Now, substituting into (H.8)
H-10
(H.8)
M. J. Roberts - 2/18/07
( ( )) x ( ) ( + f )d = x ( f ) .
F X t =
The duality property is
()
( )
( )
F
X t x f
or
( )
( )
F
X jt 2 x ( )
( )
F
and X t x f
( )
( )
F
and X jt 2 x .
( )
F
x f is similar.
The proof of X t H.12Total-Area Integral Using Fourier Transforms
The total area under a time- or frequency-domain signal can be found by
evaluating its CTFT or inverse CTFT with an argument of zero.
j 2 ft
X 0 = x t e
dt = x t dt
f 0 ()
()
()
and
+ j 2 ft
x 0 = X f e
df = X f df
t0 ()
( )
( )
or
X 0 = x t e j t dt = x t dt
0 ()
()
()
and
1
x 0 =
2
()
1
+ j t
X j e d = 2
t0
( )
X ( j ) d
H.13Integration
( ) x ( ) d .
Let y t =
t
We know that
t
( ) ( ) x ( ) u (t ) d = x ( ) d .
x t u t =
Using multiplication-convolution duality,
H-11
M. J. Roberts - 2/18/07
() () ()
( )
( ) ( )
F
y t = x t u t Y j = X j U j
Using the Fourier pair
()
( )
F
u t 1 / j + from (H.9)
( ) + X j
( )
Y j =
j
( ) X ( j ) =
( ) + X 0 () ( )
j
X j
Making the change of variable, = 2 f the cyclic frequency form is
( )
Y f =
( ) + 1X 0 f .
() ( )
j2 f 2
X f
Therefore the Fourier pairs are
t
()
F
x d and
t
x ( ) d F
( ) + 1X 0 f
() ( )
j2 f 2
X f
( ) + X 0 () ( )
j
X j
H-12
.
(H.9)