77
Chap 5. Continuous-Time Fourier Transform and Applications
5.1 Illustrative Definition of Fourier Transform
In this chapter, we will develop the basis for Fourier analysis of non-periodic signals,
which is the only group of signals meaningful in engineering and real- life applications.
Traditionally, Fourier analysis is presented by giving the definitions as we did for Laplace
transforms in Chapter 3. Here, we will approach it as a limiting behavior of Fourier series analysis
for periodic signals. To do that, let us consider an aperiodic (non-periodic) signal x (t ) and form
its periodic extension xT (t ) by repeating it every T seconds, where T is the period.
x(t)
x T(t)
Guard Interval
.....
0
.....
t
-T
0
T
2T
It is clear from above that the expansion signal can be written as:
if 0 ≤ t < T
x (t )

xT (t ) = 
 x(t ) = x (t + nT )
(5.1)
for every int eger n
The Fourier series representation for this new periodic signal is simply:
Synthesis Equation : x T (t ) =
∞
∑ Fk .e jkw0t
(5.2a)
k = −∞
1
. ∫ xT (t ).e − jkw0 t dt
(5.2b)
T <T >
2π
In the limit as T ⇒ ∞ , we observe that w0 =
⇒ dw , which is an infinitesimally small
T
quantity in frequency-domain and it implies that kw 0 ⇒ w , a continuous variable and
1
dw
⇒
= df . Finally, with this limiting behavior, the summation in (5.2a) becomes an
T
2π
Fk =
Analysis Equation :
integral as follows:
Fk =
dw ∞
. ∫ x T (t ).e − jwt dt ⇒
2π − ∞
Fk
1
=
dw 2π
∞
− jwt
∫ xT (t ).e dt =
−∞
1
2π
∞
∫ x (t ).e
− jwt
dt
(5.3)
−∞
and
x (t ) =
∞
∞
∫ ( ∫ x (t ).e
−∞ −∞
− jwt
dt ).e
jwt
dw 1
=
2π 2π
∞
∫ X ( w).e
jwt
dw
−∞
These notes are © Huseyin Abut, August 1998
(5.4)
78
With this result, we have implicitly defined the Fourier transform relationships for x(t).
Analysis Equation : X ( w ) =
∞
∫ x(t ).e
−∞
1
Synthesis Equation : x (t ) =
2π
∞
− jwt
dt = F {x (t )}
∫ X ( w).e
jwt
(5.5a)
dw = F −1{ X ( w)}
(5.5b)
−∞
These two equations are called Fourier transform pair and normally shown by: x (t ) ⇔ X ( w) .
In general, X(w) is a complex function of the real-valued frequency variable w and it is written in
terms of magnitude and phase terms:
X ( w) = X ( w) .e jφ (w)
(5.6)
Therefore, we plot one curve for the magnitude and another one for the phase of a given
expression.
φ ( w)
|X(w)||
w
w
φ(
w
|X(w)||
)
w
Dirichlet Conditions: Fourier transform exists if:
∞
1. x(t) is absolutely integrable:
∫ x (t ) dt < ∞
−∞
2. x(t) is a well-behaving function. That is, only a finite number of jumps of finite size, minima
and maxima occur within any finite interval: t1 < t < t 2 .
Example 5.1 Find the Fourier transform of a rectangular pulse (gate function, rectangular timewindow).
-τ/2
x(t)
X(w)
1.0
τ
τ/2
t
−4π/τ
−2π/τ
These notes are © Huseyin Abut, August 1998
2π/τ
4π/τ
6π/τ
8π/τ
79
∞
e − jwt
− jwt
− jwt
X ( w ) = ∫ x( t ).e
dt = ∫ 1.e
dt =
− jw
−∞
−τ / 2
=
•
•
•
•
τ /2
τ /2
=
t= −τ / 2
1
.[e − jwt / 2 − e + jwt / 2 ]
− jw
1
wτ
2
wτ
wτ
wτ
.[ −2 j.Sin( )] = .Sin( ) = τ .Sinc( ) = τ .Sa ( )
− jw
2
w
2
2π
2
(5.7)
It is easy to observe that the resulting spectrum is real and symmetrical for all values of
frequency w.
This implies that the phase response is zero: φ ( w) = 0 for all w.
The magnitude is decreasing as a function of 1 / w .
Approximately, 90% of energy content of this spectrum is under the main lobe of the plot:
− 2π / τ < w < 2π / τ . In other words, the energy under all the tail lobes is about 10%.
Example 5.2 Find the Fourier transform of a delta function. We will solve this problem using the
Sifting Theorem definition of a delta function:
∞
F {V0 .δ (t )} = V0 ∫ δ (t ).e − jwt dt = V0
⇒
F {δ (t )} = 1 ⇒ δ (t ) ⇔ 1
(5.9)
−∞
Similarly, we can use the inversion formula to have:
1
δ (t ) =
2π
∞
∫ 1.e
jwt
dw
(510)
−∞
From this final result we can deduce that the Fourier transform of a constant is a delta function in
frequency-domain:
F {V0 } = 2π .V0 .δ ( w )
(5.11)
Example 5.3 Find the Fourier transform of a complex exponential harmonic function:
F{e jw0t } =
∞
∞
−∞
−∞
jw t − jwt
∫ e 0 .e dt =
∫e
− j ( w− w0 ) t
dt = 2πδ ( w − w0 )
e jw0t ⇔ 2πδ ( w − w0 )
(5.12)
On the other hand, the Fourier transform of a one-sided decaying exponential function is:
x (t ) = e − at u (t ) and
X ( w) =
∞
∫e
− at
.u ( t ).e
−∞
a>0
− jwt
∞
dt = ∫ 1.e −( a + jw ) t dt =
0
1
a + jw
Example 5.4 Find the Fourier transform of a periodic signal x(t) with a period T0 =
(5.13)
2π
. As we
w0
know from the previous chapter that the periodic functions have Fourier series representation:
x (t ) =
∞
∑ Fk .e jkw0t
k = −∞
Let us take Fourier transform of this equation term-by-term:
These notes are © Huseyin Abut, August 1998
80
X ( w) = F{ x(t )} =
∞
∑ Fk .F{e
jkw0t
k = −∞
}=
∞
∞
k =−∞
k = −∞
2π
∑ 2π .Fk .δ ( w − kw0 ) = ∑ 2π .Fk .δ ( w − k T
)
0
(5.14)
Therefore, the Fourier transform is a sequence of impulse functions regardless of the actual shape
of the signal x(t). In other words, the Fourier transform of all periodic functions is a family of
impulses. What make them different for various x(t) shapes are the values of the coefficients {Fk}.
Example 5.5 Using the results from the previous example, we are asked to find the Fourier
transform of an impulse train.
x (t ) =
X ( w) =
∞
∞
∑ δ (t − kT0 )
k = −∞
∑ 2π .Fk .δ ( w − kw0 ) =
k = −∞
∞
∑ 2π .Fk .δ ( w − k
k = −∞
2π
)
T0
(5.15)
1
1
1
Fk = . ∫ x (t ).e − jkw0t = . ∫ δ (t − kT0 ).e − jkw0t =
T0 <T >
T0 <T >
T0
0
0
Let us substitute these coefficient values in to the result above to obtain:
X ( w) =
∞
∑
k = −∞
2π .
1
2π
2π ∞
2π
.δ ( w − k ) =
. ∑ .δ ( w − k
)
T0
T0
T0 k = −∞
T0
(5.16)
Thus, we conclude that the Fourier transform of an impulse train is another impulse train in the
frequency-domain with different strengths in the coefficient set.
x(t):
X(w):
2π /T 0
1
1
1
1
.....
2π /T 0
.....
2π / T
T0
t
0
T0
-2 π /T 0
2T 0
2π /T0
.....
.....
-T 0
2π /T 0
0
0
2π /T 0
4 π /T 0
w
5.2 Properties of Fourier Transforms
1. Linearity: The Fourier transform is a linear transform.
a. f (t ) + bg (t ) ⇔ a.F ( w) + bG ( w)
(5.17)
2. Symmetry: If x(t) is a real signal then X ( − w) = X ∗ ( w ) or in polar form:
X ( w) = X ( w) .e jφ (w )
and
X ∗ ( w) = X ( − w) = X ( w) .e − jφ ( w)
(5.18)
Therefore, we have an even-symmetry of the amplitude spectrum and an odd-symmetry for the
phase spectrum.
These notes are © Huseyin Abut, August 1998
81
Table 5.1 Fourier Transform Pairs for Selected Signals
Example 5.6 Using the properties of real signals we can turn Fourier transform into a sine or a
cosine transform. In particular, discrete version of the cosine transform has found a neat
application area in image compression based on JPEG and MPEG techniques. These compression
techniques and their derivatives attempt to exploit the symmetry properties of imagery in the
frequency-domain.
These notes are © Huseyin Abut, August 1998
82
X ( w) =
∞
∫ x(t ).e
−∞
∞
∫
=
−∞
=
− jwt
∞
dt = ∫ x(t ).[Cos( wt ) − jSin( wt )]dt
−∞
∞
x( t )Cos ( wt ) dt − j ∫ x (t ) Sin( wt ) dt
−∞
(5.19)
∞
∫ x(t )Cos( wt) dt + 0
( Sin is odd fn.)
−∞
∞
= 2 ∫ x (t ).Cos ( wt )dt
0
As we see from this last result that the Fourier transform turns into a Fourier Cosine transform for
real signals.
3. Time-Shifting (Delays): Given: x (t ) ⇔ X ( w) then
x (t − T ) ⇔ X ( w).e − jwT = X ( w) .e j (φ − wT )
(5.20)
Notes:
• Amplitude spectrum: X (w) is unaffected by a delay (time-shift) of T units.
•
However, the phase spectrum is linearly shifted by − wT radians from the phase value φ of
the original spectrum X(w). This linear phase-shift property makes the time-delay issue very
useful in many applications.
4. Time-Scaling (Resolution Change): Given the Fourier pair: x (t ) ⇔ X ( w) then a timescale change will correspond to:
x ( at)
⇔
1
w
.X ( )
a
a
(5.21)
Notes:
• For values of a larger than "1" this works as a compressor. That is, it counts slower than one
time- unit at a time.
• On the other hand, for a smaller than "1" it works as an expander to increase the resolution in
time-domain. The resulting behavior in frequency-domain is just the opposite.
Example 5.7 Using the above property find the Fourier transform of x (t ) = a.rect (
wT
) and the resolution change we get:
2π
wT
wT
F {a.rect (
) = T . sin c (
)
2a π
2a π
at
)
T
t
T
Recalling F {rect ( )} = T .Sinc(
These notes are © Huseyin Abut, August 1998
(5.22)
83
5. Derivatives and Integrals:-Scaling: Given the Fourier pair: x (t ) ⇔ X ( w) then we have:
a.
b.
d
x (t ) ⇔ jw. X ( w)
dt
dn
x (t ) ⇔ ( jw ) n . X ( w )
n
dt
t
∫
c.
x (τ ) dτ
(5.23a)
(5.23b)
⇔ π .X (0 ).δ ( w) +
−∞
d.
1
.X ( w)
jw
If there is no D.C. term, i.e., X ( 0) = 0 then
t
∫
(5.23c)
x (τ ) dτ
−∞
⇔
1
. X ( w)
jw
(5.23d)
Example 5.8 Using the above property find the Fourier transform of a unit-step function: u(t).
u(t)
Constant
=
1.0
sgn(t )
+
0.5
t
t
0.5
-0.5
t
As we see from the above figure, the unit-step function can be written as:
1 1
+ .Sgn(t )
2 2
u (t ) =
(5.24)
But we also know that:
d 1
{ .Sgn(t )} = δ (t )
dt 2
1
equivalently : ( jw ).F { .Sgn(t )} = 1
2
(5.25)
which results in:
1
1
F { .Sgn(t )} =
2
jw
and
F {u (t )} = π .δ ( w) +
1
jw
1
1
F { } = .2π .δ ( w)
2
2
(5.26)
(5.27)
6. Energy in Aperiodic Signals (Parseval's Theorem): Recalling energy in time-domain:
∞
∞
∞
1
E = ∫ x (t ) dt = ∫ x (t ).x (t )dt = ∫ x (t ).[
2π
−∞
−∞
−∞
1
=
2π
2
∞
∫
−∞
∗
∗
∞
X ( w).[ ∫ x (t ).e
−∞
− jw
1
dt ]dw =
2π
∞
∞
∫X
∗
( w).e − jw dw]dt
−∞
1
∫ X ( w). X ( w)dw = 2π
−∞
∗
∞
∫ X ( w)
2
dw
−∞
which implies that the energy is conserved during Fourier transform and it can be summarized
by:
These notes are © Huseyin Abut, August 1998
84
∞
∞
1
E = ∫ x (t ) dt = ∫ x( t ). x ( t ) dt =
2π
−∞
−∞
2
∗
∞
1
X
(
w
).
X
(
w
)
dw
=
∫
2π
−∞
∞
∗
∫
2
X ( w) dw
(5.28)
−∞
Similarly, the energy content of a signal within a finite band of frequency: w1 ≤ w ≤ w2 will be:
1
∆E x =
2π
w2
∫
2
X ( w) dw
(5.29)
w1
Example 5.9 Compute the portion of the energy for an exponentially decaying signal in the
frequency interval: − 4 ≤ w ≤ 4 . We know the Fourier transform for this signal both from a
previous example and the Fourier Tables 5.1 that:
1
1
2
x (t ) = e −t .u (t ) ⇒ X ( w) =
⇒ X ( w) =
1 + jw
1+ w 2
When we substitute this into (5.28) and (5.29) we get:
1
E=
2π
∞
1
∫
−∞ 1 + w
1
∆E x =
2π
dw =
2
4
1
2
⇒ From int egration tables.
14 1
1
4
dw
=
.
arctan(
w
)
≅ 0 .422
∫ X ( w) dw = π ∫
0
2
π
1
+
w
−4
0
2
Therefore, the portion of the energy in this finite band is 84.4%.
7. Convolution Operation:
x(t)
Input signal
Forcing function
X(w)
System
h(t)
H(w)
y (t ) = x (t ) ∗ h (t ) = h (t ) * x (t ) =
y(t)
Response
Y(w) Output Signal
∞
∫ x (τ )h (t − τ ) dτ
−∞
jφ ( w)
Y ( w) = X ( w).H ( w) = Y ( w) .e
Y ( w) = X ( w) .Y ( w) and ∠Y ( w) = ∠X ( w) + ∠H ( w)
(5.30a)
(5.30b)
(5.30c)
Notes:
• Output magnitude spectrum is the product of magnitude spectrum of the input and the
frequency response of the system.
• The phase spectrum is the sum of the phase spectra of the input and the system.
Example 5.10 Compute the unit-step response to a system with: h (t ) = e − at .u (t ) .
Y ( w) = X ( w).H ( w) = F {x (t )}.F{h( t )}
These notes are © Huseyin Abut, August 1998
85
Y ( w) = [π .δ ( w) +
1
1
π .δ ( w) 1
1
].
=
+
jw a + jw a + jw jw a + jw
Let us note that δ ( w) ≠ 0
directly from Fourier Tables
only when w = 0. Therefore, the first term in the last expression is
π
simply: 1 st Term = δ ( w) . Using this and partial fraction expansion results in:
a
Y ( w) =
π
A
B
δ ( w) +
+
a
jw a + jw
Bjw + A( a + jw) = 1 ⇒
A=
1
a
and
B=−
1
a
Finally, the output in the frequency-domain will be:
Y ( w) =
π
1/ a
1/ a
δ ( w) +
−
a
jw a + jw
and the inverse Fourier transform yields the answer:
y (t ) =
1
.{1 − e − at }.u ( t )
a
8. Modulation Theore m:
x(t)
y(t)
X
X(w)
Y(w)
m(t)
y (t ) = x (t ).m( t )
Y ( w) =
1
2π
∞
⇔
Y ( w) =
M(w)
1
[ X ( w) * M ( w)]
2π
∫ X (ψ ).H (ψ − w)dw
(5.31)
(5.32)
−∞
In other words, to replace a time-domain multiplication, we must perform a frequency-domain
convolution. As expected, we will attempt to perform time-domain multiplication as much as we
can to tackle modulation tasks.
Example 5.11 Given a signal with a base-band spectrum (frequency content of the signal in its
natural habitat), modulate it with an impulse train of period: T0 = 2π / w0 .
These notes are © Huseyin Abut, August 1998
86
X(w)
p(t) impulse train
Baseband Signal
1.0
-w 1
w1 w
-3T 0
-2T0
-T 0
T0
2T0
3T0
4T0 t
The output of the modulator (mixer) is simply:
x S (t ) = x(t ). p (t ) = x (t ).
∞
∑ δ (t − nT0 ) =
n =−∞
∞
∑ x( nT0 ).δ (t − nT0 )
(5.33)
n =−∞
Here, the impulse train acts like a sampling function, i.e., isolating the signal at a given point.
Equivalently, in the frequency-domain:
2π ∞
2π
. ∑ δ (w −
k)
(5.34)
T
T
n= −∞
0 w= −∞
0
∞
1
1 2π
2π
1 ∞
2π
X S ( w) =
X ( w) * P( w) =
. . ∑ [ X ( w) * δ ( w −
k )] = . ∑ X ( w −
k)
2π
2π T0 w= −∞
T0
T0 w= −∞
T0
P( w) = F{ p (t )} = F{
∞
∑ δ (t − nT0 )} =
(5.35)
It is clear from this result that the output spectrum is a periodic family of the input spectra shifted
to the neighborhood of ± kw0 = ± k
2π
.
T0
T 0. X s (w)
Bandpass (modulated) signal
-w 0
-w 0 /2
-w 1
w 1 w 0 /2
These notes are © Huseyin Abut, August 1998
w0
w
87
Table 5.2 Properties of Continuous -Time Fourier Pairs
These notes are © Huseyin Abut, August 1998