Digital Signal Processing Reference
In-Depth Information
Figures 5.11(c) and (d) are sketches of the magnitude and phase frequency spectrum, respec-
tively, of the signum function.
■
The time-differentiation property of the Fourier transform
EXAMPLE 5.11
The frequency spectrum of the signal shown in Figure 5.12(a) will be found. The figure shows
a waveform for which we have not previously determined a Fourier transform. The dif-
ferentiation property of the Fourier transform can be used to simplify the process. Figure 5.12(b)
shows the first derivative of with respect to time. This waveform can be described
as a set of three rectangular pulses; however, the problem can be simplified even further by
taking a second derivative with respect to time to get
w(t)
x(t),
w(t)
y(t),
the result shown in Figure 5.12(c).
An equation for this waveform can easily be written as
A
b - a
d(t + b) -
bA
a(b - a)
d(t + a)
y(t) =
bA
a(b - a)
d(t - a) -
A
(b - a)
d(t - b),
+
and the Fourier transform found from Table 5.2:
A
b - a
e
jvb
A
b - a
e
-jvb
Y(v) =
-
bA
a(b - a)
e
jva
+
bA
a(b - a)
e
-jva
.
-
Using Euler's identity, we can rewrite this as
j2A
b - a
sin(bv) -
j2bA
a(b - a)
sin(av)
Y(v) =
jv2Ab
b - a
sin(bv)
jv2bA
b - a
sin(av)
=
-
.
bv
av
Because
y(t) = d[x(t)]/dt
, the time differentiation property yields
Y(v) = jvX(v)
or
1
jv
Y(v) + kd(v),
X(v) =
(5.22)
where, because the time-averaged value of
x(t)
is seen by inspection to be zero,
k = 0.
Thus,
2Ab
b - a
[sinc(bv) - sinc(av)].
X(v) =
Similarly,
1
jv
X(v) + kd(v),
W(v) =
(5.23)
