Civil Engineering Reference
In-Depth Information
+∞
∫
xt
()d
t
= ∞
,
−∞
which means that the Fourier transform according to Equation
(1.13)
does not exist.
2.0
0.0
-2.0
-4.0
4.0
2.0
0.0
-2.0
-4.0
4.0
2.0
0.0
-2.0
-4.0
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
0.20
Time
t
(s)
Figure 1.9
Examples of time functions of a stochastic signal. The signals (oscillations) are limited to a
frequency interval of 10-250 Hz.
Obviously, we shall not be able to measure over an infinite time in any case. To
perform an analysis the idea is to define a new time function
x
T
(
t
), equal to the original
function
x
(
t
) in a time interval
T
but equal to zero otherwise. We then get an estimate of
X
(
f
) by calculation of a Fourier transform over the time interval
T
T
=
∫
−
j2
π
ft
X
()
f
=
Xf T xt
(,)
()
e
d.
t
(1.17)
T
T
0
A finite transform as shown here will always exist for a time segment of a stationary
stochastic function. Taking Equation
(1.6)
into consideration we may see that for the
X
(
f
k
,T
)
=
T X
k
with
k
=
± 1, ± 2, ± 3,...
This means that when performing the transform and letting the frequency
f
just to take on
the discrete values
f
k
we get a Fourier series with period
T
. This is in fact the method
used when processing data digitally. Before taking up that theme we shall introduce an
