Building Acoustics

Oscillating systems 13

xt t()d

+∞

−∞

∫ =∞,

which means that the Fourier transform according to Equation (1.13) does not exist.

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)

-4.0

-2.0

0.0

2.0

4.0

-4.0

-2.0

0.0

2.0

4.0

x(

t)

-4.0

-2.0

0.0

2.0

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 xT(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

j2

0

() (,) () d.

T

ft

XTTfXfT xte t

==− π

∫ (1.17)

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
discrete frequency components fk =k/T the transform in Equation (1.17) will give

X(fk,T) = T Xk with k = ± 1, ± 2, ± 3,...

This means that when performing the transform and letting the frequency f just to take on
the discrete values fk 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