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2A.5 Fourier Transform
The Fourier transform, Y( f ), of a signal, y(t), relates the time domain to the frequency domain.
Specifically,
Y ð f Þ ¼
ðþ1
21
yðtÞexpð2j2p ftÞdt ¼
ðþ1
21
yðtÞe2jvt dt ð2A:13Þ
Using the Fourier operator, F, terminology:
Y ð f Þ ¼ FyðtÞ ð2A:14Þ
Note that if yðtÞ ¼ 0 for t , 0; as in the conventional definition of system excitations and responses, the
Fourier transform is obtained from the Laplace transform by simply changing the variable according to
s ¼ j2p f or jv: The Fourier is a special case of the Laplace, where, in Equation 2A.2, s ¼ 0:
Y ð f Þ ¼ Y ðsÞjs¼j2p f ð2A:15Þ
or
Y ðvÞ ¼ Y ðsÞjs¼jv ð2A:16Þ
The (complex) function Y( f) is also termed the (continuous) Fourier spectrum of the (real) signal, y(t).
The inverse transform is given by:
yðtÞ ¼
ðþ1
21
Y ð f Þexpðj2p ftÞdf ð2A:17Þ
or
yðtÞ ¼ F21Y ð f Þ
Note that according to the definition given by Equation 2A.13, the Fourier spectrum, Y( f ), is defined for
the entire frequency range f(2 1, þ1) which includes negative values. This is termed the two-sided
spectrum. Since, in practical applications it is not possible to have “negative frequencies,” the one-sided
spectrum is usually defined only for the frequency range f(0, 1).
In order that a two-sided spectrum has the same amount of power as a one-sided spectrum, it is
necessary to make the one-sided spectrum double the two-sided spectrum for f . 0.
If the signal is not sufficiently transient (fast-decaying or damped), the infinite integral given by
Equation 2A.13 might not exist, but the corresponding Laplace transform might still exist.
2-50 Vibration and Shock Handbook
© 2005 by Taylor & Francis Group, LLC
2A.5.1 Frequency-Response Function (Frequency Transfer Function)
The Fourier integral transform of the impulse-response function is given by
Hð f Þ ¼
ð1
21
hðtÞexpð2j2p ftÞdt ð2A:18Þ
where f is the cyclic frequency (measured in cycles/sec or hertz). This is known as the frequency-response
function (or frequency transfer function) of a system. Fourier transform operation is denoted as FhðtÞ ¼
Hðf Þ: In view of the fact that hðtÞ ¼ 0 for t , 0; the lower limit of integration in Equation 2A.18 can be
made zero. Then, from Equation 2A.9, it is clear that H( f) is obtained simply by setting s ¼ j2p f in H(s).
Hence, strictly speaking, we should use the notation Hðj2p f Þ and not H( f). However, for notational
simplicity, we denote Hðj2p f Þ by H( f). Furthermore, since the angular frequency v ¼ 2p f , we can
express the frequency-response function as HðjvÞ; or simply by HðvÞ for notational convenience. It
should be noted that the frequency-response function, like the Laplace transfer function, is a complete
representation of a linear, constant-parameter system. In view of the fact that both uðtÞ ¼ 0 and yðtÞ ¼ 0
for t , 0, we can write the Fourier transforms of the input and the output of a system directly by setting
s ¼ j2p f ¼ jv in the corresponding Laplace transforms.
Then, from Equation 2A.8, we have
Y ð f Þ ¼ Hð f ÞUð fÞ ð2A:19Þ
Note: Sometimes, for notational convenience, the same lowercase letters are used to represent the
Laplace and Fourier transforms as well as the original time-domain variables.
If the Fourier integral transform of a function exists, then its Laplace transform also exists. The
converse is not generally true, however, because of the poor convergence of the Fourier integral in
comparison to the Laplace integral. This arises from the fact that the factor expð2stÞ is not present in the
Fourier integral. For a physically realizable, linear, constant-parameter system, H( f ) exists even if U( f )
and Y( f) do not exist for a particular input. The experimental determination of H( f ), however, requires
system stability. For the nth-order system given by Equation 2A.10, the frequency-response function is
determined by setting s ¼ j2p f in Equation 2A.12 as
Hð f Þ ¼
b0 þ b1j2p f þ · · · þ bmðj2p f Þm
a0 þ a1j2p f þ · · · þ anðj2p f Þn ð2A:20Þ
This, generally, is a complex function of f, which has a magnitude denoted by lHð f Þl and a phase angle
denoted by /Hðf Þ:
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