082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

766 Chapter 24 The Fourier transform
5 Given F{u(t)e −t } = 1 ,use the second shift
1+jω
theorem to find
F{u(t +4)e −(t+4) }
Verify your result bydirect integration.
6 Find,usingthe secondshift theorem,
{ }
F −1 6e −4jωsin2ω
ω
Solutions
2
4cosω
−ω 2 + 4sinω
ω 3
(a)
(b)
−4cos(ω −3) 4sin(ω −3)
(ω −3) 2 +
(ω −3) 3
−4cos(ω −j) 4sin(ω −j)
(ω −j) 2 +
3 (a) u(t)e −t e −7jt (b) e jt u(t)e −t/2
e 4jω
5
1+jω
6 3 for2 t 6,0otherwise
(ω −j) 3
24.4 SPECTRA
IntheFourieranalysisofperiodicwaveformswestatedthatalthoughawaveformphysicallyexistsinthetime(orspatial)domainitcanberegardedascomprisingcomponents
with a variety of temporal (or spatial) frequencies. The amplitude and phase of these
components are obtained from the Fourier coefficients a n
and b n
. This is known as a
frequency domain description. Plots of amplitude against frequency and phase against
frequency are together known as the spectrum of a waveform. Periodic functions have
discrete or line spectra; that is, the spectra assume non-zero values only at certain frequencies.
Only a discrete set of frequencies is required to synthesize a periodic waveform.Ontheotherhandwhenanalysingnon-periodicphenomenaviaFouriertransform
techniqueswefindthat,ingeneral,acontinuousrangeoffrequenciesisrequired.Instead
ofdiscretespectrawehavecontinuousspectra.ThemodulusoftheFouriertransform,
|F(ω)|,givesthespectrumamplitudewhileitsargumentarg(F(ω))describesthespectrum
phase.
Example24.8 InExample 24.2 the Fourier transformof
f(t)=
{ 1 |t|1
0 |t|>1
wasfoundtobeF(ω) = 2sinω . Sketch the spectrum of f (t).
ω
Solution F(ω) is purely real. The spectrum of f (t) is depicted by plotting |F(ω)| against ω as
sinω
illustrated in Figure 24.4. Note that lim = 1. (See Review exercises in
ω→0 ω
Chapter 18.)