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

768 Chapter 24 The Fourier transform
uX(v)u
uF(v)u
(a)
v
(b)
–v c
Figure24.6
Amplitude modulation. (a)Spectrum ofthe modulation signal; (b)spectrum ofthe
amplitude-modulated signal.
Taking the Fourier transformand using the first shifttheorem yields
{ x(t)(e
jω t c +e −jω c t }
)
F{φ(t)} = (ω) = F
2
{ e
jω t } {
c x(t) e
−jω t }
c x(t)
= F + F
2 2
v c
v
= 1 2 (X(ω−ω c )+X(ω+ω c ))
whereX(ω) = F{x(t)}, the frequency spectrum of the modulation signal.
Let us consider the case where the frequency spectrum, |X(ω)|, has the profile
shown in Figure 24.6(a). The frequency spectrum of the amplitude-modulated signal,
|(ω)|, is shown in Figure 24.6(b). All of the frequencies of the amplitudemodulatedsignalaremuchhigherthanthefrequenciesofthemodulationsignalthus
allowing a much smaller antenna to be used to transmit the signal. This method of
amplitude modulation is known as suppressed carrier amplitude modulation because
the carrier signal is modulated to its full depth and so the spectrum of the
amplitude-modulated signalhas no identifiable carrier component.
EXERCISES24.4
1 Show thatthe Fourier transformofthe pulse
{ t+1 −1<t<0
f(t)=
1−t 0<t<1
can be written as
Plotagraph ofthe spectrum of f (t)for
−2π ω 2π.Writedown an integral expression
forthe pulse whichwould result ifthe signal f (t)
were passed through a filterwhicheliminatesall
angularfrequenciesgreaterthan2π.
Solutions
F(ω) = 2 ω 2 (1−cosω)
1
∫
1 2π
2(1−cosω) ejωt
2π −2π ω 2 dω
24.5 THEt−ω DUALITYPRINCIPLE
We have, fromthe definition ofthe Fourier integral,
f(t)= 1 ∫ ∞
F(ω)e jωt dω (24.6)
2π
−∞