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54
Signals and Spectra Chap. 2
Example 2–5 SPECTRUM OF A SINUSOID
Find the spectrum of a sinusoidal voltage waveform that has a frequency f 0 and a peak value of A
volts. That is,
v(t) = A sin v 0 t where v 0 = 2pf 0
From Eq. (2–26), the spectrum is
q
V(f) = Aa ejv0t - e -jv 0t
b e -jvt dt
L 2j
-q
q
= A 2j L-qe -j2p(f-f0)t dt - A 2j L
By Eq. (2–48), these integrals are equivalent to Dirac delta functions. That is,
V(f) = j A 2 [d(f + f 0) - d(f - f 0 )]
Note that this spectrum is imaginary, as expected, because v(t) is real and odd. In addition, a
meaningful expression was obtained for the Fourier transform, although v(t) was of the infinite
energy type and not absolutely integrable. That is, this v(t) does not satisfy the sufficient (but not
necessary) Dirichlet conditions as given by Eqs. (2–31) and (2–32).
The magnitude spectrum is
|V(f)| = A 2 d(f - f 0) + A 2 d(f + f 0)
where A is a positive number. Since only two frequencies (f =;f 0 ) are present, u( f ) is strictly
defined only at these two frequencies. That is, u(f 0 ) = tan -1 (-1/0) = -90° and
u(-f 0 ) = tan -1 (1/0) =+90°. However, because ƒV(f)ƒ = 0 for all frequencies except f =;f 0
and V(f) = |V(f)| e ju(t) , u( f ) can be taken to be any convenient set of values for f Z;f 0 . Thus,
the phase spectrum is taken to be
q
-q
e -j2p(f # f0 )t dt
|V(f)|
¨(f)
A
–– 2
f 0
Weight
is A/2.
+90°
– f 0
f f
–90°
(a) Magnitude Spectrum
Figure 2–4
(b) Phase Spectrum (¨0=0)
Spectrum of a sine wave.