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114
Signals and Spectra Chap. 2
Solution.
and
Note: The peak instantaneous power is
P = V 2 rms /R = (1.79) 2 /600 = 5.32 mW
P
5.32 * 10-3
10 log a b = 10 log a
-3
10 10 -3 b = 7.26 dBm
max [p(t)] = max [v(t) i(t)] = max [v 2 (t)/R]
= (e)2
600
= 12.32 mW
SA2–3 Evaluation of Spectra by Superposition
w(t) =ßa t - 5
10
Find the spectrum for the waveform
b + 8 sin (6pt)
Solution.
The spectrum of w(t) is the superposition of the spectrum of the rectangular pulse and the spectrum
of the sinusoid. Using Tables 2–1 and 2–2, we have
cßa t - 5
10
bd = 10
sin (10pf)
10pf
e -j2pf5
and using the result of Example 2–5, we get
[8 sin (6pt)] = j 8 [d(f + 3) - d(f - 3)]
2
Thus,
W(f) = 10
sin (10pf)
e -j10pf + j4[d(f + 3) - d(f - 3)]
10pf
SA2–4 Evaluation of Spectra by Integration
Find the spectrum for w(t) = 5 - 5e -2t u(t).
Solution.
or
q
W(f) = w(t)e -jv t dt
L
-q
q
q
= 5e -j2pft dt - 5 e -2t e -jv t dt
L L
-q
= 5d(f) - 5 `
w(f) = 5d(f) -
0
e -2(1+jpf) t q
-2(1 + jpf) `
0
5
2 + j2pf