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Sec. 2–11 Study-Aid Examples 115
SA2–5 Evaluation of FT by Superposition
Fig. 2–26. Find the Fourier transform of w(t).
Assume that w(t) is the waveform shown in
Solution. Referring to Fig. 2–26, we can express w(t) as the superposition (i.e., sum) of two
rectangular pulses:
Using Table 2–2 and Table 2–1, we find that the FT is
or
w(t) =ßa t - 2
4
b + 2ßa t - 2
2
W(f) = 4Sa(pf4)e -jv2 + 2(2)Sa(pf2)e -jv2
W(f) = 4[Sa(4pf) + Sa(2pf) e -j4pf
SA2–6 Orthogonal Functions Show that w 1 (t) =ß(t) and w 2 (t) = sin 2pt are orthogonal
functions over the interval -0.5 6 t 6 0.5.
Solution.
b
b
0.5
w 1 (t) w 2 (t) dt = 1 sin 2ptdt = - `
La
3
-0.5
cos 2pt
2p
= -1 [cos p - cos (-p)] = 0
2p
0.5
`
-0.5
Because this integral is zero, Eq. (2–77) is satisfied. Consequently, ß(t) and sin 2pt are orthogonal
over -0.5 6 t 6 0.5. [Note: ß(t) and sin 2pt are not orthogonal over the interval
0 6 t 6 1, because ß(t) is zero for t 7 0.5. That is, the integral is 1/ p (which is not zero).]
SA2–7 Use FS to Evaluate PSD Find the Fourier series and the PSD for the waveform shown
in Fig. 2–25. Over the time interval 0 6 t 6 1, v(t) is described by e t .
w(t)
3
2
1
1 2 3 4
Figure 2–26
t