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Sec. 6–2 Power Spectral Density 427
t 1 =–T/2
=T/2-t 1
t 1
T
1
d
T/2
3
dt 1
–T/2 0
T/2
2 d
=–T/2-t 1
–T/2
t 1 =T/2
–T
Figure 6–4 Region of integration for Eqs. (6–47) and (6–48).
As seen in the figure, this is accomplished by covering the total area by using 2 when t 6 0
and 3 when t 0. Thus Eq. (6–47) becomes
0 t 1 = T/2
ƒ X T (f) ƒ 2 = c
L L
2
3
(6–48)
Now, assume that x(t) is stationary, so that R x (t 1 , t 1 t) = R x (t), and factor R x (t) outside the
inner integral. Then
0
ƒ X T (f) ƒ 2 = R x (t)e -jvt ct 1 -
L
-T
0
T
= (T + t)R x (t)e -jvt dt + (T - t)R x (t)e -jvt dt
L L
-T
-T
T t 1 = T/2-t
+ c R x (t 1 , t 1 + t)e -jvt dt 1 d dt
L L
0
t 1 = -T/2- t
⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
t 1 = -T/2
⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
T/2
T/2-t
R x (t 1, t 1 + t)e -jvt dt 1 d dt
T
d dt + R(t)e -jvt ct 1`
L
0
0
T/2-t d dt
-T/2