5.15. Let X (t) = A sin (ω
0
t + Φ) , where ω
0
is a constant and A and Φ are random variables with
marginal probability density functions given by
f
A
(a) = W
0
/2 for −W
0
≤ a ≤ W
0
f
Φ
(φ) = 1/2π for 0 < φ ≤ 2π.
Find the autocorrelation function and the spectral density, R
XX
(τ) and S
XX
(ω) .
Solution: The autocorrelation function is given by
R
XX
(τ) = E
$
A
2
sin (ω
0
t + Φ) sin (ω
0
(t + τ) + Φ)
%
.
Assuming that A and Φ are independent,
R
XX
(τ) =
W
0
−W
0
2π
0
a
2
sin (ω
0
t + φ) sin (ω
0
(t + τ) + φ)
W
0
2
1
2π
dφda
=
W
4
0
6
cos ω
0
τ.
Then
S
XX
(ω) =
1
2π
∞
−∞
R
XX
(τ) exp (−iωτ) dτ
=
1
2π
∞
−∞
W
4
0
6
cos ω
0
τ · exp (−iωτ) dτ
=
W
4
0
12
[δ(ω −ω
0
) + δ(ω + ω
0
)].
This integral was evaluated by replacing the cos ω
0
τ by complex exponential functions and then integrat-
ing, with Dirac delta functions as the result.
150
Section 5.6: Power Spectra
5.16. Complete the alternate derivation of the Wiener-Khinchine theorem.
Solution: Let
X(t
1
) =
∞
−∞
1
T
1
T/2
−T/2
X(η)e
−iω
m
η
dη
2
e
iω
m
t
1
X(t
2
) =
∞
−∞
1
T
1
T/2
−T/2
X(ν)e
−iω
n
ν
dν
2
e
iω
n
t
2
.
Then, the correlation of X(t
1
) and X(t
2
) is given by
R
XX
(τ) = E {X(t
1
)X(t
2
)} = E
∞
−∞
∞
−∞
1
T
2
T/2
−T/2
T/2
−T/2
X(η)X(ν)e
−i[ω
m
η+ω
n
ν]
dηdυe
i[ω
m
t
1
+ω
n
t
2
]
Setting υ = η + τ , t
1
= t, and t
2
= t + τ and rearranging the order of operation produces
R
XX
(τ) =
∞
−∞
∞
−∞
1
T
2
T/2
−T/2
T/2−η
−T/2−η
E {X(η)X(ν)}e
−i[ω
m
η+ω
n
(η+τ )]
dηdτ e
i[ω
m
t+ω
n
(t+τ)]
.
Using E {X(η)X(ν)} = R
XX
(τ), and let ω
mn
= ω
m
+ ω
n
for convenience. Rearrange the integration as
follows
R
XX
(τ) =
∞
−∞
∞
−∞
1
T
2
T/2
0
T/2−η
−T/2−η
R
XX
(τ)e
−i[ω
m
η+ω
n
(η+τ)]
dηdτe
i[ω
m
t+ω
n
(t+τ)]
+
1
T
2
0
−T/2
T/2−η
−T/2−η
R
XX
(τ)e
−i[ω
m
η+ω
n
(η+τ)]
dηdτe
i[ω
m
t+ω
n
(t+τ)]
.
Evaluation of the above integral produces
R
XX
(τ) =
∞
−∞
∞
−∞
1
T
2
sin(ω
mn
T/2)
ω
mn
T
−T
R
XX
(τ)e
−iω
n
τ
[e
iω
mn
τ
+ 1]dτe
i[ω
m
t+ω
n
(t+τ)]
.
Let
1
T
2
=
∆ω
m
∆ω
n
(2π)
2
to produce
R
XX
(τ) =
∞
−∞
∞
−∞
∆ω
m
∆ω
n
(2π)
2
sin(ω
mn
T/2)
ω
mn
T
−T
R
XX
(τ)e
−iω
n
τ
[e
iω
mn
τ
+ 1]dτe
i[ω
m
t+ω
n
(t+τ)]
.
As T → ∞, let ω
m
→ ω
1
and ω
n
→ ω
2
. Also, we need δ(ω) = lim
T →∞
sin(ωT )
πω
where δ is the Dirac delta
function. This produces
R
XX
(τ) =
∞
−∞
∞
−∞
1
4π
δ(ω
1
+ ω
2
)
∞
−∞
R
XX
(τ)e
−iω
2
τ
[e
iω
2
τ
+ 1]dτe
i(ω
1
+ω
2
)τ
dω
1
dω
2
.
Evaluation of the integral in either ω
1
or ω
2
produces
R
XX
(τ) =
∞
−∞
1
4π
∞
−∞
2R
XX
(τ)e
−iω
1
τ
dτ
e
iω
1
τ
dω
1
.
Let ω
1
= ω, and simplifying yields
R
XX
(τ) =
∞
−∞
1
2π
∞
−∞
R
XX
(τ)e
−iωτ
dτ
e
iωτ
dω.
The term in the bracket is the spectral density, and thus the Wiener-Khinchine theorem has been proven.
151
5.17. Discuss whether the function
R
XX
(τ) =
R
o
for |τ| < τ
o
0 elsewhere
can be an autocorrelation function. Explain.
Solution: The spectral density is the Fourier transform of R
XX
(τ) , or
S
XX
(ω) =
1
2π
∞
−∞
R
XX
(τ) exp (−iωτ) dτ
=
1
2π
τ
0
−τ
0
R
o
exp (−iωτ) dτ
=
1
2π
τ
0
−τ
0
R
o
cos ωτ dτ
=
R
o
2π
sin ωτ
ω
∞
τ=−∞
= R
0
δ(ω).
where the last simplification occurs since R
XX
(τ) = R
o
is an even function. Then, the spectral density is
given by
S
XX
(ω) = R
o
sin ωτ
o
ωπ
.
This expression permits S
XX
(ω) to be negative for some frequency ranges.
152
5.18. Sketch the following:
(a) a typical autocorrelation function for a stationary random process
(b) a typical spectral density for a stationary random process
(c) a function that cannot be a spectral density
(d) a Gaussian function for a nonstationary random process
(e) how an ensemble average for the correlation function is performed for a general random process.
Solution:
(a) A symmetric function about τ = 0 decays to µ
2
as t → ±∞ as shown in Figure 5.5.
(b) A non-negative symmetric function about ω = 0 with area equal to σ
2
.
(c) A spectral density cannot be negative or imaginary.
(d) The shape of the density function and the location of mean value changes with time.
(e) See Figure 5.3.
153
5.19. Consider a random signal with a mean value of 2.0 cm and a constant spectral density,
W (f) = 0.002 cm
2
/Hz
40 ≤ f ≤ 1600 Hz,
and zero outside this frequency range. Find the variance of the response.
Solution: The mean square value is found from
E(X
2
) =
∞
0
W (f)df
=
1600
40
0.002df = 3.12 cm
2
.
The variance is
σ
2
X
= E(X
2
) −µ
2
X
= 3.12 − 2
2
= −0.88 cm
2
.
This is clearly impossible because variance is always positive.
This result can be used in a design procedure where, for example, X has its realizations in the range
µ
X
±2σ
X
.
154
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