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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