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EXAMPLES (CORRELATION FUNCTION AND PSD)
OUTLINE
- 7-2 Relation of Spectral Density to Fourier Transform
- 7-6 Mean-Square values from Spectral Density
- 6-7 Crosscorrelation Functions
- 6-8 Examples of Crosscorrelation Functions
- 6-9 Correlation Matrices
Reading: G. R. Cooper & C. D. McGillem Chapters 6 and 7
EE/STAT 322, #21 1
Question: (modified: 7-1.1) A random process: X(t) = M , |t| ≤ T. M is
uniformly distributed in (0, 10).
(a) Find the mean value of X(t); (b) Find Fourier transform of X(t),
F
X
(ω); (c) Find E[F X
(ω)].
Solution: X(t) = M rect(
t
2 T
(a) X(t) =
E[M ] = 5 −T ≤ t ≤ T
0 elsewhere
(b) F X
(ω) =
T
−T
M e
−jωt
dt = 2T M sinc(2T f )
= 2T M sin(2πT f )/(2πT f ) = sin(2πT f )
M
πf
(c) E[F X
(ω)] = 2T E[M ] sinc(2T f ) = 10T sinc(2T f ).
Question: (modified: 7-2.1) Use the Parsevel’s theorem to evaluate the
following integrals:
(a)
∞
−∞
sin(4ω)
4 ω
sin(ω)
ω
dω;
Solution: (a) Parseval’s theorem:
If F{f (t)} = F (ω) and F{g(t)} = G(ω), respectively, then
∞
−∞
f (t)g(t)dt =
2 π
∞
−∞
F (ω)G(−ω)dω.
F (ω) =
sin(4ω)
4 ω
= (sinc(8T f )) ⇒f (t) = F
− 1 {F (ω)} =
1
8
, |t| ≤ 4.
EE/STAT 322, #21 3
In general, F (ω) = sinc(2T f ) ⇒f (t) =
1
2 T
, |t| ≤ T.
Proof:
T
−T
1 · e
−jωt
dt =
e
−jωt
jω
0
T
e
−jωt
jω
−T
0
1 −e
−jωT
jω
e
jωT − 1
jω
= 2T
e
jωT −e
−jωT
2 T jω
= 2T
sin(T ω)
T ω
Thus F
− 1
sin(T ω)
T ω
= 1/(2T ), |t| < T.
So we can show that f (t) =
1
8
for − 4 ≤ t ≤ 4.
Further G(−ω) = G(ω) =
sin(ω)
ω
⇒g(t) =
1
2
, |t| ≤ 1.
Thus,
∞
−∞
F (ω)G(−ω)dω = 2π
∞
−∞
f (t)g(t)dt = 2π
1
− 1
1
8
1
2
dt = π/ 4.
Question: (modified: 7-3.1) Decide whether each of the following functions
can be a valid PSD. State why.
(a)
1
ω
2 +4ω+
(b)
ω
2
ω
4 +6ω
2
(c) δ(ω) +
ω
3
ω
4
Solution: A valid PSD function S(ω) must be an even function (symmetric
around ω = 0 ⇒S(ω) = S(−ω)), and non-negative S(ω) ≥ 0 , for any ω.
(a) No. Because
1
ω
2 +4ω+
1
(ω+2)
2 − 3
1
3
< 0 , when ω = − 2.
(b) Yes.
ω
2
ω
4 +6ω
2
ω
2
(ω
2 +3)
2
≥ 0 and it is even.
(c) No. Because
ω
3
ω
4
< 0 when ω < 0.
EE/STAT 322, #21 7
Question: (modified: 7-3.2) A stationary process
X(t) = M + 2 cos(3t + θ 1
) + sin(4t + θ 2
where M is uniformly distributed in (-1,3),
θ 1
and θ 2
are independent and uniformly distributed in (0, 2 π), respectively.
(a) Find the mean, mean-square value and variance of X(t);
Solution: (a) E[X(t)] = E(M ) = 1.
E[X
2
(t)] = E[M
2
] + E[4 cos
2
(3t + θ 1
)] + E[sin
2
(4t + θ 2
)]
3
− 1
y
2 / 4 dy + 4 · 0 .5 + 0.5 = 2.33 + 2.5 = 4. 83.
var {X(t)} = E[X
2
(t)] − X(t)
2
2
= 3. 83.
Question (b) Find R X
(τ );
Solution: RX (τ ) = E[X(t)X(t + τ )]
= E[M
2
+4 cos(3t+θ 1
) cos(3(t+τ )+θ 1
)+sin(4t+θ 2
) sin(4(t+τ )+θ 2
)].
Note that E[M cos(3t + θ 1
)] = 0, and
E[cos(3t + θ 1 ) sin(4(t + τ ) + θ 2 )]
1
(2π)
2
2 π
0
2 π
0
cos(3t + θ 1
) sin(4(t + τ ) + θ 2
)dθ 1
dθ 2
E[cos(3t + θ 1
) cos(3(t + τ ) + θ 1
)] = cos(3τ )/ 2.
E[sin(4t + θ 2
) sin(4(t + τ ) + θ 2
)]
= E[
1
2
{cos(4τ ) − cos(4(2t + τ ) + 2θ 2
)}] = cos(4τ )/ 2.
⇒R
X
(τ ) = E[M
2
] + 4 cos(3τ )/2 + cos(4τ )/ 2
= 2.33 + 2 cos(3τ ) + cos(4τ )/ 2.
EE/STAT 322, #21 9
Question (c) Find PSD S X
(ω); find X
2 (t) using the PSD; list the frequency
components of X(t).
Solution: Some transform results:
- For constant C, F{C} = 2πCδ(ω), where δ(ω) is the Dirac delta
function.
Proof: F
− 1 { 2 πCδ(ω)} =
1
2 π
∞
−∞
2 πCδ(ω)e
jωτ dτ =
1
2 π
2 πC = C.
τ )} = π[δ(ω + ω 0
) + δ(ω − ω 0
)].
Proof: F
− 1
{π[δ(ω + ω 0
) + δ(ω − ω 0
)]}
1
2 π
∞
−∞
π[δ(ω + ω 0
) + δ(ω − ω 0
)]e
jωτ dτ
1
2 π
π[e
jω 0
τ
−jωτ
] = cos(ω 0
τ ).
Question (b) Find the autocorrelation function of V (t) = X(t) − Y (t);
Solution:
(b) R V
(τ ) = E[V (t)V (t + τ )] = E[{X(t) − Y (t)}{X(t + τ ) − Y (t + τ )}]
= R
X
(τ ) + R Y
(τ ) = 5e
−|τ |
cos 10πτ + sinc(10τ ).
Question (c) Find the cross-correlation function of U (t) and V (t).
Solution: R U V
(τ ) = E[U (t)V (t + τ )]
= E[{X(t) + Y (t)}{X(t + τ ) − Y (t + τ )}]
= R
X
(τ ) − R Y
(τ ) − R XY
(τ ) + R Y X
(τ )
= R
X
(τ ) − R Y
(τ )
= 5e
−|τ | cos 10πτ − sinc(10τ ).
EE/STAT 322, #21 13
Question (d) Find the autocorrelation function of Z(t) = X(t)Y (t);
Solution: R Z
(τ ) = E[Z(t)Z(t + τ )]
= E[X(t)Y (t)X(t + τ )Y (t + τ )]
= E[X(t)X(t + τ )] · E[Y (t)Y (t + τ )]
= R
X
(τ )R Y
(τ ) = 5e
−|τ | cos(10πτ ) · sinc(10τ ).
Question (e) Find the maximum value of R U V
(τ ) using the upper bound
of eq. (6-29). Compare this with the actual maximum value.
Solution: Using (6-29), |R XY
(τ )| ≤ [R X
(0)R
Y
(0)]
1 / 2
,
⇒|R
U V
(τ )| ≤ [R U
(0)R
V
(0)]
1 / 2
= [(5 + 1) · (5 − 1)]
1 / 2
=
The actually maximum value for |R U V
(τ )| is when τ = 0,
⇒ 5 e
−|τ |
cos 10πτ − sinc(10τ ) = 5 − 1 = 4.
Question: (modified: 6-8.1)
X(t) = 0.1 sin(2t + θ), and θ is uniformly distributed in (−π, π). Observed
signal is Y (t) = X(t) + N (t), and R N
(τ ) = 5e
− 10 |τ | .
(a) Find R Y
(τ ) for τ = 0.
(b) Find the smallest positive value of τ for which |R Y
(τ )| ≥ 10 R N
(τ ).
Solution (a) R X
(τ ) =
- 1
2
2
cos(2τ ).
So R Y
(τ ) = R X
(τ ) + R N
(τ ) = 0.005 cos(2τ ) + 5e
− 10 |τ | .
(b) |R Y
(τ )| ≥ 10 R N
(τ ) ⇒| 0 .005 cos(2τ ) + 5e
− 10 |τ | | ≥ 10 · 5 e
− 10 |τ |
⇒ 0 .005 cos(2τ ) ≥ 45 e
− 10 |τ |
, ⇒τ =?.
EE/STAT 322, #21 15
Question: (modified: 6-9.1) A stationary process X(t) with
R
X
(τ ) = 3e
−|τ |
- 3 is sampled every 0.5 seconds for three consecutive
samples. (a) Find the correlation matrix R X
of X(t) ;
Solution: (a) ∆t = 0. 5 , N = 3.
R
X
R
X
(0) R
X
(0.5) R
X
R
X
(0.5) R
X
(0) R
X
R
X
(1) R
X
(0.5) R
X
where R X
(0) = 3e
− 0
R
X
(0.5) = 3e
− 0. 5
R
X
(1) = 3e
− 1
(b) Repeat part (a) if the sample size is 9.
Solution: When n = 9, we can use Student’s t distribution with
v = n − 1 = 8 degree of freedom.
Let t =
ˆ ¯ X−
¯ X
˜ S/
√
n
Let the confidence interval be denoted by [ ¯X − k
S/
n,
X + k
S/
n].
F
T
(k) − F T
(−k) = 2F T
(t) − 1 = 0. 9 ,⇒F T
(k) = 0. 95 and v = 8
⇒k = 1. 86.
Thus, the limits are
[ ¯X − k
S/
n,
X + k
S/
n]
= [100 − 1. 86 ∗ 5 /
9] = [96. 9 , 103 .1].
EE/STAT 322, #21 19
(c) Find the confidence interval of 90% (one-sided test) for sample mean
with n = 9.
Solution:
Let t =
ˆ¯ X−
¯ X
˜ S/
√
n
X = 100,
S = 5, n = 9.
The confidence interval:
X ∈ [ ¯X − k
S/
n, ∞) ⇒t ∈ [−k, ∞).
F
T
(∞) − F
T
(−k) = 1 − F T
(−k) = F T
(k) = 0. 9 ,⇒k = 1. 397.
Thus, the limits are
[ ¯X − k
S/
n, ∞) = [100 − 1. 40 ∗ 5 /
9 , ∞] = [96. 67 , ∞].
(d) If somebody claims that the RVs have a mean of at least of 105. With
20 samples, sample mean is 102 and standard deviation of is 5, is this
claim justified with 95% of confidence?
Solution:
Confidence interval for
X: [
X − t c
S/
n, ∞).
Define t =
ˆ¯ X−
¯ X
˜ S/
√
n
, and t follows Student’s t distribution with v = n−1 =
19 degrees of freedom.
Thus, the limits for t with valid hypothesis are [−t c
F
T
(∞) − F
T
(−t c
) = 1 − F
T
(−t c
) = F
T
(t c
⇒F
T
(t c
) = 0. 95 with v = 19 ⇒t c
The lower limit is X c
= ¯X − k
S/
n = 105 − 1. 729 ∗ 5 /3 = 102. 1183.
Since
X = 102 < X
c
, ⇒the claim is not valid.
EE/STAT 322, #21 21
Example: If somebody claims that the independent RVs have a mean of at
least of 90. With 10 samples, sample mean is 85 and standard deviation of
is 10, is this claim justified with 90% of confidence?
v & F 0.90 0.95 0.
Solution:
X = 90, σ x
Confidence interval for
X: [ ¯X − t c
S/
n, ∞).
Define t =
ˆ ¯ X−
¯ X
˜ S/
√
n
, then v = n − 1 = 9.
Thus, the limits for t with valid hypothesis are [−t c
F
T
(∞) − F
T
(−t c
) = 1 − F
T
(−t c
) = F
T
(t c
) = 0. 9 , with v = 9 ⇒t c
