Civil Engineering Reference
In-Depth Information
Similarly, given a zero mean variable
()
(
)
x t
a
sin
t
2
T
=⋅
ω
, where
ωπ
=
. Its variance is
0
0
0
then given by
⎡
2
⎤
T
0
2
1
⎡
⎛ ⎞
2
⎤
a
T
π
2
⎢
⎥
lim
na
∫
sin
t
t
σ
=
⋅
⋅
=
⎢
⎜ ⎟
⎥
x
nT
⎢
⎥
2
⋅
n
→∞
⎢
⎥
⎣
⎝ ⎠
⎦
0
0
0
⎣
⎦
Given a second zero mean variable comprising two cosine functions with different amplitudes and
frequencies, i.e.:
()
(
)
(
)
x t
a
cos
t
a
cos
t
=⋅
ω
+⋅
ω
1
1
2
2
2
T
2
T
TT
=
12
1
where
ωπ
=
and
ωπ
=
. It is easily seen that if
then
1
1
2
2
() (
)
(
)
x t
aa
cos
t
=+⋅
ω
1
2
1
and thus, the calculation of its variance is identical to the solution given above, i.e.:
(
2
)
aa
+
2
1
2
σ
=
x
2
()
If
TT
≠
12
1
then the variance of
x t
is given by
T
1
2
2
2
(
)
(
)
(
)
2
2
(
)
∫
⎡
⎤
σ
=
lim
a
cos
ω
t
+
2
a a
cos
ω
t
cos
ω
t
+
a
cos
ω
t
dt
x
⎣
1
1
1
2
1
2
2
2
⎦
T
T
→∞
0
Substituting
TnT
=⋅
into the integration of the first two terms and
TnT
=⋅
into the third,
11
22
where
n
and
n
are integers, then
T
T
1
1
1
1
2
2
2
(
)
(
)
(
)
lim
na
∫
cos
t t
lim
n aa
∫
2
cos
t
cos
t t
σ
=
⋅
ω
+
⋅
ω
ω
x
11
1
1
12
1
2
nT
nT
n
n
→∞
→∞
1
1
11
11
0
0
T
2
1
2
2
(
)
lim
na
∫
cos
t t
+
⋅
ω
2
2
2
nT
n
→∞
2
22
0
It is seen that the first and the third integrals are identical to the integral of a single cosine squared
shown above, and thus, they are equal to
2
1
2
2
a
, respectively. The second integral,
containing the product of two cosine functions, may most effectively be solved by the substitution
(
a
2
and
ˆ
)
t
==
ωπ
t
2
Tt
, in which case it is given by
1
1
⎡
⎤
⎛
T
⎞
⎛
T
⎞
sin 2
1
1
sin 2
1
1
π
−
π
+
⎢
⎥
⎜
⎟
⎜
⎟
⎡
⎤
2
π
T
T
2
aa
T
⎛ ⎞
T
aa
⎝
⎠
⎝
⎠
()
⎢
⎥
ˆ
ˆ
ˆ
2
2
12
1
1
12
⋅
⎢
∫
cos
t
⋅
cos
t dt
⎥
=
+
⎜ ⎟
⎢
⎥
T
2
T
⎢
π
⎥
π
⎛
T
⎞
⎛
T
⎞
⎝ ⎠
1
2
⎣
0
⎦
21
1
21
1
⎢
−
+
⎥
⎜
⎟
⎜
⎟
T
T
⎢
⎥
⎝
⎠
⎝
⎠
⎣
⎦
2
2
⎛
T
⎞
sin 2
1
π
⎜
⎟
T
0 if
TT
/
is an integer unequ
al to 1
aa
⎧
⎪
⎨
=
⎝
⎠
2
12
12
=
which is
TT
TT
≠
0 if
TT
/
is not an integer
π
⎪
⎩
1
2
−
12
2
1
