3 54
Problems and Solutions Section 3.5 (3.39 through 3.42)
3.39
Calculate the meansquare response of a system to an input force of constant PSD,
S
0
,
and frequency response function
H
!
(
)
=
10
3
+
2
j
!
(
)
Solution:
Given:
S
ff
=
S
0
and
H
!
(
)
=
10
3
+
2
j
!
The mean square of the response can be found from Eqs (3.66) and (3.68):
x
2
=
E
x
2
!
"
#
$
=
%&
&
’
H
(
(
)
2
S
ff
(
(
)
d
(
x
2
=
S
0
%&
&
’
10
3
+
2
j
(
2
d
(
Using Eq. (3.67) yields
x
2
=
50
!
S
0
3
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3 55
3.40
Consider the base excitation problem of Section 2.4 as applied to an automobile model of
Example 2.4.1 and illustrated in Figure 2.16. In this problem let the road have a random
stationary cross section producing a PSD of
S
0
. Calculate the PSD of the response and
the meansquare value of the response.
Solution:
Given:
S
ff
=
S
0
From example 2.4.1:
m
=
1007 kg,
c
=
2000 kg/s,
k
=
40,000 N/m
!
=
c
2
km
=
2000
2
40000
i
1007
=
0.157
(underdamped)
So,
H
!
(
)
=
1
k
"
m
!
2
+
jc
!
=
1
4
#
10
4
"
1007
!
2
+
2000
j
!
H
!
(
)
2
=
1
4
#
10
4
"
1007
!
2
(
)
2
+
2000
(
)
2
j
!
2
H
!
(
)
2
=
1
1.01
#
10
6
!
4
"
4.06
#
10
7
!
2
+
1.6
#
10
9
The PSD is found from equation (3.62):
S
xx
!
(
)
=
H
!
(
)
2
S
ff
!
(
)
S
xx
!
(
)
=
1
1.01
"
10
6
!
4
#
8.46
"
10
7
!
2
+
1.6
"
10
9
The mean square value is found from equation (3.68):
x
2
=
E
x
2
!
"
#
$
=
%&
&
’
H
(
(
)
2
S
ff
(
(
)
d
(
x
2
=
S
0
%&
&
’
1
4
)
10
4
%
1007
(
2
+
2000
j
(
2
d
(
Using equation (3.70) yields
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 Spring '09
 J.G.Lee
 Signal Processing, Impulse response, mean square, Eqs, frequency response function

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