|
x t
e
x
v
t
a
x
t
a
t
a
ζω
ζω
ω
ω
ζω
ω
ζω
ω
ω
ω
−
−
=
=
=
⎛
⎞
= −
+
+
⎜
⎟
⎝
⎠
⎛
⎞
∴
=
=
+ −
⎜
⎟
⎝
⎠
⎧
⎫
∴
=
+
−
+
−
≥
⎨
⎬
⎩
⎭
Ch. 3: Forced Vibration of 1-DOF System
Arbitrary Excitation
Ideally, arbitrary excitation can be expressed as linear
combinations of simpler excitations. The simpler
excitations are simple enough that the response
is readily available. This concept is exactly used by
Fourier.
Now, the idea is to regard the arbitrary excitation as a
superposition of impulses of varying magnitude and
applied at different times. It is used when the excitation
can be easily described in time domain.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Consider the excitation F(t). We can imagine that it is
constructed from infinite impulses at different times.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
Convolution integral theorm
3.5 Non-periodic Excitation
( )
( )
(
)
( )
Focus on the time interval
, at which
the impulse of magnitude
is acting. This
shifted impulse can be written as
.
The response of the LTI system to this particular
impulse is
,
t
F
F
t
x t
F
τ
τ
τ
τ τ
τ τδ
τ
τ
< < + Δ
Δ
Δ
−
Δ
=
( )
(
)
( )
( )
(
)
( )
( )
( )
(
)
( )
( ) (
)
0
Since by sampling property
,
and the system is linear, the response to
is
In the limit as
0,
.
t
h t
F t
F
t
F t
x t
F
h t
x t
F
h t
d
τ
τ
τ τ
τ
τ τδ
τ
τ τ
τ
τ
τ
τ τ
Δ
−
=
Δ
−
=
Δ
−
Δ →
=
−
∑
∑
∫
Ch. 3: Forced Vibration of 1-DOF System
Convolution integral theorm
The response of the arbitrary excitation is the
superposition of shifted impulse responses.
3.5 Non-periodic Excitation
Interpretation
for the whole
range of time; t
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
(
)
( )
To obtain
from
, we need to carry out
two operation; shifting and folding. This is another interpretation
for the specific time t. The figures show the steps in evaluating the convolution.
If we
h t
h
τ
τ
−
( )
(
) ( )(
)
(
) ( )
( )
( )
0
0
define a new variable
, then
and
. With the change of the integration limits,
That is the convolution is symmetric in
and
.
To decide which formula to u
t
t
t
t
d
d
x t
F t
h
d
F t
h
d
F t
h t
λ
τ
τ
λ
τ
λ
λ
λ
λ
τ
τ τ
= −
= −
= −
=
−
−
=
−
∫
∫
( )
( )
( )
( )
se depends on the nature of
and
.
It is obvious that if the excitation
or the impulse response
is too complicated, we may be unable to evaluate the closed form
solution of the convolution inte
F t
h t
F t
h t
gral. The excitation may not at all
be written as functions of time. In these cases, the integration must
be carried out numerically.
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
Interpretation for the specific time; t
Ch. 3: Forced Vibration of 1-DOF System
Ex.
Determine the response of the underdamped MBK
to the unit step input.
3.5 Non-periodic Excitation
u(t)
1
0
t
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
( )
( ) (
)
( )
( )
(
)
( ) (
)
( ) (
)
(
)
( )
( )
(
)
0
0
and
is the system impulse response
shifted by and mirrored about the vertical axis.
If
0,
0 because of no overlap
If 0,
0, 0
, 0
Let
t
t
x t
F
h t
d
F
u
h t
t
t
F
h t
t
F
h t
h t
x t
t
x t
h t
d
t
t
τ
τ τ
τ
τ
τ
τ
τ
τ
τ
τ
τ τ
=
−
=
−
<
−
=
>
−
=
−
∴
=
<
∴
=
−
>
−
∫
∫
( )
( )
( )
0
0
. Hence
1
sin
Substitute sin
and use
2
1
1
cos
sin
,
0
n
d
d
n
t
t
d
d
i
i
ax
ax
d
t
n
d
d
d
d
d
x t
h
d
e
d
m
e
e
e
e dx
c
i
a
x t
e
t
t
t
k
ζω λ
ω λ
ω λ
ζω
τ λ
τ
λ
λ λ
ω λ λ
ω
ω λ
ζω
ω
ω
ω
−
−
−
=
= −
∴
=
=
−
=
=
+
⎡
⎤
⎛
⎞
∴
=
−
+
>
⎢
⎥
⎜
⎟
⎝
⎠
⎣
⎦
∫
∫
∫
Ch. 3: Forced Vibration of 1-DOF System
Ex.
Find the undamped response for the sinusoidal
pulse force shown using zero i.c.
3.5 Non-periodic Excitation
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
( )
( ) (
)
( )
( )
(
)
( ) (
)
( )
0
0
0
0
0
2
sin
sin
2
1
sin
is the system impulse response shifted by
and mirrored about the vertical axis.
If
0,
0 because of no overlap
0, 0
t
n
n
x t
F
h t
d
F
F
F
T
T
h
m
h t
t
t
F
h t
x t
t
τ
τ τ
π
π
τ
τ
τ
τ
ω τ
ω
τ
τ
τ
=
−
⎛
⎞
⎛
⎞
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
=
−
<
−
=
∴
=
<
∫
Ch. 3: Forced Vibration of 1-DOF System
3.5 Non-periodic Excitation
( ) (
)
(
)
( )
( ) (
)
(
)
(
)
(
)
( )
(
)
(
)
0
0
0
0
0
0
0
0
0
2
1
If 0
,
sin
sin
sin
sin
1
using the relation sin
sin
cos
cos
and some arrangements
2
sin
sin
, 0
where
1
n
n
t
t
n
n
n
t
T
F
h t
F
t
T
m
F
x t
F
h t
d
t
d
m
T
F
x t
t
r
t
t
T
k
r
π
τ
τ
τ
ω
τ
ω
π
τ
τ τ
τ
ω
τ τ
ω
α
β
α β
α β
ω
ω
⎛
⎞
< <
−
=
×
−
⎜
⎟
⎝
⎠
⎛
⎞
=
−
=
−
⎜
⎟
⎝
⎠
=
−
−
+
⎡
⎤
⎣
⎦
∴
=
−
< <
−
∫
∫
( )
( ) (
)
( ) (
)
(
) ( )
( )
(
)
[
]
(
)
(
)
{
}
0
0
2
0
0
0
0
0
0
0
0
2
, ,
If ,
sin
sin
sin
sin
,
1
superposition of the out-of-phase shifted sine trains
n
n
T
t
t
t T
n
n
r
k
m
T
t
T
x t
F
h t
d
F
h t
d
F t
h
d
F
x t
t
r
t
t T
r
t T
t
T
k
r
π
ω
ω
ω
ω
τ
τ τ
τ
τ τ
τ
τ τ
ω
ω
ω
ω
−
=
=
=
>
=
−
=
−
=
−
∴
=
−
−
−
−
−
>
⎡
⎤
⎣
⎦
−
∫
∫
∫
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