When the external force F(t) is periodic with period = 2/, it can be expanded in a Fourier series
tjbtjaa
tFj
i
j
i sincos2 11
0
2,... 1, 0,,cos2
0 jtdtjtFai
2,... 1, ,sin2
0 jtdtjtFbi
The equation of motion of the system can be expressed as
Using the superposition principle, the steady state solution is the sum of the steady state solutions of
tjbtjaa
kxxcxmj
i
j
i sincos2 11
0
2
0akxxcxm
tjakxxcxm j cos
tjbkxxcxm j sin
The solutions;
k
atxp
2
0
j
j
p tj
jrrj
katx
cos
21
/
2222
j
j
p tj
jrrj
kbtx
sin
21
/
2222
22
1
1
2tan
rj
jrj
The complete steady state solution;
12222
12222
0
sin
21
/
cos
21
/
2
j
j
j
j
j
j
p
tj
jrrj
kb
tj
jrrj
ka
k
atx
22
1
1
2tan
rj
jrj
In the study of vibrations of valves used in hydraulic control systems, the valve and its elastic stem are modeled as a damped spring-mass system, as shown in the next figure. In addition to the spring force and damping force, there is a fluid pressure force on the valve that changes with the amount of opening or closing of the valve. Find the steady-state response of the valve when the pressure in the chamber varies as indicated. Assume k=2500 N/m, c=10 N-s/m, and m=0.25 k.g
In some cases, the force acting on a system may be quite irregular and may be only determined only experimentally.
Using numerical integration procedure (trapezoidal rule)
N
i
iFN
a1
0
2...3,2,1 ,
2cos
2
1
jtj
FN
a iN
i
ij
...3,2,1 ,2
sin2
1
jtj
FN
b iN
i
ij
12222
12222
0
sin
21
/
cos
21
/
2
j
j
j
j
j
j
p
tj
jrrj
kb
tj
jrrj
ka
k
atx
Some of the methods that can be used to find the response of a system to an arbitrary excitation are; Representing the excitation by a Fourier integral
Method of convolution integral
Method of Laplace Transform
First approximating F(t) by a suitable interpolation model and then using a numerical procedure
Numerically integrating the equation of motion
A nonperiodic exciting force usually has a magnitude that varies with time; it acts for a specified period of time and then stop.
The simple form is the impulsive force
A force that has a large magnitude F and acts for a very short period of time t
From dynamics:
12 Impulse xmxmtF
tt
tFdtF
~
By designating the magnitude of the impulse;
A unit impulse (f) is defined as
1lim~
0
FdtFdtf
tt
tt
Response to an Impulse
We first consider the response of SDOF system to an impulse excitation
For an underdamped system, the solution of the equation of motion
is
0 kxxcxm
txx
txetx dd
nd
tn
sincos 000
nm
c
2
2
2
21
m
c
m
knd
m
kn
From the impulse-momentum,
and the initial conditions,
01~
impulse xmf
tm
etgtx d
d
tn
sin
mx
10 00 x
0for t 0 xx
Impulse Response Function
If the magnitude is F instead of 1, the initial velocity is F/m and the response
tFgtm
Fetx d
d
tn
sin
tFgtx
If the impulse F is applied at arbitrary time t =
Response to a General Forcing Conditions The arbitrary external force may be assumed to be made
up of a series of impulse of varying magnitude
tgFtx
tgFtx
dtgFtxt
0
dteF
mtx d
tt
d
n sin1
0
The Laplace transform method can be used to find the response of a system under any type of excitation, including the harmonic and periodic types. This method can be used for the efficient solution of linear differential equations, particularly those with constant coefficients. It permits the conversion of the differential equations into algebraic ones, which are easier to manipulate. The major advantages of the method are that it can treat discontinuous functions without any particular difficulty and it automatically takes into account the initial conditions.
The Laplace Transform of a function x(t) is defined as,
The integration is with respect to t, the transformation gives a function of s
Steps; Write the equation of motion of the system.
Transform each term of the equation, using known initial conditions.
dttxetxsx st
0
L
Steps;
Solve for the transformed response of the system
Obtain the desired solution by using inverse Laplace transformation.
To solve the forced vibration equation,
tdt
dxtx
tFkxxcxm
tdt
xdtx
2
2
00
xsxsdttdt
dxet
dt
dx st
L
0022
2
02
2
xsxsxsdttdt
xdet
dt
xd st
L
dttFetFsF st
0
L
tFtxktxctxm LLLL
0)0(2 xcmsxmsFsxkcsms
Ignoring the homogeneous solution of the differential equation;
kcsmssx
sFsZ 2
0000 xx
222 2
11
nnssmkcsmssF
sxsY
sFsYsx sFsYsxtx 11 LL
Transfer Function
Inverse Laplace transform
Considering the generals solution of the differential equation; 00 00 xxxx
02202222 21
2
2
2x
ssx
ss
s
ssm
sFsx
nnnn
n
nn
dteFm
tex
tex
tx
d
tt
d
d
t
d
d
t
n
nn
sin1
sinsin1
0
012/12
0
1L
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