Response of linear SDOF to arbitrary
excitation
Objective: Learn how to find the response of
a linear SDOF system to a given input
(excitation)
Preliminary definitions:
Response = natural response + forced
response
Natural response: solution of equation of
motion of the system when the excitation is
zero. The expression for natural response
contains constants.
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Forced response: any solution of equation of
motion of the system for non zero excitation.
If the natural response tends to zero when
time tends to infinity and the limit of the
forced response as time goes to infinity
exists and is bounded (not infinite), then the
limit is called steady state response.
Transient response: Process of going from
initial state to steady state.
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Steady
Transient response
state
Transient response is due to both the
application of the force and the non zero
initial conditions
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Outline of this chapter
1. Impulse response function
2. Response to arbitrary excitation
3. Shock spectrum
4. Numerical calculation of response
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1. Impulse response function
Impulse function: idealization of shortduration force applied suddenly
Force
Force
Idealization
Fˆ (t )
Area= F̂
time
time
 (t ) : unit impulse applied at t  0
F̂ : impulse of force  force  duration
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Impulse response
k
c
m
δ(t)
h(t)
Find response of single degree of freedom
system to a unit impulse:
mx(t )  cx (t )  kx(t )   (t ),
x (0 )  0,
.
x (0 )  0
Case 1: <1
 1 n t
e
sin( d t ) for t  0

x (t )  h(t )   m d
 0
for t  0
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If impulse was applied at time, τ, then
 1 n (t  )
e
sin[ d (t   )] for t  

x(t )  h(t   )   m d

0
for t  
Impulse responses of two systems with natural
frequency 6.28 rad/sec and damping ratios 0.1
and 0.8.
.
Impulse reponses two systems
displacement
Slope is 1/m here
time (sec)
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Observations:
 Transient response dies out faster when
damping increases.
 Displacement overshoot decreases with
damping.
 Maximum displacement does not occur
exactly at time equal to one fourth of a
period, unless damping is zero.
 Slope just after impulse has been
applied (i.e. at t=0+) is 1/m.
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Case 2: Overdamped system, >1
1
1
 nt  2 1
 nt  2 1


t
n
h( t ) 
e
[e
e
]
2mn  2  1
Case 3: Critically damped system, =1
h( t ) 
1  n t
te
m
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2. Response to arbitrary excitation
First, assume that system is at rest at t=0.
Idea: Use superposition principle. Split
excitation to sum of impulses. Find
response to each impulse and sum up the
responses.
t
x(t )   F ( )h(t   )d
0
Equivalent equation for response:
t
x (t )   F (t   )h( )d
0
The above are called convolution integrals.
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If system is not at rest at t=0, then the
response is the free vibration response due
the non zero initial conditions plus the above
convolution integral
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Step response
Unit step function
1
t
Consider underdamped systems only.
Response to unit step:
x (t )  U (t ) 
1
mn 2
1
[1 
1 2
e  n t cos(d t   )] 
1
1
[1 
e  n t cos(d t   )]
k
1 2
where :   tan -1 (

1
2
)
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Overshoot
Steady state (quasi static response)
Time to peak, tp=half damped
natural period
Slope is zero here
Observations:

Response oscillates about quasi static
response with frequency, ωd.

Response converges to quasi static
response as time tends to infinity. This
response is the steady state.
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
At time t=0, the velocity is zero. (note
difference with slope at time zero of
impulse response)

Time to peak is equal to half period.
(note difference with impulse response)
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3. Shock spectrum
Shock: Sudden application of force resulting
in transient response.
Shock spectrum: maximum response vs.
normalized frequency
Usually normalized frequency is the ratio of
the shock duration divided by the natural
period of the system.
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4. Numerical simulation of response
It is often difficult to calculate the
convolution integral or solve the differential
equation of motion. We could use
numerical simulation in this case. There are
two approaches for numerical simulation:
1)
Solve the differential equation
of motion
2)
Calculate numerically the
convolution integral
1)
Numerical solution of differential
equation of motion:
Most computer programs can only solve first
order differential equations. We can convert
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the equation of motion to a system of two
first order differential equations as follows.
Start with the original equation:
mx  cx  kx  F (t ) 
F (t )
x  2n x  n 2 x 
m
Let
x  x2 , and x  x1.
Then the above equation
of motion becomes:
x1 (t )  x2 (t )
x 2 (t )  2n x2 (t )   n 2 x1 (t ) 
F (t )
m
We can solve the above equations
numerically using Mathematica, Mathcad or
Matlab.
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