3.5 Random Vibrations
Random Vibrations
• So far our excitations have been harmonic,
periodic, or at least known in advance
• These are examples of deterministic excitations,
i.e., known in advance for all time
– That is given t we can predict the value of F(t) exactly
• Responses are deterministic as well
• Many physical excitations are nondeterministic, or
random, i.e., can’t write explicit time descriptions
– Rockets
– Earthquakes
– Aerodynamic forces
– Rough roads and seas
• Stationary signals are those whose statistical
properties do not vary over time
• Functions are described in terms of probabilities
– Mean values*
– Standard deviations
• Random outputs related to random input via
system transfer function
*ie given t we do not know x(t) exactly, but rather
we only know statistical properties of the response
such as the average value
• The responses x(t) are also nondeterministic
1
Autocorrelation function and
power spectral density
HARMONIC
∫
T
0
∫
∞
−∞
RANDOM
Signal
Arms
x(t)
time
x(t)
time
-A
T
x(t) x(t + τ )dτ
Rxx (τ )e
− jωτ
A2rms
Autocorrelation
A2/2
τ time
shift
Rxx(τ)
The Power Spectral Density describes the power in a signal
as a function of frequency and is the Fourier transform of the
autocorrelation function.
1
Sxx (ω ) =
2π
Examples of signals
T
A
The autocorrelation function describes how a signal is
changing in time or how correlated the signal is at two
different points in time.
1
Rxx (t) = lim
T →∞ T
2
Rxx(τ)
τ time
shift
-A2/2
Power Spectral Density
dτ
Sxx(ω)
Sxx(ω)
Frequency
(Hz)
FFT is a new way for calculating Fourier Transforms
Frequency
(Hz)
1/T
3
4
Expected Value
(or ensemble average)
More Definitions
Average:
T
T
1
x = lim ∫ x(t)dt
T →∞ T
0
Mean-square:
1 2
x (t)dt
T →∞ T ∫
0
rms:
(3.48)
The expected value is also given by
∞
T
xrms
(3.63)
The Probability Density Function, p(x), is the probability
that x lies in a given interval (e.g. Gaussian Distribution)
T
x 2 = lim
x(t)
dt = x
The expected value = E[x(t)] = Tlim
→∞ ∫ T
0
(3.47)
1
= x = lim ∫ x 2 (t)dt
T →∞ T
0
2
E[x] =
(3.49)
∫ xp(x)dx
(3.64)
−∞
5
Recall the Basic Relationships for
Transforms:
6
What can you predict?
The response of SDOF with f(t) as input:
Recall for SDOF:
transfer function : G(s) =
Deterministic Input:
1
2
ms + cs + k
X(s) = G(s)F(s)
Random Input:
Sxx (ω ) = H (ω ) S ff (ω )
2
t
frequency response function : G ( jω ) = H (ω ) =
unit impulse response function : h ( t ) =
L[h(t)] =
x(t) = ∫ h(t − τ ) f (τ )dτ
1
k − mω 2 + cωj
1
mω d
0
∞
E[x 2 ] =
∫
H (ω ) S ff (ω )dω
2
−∞
In a Lab, the PSD function of a random input and the output can be measured simply
in one experiment. So the FRF can be computed as their ratio by a single test, instead
of performing several tests at various constant frequencies.
e −ζω t sin ω d t
n
1
= G(s)
ms + cs + k
2
And the Fourier Transform of h(t) is H(ω)
7
Here we get an exact
time record of the output
given an exact record of the
input.
Here we get an expected
value of the output given
a statistical record of the
input.
8
Example 3.5.2
Example 3.5.1 PSD Calculation
Consider mx + cx + kx = F (t ), where the PSD of F (t ) is constant S 0
Consider the system of Example 3.5.1 and compute:
The corresponding frequency response function is:
H (ω ) =
1
k − mω + cω j
⇒ H (ω ) =
2
2
1
k − mω + cω j
1
=
2 2
( k − mω ) + c 2ω 2
2
=
1
1
i
k − mω + cω j k − mω 2 − cω j
2
= S0
From equation (3.62) the PSD of the response becomes:
Sxx = H (ω ) S ff =
2
2
∞
1
−∞
k − mω n2 + cω j
E ⎡⎣ x 2 ⎤⎦ = S0 ∫
(2.59)
2
Mean Square Calculation
πm
kcm
=
dω
π S0
kc
Here the evaluation of the integral is from a tabulated value
See equation (3.70).
S
(k − mω ) + (cω )2
0
2 2
9
10
Using the convolution equation as a tool,
compute the maximum value of the response
Section 3.6 Shock Spectrum
Arbitrary forms of shock are probable
(earthquakes, …)
The spectrum of a given shock is a plot of the
maximum response quantity (x) against the
ratio of the forcing characteristic (such as
rise time) to the natural period.
Maximum response gives maximum stress.
Recall the impulse response function undamped system:
h(t − τ ) =
1
sin ω n (t − τ )
mω n
(3.73)
⇒
x(t)max =
1
mω n
∫
t
0
F(τ )sin (ω n (t − τ ))dτ
(3.74)
max
Such integrals usually have to be computed numerically
x(t) =
∫
t
0
F(τ )h(t − τ )dτ
(3.71)
12
Example 3.6.1 Compute the response spectrum
for gradual application of a constant force F . Assume zero
initial conditions
mx(t ) + kx(t ) = F (t )
0
t1=infinity, means static loading
F0
t
t1
0
⎧
⎪
F2 (t) = ⎨ t − t1
⎪−( t )F0
1
⎩
x1 (t) =
ω n t F0τ
k
∫
0
t1
sin ω n (t − τ )dτ =
F0 ⎛ t sin ω nt ⎞
−
k ⎜⎝ t1
ω nt1 ⎟⎠
0 < t < t1
(3.75)
F(t) = F1 (t) + F2 (t)
t
F1 (t) = F0
t1
time shift and negative, like half sine problem
F(t)
Split the solution into two parts
and use the convolution integral
For x2 apply
x2 (t) = −
time shift t1
0 < t < t1
t ≥ t1
x(t) = x1 (t) + x2 (t) =
F0 ⎛ t − t1 sin ω n (t − t1 ) ⎞
−
⎟⎠ , t ≥ t1
k ⎜⎝ t1
ω nt1
(3.76)
F0 ⎛ t sin ωnt ⎞ F0 ⎛ t − t1 sin ωn (t − t1 ) ⎞
−
⎜ −
⎟− ⎜
⎟Φ(t − t1 )
k ⎝ t1 ωn t1 ⎠ k ⎝ t1
ω nt 1
⎠
The characteristic time of the input
(3.77)
13
Next find the maximum value
of this response
14
Solve for t at xmax, denoted tp
− cos ω nt + cos ω n (t − t1 ) t =t = 0
To get max response, differentiate x(t).
p
In the case of a harmonic input (Chapter 2) we computed this
by looking at the coefficient of the steady state response,
giving rise to the Magnitude plots of figures 2.7, 2.8, 2.13.
Need to look at two cases 1) t < t1 and 2) t > t1
For case 2) solve: (what about case 1? Its max is Xstatic)
cos ω nt p = cos ω nt p cos ω nt1 + sin ω nt p sin ω nt1
⎛ 1 − cos ω n t1 ⎞
⇒ ω n t p = tan −1 ⎜
⎝ sin ω nt1 ⎟⎠
sin 2 ω nt1 + (1 − cos ω nt1 )2
= 2 (1 − cos ω nt1 )
⎤
d ⎡ F0
(ω nt1 − sin ω nt + sin ω n (t − t1 ))⎥ = 0 ⇒
⎢
dt ⎣ kω nt1
⎦
1− cos ωn t1
ωn t p
sin ωnt1
15
16
From the triangle:
1
(1− cos ωnt1 )
2
−sin ωnt1
cos ωn t p =
2(1− cos ωnt1 )
sin ωnt p = −
Response Spectrum
Minus sq root taken as + gives a negative magnitude
Substitute into x(tp) to get nondimensional Xmax :
xmax k
1
= 1+
2(1− cos ωnt1 )
F0
ωn t1
•Indicates how normalized max output
changes as the input pulse width increases.
•Very much like a magnitude plot.
•Shows very small t1 can increase
the response significantly: impact, rather than
smooth force application
•The larger the rise time, the smaller the peaks
•The maximum displacement is minimized if
rise time is a multiple of natural period
•Design by MiniMax idea
1st term is static, 2nd is dynamic. Plot versus:
t 1 ω nt 1
=
T 2π
x
k
X = max
F0
Input characteristic time
System period
18
17
Comparison between impulse and
harmonic inputs
Impulse Input
Transient Output
Max amplitude versus
normalized pulse
“frequency”
Review of The Procedure for Shock Spectrum
Harmonic Input
Harmonic Output
Max amplitude versus
normalized driving
frequency
1.
2.
3.
4.
5.
Find x(t) using convolution integral
Compute its time derivative
Set it equal to zero
Find the corresponding time
Evaluate the max possible value of x
(be careful about points where the
function does not have derivative!!)
6. Plot it for different input shocks
19
20
3.7 Measurement via Transfer Functions
• Apply a sinusoidal input and measure
the response
• Do this at small frequency steps
• The ratio of the Laplace transform of
these to signals then gives and
experiment transfer function of the
system
Several different signals can be
measured and these are named
receptance:
mobility:
inertance:
21
The magnitude of the compliance transfer
function yields information about the systems
parameters
H ( jω ) =
H( j
1
(k − mω )2 + (cω )2
k
1
)=
m
cω n
H (0) =
1
k
(3.90)
(3.89)
H ( jω )
(3.91)
23
X(s)
1
=
2
F(s) ms + cs + k
sX(s)
s
=
2
F(s) ms + cs + k
s2
s 2 X(s)
=
ms 2 + cs + k
F(s)
(3.86)
(3.87)
(3.87)
22