SINE
xs(x)
SIgnals and NEtworks
1
Atomic Norm Minimization for Modal Analysis with Random Spatial Compression
Shuang Li, Dehui Yang and Michael B. Wakin
Electrical Engineering and Computer Science, Colorado School of Mines
Introduction
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Physical structures:
Mississippi River Bridge (2007)
Sampoong Department Store (1995)
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How to detect the damage that can be caused over
time by continuous use?
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Natural frequencies
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Mode shapes
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Damping ratios
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Uniform sampling
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Synchronous random sampling
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Asynchronous random sampling
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Random temporal compression
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Random spatial compression
Modal Expansion Theorem
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Second-order equations of motion for an N degree of
freedom linear system:
[M ]
{ ¨
x(t)
} + [C]{ ˙x(t)} + [K]{x(t)} = {f(t)}
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Modal expansion with K active modes:
{x(t)} = [Ψ]{q(t)} =
K
X
k=1
{ψ
k
}q
k
(t)
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Free vibration & no damping:
q
k
(t) = A
k
e
j2πf
kt
Problem Formulation
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Analytic signal:
{x(t)} =
K
X
k=1
{ψ
k
}A
k
e
j2πf
kt
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Taking Nyquist samples at
T =
{t
1
, t
2
,
· · · , t
M
} = {0, T
s
,
· · · , (M − 1)T
s
}.
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Data matrix:
[
X] = [x(t
1
), x(t
2
),
· · · , x(t
M
)]
=
K
X
k=1
A
k
{ψ
k
}a(f
k
)
>
∈ C
N
×M
with
a(f
k
) := [e
j2πf
kt
1, e
j2πf
kt
2,
· · · , e
j2πf
kt
M]
>
.
Randomized Spatial Compression
y
m
=
h[X](:, m), b
m
i
=
h[X]e
m
, b
m
i
=
h[X], b
m
e
H
m
i
1
≤ m ≤ M,
Atomic Norm Minimization
min
[ ˆ
X]
k[ ˆ
X]
k
A
s. t. y
m
=
h[ ˆ
X], b
m
e
H
m
i, 1 ≤ m ≤ M.
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Atomic set:
A = {ha(f)
>
:
khk
2
= 1
}
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Atomic norm:
k[X]k
A
= inf
{t > 0 : [X] ∈ t conv(A)}
= inf
(
X
k
c
k
: [
X] =
X
k
c
k
h
k
a(f
k
)
>
, c
k
≥ 0
)
.
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Dual polynomial:
Q(f) = Ya(f)
Theoretical Guarantee
Theorem 1
[Yang, 2016]
Suppose we observe the data matrix
[
X]
with the above random
spatial compression scheme. Assume that the random vectors
b
m
are
i.i.d samples from an distribution with the isotropic and
µ
-incoherent
properties. Assume that the signs
{ψ
k}(n)A
k|{ψ
k}(n)A
k|
are drawn i.i.d. from
the uniform distribution on the complex unit circle, and assume the
minimum separation
∆
f
= min
k
6=j
|(f
k
− f
j
)T
s
| ≥
M
4
−1
. Then there exists
a numerical constant
C
such that
M
≥ CµKN log
M KN
δ
log
2
M N
δ
is sufficient to guarantee that we can recover
[
X]
via ANM and
localize the frequencies with probability at least
1
− δ
.
ANM-based Strategy vs. SVD-based Strategy
m1 m2 m3 m4 m5 m6
k1 k2 k3 k4 k5 k6 k7
Figure 1: Undamped box car system with m1 = 1, m2 = 2, m3 = 3, m4 = 4, m5 = 5, m6 = 6 kg, and the stiffness values are k1 = k7 = 500, k2 = k6 = 150, k3 = k4 = k5 = 100 N/m.
0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency(Hz) jj Q (f ) jj2
Figure 2: Uniform sampling: frequency localization from dual polynomial Q(f) in the box cars system.
Note that the true mode shapes here are not orthogonal to each other. Therefore, it’s obvious to see the outperformance of ANM based algorithm from Fig. 4. Moreover, the MAC for AMN is (1, 1, 1, 1, 1, 1), while the MAC for SVD is (0.8860, 0.6663, 0.7646, 0.9557, 0.9629, 0.9889), which verifies that the SVD based algorithm fails to recover the true mode shapes when the mode shapes are not orthogonal. With some simple calculations, we can get the minimum separation ∆f = 0.0054. Theorem 3.1 indicates that we need M ≥ max{∆4
f + 1, 257} = max{741, 275} to get perfect recovery. However, this simulation does indicate
that our Theorem 3.1 is too strict and we do not need to use as many as 275 uniform samples to get perfect recovery.
For convenience, we will use random mode shapes to test our ANM based algorithms in the following experiments.
4.2
Asynchronous v.s. synchronous random sampling
In this experiment, we compare the performance of asynchronous and synchronous random sampling with respect to correlated mode shapes, which are shown in Fig. 5 (Only the first two mode shapes are correlated). The true frequencies are set as 2, 3 and 10 Hz. We use Nyquist rate to get M = 100 uniform samples from each sensor. However, only M0 samples are randomly observed from each sensor according to the asynchronous scheme or synchronous scheme. M0 ranges from 2 to 20. Note that M0 = |ΩS| in the synchronous random sampling. However, in the asynchronous random sampling, we use M0 to denote the average number of observed measurements from each sensor, i.e., M0N = |ΩA|. Other parameters are set same as in Section
4.1. 100 trials are performed in this experiment. Fig. 6 shows that when compared with synchronous
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Mass:
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Orthogonal: m
1= m
2= m
3= m
4= m
5= m
6= 1 kg.
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Non-orthogonal: m
1= 1, m
2= 2, m
3= 3, m
4= 4, m
5= 5, m
6= 6 kg.
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Stiffness:
k
1= k
5= 200, k
2= k
6= 150, k
3= 100, k
4= 50, k
7= 200 N/m.
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M = 150, N = 6.
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# of measurements:
SVD(M
× N), ANM(M).
ANM-based Strategy vs. SVD-based Strategy
0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency(Hz) jj Q (f ) jj2
Frequency localization with dual polynomial.
2 4 6 0.2 0.4 0.6 M od e Sha pe 1 2 4 6 0.2 0.4 M od e Sha pe 2 2 4 6 0.2 0.4 0.6 M od e Sha pe 3 2 4 6 0.2 0.4 0.6 M od e Sha pe 4 2 4 6 0.2 0.4 0.6 Node M od e Sha pe 5 2 4 6 0.2 0.4 0.6 0.8 Node M od e Sha pe 6 True ANM SVD
Orthogonal mode shapes.
0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency(Hz) Am pl itud e True Frequencies Estimated Frequencies
The estimated frequencies.
2 4 6 0.2 0.4 0.6 M od e Sha pe 1 2 4 6 0.2 0.4 0.6 M od e Sha pe 2 2 4 6 0.2 0.4 0.6 M od e Sha pe 3 2 4 6 0.2 0.4 0.6 M od e Sha pe 4 2 4 6 0.2 0.4 0.6 Node M od e Sha pe 5 2 4 6 0.2 0.4 0.6 0.8 Node M od e Sha pe 6 True ANM SVD
Non-orthogonal mode shapes.
M vs. K and N
Sparsity (K) N um ber o f m eas ur em en ts (M )Percentage of successful recovery
2 4 6 8 10 10 20 30 40 50 60 70 80 90 100 0 0.2 0.4 0.6 0.8 1
number of sensors is set as N = 5
Number of sensors (N) N um ber o f m eas ur em en ts (M )
Percentage of successful recovery
2 4 6 8 10 10 20 30 40 50 60 70 80 90 100 0 0.2 0.4 0.6 0.8 1
