MODAL ANALYSIS - the Dept. of Mechanical Engineering at - Vrije ...

3.1.3 Normal Equations
Many estimators used in modal analysis form the normal equations explicitly, i.e. they compute
Re( J J)
H
explicitly. Note that the real part of J J
H has to be taken because the coefficients are real.
Deriving the estimates directly from the Jacobian matrix leads to a better-conditioned problem.
However, forming the normal equations can result in a faster implementation, as will be the case
here too. The normal equations can be written as
⎡R1
⎢
0
⎢
⎢ �
⎢
T
⎢S1
⎣
S
0
R
2
T
2
�
�
�
S1
⎤ ⎧
⎪
S
⎥
2 ⎥ ⎪
� ⎥ ⋅ ⎨
⎥ ⎪
Tk
⎥ ⎪
⎦ ⎩⎪
NoN
i
∑
k=
1
�
1
2
N N
o
i
⎫
⎪
⎪
⎬ ≈ 0
⎪
⎪
⎪⎭
H
H
H
with R k = Re( X k X k ) , S k = Re( X k Yk
) , and T k = Re( Yk
Yk
) . The entries of these matrices equal
⎛
R ⎜
k ( r,
s)
= Re
⎜
⎝
⎛
T ⎜
k ( r,
s)
= Re
⎜
⎝
N
f
∑
f = 1
⎛
S
⎜
k ( r,
s)
= − Re
⎜
⎝
N
N
f
∑
f
∑
f = 1
W
f = 1
k
W
k
( ω )
k
f
f
f
⋅Ω
k
H
r−1
W ( ω ) H ( ω )
2
( ω )
2
k
f
( ω ) Ω
f
f
⋅Ω
s−1
H ( ω ) ⋅Ω
2
H
r−1
H
r−1
⎞
( ω ⎟ f )
⎟
⎠
( ω ) Ω
( ω ) Ω
f
f
s−1
s−1
⎞
( ω ⎟ f )
⎟
⎠
⎞
( ω ⎟ f )
⎟
⎠
If a discrete time-domain model is used, i.e. Ω j ( ω f ) = exp( −iω
f Ts ⋅ j)
, and if the frequencies are
uniformly distributed (i.e. ω f = f ⋅ Δω
, f N f , , 1 � = , with summations can be rewritten as
Δ ω = 2π
NTs
), then, the above
⎛
R ⎜
k ( r,
s)
= Re
⎜
⎝
⎛
T ⎜
k ( r,
s)
= Re
⎜
⎝
N
f
∑
f = 1
⎛
S
⎜
k ( r,
s)
= − Re
⎜
⎝
N
N
f
∑
f
∑
f = 1
W
f = 1
k
W
k
( ω )
k
f
f
f
⋅ e
k
i2π
( r−s
) f N
W ( ω ) H ( ω )
2
( ω )
2
k
f
f
⋅ e
⎞
⎟
⎠
H ( ω ) ⋅e
2
i2π
( r−s
) f N
i2π
( r−s
) f N
⎞
⎟
⎠
⎞
⎟
⎠
One can readily verify that the above matrices have a Toeplitz structure and that their entries can be
time-efficiently computed with the Fast Fourier Transform (FFT) algorithm.
3.1.4 Reduced Normal Equations
Although the number of rows of the normal matrix in (29) is much smaller than the number of rows
of the Jacobian matrix (28), its size is still quite huge (i.e. (n+1)(No Ni +1) rows and columns). As
we are mainly interested in a fast and stable method to construct a stabilization chart (see next
section), only the denominator coefficients (i.e. the poles) are in fact required. Elimination of the
numerator coefficients
yields
⎡
⎢
⎣
k
−1
= −R
⋅S
⋅
NoN
i
∑
k=
1
k
k
T −1
⎤
Tk − Sk
⋅ R k ⋅ Sk
⎥ ⋅ ≈ 0
⎦
(29)
(30)
(31)
(32)
(33)