[PPT] - On a rank factorisation problem arising in vibration analysis A. PowerPoint Presentation

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On a rank factorisation problem arising in vibration analysis

A. Quadrat

Inria Paris, Ouragan, IMJ–PRG, Sorbonne Universit´ e In collaboration with Elisa Hubert (Univ. Lyon), Axel Barrau (Safran Tech), Yacine Bouzidi (Inria Lille), Roudy Dagher (Inria Chile)

Journ´ ees Nationales de Calcul Formel Luminy, 02/03/2020

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Motivation: Gear box vibration analysis

Elisa Hubert CIFRE PhD’s thesis⋆ (Safran Tech) Gearbox health monitoring Up to date surveillance Reduction of damage frequency Optimization of maintenance intervention

Say VRAOUMMM*!

⋆ D´

emodulation d’amplitude et de phase de signaux multi-porteuses : Application aux signaux vibratoires d’engrenage, Univ. Lyon, 2019

A. Quadrat

On a rank factorisation problem arising in vibration analysis

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Demodulation problem

K: field (e.g. Q, R, C),

M ∈ Kn×m, D1, . . . , Dr ∈ Kn×n Jn =   

. . . 1 . . . 1 . . . 1 . . .

   ∈ Kn×n

Definition: M ∈ Cn×m is centrohermitian if Jn M Jm = M

CHn,m(K): set of all the centrohermitian matrices of Kn×m · Frob: Frobenius norm, i.e. AFrob =

trace(A⋆A)
The demodulation problem: M ∈ CHn,m(C)

Find u ∈ CHn,1(C), vi ∈ CH1,m(C) : r

i=1 Di u vi = M

arg minu∈CHn,1(C), vi∈CH1,m(C) r

i=1 Di u vi − MFrob

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

A. Quadrat

On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

Gearing vibration produced by 2 wheels W1 and W2 (Capdessus 92)

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

Gearing vibration produced by the teeth of 2 wheels W1 and W2

Ni denotes the number of teeth of the wheel Wi

fe = N1 f1 = N2 f2

s static transmission error producing the vibration (Mark 78)

s(t) = se(t) (F + s0(t) + s1(t) + s2(t)) se and s0: periodic signals of frequency fe si: periodic signal of frequency fi, i = 1, 2 ⇒ Amplitude modulation (Capdessus 92) ⇒ s(t) = sc(t) sm(t) sm: modulation fm = T −1

m

(low-frequency) sc: carrier fc = T −1

c

= N fm, N ∈ N≥0 (high-frequency)

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

p : R → R: T-periodic integrable function, a ∈ R:

p0(t) = p(t), a ≤ t < a + T, 0, else, ⇒ p(t) = p0(t) ⋆

j∈Z

δ(t − j T) =

j∈Z

p0(t − j T)

Fourier transform & Poisson’s formula yield

ˆ p(ν) = ˆ p0(ν) 1 T

j∈Z

δ

ν − j

T

=
j∈Z

1 T ˆ p0 j T

δ
ν − j

T

Inverse Fourier transform p(t) =

j∈Z cj(p) e

2πijt T

(Fourier series) cj(p) = 1 T ˆ p0 j T

= 1

T a+T

a

p(t) e− 2πijt

T dt

cj(p) = c−j(p), j ∈ Z. p is real ⇒ c−j(p) = cj(p), j ∈ Z
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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

s(t) = sc(t) sm(t),

T −1

c

= N T −1

m

ˆ sm(ν) =

j∈Z(cm)j δ

ν −

j Tm

,

(cm)j = 1 Tm ˆ sm0 j Tm

,

ˆ sc(ν) =

k∈Z(cc)k δ

ν − k

Tc

,

(cc)k = 1 Tc ˆ sc0 k Tc

,

ˆ s(ν) =

l∈Z(cs)l δ

ν −

l Tm

,

(cs)l = 1 Tm ˆ s0 l Tm

,

ˆ s(ν) = (ˆ sm ⋆ ˆ sc)(ν) =

j∈Z
k∈Z

(cm)j (cc)k δ

ν −

j Tm − k Tc

=
j∈Z
k∈Z

(cm)j (cc)k δ

ν − j + k N

Tm

⇒ (cs)l =
{(j,k)∈Z2 | j+k N=l}

(cm)j (cc)k

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

          . . . (cs)−1 (cs)0 (cs)1 . . .           =           . . . . . . . . . . . . . . . . . . . . . . . . (cm)2N−1 (cm)N−1 (cm)−1 (cm)−N−1 (cm)−2N−1 . . . . . . (cm)2N (cm)N (cm)0 (cm)−N (cm)−2N . . . . . . (cm)2N+1 (cm)N+1 (cm)1 (cm)−N+1 (cm)−2N+1 . . . . . . . . . . . . . . . . . . . . . . . .                     . . . (cc)−1 (cc)0 (cc)1 . . .           .

A. Quadrat

On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

Hypotheses on the signals

sc: finite number of non-zero harmonics

ˆ sc(ν) =

0≤|k|≤r

(cc)k δ

ν − k

Tc

⇒ (cs)l =
0≤|k|≤r

(cm)l−k N (cc)k

sm does not induce overlaps in ˆ

s = ˆ sm ⋆ ˆ sc ∀ k ∈ Z, k N Tm + |j| Tm < (k + 1) N Tm − |j| Tm ⇔ 2 |j| < N

k N

Tm

(k + 1) N

Tm

k N

Tm + |j| Tm

(k + 1) N

Tm − |j| Tm

(cm)j = 0 for all j / ∈ {j ∈ Z | 2 |j| < N} = {−p, . . . , p}, where p = N − 1 2 , if N odd, N 2 − 1, if N even

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

|l| ≤ t := p + r N

cs :=                 (cs)−t . . . (cs)−1 (cs)0 (cs)1 . . . (cs)t                 , cc :=                 (cc)−r . . . (cc)−1 (cc)0 (cc)1 . . . (cc)r                 , cm :=                 (cm)−p . . . (cm)−1 (cm)0 (cm)1 . . . (cm)p                

N odd : A =        cm . . . cm ... . . . . . . ... ... . . . cm        = cm ⊗ I2 r+1 cs = A cc

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On a rank factorisation problem arising in vibration analysis

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Gear box vibration analysis

The vectorization satisfies

vec(U V W ) =

W T ⊗ U
vec(V )

U = I2 r+1, W = cmT ∈ C1×(2 p+1), V = cc ∈ C(2 r+1)×1, ⇒ cc cmT = vec−1(cs) c−j(s•) = cj(s•) ⇒ cc, cm, vec−1(cs) centrohermitian

Amplitude demodulation problem: Ms ∈ CH(2 r+1),(2 p+1)(C).

Find cc ∈ CH(2 r+1),1(C) and cc ∈ CH1,(2 p+1)(C) such that cc cmT = Ms

Phase and amplitude demodulation problem:

cc cm1

T + D cc cm2 T = Ms,

D := 2 π i fc diag(−r, . . . , r)

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On a rank factorisation problem arising in vibration analysis

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The rank factorization problem

The rank factorization problem: Determine u ∈ CHn,1(C) and

v1, . . . , vr ∈ CH1,m(C) − if they exist − such that M =

r

i=1

Di u vi (bilinear polynomial system)

A system-theoretical approach:

A(u) := (D1 u . . . Dr u) ∈ Kn×r, v := (vT

1

. . . vT

r )T ∈ Kr×m

⇒ A(u) v = M

Necessary condition: rankK(M) ≤ r
If rankK(M) = r then v has full row rank since

λ v =

r

i=1

λi vi = 0, λk = 0 ⇒ M =

1≤i=k≤r
Di − λi λ−1

k

Dk

u
vi

∀ λ ∈ K \ {0} : (u, v) → (λ u, λ−1 v)

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On a rank factorisation problem arising in vibration analysis

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The rank factorization problem

(D1 u . . . Dr u)

A(u)

v = M, l := rankK(M) ≤ r

L ∈ Kp×n: full row rank matrix defining a basis of

kerK(.M) = {λ ∈ K1×n | λ M = 0}, p := dimK(kerK(.M)) A(u) v = M ⇒ L A(u) v = L M = 0 ⇒ L A(u) = 0 (v full row rank) ⇔      L D1 u = 0, . . . L Dr u = 0, (V) ⇔ u = Z ψ, ∀ ψ ∈ Kd×1 d = dimK(V) & Z ∈ Kn×d full column rank defining a basis of V ⇒ i = 1, . . . , r, L Di Z = 0

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On a rank factorisation problem arising in vibration analysis

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The rank factorization problem

X ∈ Kn×l: full column rank matrix defining a basis of imK(M.)

⇒ ∃! Y ∈ Kl×m full row rank : M = X Y

kerK(.M) = imK(.L) ⇒ kerK(L.) = imK(M.) = imK(X.)

i = 1, . . . , r, L Di Z = 0 ⇒ ∃ Wi ∈ Kl×d : Di Z = X Wi A(Z ψ) = (D1 Z ψ . . . Dr Z ψ) = X (W1 ψ . . . Wr ψ)

Let u = Z ψ and B(ψ) := (W1 ψ . . . Wr ψ) ∈ Kl×r

A(Z ψ) v = X B(ψ) v = M = X Y ⇒ B(ψ) v = Y (X full colum rank)

Example: If l := rankK(M) = r, then B(ψ) ∈ Kr×r
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On a rank factorisation problem arising in vibration analysis

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The rank factorization problem

Theorem: Let S ∈ Kr×s and y ∈ Kr×t. Let the rows of

T ∈ Kq×r form a basis of kerK(.S), q = dimK(kerK(.S)). Then, S x = y admits a solution x ∈ Ks×t iff y satisfies the compatibility condition T y = 0 All the solutions x ∈ Ks×t of S x = y are then given by ∀ z ∈ Ku×t, x = E y + C z, where E ∈ Ks×r is a generalized inverse of S, i.e., S E S = S, and C ∈ Ks×u is a matrix whose columns form a basis of kerK(S.) := {η ∈ Ks×1 | S η = 0}

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On a rank factorisation problem arising in vibration analysis

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The rank factorization problem

B(ψ) := (W1 ψ . . . Wr ψ) ∈ Kl×r, B(ψ) v = Y , kerK(.Y ) = 0

Corollary: The problem is solvable iff1

∃ ψ ∈ Kd×1 \ {0} : rankK(B(ψ)) = l ⇔ ∃ ψ ∈ Kd×1 \ {0} : B(ψ) admits a right inverse E(ψ) If so, then the solutions of the problem are defined by u = Z ψ, ∀ z ∈ K(r−l)×m, v = E(ψ) Y + C(ψ) z where C(ψ) ∈ Kr×(r−l) defines a basis of kerK(B(ψ).). Finally, if l := rankK(M) = r and det(B(ψ)) = 0, then u = Z ψ, v = B(ψ)−1 Y

1Hubert, Barrau, Bouzidi, Dagher, Q., On a rank factorisation problem

arising in gearbox vibration analysis, IFAC World Congress, 2020

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On a rank factorisation problem arising in vibration analysis

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Example: amplitude demodulation

If d = 1, then ψ ∈ K and

B(ψ) := (W1 ψ . . . Wr ψ) = (W1 . . . Wr) ψ Pb is solvable iff W := (W1 . . . Wr) ∈ Kl×r admits a right inverse.

Example: r = 1 and D1 = In. Then, W1 = 0

⇒ Pb solvable is always solvable (SVD2)

2Hubert, Barrau, El Badaoui, New multi-carrier demodulation method

applied to gearbox vibration analysis, International Conference on Acoustics, Speech, and Signal Processing, 2018

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On a rank factorisation problem arising in vibration analysis

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Defect of exactness

A(u) v = M ⇒ L A(u) = 0 ⇒ u = Z ψ, ψ ∈ Kd×1

⇒ solvability of A(Z ψ) v = M?

By construction, u = Z ψ satisfies L A(Z ψ) = 0 for all ψ ∈ Kd×1

⇒ ∀ ψ ∈ Kd×1, kerK(.M) = imK(.L) ⊆ kerK(.A(Z ψ))

A(Z ψ) v = M is solvable iff L′ M = 0, where

kerK(.A(Z ψ)) = imK(.L′) i.e. iff kerK(.A(Z ψ)) = imK(.L′) ⊆ kerK(.M) Hence, the problem is solvable iff ∃ ψ ∈ Kd×1 : kerK(.A(Z ψ))/ kerK(.M) = 0

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On a rank factorisation problem arising in vibration analysis

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Example: phase and amplitude demodulation

∃ ψ ∈ Kd×1 : kerK(.A(Z ψ))/ kerK(.M) = 0 ⇔ dimK(kerK(A(Z ψ).)) = r − rankK(M) = r − l

Case l = rankK(M) = r = 2 and D1 = In
L u = 0,

L D2 u = 0, ⇒

u ∈ kerK(L.),

dimK(kerK(L.)) = rankK(M) = 2 ⇒ d ≤ 2 kerK(A(u).) = {(α1 α2)T ∈ K2×1 | u α1 + D2 u α2 = 0}

Corollary: Pb is solvable iff

∃ ψ ∈ Kd×1 \ {0} : u = Z ψ is not an eigenvalue of D2

Example: D2 = diag(d1, . . . , dn), di = dj, pb solvable iff

∃ ψ ∈ Kd×1 \ {0} : u = Z ψ / ∈ K ei, i = 1, . . . , n, where {ei}i=1,...,n is the standard basis of Kn×1

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On a rank factorisation problem arising in vibration analysis

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Centrohermitian matrices

Definition: A matrix Q ∈ Kn×m is J-real if Jn Q = Q

Q2t = 1 √ 2

It

i It Jt −i Jt

, Q2t+1 =

1 √ 2    It i It √ 2 Jt −i Jt    (unitary)

Theorem (Lee 80): T ∈ Kn×n, U ∈ Km×m non-singular J-real
matrices. Then,

ϕ : Kn×m − → Kn×m M − → T −1 M U bijectively maps CHn,m(K) onto Rn×m: ∀ M ∈ CHn,m(K), T −1 M U ∈ Rn×m If m = n, then ϕ isomorphically maps the ring CHn(K) onto Rn×n M =

r

i=1

Di u vi ⇔ ϕ(M) =

r

i=1

ϕ(Di) uϕ vi ϕ, uϕ ∈ Rn×1, vi ϕ ∈ R1×m

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On a rank factorisation problem arising in vibration analysis

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Example

M =    9 + 18 i −225 9 + 198 i 9 − 198 i −225 9 − 18 i    , rankK(M) = 2, D1 = I3, D2 =    −i i    Q3 = 1 √ 2    1 i √ 2 1 −i    , Q−1

3

= Q⋆

3

Mϕ := ϕ(M) = Q⋆

3 M Q3 =

   18 −225 √ 2 180 216    ∈ R3×3

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On a rank factorisation problem arising in vibration analysis

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Example

D2ϕ := ϕ(D2) = Q⋆

3 D2 Q3 =

   −1 1    ∈ R3×3 Lϕ = (0 1 0), dϕ = 2, Zϕ =    1 1    , Xϕ =    18 180 216    , Yϕ =   1 −5 4 √ 2 1   W1ϕ =    1 216 − 1 2160 1 180    , W2ϕ =    − 1 216 1 180 1 2160   

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On a rank factorisation problem arising in vibration analysis

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Example

Bϕ(ψ) = (W1ϕ ψ W1ϕ ψ), det(Bϕ(ψ)) = c(ψ2

1 + ψ2 2)

∀ ψ = (ψ1 ψ2)T ∈ R2×1 \ {0} : uϕ = Z ψ = (ψ2 ψ1)T vϕ = B−1

ϕ Yϕ

= (ψ2

1 + ψ2 2)−1

18 (12 ψ1 + ψ2)

−225 √ 2 ψ2 180 ψ2 18 (ψ1 − 12 ψ2) −225 √ 2 ψ1 180 ψ1

u = Q3 uϕ =

√ 2 2 (ψ2 + i ψ1 ψ2 − i ψ1)T v = vϕ Q⋆

3 =

√ 2 (ψ2

1 + ψ2 2)−1

9 (12 ψ1 + ψ2) − 90 i ψ2

−225 ψ2 9 (12 ψ1 + ψ2) + 90 i ψ2 9 (ψ1 − 12 ψ2) − 90 i ψ1 −225 ψ1 9 (ψ1 − 12 ψ2) + 90 i ψ1

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On a rank factorisation problem arising in vibration analysis

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Centrohermitian solutions

M, D1, . . . , Dr: centrohermitian, v = (vT

1

. . . vr)T S = {(u, v1, . . . , vr) ∈ CHn,1(C) × CH1,m(C)r | A(u) v = M} A(u) = (D1 u . . . Dr u) = (Jn D1 Jn u . . . Jn Dr Jn u) = Jn A(Jn u) A(u) v = M ⇔ A(u) v = M = Jn M Jm ⇔ Jn A(u) v Jm = A(Jn u) (v Jm) = M ⇒ T : S − → S (u, v) − → (Jn u, v Jm) T 2 = idS ⇒ ∃ R ∈ Kd×d : Z = Jn Z R, R R = R R = Id R: coninvolutory matrix

u = Z ψ. Then

Jn u = u ⇔ Z R ψ = Z ψ ⇔ R ψ = ψ

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On a rank factorisation problem arising in vibration analysis

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Centrohermitian solutions

R := {ψ ∈ Kd×1 | R ψ = ψ},

R R = R R = Id

ψ = ψr + i ψi,

ψr, ψi ∈ Rd×1

R = Rr + i Ri,

Rr, Ri ∈ Rd×d, R2

r + R2 i = Id,

Rr Ri = Ri Rr H =

Rr

Ri Ri −Rr

∈ R2d×2d,

H2 = I2d H: involutory matrix, 1 ∈ spec(H) ⊆ {−1, 1} ψ ∈ R ⇔ H

ψr

ψi

=
ψr

ψi

⇒ R = 0
kerK((H − I2d).) = imR(G), G = (G T

1

G T

2 )T ∈ R2d×e

ψr

ψi

= G γ,

∀ γ ∈ Re×1 ⇒ R = (G1 + i G2)

Z ′

Re×1

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On a rank factorisation problem arising in vibration analysis

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Centrohermitian solutions

∀ ψ ∈ Z ′ Re×1: u = Z ψ centrohermitian
Corollary: Pb solvable iff there exists γ ∈ Re×1 \ {0} such that

B

Z ′ γ
= (W1 Z ′ γ . . . Wr Z ′ γ)

admits a right inverse. If so, then    uJ = (Z Z ′) γ, vJ = v + v Jm 2 , is a centrohermitian solution, where v = E(γ) Y + C(γ) z, z ∈ K(r−l)×1 E(γ) right inverse of B(Z ′ γ) C(γ) ∈ Kr×(r−l) is a full column rank matrix such that kerK(K(γ).) = imK(C(γ).)

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On a rank factorisation problem arising in vibration analysis

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Example

M =    9 + 18 i −225 9 + 198 i 9 − 198 i −225 9 − 18 i    , rankK(M) = 2, D1 = I3, D2 =    −i i    X =    9 + 18 i −225 9 − 198 i −225    , Y =

1

1 1 − 4

5 i

,

L = (0 1 0) , d = 2, Z =    1 1    , R =

1

1

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On a rank factorisation problem arising in vibration analysis

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Example

R = {ψ ∈ C2×1 | R ψ = ψ} H =        1 1 −1 −1        ∈ R4×4, spec(H) = {−1, 1} the eigenspace associated with 1 is defined by G = (G T

1

G T

2 )T

G1 =

1

1

, G2 =
−1

1

⇒ Z ′ = G1 + i G2 =
−i

1 i 1

,

R = Z ′ R2×1

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On a rank factorisation problem arising in vibration analysis

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Example

W1 =    i 216 − i 216 − 1 2700 + i 5400 − 11 2700 − i 5400    , W2 =    − 1 216 − 1 216 − 1 5400 − i 2700 − 1 5400 + 11 i 2700    ⇒ det(B(Z ′ γ)) =

W1 Z ′ γ

W2 Z ′ γ

= c (γ2

1 + γ2 2)

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On a rank factorisation problem arising in vibration analysis

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Example

∀ γ ∈ R2×1 \ {0}, uJ = Z Z ′ γ =    i γ1 + γ2 −i γ1 + γ2    v = B(Z ′ γ)−1 Y = (γ2

1 + γ2 2)−1

9 (12 γ1 + (1 − 10 i) γ2)

−225 γ2 9 (12 γ1 + (1 − 10 i) γ2) + 180 i γ2 9 (−12 γ2 + (1 − 10 i) γ1) −225 γ1 9 (−12 γ2 + (1 − 10 i) γ2) + 180 i γ1

vJ = v + v J3

2 = (γ2

1 + γ2 2)−1

9 (12 γ1 + (1 − 10 i) γ2)

−225 γ2 9 (12 γ1 + (1 + 10 i) γ2) 9 (−12 γ2 + (1 − 10 i) γ1) −225 γ1 9 (−12 γ2 + (1 + 10 i) γ2)

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On a rank factorisation problem arising in vibration analysis

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Conclusion

Study a rank factorization pb coming from gear box vibration

⇒ (phase and) amplitude demodulation pb in signal processing

To be continued:

Understand the geometric aspects (multiprojective space?) Understand the topological aspects (vector bundle?) Continue the study of the non-exact case, i.e. arg min

u∈CHn,1(C), vi∈CH1,m(C)

r
i=1

Di u vi − M

Frob

U, V unitary ⇒ U A V Frob = AFrob =

trace(A⋆A)

minu∈CHn,1(C), vi∈CH1,m(C) r

i=1 Di u vi − MFrob

= minuϕ∈Rn×1, viϕ∈R1×m r

i=1 ϕ(Di) uϕ viϕ − ϕ(M)Frob

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On a rank factorisation problem arising in vibration analysis