Lecture 1: Getting started with problematic inversions Denis - PowerPoint PPT Presentation

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Lecture 1: Getting started with problematic inversions Denis Maillet, Yvon Jarny, Daniel Petit, Olivier Fudym, Philippe Le Masson LEMTA Nancy - LTN Nantes - Institut P’ Poitiers - Ecole Mines Albi - LIMATB Lorient Example 1: Square system of linear equations Example 2: Different inverse problems fo steady state 1D heat transfer through a wall 1

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Example 1: Square system of linear equations − = 10 x 21 x 9 1 2 − = 39 x 81 x 1 1 2 scalar relationship → vector −       10 21 3 9 S x x y S x = = = ⇒ = = exact exact       − mo       39 81 1 36 Direct problem : input (known) output (calculated) Model : y mo = η η ( x ) η η Structure of model 2

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 −     10 21 9 = = = S y S x exact     − mo     39 81 36 inverse problem : data (known) unknown   3 Solution with exact data y mo : No problem ! x S − y x = = = 1 exact   mo   1   9 . 1   0 . 1 ≈ 1 % of y mo 1 y y Solution with noisy data y : = + = ε = ε     mo −  35 . 7   0 . 3  ≈ 1 % of y mo 2 Noise amplification ! noise −       10 21 9 , 1 1 . 40 53 % error for x 1 x x x e = = + = exact ˆ       x −       39 81 35 . 7 0 . 233 77 % error for x 2 3 estimate estimation error

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 1 / 2   = ∑ 2   u u  2 Euclidian distance (L2 norm) :  j   = j 1 e x x x = − exact ˆ = y − y ε estimation error mo Measurement error (noise) S − 1 ε e 1 . 774 x = = = = ε S = k a ( ) 5 . 61 absolute det ( ) 9 ε ε 0 . 316 coefficients of S ≤ = ε k ( ) cond ( ) 958 maximum: amplification of r measurement relative S − S − y e x 1 ε 1 exact / / error (leverage) 1 . 774 / 3 . 16 x mo = = = = ε k ( ) 65 . 8 4 y y r ε ε / / 0 . 316 / 37 . 11 mo mo

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Example of inverse heat conduction problem (IHCP) Direct problem Computed Model( s ) Heat flux Temperatures method e s r e v n i Inverse problem Retrieved heat flux ? Measured temperatures 5

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Example 2: Different inverse problems fo steady state 1D heat transfer through a wall Physical system with sensors Thermal model • plane wall for exact output of sensor T s • 2 temperature sensors: 1 on rear face (exact measurement T e ) • homogeneous material 1 inside (x = x s ; noisy measurement y) (conductivity λ ) • steady state • stimulation q (x = 0) • 1D heat transfer • no internal source • Fourier law Possible objectives (types of inverse problems) • flux q entering the wall (x = 0) ? • front face temperature T 0 ? • internal temperature distribution T (x) ? • conductivity λ ? 6

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 ∂ ∂ 2 T T = − λ = = 0 with q and T T State equations: T ? ∂ = e ∂ 2 x e x x = x 0 model structure Assumption: λ known, no error for T e = = η λ ≡ − λ y T ( x ; q , T , ) T q x / no error for e and x s mo x 1 0 0 Objective: find q, T 0 , T x parameters dependent variable = η λ Output equation: T ( x ; q , T , ) or explained s 1 s 0 variable     x x x ≡ η ≡  η  = +   s s s , T , T ( x /e, T ) 1 - T T 2 2 s 0 e 0 0 e     e e e Measurement: = + ε y T Random variable s p.d.f: E ( ε ) = 0 E ( ε 2 ) = σ 2 measurement noise exact unknown temperature σ ε 0 Estimation = exact matching: estimate   x = = η η ˆ ˆ   s 7 , T , T y T ( x /e, T ) s 2 2 s 0 0 e   e

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Solution of inverse problem: estimation of T 0 * 1 x = − * = ˆ s T y T with x x / e 0 e s s − * − * 1 x 1 x s s = ε − * ⇒ = σ = σ − * e / ( 1 x ) E ( e ) 0 a n d / ( 1 x ) Estimation error: T 0 s T 0 0 s Good estimation of T 0 for shallow measurement Estimation of T (x) : * * − * x x 1 - x * = η ˆ = ˆ = + s T ( x ) ( x , T ) T y T recalc 2 0 x e * * − − 1 x 1 x s s * = with x x / e Estimation error: * 1 - x = ε ⇒ σ = σ = e K K with K Tx Tx * − 1 x s 8

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Two regions for estimation of T x * 1 - x ∈ • in between measurements points x [ x , e ] σ = σ = K wit h K s Tx − * 1 x interpolation = attenation of error s ≤ K 1 well-posed problem (Hadamard, 1902): - solution exists - it is unique - it depends continuously of the data [ [ ∈ • outside measurements points interval x 0 , x s extrapolation = attenation of error > K 1 ill-posed problem (Hadamard, 1902) → ⇒ → ∞ ∀ ≠ x s e K x e Very bad design ! 9

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 − y T = λ Estimation of flux q : e ˆ q − e x s λ λ 1 1 = ε ⇒ σ = σ ⇒ σ = Estimation error: e / q q − q − q − * e x e x SNR 1 x s s s = e − σ SNR ( T T )/ 0 signal over noise ratio Numerical application: e = 0.2 m - λ = 1 W.m -1 .K -1 - T 0 – T e = 30° C x s = 0.18 m - σ = 0.3° C ⇒ = σ = SNR 100 and / q 10 % q mid-slab measurement: = = σ = ⇒ x e / 2 0 . 10 m / q 2 % s q 10

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Errors for parameters "assumed to be known" Assumption: λ known, no error for T e no error for e error for x s Objective: find q, T 0 , T x = + δ nom x x s s nominal (« a priori ») location location of sensor error (deterministic) (random) exact location (random) σ p.d.f: E ( δ ) = 0 E ( δ 2 ) = 2 pos σ pos δ 0 11

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 = η + ε = η + ε nom y ( x / e , T , T ) ( x / e , T , T ) ' Signal and model : 2 s 0 e 2 s 0 e ε = δ − + ε with ' ( T T )/ e 0 e ( ) σ = ε = σ + − 2 σ 2 2 2 Equivalent temperature noise : ' var ( ' ) ( T T ) / e 0 e pos ( ) = σ + 2 2 2 1 SNR / R pos = σ with R e / pos pos 1 1 σ = Estimation error: / q q − * 1 x SNR ' s = − σ SNR ' ( T T )/ ' e 0 Numerical application: e = 0.2 m - λ = 1 W.m -1 .K -1 - T e – T 0 = 30° C x s = 0.18 m - σ = 0.3° C - σ pos = 2 mm ⇒ SNR = 100 = σ = = R e / 200 / 2 100 pos pos ⇒ σ = / q 14 . 1 % q 12

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Assumption: no error for T e λ = λ + nom exact e λ no error for e no error for x s error for λ λ λ λ nominal (« a priori ») conductivity Objective: find q, T 0 , T x conductivity error (deterministic) (random) exact conductivity Estimation of flux q : (random) − λ + exact y T e ( ) σ 2 = λ = λ − + ε nom e p.d.f: E ( e λ ) = 0 E ( e λ 2 ) = ˆ q T T λ − − s e e x e x s s   λ − ε exact  +  ( T T ) e σ λ λ   = +   s e 1 1   − λ −  exact  e x  T T  s s e Estimation error: δ 0 λ exact assumptions: - small e / λ - large SNR   ε e  λ  + = + + exact exact q e q 1   Numerical application: q − λ exact  T T  s e e = 0.2 m - λ = 1 W.m -1 .K -1 - T e – T 0 = 30° C ε e e 1 x s = 0.18 m - σ = 0.3° C - σ pos = 0 mm ⇒ = λ + q σ λ exact exact SNR σ λ = 0.1 W.m -1 .K -1 q ⇒ σ = / q 10 . 1 % 1 / 2   q σ σ 2 1   ≈ λ + q ( )   exact 2 2 13 λ q SNR exact  

Eurotherm Advanced School – Metti 5 – Roscoff – June 13-18, 2011 Thank you for your attention ! 14