Interval and p-Box Techniques for Model Validation: on the Example - PDF document

1 Interval and p-Box Techniques for Model Validation: on the Example of the Thermal Challenge Problem Vladik Kreinovich Department of Computer Science University of Texas at El Paso 500 W. University El Paso, Texas, 79968, USA office phone (915) 747-6951 email vladik@utep.edu http://www.cs.utep.edu/vladik University of Texas at El Paso

2 Realistic Measurement Situations • Often, the measurement result z depends: – not only on the measured value x , but also – on the parameters s of the experiment’s setting – and on the values of some auxiliary quantities y . • The dependence z = f ( x, s, y ) is usually known. • Ideal case: we know y , so we find x . • Real case: we know y with some uncertainty. • Usually: uncertainty in y leads to extra measurement error in x . • Good news: often, we can combine multiple measure- ment results and decrease influence of y ’s uncertainty. • We get sub-noise measurement accuracy: better than the accuracy with which we know y . University of Texas at El Paso

3 Example: Multi-Spectral Imaging � • We measure I ( f, � p ) = I ( f, � p ) + D ( f, � p ), where: • I ( f, � p ) = C ( f ) · I ( � p ) is the intensity of the source on frequency f at point p ; • D ( f, � p ) is the intensity of dust radiation. • Often, D ≫ I , so we cannot determine the object’s structure. p ) · f α . • We know how D depends on f : D ( f, � p ) = D ( � • Here, x = I , s = f , y = D , and z = f ( x, s, y ) = C ( s ) · x + y · s α . • Based on two observations z i = C ( s i ) · x + y · s α i , we can apply linear algebra ideas to eliminate y : z 1 · s α 2 − z 2 · s α 1 = x · ( C ( s 1 ) · s α 2 − C ( s 2 ) · s α 1 ) . • Result: we uncover previously unseen spiral and ring- like structures in distant galaxies. University of Texas at El Paso

4 VLBI Astrometry • Very Large Baseline Interferometry (VLBI): we si- multaneously observe a distant radiosource by two (or more) radioantennas i , j . • Ideal case: time delay between the two antennas τ i,j,k = 1 c · � b i,j · � s k . • Synchronization is not perfect (∆ t i � = 0), hence τ i,j,k = 1 c · � b i,j · � s k + ∆ t i − ∆ t j . s k , y = ( � • Here, z = τ , x = � b i,j , ∆ t i ). • Measurement error in τ corresponds to accuracy ≈ 0 . 001 ′′ , but inaccuracy in ∆ t i is much worse. • Differential astrometry: ∆ τ i,j,k,l = 1 c · � b i,j · ∆ � s k,l , def where ∆ τ i,j,k,l = τ i,j,k − τ i,j,l , drastically improves the accuracy. University of Texas at El Paso

5 VLBI Astrometry: Arc Method • To get rid of baseline vectors, we need 4 antennas: ∆ τ 1 , 2 ,k,l = 1 s k,l ; ∆ τ 2 , 3 ,k,l = 1 c · � c · � b 1 , 2 · ∆ � b 2 , 3 · ∆ � s k,l , ∆ τ 3 , 4 ,k,l = 1 c · � b 3 , 4 · ∆ � s k,l . B i,j · 1 c · � • For the dual basis � b i,j = δ ( i,j ) , ( i ′ ,kj ′ ) , we get s k,l = ∆ τ 1 , 2 ,k,l · � B 1 , 2 + ∆ τ 2 , 3 ,k,l · � B 2 , 3 + ∆ τ 3 , 4 ,k,l · � � B 3 , 4 . • Express � B i,j as a linear combination of � s 1 , 2 , � s 1 , 3 , � s 1 , 4 . • For any other source k , we have a similar expression s 1 = ∆ τ 1 , 2 ,k, 1 · � B 1 , 2 +∆ τ 2 , 3 ,k, 1 · � B 2 , 3 +∆ τ 3 , 4 ,k, 1 · � � s k, 1 = � s k − � B 3 , 4 . • Hence, � s k is a linear combinations of � s 1 , 2 , � s 1 , 3 , � s 1 , 4 . • We have a linear transformation T between the actual and the observed values � s k . • Since � � s k � = 1, T is rotation. • So, we can determine positions modulo rotation. University of Texas at El Paso

6 VLBI Imaging • Problem: find the image I ( � p ). • Solution: find Fourier transform F ( � b ) of I ( � p ). • Ideal case: the phase shift ϕ i,j between the signals � observed by antennas i and j is equal to the phase ϕ i,j of F ( � b ij ). • In reality: due to synchronization errors ∆ ϕ i , ϕ i,j = ϕ i,j + ∆ ϕ i − ∆ ϕ j . � • Here, z = ϕ i,j , x = ϕ i,j , y = ∆ ϕ i . � • Closure phase method eliminates the effect of the auxiliary parameters by considering the “closure phase” ϕ ij + � ϕ jk + � ϕ ki for which: � ϕ ij + � ϕ jk + � ϕ ki = ϕ ij + ϕ jk + ϕ ki . � University of Texas at El Paso

7 Image Georeferencing • Problem: find the relative orientation of geospatial images I 1 ( � p ) and I 2 ( � p ). • Problem reformulated: find shift, rotation angle, and scaling between the images. • Difficulty: to find an angle with accuracy of 1 ◦ , we need 360 tests; we need 4 parameters, so we need 360 4 ≈ 10 9 tests – practically impossible. • Idea: separate the problem – find rotation angle and scaling separately from finding the shift. • Fact: in Fourier domain, when I 2 ( � p ) = I 1 ( � p + � a ), then F 2 ( � ω ) = F 1 ( � ω ) · exp(i · � ω · � a ). • Here, x = F ( � ω ), y = � a . • Solution: the shift-independent combination is the absolute value | F i ( � ω ) | . University of Texas at El Paso

8 Measuring Strong Electric Currents • Problem: measuring the cable current I at an alu- minum plant. • Specifics: I is difficult to measure directly. • Specifics: I is measured by its magnetic field E . • Ideal case (single cable): E = I/r , where r is the distance between the sensor and the cable’s axis. • Real plants: there is often an auxiliary nearby cable. • Here, z = E , x = I , s = sensor locations, y = location and current in the auxiliary cable. • Difficulty: z = f ( x, s, y ) non-linearly depends on the (unknown) location of the auxiliary cable. • Solution: combining the measurements from different sensors eliminates the influence of the auxiliary cable. University of Texas at El Paso

9 Ultrasonic Non-Destructive Testing (in brief) • Problem: find the location and orientation of hidden faults in a plate. • Related active measurements: – send ultrasonic Lamb waves to the plate; – measure the waves that propagated along the plate. • Difficulty: the resulting signals depend both on the location and on the orientation of the fault. • Idea: separate the effects of location and orientation. • Solution: by appropriately combining sensor read- ings, we can minimize the effect of location. • Thus, we can easily determine the fault’s orientation. University of Texas at El Paso

10 Formulation of the General Problem • General problem: • Objective: we are interested in n x scalar parame- ters that form x . • Measurement situation: each n z -component mea- surement result z depends not only on x , but also on n y components of the auxiliary quantity(-ies) y : z = f ( x, s, y ). • Desirable objective: determine x without knowing y precisely. • Two possible situations: • y is fixed (cannot be varied), but we can change s . Example: multi-spectral imaging. • We cannot change the settings s , but we can use different values of y . Example: VLBI astrometry. University of Texas at El Paso

11 Variable Settings: Analysis of the Problem • Situation: after we performed the measurement in N s different settings s 1 , . . . , s N s , we get N s measurement results z 1 , . . . , z N s . • Situation: we do not know y . • Conclusion: select N s so that we will be able to uniquely determine both x and y . • After N s measurements, we have N s n z -component equations z i = f ( x, s i , y ) to determine n x unknown components of x and n y unknown components of y . • Fact: # of equations must be ≥ # of unknowns. • We have N s · n z scalar equations for n x + n y unknowns. • Recommendation: perform the measurements in at least N s ≥ ( n x + n y ) /n z different settings. University of Texas at El Paso

12 Practical Question: How to Solve the System of Equations? • Difficulty: in general, the dependence z = f ( x, y ) is non-linear. • So, we have a system of non-linear equations. • What helps: often, we know good approximations x (0) and y (0) to x and y . • How it helps: = x − x (0) and – We only need to find ∆ x def ∆ y def = y − y (0) . – Usually, ∆ x and ∆ y are small. – So, we can expand f ( x, y ) in Taylor series in ∆ x and ∆ y and ignore 2nd and higher order terms. – As a result, to find ∆ x and ∆ y , we get an easier- to-solve system of linear equations. University of Texas at El Paso

13 Variable Settings: Example • Case study: multi-spectral astronomical imaging. p ) · f α . � • Reminder: I ( f, � p ) = C ( f ) · I ( � p ) + D ( � � • Here, z = I , x = I , s = f , y = D , and z = f ( x, s, y ) = C ( s ) · x + y · s α . • Specifics: n z = 1, n x = 1, and n y = 1. • General recommendation: we must have at least ( n x + n y ) /n z = (1 + 1) / 1 = 2 settings. • Confirmation: we have shown that, based on mea- surements in two different settings z 1 = C ( s 1 ) · x + y · s α 1 , z 2 = C ( s 2 ) · x + y · s α 2 , we can uniquely determine the desired value x : z 1 · s α 2 − z 2 · s α 1 = x · ( C ( s 1 ) · s α 2 − C ( s 2 ) · s α 1 ) . University of Texas at El Paso