Laboratory Exercise System Identification. Laboratory Experiment AS-PA6. "Design of Experiments"

Transcription

1 Formel-Kapitel Abschnitt FACHGEBIET Systemanalyse Laboratory Exercise System Identification Laboratory Experiment AS-PA6 "Design of Experiments" Responsible professor: Dr.-Ing. Thomas Glotzbach Responsible for lab experiment: Dr.-Ing. M. Eichhorn Raum: Zuse-Bau 087 Stand 04/7 TECHNISCHE UNIVERSITÄT ILMENAU

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3 Contents. Aim. Theoretical Basics 3.. The Box-Wilson-Method First Step Setting up a simple linear model and determination of its parameters Second step Determination of the gradient based on the obtained linear model and execution of steps in this direction Third step Determination of the second order model parameters Quality evaluation Adequacy test of the model approach Significance test of the parameters Coefficient of determination Using MATLAB Pre-assignments 3.. First pre-assignment Second pre-assignment Experimental procedure 4.. First task Second task Third task Appendix 3

4 . Aim One of the most common tasks in chemical and process industries is to determine the optimal conditions, such as temperature, concentration, pressure and flow for the underlying process. One way to accomplish this is to use Experimental Design and the Box-Wilson method also known as the Response Surface Method (RSM) to detect a point within the area of operation with a minimum number of trials where the response surface has its maximum. The Box- Wilson method uses a combination of experimental design, regression techniques and a gradient approach for optimization to determine the parameters of the process models.

5 . Theoretical Basics.. The Box-Wilson-Method... First Step Setting up a simple linear model and determination of its parameters Since at the beginning of the investigations it can be assumed that the optimum is far away, the so-called quasi-stationary area, a linear model can be postulated in the search area. The parameters are estimated using direct regression as in the following equation: T T aˆ = U U U y () In case of using an orthogonal experimental design, the simplified estimation rule applies as follows: or rather T aˆ = U y ( n ˆ Number of independent experiments) () n n+ n0 aˆ = u y + ( n 0 ˆ Number of zero point experiments, u0j=) (3) 0 0 j n n0 j= j = n â uy i ij j n j= i=,,m (4) The parameter â 0 is the coefficient of the operating point and the parameters â i are the linear coefficients of the individual inputs ui. For a system with two inputs, the following equation applies: yˆ = aˆ + au ˆ + au ˆ (5) 0... Second step Determination of the gradient based on the obtained linear model and execution of steps in this direction For the following equation: k y = a i u, (6) i= i the gradient of the goal function y(u) is: 3

6 . Theoretical Basics y y y y = η + η ηk = aη + aη u u uk where η ηκ are unit vectors, which are perpendicular to each other a η k k (7) For the search steps in this direction always applies: u : u :...: u = a : a :...: a (8) k k This means that with the definition of one increment in a coordinate direction the other increments are likewise defined: aν u ν = uµ ; ν = µ =,,..., k (9) a µ After the definition of the increment vector, search steps in gradient direction will be undertaken starting from the defined initial trial point P 0, 0 u + u u u u = + = = u P P u 0 uk + uk uk (0) until the value y cannot be improved any more, which means that the minimum or maximum has been achieved...3. Third step Determination of the second order model parameters The best target function value determined in the direction with the steepest gradient will now be the center point of the second order experimental design. The resulting model has the following form: i= k y = a + a u + a u u + a u () 0 k i i k ij i j i= j= i+ i= As the experimental design, a composite rotatable design by Box and Hunter is selected. It consists of a k - factorial design and a star design with axes and center points. Rotability implies that the variances var{a0}, var{ai}, var{aij} und var{aii} for points in the experimental region remain constant at the same distance from the origin. Figure. shows the arrangement of the experimental design points. k ii i 4

7 .. The Box-Wilson-Method U P3 P - Center point U P4 - P Figure.: Experimental design The structure and the matrix U are shown in Table.. j u0 u u uu u u α 0 0 α 0 6 -α 0 0 α α 0 0 α 8 0 -α 0 0 α Table.: Structure of the experimental design The experimental design consists of the following parts: Design core (first order design)...4 nc = 4 Star points na = 4 α = Center points n0 = 5 The parameters are estimated analog to the design of the linear model using direct regression. The following figure shows the individual steps of the Box-Wilson method for a quadratic function. 5

8 . Theoretical Basics. Step. Step 3. Step Figure.: The several experimental steps of the Box-Wilson-method [Photo credit: Dr.-Ing. Winfried Winkler] 6

9 .. Quality evaluation.. Quality evaluation... Adequacy test of the model approach Compare the variance of the experimental error se with the residual standard variance sr using the F-test: R e s Fˆ = () s The variance of the experimental error will be calculated from the zero point trials (see Table.) and is a measure of the variance caused exclusively by the disturbance: s e = (3) n u u y 0 n0 ( yi0 y 0 ) i= n0 zero point trials Table.: Determination of the variance of the experimental design The residual standard variance is obtained by the residual sum of squares of the model error: n s ˆ R y yi (4) n m i where n =ˆ Total number of trials (n=nc+na+n0) m =ˆ Number of parameters It results from the error, which depends on the degree to which the chosen model approach corresponds to the system structure as well as on the magnitude of the disturbances. The model is adequate, if ˆF with ν = n-m - (5) F αν ν,, ν = n 0 The residual standard variance then only differs by chance from the variance of the experimental error. This means that the error results exclusively from the disturbances and not from the chosen model approach. 7

10 . Theoretical Basics... Significance test of the parameters The estimated parameters are significant, if: is valid. With { } inverse of the matrix { } â t var a (6) i α/, ν i var a = s p (The parameters pii are elements on the main diagonal of the i R ii T U U T P= U U ) results: â t s p. (7) i α/, ν R ii Thereby a hypothesis will be assumed, that the parameter ai = 0. The confidence interval of this parameter which is generally defined as: thus become: a a = t s p (8) ˆi i α/, ν R ii a = t s p. (9) ˆi α/, ν R ii In case of a ˆi < tα/, ν s p, the hypothesis will be accepted and ai is statistically equal to R ii zero, with an error probability of α. When equation (7) is met, the hypothesis is rejected and the parameter ai is statistically not equal to zero and therefore significant. s For orthogonal experimental design the variance of the parameter ai is var{ a } = R i and this n results in the following simplification: sr âi tα/, ν. (0) n The parameter n corresponds to the total number of experiments. The value of the Student s t- distribution tα/,ν is called t-value and is determined by the significance level α and the degrees of freedom ν=n-m-, whereby n corresponds to the number of experiments and m to the number of parameters. For the choice of the significance level α, a two-sided confidence interval is assumed, which is why the index of t is α/. This two-sided confidence interval is usually used for hypothesis testing of parameters and measured values, because the estimated values could never achieve an infinitely large negative or positive value, which is the case when using a one-sided confidence interval. The value for tα/,ν can be directly read from a table (see appendix) or determined computationally (see section.3). The values for α/ and -α correspond to the surface area of the shaded sections in Figure.. The possible range for the true parameter ai is inside the confidence interval and is determined using the following equation: aˆ t s p a aˆ t s p () i /, R ii i i /, R ii If a confidence interval includes a zero-value, the parameter ai is not significant. This means that âi tα/, ν sr pii must apply (see equation (7)), so that both bounds of the confidence interval have the same sign and thus not include the zero-value. 8

11 .3. Using MATLAB p(x) -α α/ α/ a t s p i /, R ii Hypothesis rejected a i Hypothesis accepted a t s p i /, R ii Hypothesis rejected x Figure.: Two-sided confidence interval of the parameters ai..3. Coefficient of determination The coefficient of determination R evaluates the quality of consistency of the calculated model output values with the measured values: R n i= = n ( yˆ y) ( yi y) The closer the value R is to, the better is the degree of consistency..3. Using MATLAB i= In the following, several useful MATLAB functions for statistical and empiric statistical parameters are presented. Thereby is y a vector. yquer = mean(y) Determines the mean of vector y yvar = var(y) Determines the variance of vector y ystd = std(y) Determines the standard deviation of vector y ydach = U*a Calculates the output vector ydach by means of the measurement matrix (regression matrix) U and the parameter vector a The coefficient of determination, equation (), can be calculated as follows: B = sum((ydach-yquer).^)/sum((y-yquer).^) The operator \ can be used to determine the parameters of a system of equations using direct regression: a = U\y The parameter y corresponds to the vector of the measured output signals. U includes the regression matrix, whereby each row represents a trial or sample. For two inputs u und u the regression matrix X for a second-order model according to equation () can be expressed as follows: U = [ones(length(y),), u, u, u.^, u.^, u.*u] i () 9

12 . Theoretical Basics The MATLAB function regress from the Statistics Toolbox can be used to determine the parameters of a linear equation system by linear regression: [a] = regress(y,u) This function can also calculate the statistical parameters: [a,aint,r,rint,stats] = regress(y,u) y n Vector of the measured output values U n p Regression matrix a p Vector of the estimated coefficients using direct regression aint p Vector that contains the 95%-confidence intervals for the coefficients ai (see equation ()) (Is a zero value located within a confidence interval of a coefficient, this coefficient is not significant) r p Residual vector, difference between the measured and modeled output (y-ydach) rint p Vector that contains the confidence interval for the residuals r stats() Coefficient of determination R (see equation ()) stats() F-statistics stats(3) p-value stats(4) Estimated error variance (corresponds to equation (4) with ) n m By default, the function regress uses a 95%- confidence interval. When using a confidence interval with another confidence level, the value alpha can be passed to the function additionally. The confidence level corresponds to: 00*(-alpha)%. [ ] = regress(y,u,alpha) To determine the value of an F-distribution F ανν with an error probability of α=5% (alpha=0.05) and the two degrees of freedom ν und ν, the function finv can be used. F=finv(-alpha,nu,nu) Thereby alpha = α/00%. For an error probability of α=5% and the degrees of freedom ν = 6 and ν =4 the function is: F=finv(0.95,6,4)=6.63 To determine the value of a t-distribution of a one-sided confidence interval of p = -α with the degree of freedom ν, the function tinv can be used. t=tinv(-alpha,nu) A two-sided confidence interval will usually be required. In this case the first parameter in function tinv has the value -α/. For an error probability α=5% and ν=4 and a one-sided confidence interval the function call is: t=tinv(-0.05,4)=.38 and for a two-sided confidence interval: t=tinv(-0.05,4)=

13 3. Pre-assignments 3.. First pre-assignment Show by using a calculation example, that for a full orthogonal design for a first-order model, the equation can be reduced to T T [ U U ] U y â = (3) T â = U y (4) n or = n â uy i ij j n j=. (5) For this example use a design with two input variables! 3.. Second pre-assignment Use the design in Table 3. Versuch u u u u y 0 0 5,45-0 0, , , , , , ,50 Table 3.: Design for the second pre-assignment to determine the parameters for a linear model approach: yˆ = aˆ + au ˆ + au ˆ (6) 0 Determine the gradient direction for this model and draw it together with the contour lines of the same quality (y=const.) for the trial points P, P, P3, P4 in a u, u diagram. Calculate the coefficient of determination and test the significance of the parameters for an error probability of α=5% for a two-sided confidence interval!

14 4. Experimental procedure The inputs u und u of the analog model are controlled according to the defined design. The input signals can be adjusted via a potentiometer in a range from 0 00, which corresponds to a voltage range of 0 0 V. Please note that, the scaling of the input potentiometers is only an estimate, therefore the exact value should be adjusted using the digital voltmeter. The output signal can be read off the digital voltmeter. This value can be inserted directly in the experimental design. A scaling is not necessary. The noise signal will be generated randomly from a low pass filtered PRBS (pseudorandom bit sequence) signal by pressing a push button. The random value will be held by a sample and hold circuit when the button is released. This allows an easy reading of the disturbed output signal. The noise generator must be set to a value of (rotary switch.0, toggle switch x ). 4.. First task Place around the operating point P0(u', u') = P0(.5V,.5V) a factorial design for inputs with a step size u ' = u' = 0.5V. Measure the associated output voltage. Due to the disturbed output signal, each value has to be measured five times in order to calculate a mean value subsequently. Determine the parameters of a linear model with the normalized input values using equation (3) und (4) manually! 4.. Second task Determine the gradient direction, carry out more trials, starting from the best (maximal) value of the design in task 4. in gradient direction with an increment of u'= 0.5 V until the maximal output value is reached. Each value has to be measured five times in order to calculate a mean value subsequently Third task Place a second-order design by Box and Hunter around the best value found in task 4. or in the immediate vicinity (see Table.). Use a step size of u' = u' =.0V. Measure the associated output voltage five times for the same input settings. Calculate the mean value of these values. Determine the model parameters by regression (Use the function regress in MATLAB to solve the equation system and to detect the statistical parameters.) Use unnormalized values for the inputs this time! Determine the coefficient of determination R, the residual standard variance sr and the parameter variances. Test the adequacy of the model as well as the significance of the parameters for an error probability of α=5% for a two-sided confidence interval! Modify your design and repeat the procedure in case of nonsignificant parameters!

15 5. Appendix t-distribution ttab(p,ν) (Student distribution) Number of p for two-sided confidence interval (p=-α) degrees of 0,5 0,75 0,8 0,9 0,95 0,98 0,99 0,998 freedom p for one-sided confidence interval (p=-α) ν 0,75 0,875 0,9 0,95 0,975 0,99 0,995 0,999,000,44 3,078 6,34,706 3,8 63,657 38,309 0,86,604,886,90 4,303 6,965 9,95,37 3 0,765,43,638,353 3,8 4,54 5,84 0,5 4 0,74,344,533,3,776 3,747 4,604 7,73 5 0,77,30,476,05,57 3,365 4,03 5, ,78,73,440,943,447 3,43 3,707 5,08 7 0,7,54,45,895,365,998 3,499 4, ,706,40,397,860,306,896 3,355 4,50 9 0,703,30,383,833,6,8 3,50 4,97 0 0,700,,37,8,8,764 3,69 4,44 0,697,4,363,796,0,78 3,06 4,05 0,695,09,356,78,79,68 3,055 3, ,694,04,350,77,60,650 3,0 3,85 4 0,69,00,345,76,45,64,977 3, ,69,97,34,753,3,60,947 3, ,690,94,337,746,0,583,9 3, ,689,9,333,740,0,567,898 3, ,688,89,330,734,0,55,878 3,60 9 0,688,87,38,79,093,539,86 3, ,687,85,35,75,086,58,845 3,55 0,686,83,33,7,080,58,83 3,57 0,686,8,3,77,074,508,89 3, ,685,80,39,74,069,500,807 3, ,685,79,38,7,064,49,797 3, ,684,78,36,708,060,485,787 3, ,684,77,35,706,056,479,779 3, ,684,76,34,703,05,473,77 3,4 8 0,683,75,33,70,048,467,763 3, ,683,74,3,699,045,46,756 3, ,683,73,30,697,04,457,750 3, ,68,67,303,684,0,43,704 3, ,679,64,99,676,009,403,678 3,6 60 0,679,6,96,67,000,390,660 3,3 70 0,678,60,94,667,994,38,648 3, 80 0,678,59,9,664,990,374,639 3, ,677,58,9,66,987,368,63 3, ,677,57,90,660,984,364,66 3, ,676,54,86,653,97,345,60 3, ,675,53,84,650,968,339,59 3, ,675,5,84,649,966,336,588 3, 500 0,675,5,83,648,965,334,586 3,07 0,674,50,8,645,960,36,576 3,090 Alternative calculation: for a one-sided confidence interval in MATLAB: = tinv(p,nu) for a one-sided confidence interval in Excel: = T.INV(p;nu) for a two-sided confidence interval in Excel: = TINV(-p;nu) = T.INV.S(-p;nu) from Office 00 3

16 F-distribution Ftab (p,ν,ν ) for p=0.95 or α=0.05 ν ν Alternative calculation: in MATLAB: in Excel: = finv(p,nu,nu) = FINV(-p;nu;nu) 4