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Unformatted text preview: Homework 5 Lectures 19 ‐ 20 Math 344 Ohio University Spring Quarter 09’‐10’ Bob Benjamin Astrom Elbert 19.1) The given data consists of 11 data points (ti , yi) as follows: t = [0, .1, .499, .5, .6, 1.0, 1.4, 1.5, 1.899, 1.9, 2.0] y = [0, .06, .17, .19, .21, .26, .29, .29, .30, .31, .31] When 11 data points are given the appropriate fit is a 10th degree polynomial to approximate the data curve. After selecting the 10th degree polynomial option from Tools > Basic Fitting, MatLab displays a warning that the representation is badly conditioned. The reason that this fit is badly conditioned is that some of the t values are very close together: t = .499 and t = .5; t = 1.899 and t = 1.9. When the “input” points are this close together, the system that MatLab must solve to find the best fit function is nearly singular. *See attached graph Figure 19.1* 19.2) If we consider the same plot of the data points given in Problem 19.1, but instead utilize a Spline Interpolant to create a best fit function, we see that Spline Interpolant creates a best fit curve that does not vary as dramatically as the 10th degree polynomial fit. For instance, between the first and second data points, the Spline Interpolant line only reaches a maximum value of less than .2, whereas the 10th degree fit has a maximum value between t0 and t1 of more than 5! While this is quite extreme, considering the overall small values of yi as well as the relatively small change between yi and yi+1, we cannot rule out any of the approximations since we do not have additional details or knowledge about the data. The cubic approximation appears to be a more accurate representation of the steady increase of the yi’s, but, without more knowledge about the data and what it represents, we are not able to determine the best fit. *See attached graphs Figure 19.2 and overlay of 19.1 and 19.2* Figure 19.1 10th Degree Ploynomial 10
10 9 8 7 6 5 y =  8e+002*x + 8.7e+003*x  4e+004*x + 1e+005*x  1.6e+005*x + 1.6e+005*x 4 3 2 9.8e+004*x + 3.5e+004*x  6.1e+003*x + 3.4e+002*x  8.4e010 5 0 Y Values 5 t=0.499 y=0.170 t=0.500 y=0.190 t=1.899 y=0.300 t=1.900 y=0.310 10 data 4 10th degree 15 0 0.2 0.4 0.6 0.8 1 T Values 1.2 1.4 1.6 1.8 2 Figure 19.2 Spline Interpolant 0.8 0.6 t=1.899 y=0.300 t=1.900 y=0.310 0.4 0.2 0 Y Values 0.2 0.4 t=0.499 y=0.170 t=0.500 y=0.190 0.6 0.8 data 4 spline 1 1.2 0 0.2 0.4 0.6 0.8 1 T Values 1.2 1.4 1.6 1.8 2 Combined Figure 19.1 and 19.2 Spline Interpolant vs. 10th Degree Ploynomial 10
10 9 8 7 6 5 y =  8e+002*x + 8.7e+003*x  4e+004*x + 1e+005*x  1.6e+005*x + 1.6e+005*x 4 3 2 9.8e+004*x + 3.5e+004*x  6.1e+003*x + 3.4e+002*x  8.4e010 5 0 Y Values 5 t=0.499 y=0.170 t=0.500 y=0.190 t=1.899 y=0.300 t=1.900 y=0.310 10 data 4 spline 10th degree 15 0 0.2 0.4 0.6 0.8 1 T Values 1.2 1.4 1.6 1.8 2 Problem 20.1) Graph 1) Absorption Coefficient vs. Frequency 1/2" Binary Amplitude Diffuser 1.1 data 1 spline 1 0.9 0.8 Absorption Coeffcient 0.7 0.6 0.5 0.4 0.3 0.2 0.1 125 160 200 250 315 400 500 630 800 1k 1.25k 1.6k 2k 2.5k 3.15k 4k Frequency Hz 5k Absorption Coefficients for 1/2" Binary Amplitude Diffuser (BAD panel) Absorption coefficients indicate the percentage of sound absorbed during a random incident reverberation testing in a diffuse sound field reverberation chamber. The testing is for a Binary Amplitude Diffuser as manufactured by RPG Incorporated. Data was obtained from Acoustic Absorbers and Diffusers Trevor Cox and Peter D'Antonio Second Addition and RPG's website rpginc.com. The graph indicates that the absorption of sound of the BAD panel is strongest (highest absorption) in the mid frequencies and lower at the high and low frequencies. Figure 11.1 Top: construction of a hybrid surface: porous absorber (left), the mask (middle), and fabric covering (right). Bottom: an application on the lower part of the wall in the rehearsal room for the Commodores, the US Navy jazz ensemble. (Acoustician: Polysonics Corporation)
1.2 1 0.8 0.6 0.4 0.2 0 100 160 250 400 630 1000 f (Hz) 1600 2500 4000 Figure 11.2 Random incidence absorption coefficient for a hybrid surface (BADTM panel) compared to mineral wool. Four different backing depths for the panel are shown: 2.5 cm BAD; 5.1 cm BAD; 7.6 cm BAD; 10.2 cm BAD; and 2.5 cm Fibreglass. Absorption coefficient Problem 20.1) Graph 2) Equal Loudness Contour data 1 spline 110 100 90 Phon (dB) 80 70 60 50 1 10 10 2 10 Frequency (Hz) 3 10 4 Equal Loudness Contour for 60dB tone at 1000Hz The Equal Loudness Contour represents the sound pressure level in phons which humans perceive single frequency tones as the same loudness level. Data produced from ISO226 standard as produced by a MATLab .m file written by Jeff Tackett and modified to produce the above plot. The data for this plot can be found in any number of books such as Fundamentals of Acoustics Kinsler and Frey Third Edition Page 272 The figure below is the Equal loudness Contour for multiple frequencies. function [spl, freq] = iso226(phon); % % Generates an Equal Loudness Contour as described in ISO 226 % % Usage: [SPL FREQ] = ISO226(PHON); % % PHON is the phon value in dB SPL that you want the equal % loudness curve to represent. (1phon = 1dB @ 1kHz) % SPL is the Sound Pressure Level amplitude returned for % each of the 29 frequencies evaluated by ISO226. % FREQ is the returned vector of frequencies that ISO226 % evaluates to generate the contour. % % Desc: This function will return the equal loudness contour for % your desired phon level. The frequencies evaulated in this % function only span from 20Hz  12.5kHz, and only 29 selective % frequencies are covered. This is the limitation of the ISO % standard. % % In addition the valid phon range should be 0  90 dB SPL. % Values outside this range do not have experimental values % and their contours should be treated as inaccurate. % % If more samples are required you should be able to easily % interpolate these values using spline(). % % Author: Jeff Tackett 03/01/05 % /\ %%%%%%%%%%%%%%%%% TABLES FROM ISO226 %%%%%%%%%%%%%%%%% % \/ f = [20 25 31.5 40 50 63 80 100 125 160 200 250 315 400 500 630 800 ... 1000 1250 1600 2000 2500 3150 4000 5000 6300 8000 10000 12500]; af = [0.532 0.506 0.480 0.455 0.432 0.409 0.387 0.367 0.349 0.330 0.315 ... 0.301 0.288 0.276 0.267 0.259 0.253 0.250 0.246 0.244 0.243 0.243 ... 0.243 0.242 0.242 0.245 0.254 0.271 0.301]; Lu = [31.6 27.2 23.0 19.1 15.9 13.0 10.3 8.1 6.2 4.5 3.1 ... 2.0 1.1 0.4 0.0 0.3 0.5 0.0 2.7 4.1 1.0 1.7 ... 2.5 1.2 2.1 7.1 11.2 10.7 3.1]; Tf = [ 78.5 68.7 59.5 51.1 44.0 37.5 31.5 26.5 22.1 17.9 14.4 ... 11.4 8.6 6.2 4.4 3.0 2.2 2.4 3.5 1.7 1.3 4.2 ... 6.0 5.4 1.5 6.0 12.6 13.9 12.3]; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Hzbands = f; % Additional code to save frequency information %Error Trapping if((phon < 0)  (phon > 90)) disp('Phon value out of bounds!') spl = 0; freq = 0; else %Setup userdefined values for equation Ln = phon; %Deriving sound pressure level from loudness level (iso226 sect 4.1) Af=4.47E3 * (10.^(0.025*Ln)  1.15) + (0.4*10.^(((Tf+Lu)/10)9 )).^af; Lp=((10./af).*log10(Af))  Lu + 94; %Return user data spl = Lp; freq = f; end x = Hzbands y = Lp
plot(Hzbands,Lp,'*') % Additional code for plot Output: >> iso226(phon) x = 1.0e+004 * Columns 1 through 7 0.0020 0.0025 0.0032 0.0040 0.0050 0.0063 0.0080 Columns 8 through 14 0.0100 0.0125 0.0160 0.0200 0.0250 0.0315 0.0400 Columns 15 through 21 0.0500 0.0630 0.0800 0.1000 0.1250 0.1600 0.2000 Columns 22 through 28 0.2500 0.3150 0.4000 0.5000 0.6300 0.8000 1.0000 Column 29 1.2500 y = Columns 1 through 7 109.5113 104.2279 99.0779 94.1773 89.9635 85.9434 82.0534 Columns 8 through 14 78.6546 75.5635 72.4743 69.8643 67.5348 65.3917 63.4510 Columns 15 through 21 62.0512 60.8150 59.8867 60.0116 62.1549 63.1894 59.9616 Columns 22 through 28 57.2552 56.4239 57.5699 60.8882 66.3613 71.6640 73.1551 Column 29 68.6308 ...
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This note was uploaded on 01/15/2011 for the course MATH 345 taught by Professor Staff during the Spring '08 term at Ohio State.
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