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rect10_vallillo_HW02

# rect10_vallillo_HW02 - Homework Problem#1 clear;clc...

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% Homework Problem #1 clear;clc disp('Problem 1 - Interpolation: Answers') volume=1:6;pressure=[2494 1247 831 623 499 416];a=interp1(volume,pressure,3.8);b=interp1(volume,pressure,3.8,'spline');c=in terp1(pressure,volume,1000); %variable definitions fprintf('a) Using linear interpolation the estimated pressure at a volume of 3.8 cubic meters is %.3f kPa.\n',a) fprintf('b) Using cubic spine interpolation the estimated pressure at a volume of 3.8 cubic meters is %.3f kPa.\n',b) fprintf('c) Using linear interpolation the estimated volume at 1000 kPa is %.3f cubic meters.\n',c) % Homework Problem #2 disp('Problem 2 - Curve Fitting: Answer in Figure 1') % calculate the coefficients of 1st, 2nd, 3rd, and 4th order coef1=polyfit(volume,pressure,1); coef2=polyfit(volume,pressure,2); coef3=polyfit(volume,pressure,3); coef4=polyfit(volume,pressure,4); x=1:.2:6; y1=polyval(coef1,x);y2=polyval(coef2,x);y3=polyval(coef3,x);y4=polyval(coef4,x); figure(1) plot(volume,pressure,'o',x,y1,x,y2,x,y3,x,y4); title('Curve Fitting of Volume vs Pressure');xlabel('Volume - cubic meters');ylabel('Pressure - kPa');legend('original data','first order','second order','third order','fourth
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