Question

B.1 Write a MATLAB user-defined function for interpolation with linear splines Name the function Yint = Linear Splines, y, rint), where the input ar- guments and y are vectors with the coordinates of the data points and rint is the coordinate of the interpolated point. The output argument Yint is the y value of the interpolated point. Also write a program in a script file that plots the data points and the curve of the Linear splines that interpolates the data.

Answer 1

The code is shown. The functions f1 f2 f3 are three splines for which function is interpolated. x and y are sample points

clc;clear all
x=[8 11 15 18];
y=[5 9 10 8];
yint=LinearSpline(x,y,12.7)
x1=8:0.01:11;
xi=12.7;
f1=(-5/3).*(x1-11)+(9/2).*(x1-8);
x2=11:0.01:15;
f2=(-9/4).*(x2-15)+(10/4).*(x2-11);
x3=15:0.01:18;
f3=(-10/3).*(x3-18)+(8/3).*(x3-15);
plot(x,y,'*',xi,yint,x1,f1,x2,f2,x3,f3)
legend('Points','Interpolated point','Spline 1','Spline 2','Spline 3')

function yint=LinearSpline(x,y,xint)
n=length(x);
for(i=1:(n-1))
if( xint>x(i) && xint<x(i+1))
a=i;
break
end
end
yint=(xint-x(a+1))/(x(a)-x(a+1))*y(a)+(xint-x(a))/(x(a+1)-x(a))*y(a+1);
end


yint = 9.4250 clc;clear all X=[8 11 15 18]; y=[5 9 10 8]; yint=Linear Spline(x,y,12.7) x1=8:0.01:11; xi=12.7; f1=(-5/3). *(x1-11)+(9/2). *(x1-8); x2=11:0.01:15; f2=(-9/4). *(x2-15)+(10/4). * (x2-11); *3=15:0.01:18; f3=(-10/3).*(x3-18)+(8/3).*(x3-15); plot(x,y,'*',xi,yint,x1, f1,x2, f2, x3, f3) legend('Points', 'Interpolated point', 'Spline 1', 'Spline 2', 'Spline 3') function yint=LinearSpline(x,y,xint) n=length(x); for(i=1:(n-1)) if( xint>x(i) && xint<x(i+1)) a=i; break end end yint=(xint-x(a+1))/(x(a)-x(a+1))*y(a)+(xint-x(a))/(x(a+1)-x(a))*y(a+1); end

Figure File Edit View Insert Tools Desktop Window Help * Points Interpolated point Spline 1 Spline 2 Spline 3 18