Course No: EEE 212

Experiment No: 03

Name of the experiment:

INTERPOLATION

Name: Md. Arifur Rahman

Date of performance: 11-03-2015 Student ID: 1206130

Date of submission: 18-03-2015 Department: EEE

Level 2 / Term 2

Section: B ( B-2 )

Partners ID: 1206121

Group No:15

EXERCISE 2 -

Plot the curve corresponding to the following table using linear Interpolation.

x f(x)

0 0

1 0.8415 2 0.9093

3 0.1411

4 -0.7568

5 -0.9589

6 -0.2794

Solution % given data points

x = [0 1 2 3 4 5 6];

y = [0 0.8415 0.9093 0.1411 -0.7568 -0.9589 -0.2794];

plot(x,y,'ko'), hold;

for i=1:length(x)-1

X(1) = x(i);

X(2) = x(i+1);

sum = poly(x(i+1))*y(i)/(x(i)-x(i+1)) -

poly(x(i))*y(i+1)/(x(i)-x(i+1));

% selecting two element

matrix p=polyval(sum, X);

plot(X,p), grid on, hold on;

end

EXERCISE 3 y = sin x; x = 0:10, x = 0:0.25:10; construct the interpolant y and plot.

Solution clear all;

close

all; clc;

% creating the data points from given

function. x1 = 0:10;

x2 = 0:0.25:10;

y1 = sin(x1);

y2 = sin(x2);

% marking the data points on plot.

subplot(2,1,1),plot(x1,y1,'o'), hold on;

subplot(2,1,2),plot(x2,y2,'o'), hold on;

% Interpolation for x1 and y1.

for i=1:length(x1)-1

X1(1) = x1(i);

X1(2) = x1(i+1);

sum1 = poly(x1(i+1))*y1(i)/(x1(i)-x1(i+1))

- poly(x1(i))*y1(i+1)/(x1(i)-x1(i+1));

p1=polyval(sum1, X1);

subplot(2,1,1), plot(X1,p1), title('Linear Interpolation for x

= 0:10'), grid on, hold on; end

% Interpolation for x2 and y2. for i=1:length(x2)-1

X2(1) = x2(i);

X2(2) = x2(i+1);

sum2 = poly(x2(i+1))*y2(i)/(x2(i)-x2(i+1))

- poly(x2(i))*y2(i+1)/(x2(i)-x2(i+1));

p2=polyval(sum2, X2);

subplot(2,1,2), plot(X2,p2), title('Linear Interpolation for x

= 0:0.25:10'), grid on, hold on; end

Plot: