Question

Using MATLAB Linear Interpolation:

The following equation illustrates how to calculate an output for any input, using linear interpolation. The input must be within the bounds of the data set. The constants come from the data set; they are the nearest data points that bound the original input.

(constants from data set--> y1, x1, y2)

(output--> y) (input--> x)

Create a MATLAB function file that calculates the linearly interpolated y-value (output) provided an x-value (input). The function should linearly interpolate to find y given any x using the air velocity and thermal resistance data below. You are required to use a for-loop: Use a for loop inside this function file that loops through the rows of the data to determines which two rows of the data an input falls between. The row that is the lower bound will become the x₁ and y₁ values; the row that is the upper bound will become x₂ and y₂.

Calculate y given x, x₁, y₁, x₂, and y₂.

Test the code by calling this user-defined function inside the for loop of a script for values from 0 to 1700 (air velocity) and plotting the results.

Example:

x= 480

x₁=400 (Lower Bound)

x₂=600 (Upper Bound)

y₁=113.6, y₂=93.1

Resulting value of y: 105.4

Answer 1

%linear interpolate . m function y=linear interpolate (air velocity, thermal resist, x) x1-0; x2-0; y1-0: y2-0; k=0 ; %lower and upper bound for k1-1: length (air velocity) xair velocity (kl) x2-aivelocity(k1); xl-air velocity (k1-1); y2-thermal resist (kl) ; y1thermal resist(k1-1);

end end sdisplay y value y ylt x-x1)((y2-y1)/ (x2-x1))) end

Smain.m x-480 air velocity I0,200,400,600,800,1000,1200,1400,1600,1800] thermal_resist-[373.0,156.1,113.6,93.1,81.5,73.7,67.2,62.4,58.3 55.01 ylinear interpolatelair velocitvtheral resist x) %plot xlabell'Air Velocity') zlabell'Thermal Resistance) elatlaiFxelositxthermal xesist);

output:

У 105. 4θ

350 300 订)() 250 200 150 1 500 50 1000 1500 2000

copyable code:

%linear_interpolate.m

function y=linear_interpolate(air_velocity,thermal_resist,x)

x1=0;

x2=0;

y1=0;

y2=0;

k=0;

%lower and upper bound

for k1=1:length(air_velocity)

if x<air_velocity(k1)

x2=air_velocity(k1);

x1=air_velocity(k1-1);

y2=thermal_resist(k1);

y1=thermal_resist(k1-1);

end

end

%display y value

y=y1+ ((x-x1)*((y2-y1)/(x2-x1)));

end

%main.m

x=480;

air_velocity=[0,200,400,600,800,1000,1200,1400,1600,1800];

thermal_resist=[373.0,156.1,113.6,93.1,81.5,73.7,67.2,62.4,58.3,55.0];

y=linear_interpolate(air_velocity,thermal_resist,x)

%plot

xlabel('Air Velocity');

ylabel('Thermal Resistance');

plot(air_velocity,thermal_resist);