# 16.1.1: Linear Interpolation with corrected format

• • Carey Smith
• Oxnard College
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This is from Serhat Beyenir, A_brief-introduction-to-engineering-computation-with-matlab-11.8.pdf

Much of the formatting has been corrected in this copy.

Linear interpolation is one of the most common techniques for estimating values between two given data points. For example, when using steam tables, we often have to carry out interpolations. With this technique, we assume that the function between the two points is linear. MATLAB has a built-in interpolation function.

## The interp1 Function

Given an x-y table, y_new = interp1(x,y,x_new) interpolates to find a y_new. Consider the following examples:

##### Example $$\PageIndex{1}$$

Example 5.1
To demonstrate how the interp1 function works, let us create an x-y table with the following
commands;
x = 0:5; y = [0,20,60,68,77,110];

To tabulate the output, we can use
[x',y']

The result is
ans = 0 0 1 20 2 60 3 68 4 77 5 110

Suppose we want to find the corresponding value for 1.5 or interpolate for 1.5.

Using y_new = interp1(x,Y,x_new) syntax, let us type in:
y_new=interp1(x,y,1.5) y_new = 40

.

##### Example $$\PageIndex{2}$$

The table we created above has only 6 elements in it and suppose we need a more detailed table.
In order to do that, instead of a single new x value, we can define an array of new x values.

The interp1 function returns an array of new y values:
new_x = 0:0.2:5; new_y = interp1(x,y,new_x);

Let's display this table
[new_x',new_y']

The result is

ans = 0 0 0.2000 4.0000 0.4000 8.0000 0.6000 12.0000 0.8000 16.0000 1.0000 20.0000 1.2000 28.0000 1.4000 36.0000 1.6000 44.0000 1.8000 52.0000 2.0000 60.0000 2.2000 61.6000 2.4000 63.2000 2.6000 64.8000 2.8000 66.4000 3.0000 68.0000 3.2000 69.8000 3.4000 71.6000 3.6000 73.4000 3.8000 75.2000 4.0000 77.0000 4.2000 83.6000 4.4000 90.2000 4.6000 96.8000 4.8000 103.4000 5.0000 110.0000

.

##### Example $$\PageIndex{3}$$

Using the table below, find the internal energy of steam at 215 ºC and the temperature if the
internal energy is 2600 kJ/kg (use linear interpolation).

First let us enter the temperature and energy values
temperature = [100, 150, 200, 250, 300, 400, 500]; energy = [2506.7, 2582.8, 2658.1, 2733.7, 2810.4, 2967.9, 3131.6]; [temperature',energy']

This returns
ans = 1.0e+003 * 0.1000 2.5067 0.1500 2.5828 0.2000 2.6581 0.2500 2.7337 0.3000 2.8104 0.4000 2.9679 0.5000 3.1316

Issue the following for the first question:
new_energy = interp1(temperature,energy,215)

returns
new_energy = 2.6808e+003

Now, type in the following for the second question:
new_temperature = interp1(energy,temperature,2600)

returns
new_temperature = 161.4210