Give an x-y table, y_new = interp1(x,y,x_new) interpolates to find y_new. Consider the following examples:
Example 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];
[x',y']
ans =
0 0
1 20
2 60
3 68
4 77
5 110
y_new = interp1(x,Y,x_new) syntax, let us type in:
y_new=interp1(x,y,1.5)
y_new =
40
Example 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);
[new_x',new_y']
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 3
Using the table below, find the internal energy of steam at 215 ˚C and the temperature if the internal energy is 2600 kJ/kg (use linear interpolation).
| Temperature [C] | Internal Energy [kJ/kg] |
|---|---|
| 100 | 2506.7 |
| 150 | 2582.8 |
| 200 | 2658.1 |
| 250 | 2733.7 |
| 300 | 2810.4 |
| 400 | 2967.9 |
| 500 | 3131.6 |
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']
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
new_energy = interp1(temperature,energy,215)
new_energy =
2.6808e+003
new_temperature = interp1(energy,temperature,2600)
new_temperature =
161.4210
