How to Perform Continuous Interpolation on Matlab

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Linear interpolation, also called simply interpolation or "lerping,"^[1] is the ability to deduce a value between two values explicitly stated in a table or on a line graph. While many people can interpolate on an intuitive basis, the article below shows the formalized mathematical approach behind the intuition.

Steps

1. 1

   Identify the value for which you want to find a corresponding value. Interpolation can be used for such things as finding a logarithm or trigonometric function value or for the corresponding gas pressure or volume for a given temperature in chemistry.^[2] Because scientific calculators have largely replaced logarithmic and trigonometric tables, we'll use as our example for finding an interpolated value that of finding the pressure of a gas for a temperature whose value isn't listed in the reference table or as a graph point.

   • For the equation we'll derive, we'll represent the value that we want to find a corresponding value for as ''x'' and the interpolated value we want to find as ''y''. (We'll use these labels, because on a graph, the value we know would be plotted on the horizontal, or x-axis, and the value we're trying to find would be plotted on the vertical, or y-axis.)
   • Our ''x'' value will be the temperature of the gas, which for this example will be 37 °C (99 °F).

2. 2

   Find the closest values below and above the value of x in the table or on the graph. Our reference table doesn't give a gas pressure for 37 °C (99 °F), but it does list values for 30 °C (86 °F) and 40 °C (104 °F). The gas pressure at 30 °C (86 °F) is 3 kilopascals (kPa) and the pressure at 40 °C (104 °F) is 5 kPa.

   • Because we're representing the temperature of 37 °C (99 °F) with ''x'', we'll represent the temperature of 30 degrees as ''x₁'' and the value of 40 degrees as ''x₂''.
   • Because we're representing the pressure were trying to find with ''y'', we'll represent the pressure of 3 kPa at 30 °C (86 °F) as ''y₁'' and the pressure of 5 kPa at 40 °C (104 °F) as ''y₂''.

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3. 3

   Find the interpolated value mathematically. The equation for finding the interpolated value can be written as y = y₁ + ((x – x₁)/(x₂ - x₁) * (y₂ - y₁))^[3]

   • Plugging in the values for x, x₁, and x_/2 in their places gives (37 – 30)/(40 -30), which reduces to 7/10 or 0.7.
   • Plugging in the values for y₁ and y₂ at the end of the equation gives (5 – 3) or 2.
   • Multiplying 0.7 by 2 gives a product of 1.4. Adding 1.4 to the value of y₁, or 3, gives a value of 4.4 kPa. Checking this against our original values, 4.4 falls between 3 kPa at 30 °C (86 °F) and 5 kPa at 40 °C (104 °F), and because 37 is closer to 40 than it is to 30, the result should be closer to 5 kPa than it is to 3 kPa.

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• Question

  What is the interpolation of 27.5?

  

  When you are doing a linear interpolation, you are approximating the function between points as a line. From there, you can find the estimated value at any point between them. Therefore, you cannot ask the interpolation of a single number - rather, you need multiple numbers to determine your line, and from there you can ask what value you would expect at a point.

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• If you're good at estimating distances on graphs, you can do a rough interpolation by eyeballing the position of a point against the x-axis to determine the corresponding y-value.^[4] If the example above had been plotted on a graph where the x-axis was marked in units of 10 °C (50 °F) and the y-axis in units of 1 kPa, you could approximate the position of 37 °C (99 °F) and then look at the y-axis to find a point not quite halfway between 4 and 5 kPa. The equation above simply formalizes the thinking process and provides a more precise value.

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• Related to interpolation is extrapolation, which is finding a corresponding value for a given value outside of the range of values listed in a table or concretely plotted on a graph.^[5]

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