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# Contents

Detailed Interpolation of a Single Point

The Interpolation Functions interp1

# Detailed Interpolation of a Single Point

Suppose we have a rocket and have recorded it’s speed during the list of times.

$\displaystyle \begin{array}{*{20}{c}} {\text{time (s)}} & {\text{speed (m/s)}} \\ 0 & 0 \\ {10} & {227.04} \\ {15} & {362.78} \\ {20} & {517.35} \\ {22.5} & {602.97} \\ {30} & {901.67} \end{array}$

And we want to know it’s speed at time t=16 s. There are a number of ways we can estimate this value or interpolate this value from the given data.

## Nearest Point Interpolation (1 data point)

Simply by looking at the data we can see that the nearest time point to t=16 s is t=15 s.

$\displaystyle \begin{array}{*{20}{c}} {\text{time (s)}} & {\text{speed (m/s)}} \\ 0 & 0 \\ {10} & {227.04} \\ {15} & {362.78} \\ {20} & {517.35} \\ {22.5} & {602.97} \\ {30} & {901.67} \end{array}$

We can take the speed at t=15 s and use it as a first hand estimate for the speed at t=16 s and this is known as single point interpolation.

St16=St15 St16=362.78

## Linear Interpolation (2 data points)

Suppose we want to be a bit more accurate we can instead take two points and calculate the equation for a line. We can then plug in our x data point and use the equation of the straight line to estimate our unknown y data point. Looking at the data for t=16 s the nearest two time points are t=15 s and t=20 s.

$\displaystyle \begin{array}{*{20}{c}} {\text{time (s)}} & {\text{speed (m/s)}} \\ 0 & 0 \\ {10} & {227.04} \\ {15} & {362.78} \\ {20} & {517.35} \\ {22.5} & {602.97} \\ {30} & {901.67} \end{array}$

The equation for a straight line is:

$\displaystyle S(t)={{a}_{0}}+{{a}_{1}}t$

For two times we can write this out in matrix form:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {S({{t}_{1}})} \\ {S({{t}_{2}})} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1 & {{{t}_{1}}} \\ 1 & {{{t}_{2}}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \end{array}} \right]$

Now we can substitute in the known values:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {362.78} \\ {517.35} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1 & {15} \\ 1 & {20} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \end{array}} \right]$

We can save the known data as variables S and T:

S=[362.78;517.35]

$\displaystyle S=\left[ {\begin{array}{*{20}{c}} {362.78} \\ {517.35} \end{array}} \right]$

T=[1,15;1,20]

$\displaystyle \text{T}=\left[ {\begin{array}{*{20}{c}} 1 & {15} \\ 1 & {20} \end{array}} \right]$

And call the unknown variable U:

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \end{array}} \right]$

We know that:

$\displaystyle \text{S}=\text{T}*\text{U}$

However we want to solve for U. We can get U on it’s own by multiplying throughout by the inverse matrix of T:

$\displaystyle {{\text{T}}^{{-1}}}\text{S}={{\text{T}}^{{-1}}}\text{T}*\text{U}$

Since a matrix multiplied by it’s inverse gives the identity matrix and a matrix multipied by the identity matrix gets unchanged:

$\displaystyle {{\text{T}}^{{-1}}}\text{S}=\text{U}$

Which can be reordered:

$\displaystyle \text{U=}{{\text{T}}^{{-1}}}\text{S}$

Thus to solve for U we need the inverse matrix of T:

$\displaystyle \text{T}=\left[ {\begin{array}{*{20}{c}} 1 & {15} \\ 1 & {20} \end{array}} \right]$

This can be calculated using the function Minv=inv(M)

Tinv=inv(T)

$\displaystyle \text{Tinv}=\left[ {\begin{array}{*{20}{c}} 4 & {-3} \\ {-0.2} & {0.2} \end{array}} \right]$

Thus we can get the unknown coefficients of U using:

U=Tinv*S

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} 4 & {-3} \\ {-0.2} & {0.2} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {362.78} \\ {517.35} \end{array}} \right]$

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {-100.93} \\ {30.914} \end{array}} \right]$

i.e. in the limits of 15<t<20 a0=-100.93 and a1=30.914

Instead of using the inverse matrix we could also use left division

U=T\S

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} 1 & {15} \\ 1 & {20} \end{array}} \right]\backslash \left[ {\begin{array}{*{20}{c}} {362.78} \\ {517.35} \end{array}} \right]$

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {-100.93} \\ {30.914} \end{array}} \right]$

Now that we have the coefficients we can calculate the S(16)

St16=[1,16]*U

$\displaystyle \text{St16}=\left[ {\begin{array}{*{20}{c}} 1 & {16} \end{array}} \right]*U=\left[ {\begin{array}{*{20}{c}} 1 & {16} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {-100.93} \\ {30.914} \end{array}} \right]$

$\displaystyle \text{St16}=393.69$

We can check the equation using the known data:

$\displaystyle \text{Stx}=\left[ {\begin{array}{*{20}{c}} 1 & {x} \end{array}} \right]*U=\left[ {\begin{array}{*{20}{c}} 1 & {x} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {-100.93} \\ {30.914} \end{array}} \right]$

$\displaystyle \begin{array}{*{20}{c}} {\text{time (s)}} & {\text{speed (m/s)}} & {\begin{array}{*{20}{c}} {\text{calculated}} \\ {\text{speed (m/s)}} \end{array}} \\ 0 & 0 & {-100.93} \\ {10} & {227.04} & {208.93} \\ {15} & {362.78} & {362.78} \\ {20} & {517.35} & {517.35} \\ {22.5} & {602.97} & {594.63} \\ {30} & {901.67} & {826.49} \end{array}$

Unsurprisingly the values at the limits t=15 and t=20 give exact matches. There is a slight deviation just outside this interval and a massive deviation substantially away from this interval.

## Quadratic Interpolation (3 data points)

Suppose we want to be a bit more accurate we can instead take three points and calculate a quadratic equation. We can then plug in our x data point and use the quadratic equation to estimate our unknown y data point. Looking at the data for t=16 s the nearest three time points are t= 10 s, t=15 s and t=20 s.

$\displaystyle \begin{array}{*{20}{c}} {\text{time (s)}} & {\text{speed (m/s)}} \\ 0 & 0 \\ {10} & {227.04} \\ {15} & {362.78} \\ {20} & {517.35} \\ {22.5} & {602.97} \\ {30} & {901.67} \end{array}$

The equation for a quadratic function is:

$\displaystyle S(t)={{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}}$

For three times we can write this out in matrix form:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {S({{t}_{1}})} \\ {S({{t}_{2}})} \\ {S({{t}_{3}})} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1 & {{{t}_{1}}} & {t_{1}^{2}} \\ 1 & {{{t}_{2}}} & {t_{2}^{2}} \\ 1 & {{{t}_{3}}} & {t_{3}^{2}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \\ {{{a}_{2}}} \end{array}} \right]$

Now we can substitute in the known values:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {227.04} \\ {362.78} \\ {515.35} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1 & {10} & {{{{10}}^{2}}} \\ 1 & {15} & {{{{15}}^{2}}} \\ 1 & {20} & {{{{20}}^{2}}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \\ {{{a}_{2}}} \end{array}} \right]$

We can save the known data as variables S and T:

S=[227.04;362.78;517.35]

$\displaystyle \text{S}=\left[ {\begin{array}{*{20}{c}} {227.04} \\ {362.78} \\ {515.35} \end{array}} \right]$

T=[1,10,10^2;1,15,15^2;1,20,20^2]

$\displaystyle \text{T}=\left[ {\begin{array}{*{20}{c}} 1 & {10} & {{{{10}}^{2}}} \\ 1 & {15} & {{{{15}}^{2}}} \\ 1 & {20} & {{{{20}}^{2}}} \end{array}} \right]$

And call the unknown variable U:

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \\ {{{a}_{2}}} \end{array}} \right]$

To solve for U we use:

U=T\S

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} 1 & {10} & {{{{10}}^{2}}} \\ 1 & {15} & {{{{15}}^{2}}} \\ 1 & {20} & {{{{20}}^{2}}} \end{array}} \right]\backslash \left[ {\begin{array}{*{20}{c}} {227.04} \\ {362.78} \\ {515.35} \end{array}} \right]$

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {12.05} \\ {17.733} \\ {0.3766} \end{array}} \right]$

i.e. in the limits of 10<t<20 a0=12.05, a1=17.733 and a2=0.3766

Now that we have the coefficients we can calculate the S(16)

St16=[1,16,16^2]*U

$\displaystyle \text{St16}=\left[ {\begin{array}{*{20}{c}} 1 & {16} & {{{{16}}^{2}}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {12.05} \\ {17.733} \\ {0.3766} \end{array}} \right]$

$\displaystyle \text{St16}=\text{392}\text{.19}$

## Cubic Interpolation (4 data points)

Suppose we want to be a bit more accurate we can instead take four points and calculate a cubic equation. We can then plug in our x data point and use the cubic equation to estimate our unknown y data point. Looking at the data for t=16 s the nearest three time points are t= 10 s, t=15 s, t=20 s and t=22.5 s.

$\displaystyle \begin{array}{*{20}{c}} {\text{time (s)}} & {\text{speed (m/s)}} \\ 0 & 0 \\ {10} & {227.04} \\ {15} & {362.78} \\ {20} & {517.35} \\ {22.5} & {602.97} \\ {30} & {901.67} \end{array}$

The equation for a cubic function is:

$\displaystyle S(t)={{a}_{0}}+{{a}_{1}}t+{{a}_{2}}{{t}^{2}}+{{a}_{3}}{{t}^{3}}$

For four times we can write this out in matrix form:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {S({{t}_{1}})} \\ {S({{t}_{2}})} \\ {S({{t}_{3}})} \\ {S({{t}_{4}})} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1 & {{{t}_{1}}} & {t_{1}^{2}} & {t_{1}^{3}} \\ 1 & {{{t}_{2}}} & {t_{2}^{2}} & {t_{2}^{3}} \\ 1 & {{{t}_{3}}} & {t_{3}^{2}} & {t_{3}^{2}} \\ 1 & {{{t}_{4}}} & {t_{4}^{2}} & {t_{4}^{2}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \\ {{{a}_{2}}} \\ {{{a}_{3}}} \end{array}} \right]$

Now we can substitute in the known values:

$\displaystyle \left[ {\begin{array}{*{20}{c}} {227.04} \\ {362.78} \\ {515.35} \\ {602.97} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 1 & {10} & {{{{10}}^{2}}} & {{{{10}}^{3}}} \\ 1 & {15} & {{{{15}}^{2}}} & {{{{15}}^{3}}} \\ 1 & {20} & {{{{20}}^{2}}} & {{{{20}}^{3}}} \\ 1 & {22.5} & {{{{22.5}}^{2}}} & {{{{22.5}}^{3}}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \\ {{{a}_{2}}} \\ {{{a}_{3}}} \end{array}} \right]$

We can save the known data as variables S and T:

S=[227.04;362.78;517.35;602.97]

$\displaystyle \text{S}=\left[ {\begin{array}{*{20}{c}} {227.04} \\ {362.78} \\ {515.35} \\ {602.97} \end{array}} \right]$

T=[1,10,10^2,10^3;1,15,15^2,15^3;1,20,20^2,20^3;1,22.5,22.5^2,22.5^3]

$\displaystyle \text{T}=\left[ {\begin{array}{*{20}{c}} 1 & {10} & {{{{10}}^{2}}} & {{{{10}}^{3}}} \\ 1 & {15} & {{{{15}}^{2}}} & {{{{15}}^{3}}} \\ 1 & {20} & {{{{20}}^{2}}} & {{{{20}}^{3}}} \\ 1 & {22.5} & {{{{22.5}}^{2}}} & {{{{22.5}}^{3}}} \end{array}} \right]$

And call the unknown variable U:

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {{{a}_{0}}} \\ {{{a}_{1}}} \\ {{{a}_{2}}} \\ {{{a}_{3}}} \end{array}} \right]$

To solve for U we use:

U=T\S

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} 1 & {10} & {{{{10}}^{2}}} & {{{{10}}^{3}}} \\ 1 & {15} & {{{{15}}^{2}}} & {{{{15}}^{3}}} \\ 1 & {20} & {{{{20}}^{2}}} & {{{{20}}^{3}}} \\ 1 & {22.5} & {{{{22.5}}^{2}}} & {{{{22.5}}^{3}}} \end{array}} \right]\backslash \left[ {\begin{array}{*{20}{c}} {227.04} \\ {362.78} \\ {515.35} \\ {602.97} \end{array}} \right]$

$\displaystyle \text{U}=\left[ {\begin{array}{*{20}{c}} {-4.254} \\ {21.2655333} \\ {0.13204} \\ {0.0054347} \end{array}} \right]$

i.e. in the limits of 10<t<22.5 a0=4.254, a1=21.2655333, a2=0.13204 and a3=0.0054347

Now that we have the coefficients we can calculate the S(16)

St16=[1,16,16^2,16^3]*U

$\displaystyle \text{St16}=\left[ {\begin{array}{*{20}{c}} 1 & {16} & {{{{16}}^{2}}} & {{{{16}}^{3}}} \end{array}} \right]*\left[ {\begin{array}{*{20}{c}} {-4.254} \\ {21.2655333} \\ {0.13204} \\ {0.0054347} \end{array}} \right]$

$\displaystyle \text{St16}=392.06$

## Comparison

We can compare our interpolation for t=16 using the four different methods. As we can see when we are interpolating with higher order polynomials the result begins to converge as expected:

$\displaystyle \begin{array}{*{20}{c}} {\text{method}} & {\begin{array}{*{20}{c}} {\text{no of data}} \\ {\text{points}} \end{array}} & {\text{St16}} \\ {\text{nearest}} & 1 & {362.78} \\ {\text{linear}} & 2 & {393.69} \\ {\text{quadratic}} & 3 & {392.19} \\ {\text{cubic}} & 4 & {392.06} \end{array}$

# Interpolation of a Data Set

As we seen it is quite involved interpolating the data for a single data point. If we want to interpolate 100’s or even 1000’s of data points manually going through them all like above it would take all day. Luckily Octave/MATLAB have inbuilt interpolation functions.

intery=interp1(originalx,originaly,xdatapoints,'Method')

We can type in the data using Octave/MATLAB.

data=[0,0;
10,227.04;
15,362.78;
20,517.35;
22.5,602.97;
30,901.67]



We could also use Excel to copy and paste in the data. In MATLAB which is more polished, we can directly go to the variable editor, right click it and create a new variable and then name accordingly. This option isn’t available in Octave however we can quickly create a variable, assign in to a scalar and then copy and paste in a matrix in the variable editor.

For instance typing in:

data=1;

Then clicking the new variable data in the workspace to open it up in the variable editor:

We can right click the top cell of data in the variable editor and we can right click Paste Table to copy the Excel Data through to Octave:

Now we have our data variable:

Next we need to make the new x data. For this I am going to make a new variable called dataint and I want the first column to go from 1 to 30. Here logical indexing is used when creating the variable dataint. The (:,1) means we are creating data in all rows but using the first column. The [1:30]' is a column vector ranging from 1 to 30 in steps of 1. If you are unsure about logical indexing see my earlier guide Introduction to Arrays.

dataint(:,1)=[1:30]'

Now we are looking to interpolate the data:

intery=interp1(originalx,originaly,xdatapoints,'Method')

Our original x data originalx is the first column of data our original y data originaly is the second column of data

originalx=data(:,1)

originaly=data(:,2)

Our x datapoints xdatapoints are the first column in dataint

xdatapoints=dataint(:,1)

We will start with Nearest Interpolation meaning instead of typing in 'Method' we type in 'Nearest'. We can now get intery:

intery=interp1(originalx,originaly,xdatapoints,'Nearest')

We can clear up our code here:

intery=interp1(originalx,originaly,xdatapoints,'Nearest')

By substituting these in directly:

originalx=data(:,2)

originaly=data(:,2)

xdatapoints=dataint(:,1)

This becomes

intery=interp1(data(:,1),data(:,2),dataint(:,1),'Nearest')

Now it is more likely we want the y data to be alongside the xdatapoints so instead of having the separate output variable intery we can add it as a second column to dataint by using:

dataint(:,2)=interp1(data(:,1),data(:,2),dataint(:,1),'Nearest')

We can press the [↑] key to get our last line of code and easily modify the line of code to print linear interpolated data to column 3. For linear interpolation the method is 'Linear':

dataint(:,3)=interp1(data(:,1),data(:,2),dataint(:,1),'Linear')

MATLAB/Octave’s interp1 function doesn’t have an inbuilt quadratic interpolation so I will go directly to cubic interpolation and here the method is 'Pchip' , this time I want the data in the 4th column.

dataint(:,4)=interp1(data(:,1),data(:,2),dataint(:,1),'Pchip')

Since we were interested at the value at t=16 which is the 16th row we can output the 16th row into the command window:

t16row=dataint(16,:)

These are very similar to the values manually calculated, the differences are likely due to rounding.

The code above can be cleared up and put in a script file.

% Data
data=[0,0;
10,227.04;
15,362.78;
20,517.35;
22.5,602.97;
30,901.67];

% Lower and Upper Limits for interpolation
m=1;
n=30;