Download interp_primer

Interpolation
Interpolation
Data Point - A data point is a value at some position.
Consider position along the horizontal axis for now.
Then the picture below shows four data points, each value
shown as a height of an x at its horizontal position.
x
x
x
x
Interpolation
Grid Line - An imaginary vertical line at some position
which may not correspond to a data point's position.
A value is associated with a grid line – we use circles
in the figure to illustrate.


x

x


x





 
 
x

Interpolation
A graph – Given a collection of grid lines, with values,
Matlab will fill the spaces between grid lines with straight
lines connecting the dots representing values. For example,
see below:


x

x


x





 
 
x

Interpolation
This example – The line below does a bad job of
representing the four data points: it is jumpy and it does
not even come close to the data points. We need a way
to find grid line values that allow Matlab to better
approximate the data.


x

x


x





 
 
x

Interpolation
Nearest Neighbor – The simplest, but hardly the best, way
to get grid line values is to use the value of the nearest
data point. For example, consider this grid line:
x
x
x
x
Interpolation
Nearest Neighbor – The simplest, but hardly the best, way
to get grid line values is to use the value of the nearest
data point. Look for the nearest data point to the grid line:
0.35
x
0.83
x
x
x
0.21
2.05
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Nearest Neighbor – The simplest, but hardly the best, way
to get grid line values is to use the value of the nearest
data point. The closest data point is the 2nd from left – the
value of the grid line is the value of that point:
x
x

x
x
Interpolation
Nearest Neighbor – The simplest, but hardly the best, way
to get grid line values is to use the value of the nearest
data point. Continue for all the grid points.


x 


x

x 



x



Interpolation
Nearest Neighbor – The simplest, but hardly the best, way
to get grid line values is to use the value of the nearest
data point. Using plot(...) Matlab shows this:
x
x
x
x
Interpolation
Variables – 1: The data will have to reside somewhere.
Matlab likes matrices so put the data in a two dimensional
array: 1st column is the x position, 2nd column holds the
data values, the number of rows is the number of data pts.
Call the matrix dpts.
x
x
dpts(1,2)
dpts(1,1)
x
x
0
Interpolation
Variables – 1: The data will have to reside somewhere.
Matlab likes matrices so put the data in a two dimensional
array: 1st column is the x position, 2nd column holds the
data values, the number of rows is the number of data pts.
Call the matrix dpts.
x
x
x
dpts(2,2)
dpts(2,1)
0
x
Interpolation
Variables – 1: The data will have to reside somewhere.
Matlab likes matrices so put the data in a two dimensional
array: 1st column is the x position, 2nd column holds the
data values, the number of rows is the number of data pts.
Call the matrix dpts.
x
x
x
dpts(3,2)
dpts(3,1)
0
x
Interpolation
Variables – 2: The position and value of grid lines will
reside somewhere. Since we will use plot(x,y) to plot the
result, it is convenient to define vectors x and y to mean
position and value, respectively. Since we can choose
the grid line positions, we can set that vector right away.
X = 0:0.2:10;
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Variables – 2: The position and value of grid lines will
reside somewhere. Since we will use plot(x,y) to plot the
result, it is convenient to define vectors x and y to mean
position and value, respectively. Since we can choose
the grid line positions, we can set that vector right away.
X = 0:0.2:10;
x
x
x
x
and the length of the y vector must match this
so we can do
y = zeros(length(x));
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Writing the code – For each grid line we need to determine
which data point is nearest, then set the value of the grid
line to the value of that data point.
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Writing the code – For each grid line we need to determine
which data point is nearest, then set the value of the grid
line to the value of that data point. Start off with a simpler
problem: find data point nearest to some grid line
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Ask questions – To find distances from grid line to a data
point I need to known the position of the grid line and the
position of the data point.
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Ask questions – To find distances from grid line to a data
point I need to known the position of the grid line and the
position of the data point. We said the grid line position
is in a vector called x. But to access an element of x I need
to provide an index (i.e. x(i)). For now, we use this and
figure out how to get i later.
x
x
x
x
x(i)
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Ask questions – To find distances from grid line to a data
point I need to known the position of the grid line and the
position of the data point. Consider the position of the 1st
data point – we said it is dpts(1,1).
x
x
dpts(1,1)
x
x
x(i)
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Ask questions – To find distances from grid line to a data
point I need to known the position of the grid line and the
position of the data point. Now to find the distance between
grid line and data point 1 I can use abs(x(i)-dpts(1,1)).
abs(x(i)-dpts(1,1))
x
x
dpts(1,1)
x
x
x(i)
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Ask questions – To find distances from grid line to a data
point I need to known the position of the grid line and the
position of the data point. But I need to do this for all the
data points. Moreover, I need to compare results and pick
the data point for which abs(...) is lowest.
abs(x(i)-dpts(1,1))
x
abs(x(i)-dpts(3,1))
x
x
abs(x(i)-dpts(2,1))
x
abs(x(i)-dpts(4,1))
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – I can use a for loop to compute all the abs(...)
terms on each interation I can compare another data point.
But to write the for loop I need to know the number of points.
abs(x(i)-dpts(1,1))
x
abs(x(i)-dpts(3,1))
x
x
abs(x(i)-dpts(2,1))
x
abs(x(i)-dpts(4,1))
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – I will figure out how to get it later. For now I
will suppose a variable called npts holds that number. So
I write:
for j=1:npts
abs(x(i) – dpts(j,1));
end
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – This is no good – I will save results (dss) and
compare with a previous best (dsx):
for j=1:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; end
end
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – Still no good – I will also save the index (bst)
of the current nearest data point:
for j=1:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; bst = j; end
end
x
x
x
x
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – Hold the phone – dsx has no value when j=1
so I pull the first abs() out of the loop and start with j=2
dsx = abs(x(i) – dpts(1,1));
bst = 1;
for j=2:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; bst = j; end
end
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – Finally, I need to set the value of the ith grid
line (this is y(i))
dsx = abs(x(i) – dpts(1,1));
bst = 1;
for j=2:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; bst = j; end
end
y(i) = dpts(bst,2);
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – Now I have to repeat this for all the grid lines
for i=1:length(y)
dsx = abs(x(i) – dpts(1,1));
bst = 1;
for j=2:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; bst = j; end
end
y(i) = dpts(bst,2);
end
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – Let's add in what we have from before
x=0:0.2:10;
y=zeros(length(x));
for i=1:length(y)
dsx = abs(x(i) – dpts(1,1));
bst = 1;
for j=2:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; bst = j; end
end
y(i) = dpts(bst,2);
end
plot(x,y);
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – The only things missing are the data
points and setting npts.
x=0:0.2:10;
y=zeros(length(x));
for i=1:length(y)
dsx = abs(x(i) – dpts(1,1));
bst = 1;
for j=2:npts
dss = abs(x(i) – dpts(j,1));
if dss < dsx dsx = dss; bst = j; end
end
y(i) = dpts(bst,2);
end
plot(x,y);
0
0.2 0.4 0.6 0.8
1.0 1.2 1.4 1.6 1.8
2.0 2.2 2.4 2.6 2.8 3.0 ...
Interpolation
Write code – Read the data from a file – suppose file's
name is one.dat and it begins with a line containing the
number of data points and each line thereafter is a data
point with position followed by value (two numbers/line).
Open the file and read the number of data points.
fid = fopen('one.dat','r');
if fid < 1
error('No such file');
end
npts = fscanf(fid,'%d',1); % number points
Interpolation
Write code – Read the data from a file – suppose file's
name is one.dat and it begins with a line containing the
number of data points and each line thereafter is a data
point with position followed by value (two numbers/line).
Create the dpts matrix so we can store the data from file
fid = fopen('one.dat','r');
if fid < 1
error('No such file');
end
npts = fscanf(fid,'%d',1); % number points
dpts = zeros(npts,2);
Interpolation
Write code – Read the data from a file – suppose file's
name is one.dat and it begins with a line containing the
number of data points and each line thereafter is a data
point with position followed by value (two numbers/line).
Now store the data points
fid = fopen('one.dat','r');
if fid < 1
error('No such file');
end
npts = fscanf(fid,'%d',1); % number points
dpts = zeros(npts,2);
for i=1:npts
dpts(i,:) = fscanf(fid,'%f',2);
end
Interpolation
Write code – Read the data from a file – suppose file's
name is one.dat and it begins with a line containing the
number of data points and each line thereafter is a data
point with position followed by value (two numbers/line).
Close the file. Merge this part with the previous part.
fid = fopen('one.dat','r');
if fid < 1
error('No such file');
end
npts = fscanf(fid,'%d',1); % number points
dpts = zeros(npts,2);
for i=1:npts
dpts(i,:) = fscanf(fid,'%f',2);
end
fclose(fid);
Interpolation
2D – Each line of the file has 2 coordinates plus a value.
Grid lines occur at mesh points – each with x,y coordinates.
Let x be an x position vector as before. Establish a y vector
for the y position of a grid line. Expand dpts so dpts(k,2)
is the y position of the data point and dpts(k,3) is the
value. Then, distance from grid line to data point is
sqrt((x(i)-dpts(k,1))^2+(y(j)-dpts(k,2))^2)
which computes distance between the (i,j) grid line and
the kth data point. Use vector v to hold interpolated values
and z to hold grid line values. Then the inner loop ends w/
z(i,j) = dpts(bst,3);
and the output loop becomes two loops (i and j).
Interpolation
1D - Inverse Distance Squared
d1
f(x1 )
x
f(x3 )
x
x
f(x2 )
d2
d3
x
d4
xp
f(x4 )
Interpolation
1D - Inverse Distance Squared
d1
f(x1 )
x
f(x3 )
x
x
f(x2 )
d2
d3
x
d4
f(x4 )
xp
f(xp ) = (f(x1 )/d12 + f(x2 )/d22 + f(x3 )/d32 + f(x4 )/d42)/(1/d12 +1/ d22 +1/ d32 +1/ d42)
Interpolation
1D - Inverse Distance Squared
d1
f(x1 )
x
f(x3 )
x
x
f(x2 )
d2
d3
x
d4
f(x4 )
xp
Use 1/(+d2) instead of 1/d2 to prevent blowup!