©1999 bg mobasseri 15/24/99 interpolation and curve fitting etter: pp. 164-184 june 16, ‘99
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©1999 BG Mobasseri 1 5/24/99
INTERPOLATION AND CURVE FITTING
Etter: pp. 164-184
June 16, ‘99
©1999 BG Mobasseri 2 5/24/99
DEFINING INTERPOLATION
Interpolation can be used in at least two ways– Subsampling (taking every other sample or
more) a dense signal and later filling it in– Generating new data where none was
previously available
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SUBSAMPLING
Say you have a large sound file taking too long to transmit or too much space to store.
One solution is to subsample it, send(store) the sparse data then fill in the gaps later
©1999 BG Mobasseri 4 5/24/99
EXAMPLE
Keep the circles and discard the rest. Later, fill in the gaps
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VIDEO COMPRESSION
Video captured at 30 frames/sec. contains a lot of redundancies
Keep only a fraction of the frames and interpolate between them later
This is implemented in the MPEG standard
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INTERPOLATION METHODS
There are 3 major interpolation techniques– linear– cubic-spline( a 3rd degree polynomial)– polynomial fitting (polynomial of arbitrary
order)
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LINEAR INTERPOLATION
Simplest of its kind, works on the following principle
error
interpolation
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HOW DOES MATLAB DO IT?
The main MATLAB’s routine for 1-D interpolation is interp1 with the following syntax– yi=interp1(x,y,xi,’method’)
(x,y) are the original coarse data. xi’s are the new finer positions to be interpolated, yi is the answer. See next
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Illustrating (xi,yi)
x xiy
yi
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WORKING WITH interp1
Want to interpolate a sinc function with samples originally located at [-4:1:4]
-4 -3 -2 -1 0 1 2 3 4-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
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Linear interpolation:Try it! x=[-4:.1:4];%x-values y=[0 0 0 0 1 0 0 0 0];%coarsely sampled data at x=-4,-3,... xfiner=[-4:0.5:4];%inerpolate at -4,-3.5,-3,... yfiner=interp1(x,y,xfiner,’linear’);%interpolate at new grid
positions plot(x,y,xfiner,yfiner,'ro',xfiner,yfiner,'r-');
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LINEAR INTERPOLATION OF SINC
Interpolated at 0.5 intervals
-4 -3 -2 -1 0 1 2 3 4-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
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DEFINING CUBIC-SPLINE
When data is interpolated by cubic spline, it means we pass a 3rd order polynomial through each pair of points
Since a polynomial is passed through a pair of pointsWe can interpolate at arbitrary fine positionsBetween the two
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Example
Let’s say we have passed y=x2 between (1,1) and (2,2).
We can read any intermediate y values by simply plugging in an x value
1 2
1
4
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CUBIC-SPLINE
A more accurate interpolation can be achieved by passing 3rd degree polynomials through coarse data
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 55
6
7
8
9
10
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INTERPOLATING AN AM SIGNAL
An amplitude modulated signal is s(t)=(1+ cos2πfmt)cos2πfct( )
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2
-1.5
-1
-0.5
0
0.5
1
1.5
2AM SIGNAL,fm=2,fc=3
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Try it!
Evaluate the AM signal with fm=2,fc=3 in the range 0<t<1 in increments of.01
Subsample it 20:1. Using the kept data points, perform
linear and cubic-spline interpolation Compare your code with mine, next
page
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My code
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Output graph
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
original
interpolated
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LINEAR VS. CUBIC
spline gets closer to the actual curve
0 0.2 0.4 0.6 0.8 1-2
-1.5
-1
-0.5
0
0.5
1
1.5
2CUBIC-SPLINE vs. LINEAR INTERPOLATION
linear
originalspline
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FINER INTERPOLATION
Smoother interpolation via spline
0 0.2 0.4 0.6 0.8 1-2
-1.5
-1
-0.5
0
0.5
1
1.5
2FINER CUBIC-SPLINT INTERP.
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DIRECT READ-OFF interp1(x,y,2.5,’spline’)
returns spline interpolated value of 6.1 at 2.5
2.5
6.1
x=0:5; y=[5 8 6 7 9 8]; xi=0:.1:5; ylin=interp1(x,y,xi); ys=interp1(x,y,xi,'spline'); point=interp1(x,y,2.5,'spline')
©1999 BG Mobasseri 23 5/24/99
LEAST SQUARES CURVE FITTING
Linear and cubic spline interpolations fit curves constrained to go through the data points
A curve fitted using least squares may not pass through any data point but it will be “close” to all of them in the “least squares” sense
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LEAST SQUARES SENSE
MSE =yk −ˆ yk( )
2
k=1
N∑
N
data
Fit a line that on average is closets to all data points
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LEAST SQUARES OBJECTIVE
Find a function,e.g. a polynomial of whatever order, that minimizes the mean square error
MATLAB does this through polynomial regression
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DIFFERENCES WITH CUBIC SPLINE
Both are polynomials but cubic splines are 3rd order polynomials.
The big difference is that cubic spline fits separate 3rd degree polynomials per segment
Least squares fits a single polynomial through all data points
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DEFINING A POLYNOMIALEtter: pp. 78-86
An Nth degree polynomial is specified by N+1 coefficients
If there are N+1 data points, an Nth degree polynomial will pass through all of them
f x( ) =a0xN +a1x
N−1 +... +aN−1x+aN
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Example of polynomials
It takes a first degree polynomial, a straight line, to connect two points
It takes a 2nd degree polynomial to connect 3 points
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polyfit FUNCTION
To fit an nth degree polynomial to (x,y) data use– p=polyfit(x,y,n)
polyfit returns a vector p of n+1 coefficients in decreasing powers of x.
So if
then p=[ao,a1,a2,a3,…,aN]
f x( ) =a0xN +a1x
N−1 +... +aN−1x+aN
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Evaluating and Plotting the Fitted Polynomial:polyval
Once polynomial is specified through the vector p, it can be evaluated directly using– y=polyval(p,x)
coarse x valuesvector of polynomial coefficients
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Example using polyfit
Let x=[0 1 2 3 4 5] and y= [5 8 6 7 9 8] be the coarse data points. This is how to fit a 4th order polynomial
p=polyfit(x,y,4);%p is the coeff. vector
now let’s evaluate the polynomial at finer positions given by xfine=[0:0.1:5]
yfine=polyval(p,xfine);%plot yfine to see
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RESULT
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 55
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
data
Fitted polynomial does not passthrough any data point
©1999 BG Mobasseri 33 5/24/99
Comparing Interpolations
In the following slides we start with the following data points– x=[0 1 2 3 4 5];– y=[5 8 6 7 9 8];
We will then interpolate along x in increments of 0.1 using progressively larger order polynomials
©1999 BG Mobasseri 34 5/24/99
APPLYING polyfit
Let’s fit a first degree y=mx+h to data. We’ll get m=0.54,h=5.8
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 55
6
7
8
9
10
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3RD DEGREE POLYNOMIAL
Coefficients are [0.037 -0.349 1.407 5.460]
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 55
6
7
8
9
10
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4th DEGREE POLYNOMIAL
coeff=[-0.2500 2.5370 -8.0278 8.5503 5.0317]
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 55
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
©1999 BG Mobasseri 37 5/24/99
5th DEGREE POLYNOMIAL
Expect the polynomial to pass through all the points (why?)
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 55
5.5
6
6.5
7
7.5
8
8.5
9
9.5
©1999 BG Mobasseri 38 5/24/99
Homework
Take the data on page 33 and interpolate it in increments of 0.1 using– Linear interpolation– Cubic spline– Polynomial of 3rd degree– Superimpose your plots and show how they
compare
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