©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

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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')

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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

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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

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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

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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

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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