Modeling of Data

It is often useful to have a simple equation for a set of data that do not follow a particular model.  There are numerous ways of modeling data.  We will consider three of these.

Model 1: Linearization (Least Squares Fit or Point-Slope)

Consider the data shown in the figure to the right.  This represents the relationship between hemoglobin saturation [Hb] and oxygen pressure ().  Clearly this curve is nonlinear.  However, in some cases we might be interested in the behavior of a system around a particular point on the curve.  For example, if blood at the capillary level is maintained in a range of 20 to 40 mm Hg, we can model this part of the curve as a straight line with only a small amount of error.  It would be ridiculous to think that a straight line could model the complete curve, but in the range of interest, the linear model (dashed line) matches the measured curve astoundingly well.

To obtain the least squares fit, it is first necessary to define an error criterion.  Qualitatively we say that the difference between the measured data and the model is as small as possible.  Mathematically, we say:

 Eq. 1

,

where  is the measured value of y (corresponding to the “Measured Curve” in the figure above),  is the value that the model provides (corresponding to the dashed line in the figure), and N is the number of data points available.  For a linear model, , so that:

 Eq. 2

This is a mathematical description of what we mean by the best fit.  The square of the difference is useful because this quantity is always positive.  If there were both positive and negative terms, it would be possible for large positive errors to cancel out large negative errors in the model.  An alternative to the square would be simply to take the absolute value.  The resulting equation would be different from the least squares equation, but, if we chose to define “best fit” this way, it would still be a valid model.

To determine the values of a and b that satisfy the error criterion, it is necessary to minimize by taking the derivative with respect to each variable and setting the result equal to zero.  In other words:

 Eq. 3

 Eq. 4

We can move the derivatives into the sum (because the derivative and sum are both linear operators) and then take the derivative to obtain:

 Eq. 5

 Eq. 6

The last sum in Eq. 5 is , and is just N times the average of all of the x’s, which we will designate as .  The last term in Eq. 6 is just Nb.  Also, since m is a constant, it can be taken out from under the sums in both equations.  Equations 5 and 6 then can be written as:

 Eq. 7

 Eq. 8

Since all of the experimental data are known, the only unknowns are m and b, so these two equations can be solved simultaneously for  m and b.  Eliminating by multiplying Eq. 8 by and subtracting gives:

 Eq. 9

,

which can be solved for m to yield:

 Eq. 10

and from Eq. 8,

 Eq. 11

Taylor Series Model

The linear regression model is similar to taking a Taylor series of the curve around the point of interest.  Recall that the Taylor series is defined as:

 Eq. 12

A simple linear model can be obtained by using only the first 2 terms of this expansion.  In other words, this method simply uses the tangent line to the curve at the point of interest as a model.  Again, this would cause huge errors, in general, for points that are far away from , but if our interest is only on places near enough to , the error will be small.

Of course, more accurate models can be obtained by taking more of the terms of the Taylor series, although use terms higher than first order will negate the advantages of using a linear model.

Power Law Model

In some cases it is useful to use a model of the form:

 Eq. 13

,

where the three parameters a, b, and a are obtained by curve fitting.  There is an interesting trick that can be used to find a, b and a, which starts with the logarithm of the model:

 Eq. 14

If a value of a is already selected, then this transformed equation is linear in  and .  That is, a least squares fit of  as a function of  will give  as the slope and  as the y-intercept.  The hemoglobin saturation curve does not fit well to this type of model.  However, some phenomena fit this model well, such as the relationship between velocity in a fluid and current from a hot film anemometer.

Fourier Series Model

The Fourier series model can be truncated to only a few terms.  For example, consider the hemoglobin saturation curve again, and take as a model:

 Eq. 15

This is a two-term Fourier series model.  There are different ways to find the values of A, B, a and j.  One way is to take the fast Fourier transform of the data sequence directly.  For example, the fft function in Matlab can be used.  Assuming you have the data in the variable Hb, you can do the following:

N=length(Hb);

PO2max = 65;

a=2*pi/PO2max;

X=fft(Hb)./N;

A=X(1);

B=abs(X(2));

Phi=angle(X(2));

An alternative method is to pick three features of the curve and fit these to the model directly.  For example, if you think of the curve as being shaped like a sinusoid, the middle of the sinusoid appears to be at about where percent saturation is 40.  We will take this as the offset of the sine wave (i.e. A=40).   The peak of the sine wave appears to be at the point (70,90), which means that the amplitude, B, should be 90-40, or 50.  Assume you want the model to fulfill, in addition, the following two criteria:

 a

 b

 Eq. 16

[Hb] = 0 at = 0

Now you have two equations and two unknowns that can be solved simultaneously.

 a

 Eq. 17

 b

From Eq. 17a, , and with this in 17b we get

 Eq. 8

.

 Eq. 18

Exercises:

1. Use Excel to generate a set of 10 discrete values of this data set from  to .  With these values, construct a linear regression model of the data around the point .  Plot the data and the regression model on the same plot.  Note that the linear regression output includes a parameter “R.”  This is an indicator of the “goodness of fit.”  A value of R close to 1 indicates an excellent fit to the data.  What is the value of R for your fit?
2. Calculate the Taylor series linear model for the function.  Compare the slope of this model to that of the linear regression model.  Plot the two models on the same plot.  How different are they?  Which is the better match to the original function within the range of interest ( to ).

The data used to generate the oxygen saturation curve above are given below.

 [Hb] [Hb] [Hb] 0 0 40 72 100 95 15 20 47 80 120 97 20 29 60 87 140 100 26 42 70 90 33 60 80 92

3.       Use Excel to plot these data and the model in Eq. 18.  Note that the match is not perfect.  What is the problem?

4. You wish to model data as a quadratic of the form:

Set up the least squares problem and solve for a, b and c in terms of the data (xi,yi).