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Suppose we calculate some variable of interest, y, as a function of some other variable x. We call y the dependent variable and x the independent variable. For example, consider the data set below, taken from a simple experiment involving a vehicle, its velocity versus time is tabulated. In this case, velocity is a function of time, thus velocity is the dependent variable and the time is the independent variable.
Time [s] | Velocity [m/s] |
---|---|
0 | 20 |
10 | 39 |
20 | 67 |
30 | 89 |
40 | 111 |
50 | 134 |
60 | 164 |
70 | 180 |
80 | 200 |
In its simplest form regression analysis involves fitting the best straight line relationship to explain how the variation in a dependent variable, y, depends on the variation in an independent variable, x. In our example above, once the relationship (in this case a linear relationship) has been estimated we can produce a linear equation in the following form:
And once an analytic equation such as the one above has been determined, dependent variables at intermediate independent values can be computed.
Regression analysis with MATLAB is easy. The MATLAB Basic Fitting GUI allows us to interactively to do "curve fitting" which is a method to arrive at the best "straight line" fit for linear equations or the best curve fit for a polynomial up to the tenth degree. The procedure to perform a curve fitting with MATLAB is as follows:
Using the data set above, determine the relationship between velocity and time.
First, let us input the variables (Workspace>New variable) as shown in the following figures.
Second, we will plot the data by typing in
plot(time,velocity)
at the MATLAB prompt. The following plot is generated, select Tools>Basic Fitting:
In the "Basic Fitting" window, select "linear" and "Show equations". The best fitting linear line along with the corresponding equation are displayed on the plot:
Now let us do another curve fitting and obtain an equation for the function. Using that equation, we can evaluate the function at a desired value with
polyval
.
The following is a collection of data for an iron-constantan thermocouple (data available for download ). Engineering Fundamentals and Problem Solving by Arvid R. Eide, Roland Jenison, Larry L. Northup, Steven K. Mikelson , McGraw-Hill Higher Education. © 2007 p.114
Temperature [C] | Voltage [mV] |
---|---|
50 | 2.6 |
100 | 6.7 |
150 | 8.8 |
200 | 11.2 |
300 | 17.0 |
400 | 22.5 |
500 | 26 |
600 | 32.5 |
700 | 37.7 |
800 | 41 |
900 | 48 |
1000 | 55.2 |
We will input the variables first:
Temp=[50;100;150;200;300;400;500;600;700;800;900;1000]
Voltage=[2.6;6.7;8.8;11.2;17;22.5;26;32.5;37.7;41;48;55.2]
To plot the graph, type in:
plot(Temp,Voltage)
We can now use the Plot Tools and Basic Fitting settings and determine the equation:
By clicking the right arrow twice at the bottom right corner on the Basic Fitting window, we can evaluate the function at a desired value. See the figure below which illustrates this process for the temperature value 1150 C.
Now let us check our answer with a technique we learned earlier. As displayed on the plot, we have obtained the following equation: y=\mathrm{0.052831}x+\mathrm{0.67202} This equation can be entered as polynomial and evaluated at 650 and 1150 as follows:
>>p=[0.052831,0.67202]
p =0.0528 0.6720>>polyval(p,1150)
ans =61.4277
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