Lab8 - Engineering 7 Introduction to Programming for...

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Engineering 7: Prof. Alexandre Bayen Introduction to Programming for Engineers Spring 2011 Lab 8: Interpolation and Series Date Assigned: 5:00pm, Friday – April 1. Date Due: 5:00pm, Friday – April 8. For the problems in this assignment, you should assume that all vector input and output arguments are column vectors. Vectors defined by x and y are always the same size. Vectors defined by X and Y are always the same size. You may assume that the values contained in x and X are unique and in ascending order. You may also assume that the all the values of X are between min(x) and max(x), inclusively. Problem 1: Write a function with header [] = myInterpPlotter(x,y, X, option) where x and y are column vectors containing x and y data points, and X is a column vector containing the coordinates for which an interpolation is desired. The input argument option should be a string either ‘linear’ , ‘spline’ , ‘nearest’ . Your function should produce a plot of the data points (x,y) marked as red circles, and the points (X,Y) where X is the input vector, and Y is the interpolation at the points contained in X defined by the input argument specified by option. The points (X,Y) should be connected by a blue line. Be sure to include title, axis labels, AND a legend. Hint: You should use the function interp1 . Test Cases: >> x = [0 .1 .15 .35 .6 .7 .95 1]; >> y = [1 0.8187 0.7408 0.4966 0.3012 0.2466 0.1496 0.1353]; >> myInterpPlotter(x,y, linspace(0,1,100), 'nearest') 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 nearest interpolation of data x y data points interpolation
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Lab8 - Engineering 7 Introduction to Programming for...

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