8 - Chapman MATLAB Programming for Engineers

Cumulative integral of humps(x)

Figure 11.4: Cumulative integral of humps(x)

An accelerometer measures acceleration and the resulting measurements can be used to estimate velocity and displacement. The acceleration is integrated to estimate velocity and the velocity in turn is integrated to estimate displacement. Suppose a vehicle starts from rest at time t = 0 and that its measured acceleration is given by the following table:

Time (s) Accel (m/s2)

00

12

24

37

411

517

624

732

841

948

10 51

A script to estimate the velocity:

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Simpson’s Rule

If the area under a curve is approximated by areas under quadratic curve, with the interval [a, b] divided into 2n equal subintervals, then the area can be approximated by Simpson’s rule:

where the xk values represent the endpoints of the subintervals and where x0 = a, x2n = b, and ∆x = (b − a)/(2n).

If the piecewise components of the approximating function are higher-degree functions, the integration techniques are referred to as Newton-Cotes integration techniques. The estimate of an integral improves as more components are used to approximate the area under a curve. An integral approximation can be made to meet a desired relative error by increasing the number of components until the error specification is met.

If the function to be integrated contains a singularity (a point at which the function or its derivatives are infinity or are not defined), a satisfactory result may not be provided by numerical techniques.

Matlab provides two quadrature functions for performing numerical function integration.

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quad(’function’,a,b) Returns the approximation of the integral of the function f(x), whose name is contained in the string ’function’, from a to b to within a relative error of 0.001 using an adaptive recursive Simpson’s rule. Function f must return a vector of output values if given a vector of input values. Inf is returned if an excessive recursion level is reached, indicating a possibly singular integral.

quad8(’function’,a,b) Returns the approximation of the integral of the function between a and b using an adaptive Newton-Cotes 8-panel rule. This function is better than the quad function at handling some functions with singularities.

For example, consider computing the integral of the humps function again:

>>quad(’humps’,-1,2) ans =

26.34497558341242

>>quad8(’humps’,-1,2) ans =

26.34496024631924

Here the quad solution achieves six significant digit accuracy and the quad8 solution achieves eight significant digit accuracy.

Two-Dimensional Integration

Two-dimensional integration is evaluated using the function dblquad, which approximates the integral

ymax xmax

f(x, y)dxdy

ymin xmin

q = dblquad(’function’,xmin,xmax,ymin,ymax): Returns the evaluation of the double integral f(x,y) using the quad quadrature function where x is the inner variable ranging from xmin to xmax, and y is the outer variable ranging from ymin to ymax. The first argument ’function’ is a string representing the integrand function. This function must be a function of two variables of the form z =f(x,y). The function must take a vector x and a scalar y and return a vector z that is the function evaluated at y and each value of x.

To use dblquad, one must first create a function that evaluates f(x, y). Consider, for example, the function func.m:

function z = func(x,y)

% FUNC(X,Y) computes an example function of two variables z = sin(x).*cos(y) + 1;

Using the commands listed below, this function can be plotted as shown in Figure 11.5:

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x = linspace(0,pi,20);

y = linspace(-pi,pi,20); [xx,yy] = meshgrid(x,y);

zz = func(xx,yy); mesh(xx,yy,zz),xlabel(’x’),ylabel(’y’)

2

1.5

1

0.5

0

4

Figure 11.5: Mesh plot of func(x,y)

The integral of this function over limits 0 ≤ x ≤ π, −π ≤ y ≤ π is given by

>> dblquad(’func’,0,pi,-pi,pi) ans =

19.73921476256606

11.2Numerical Di erentiation

The derivative of a function f(x) is defined as a function f (x) that is equal to the rate of change of f(x) with respect to x

f (x) = df(x) dx

Di erentiation and integration are complementary operations. For example, if

f(x) = g(x)dx

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then g(x) is the derivative of f(x) with respect to x. This relationship is written as

g(x) = df(x) = f (x) dx

The derivative f (x) can be interpreted geometrically as the slope of the function at (x, f(x)), where the slope of f(x) is the slope of the tangent line to the function at x, as shown in Figure 11.6.

Di erence Expressions

Numerical di erentiation techniques estimate the derivative of a function at a point xk by approximating the slope of the tangent line at xk using values of the function at points near xk. The approximation of the slope of the tangent line can be done in several ways, as shown in Figure 11.7.

The backward di erence approximation of the derivative at xk is the slope of the line between f(xk−1) and f(xk), as shown in Figure 11.7a

f (xk) = f(xk) − f(xk−1) xk − xk−1

If the samples of x are uniformly spaced so that xk − xk−1 = ∆x

f (xk) = f(xk) − f(xk−1) ∆x

The forward di erence approximation of the derivative at xk is the slope of the line between f(xk) and f(xk+1), as shown in Figure 11.7b

f (xk) = f(xk+1) − f(xk) xk+1 − xk

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For uniform spacing ∆x

f (xk) = f(xk+1) − f(xk) ∆x

Note that the forward di erence at xk is the same as the backward di erence at xk+1.

Figure 11.7: Techniques for approximating f (xk).

The quality of both of these derivative approximations depends heavily on two factors: the spacing of the samples and the scatter in the data due to measurement error. The greater the spacing, the more di cult it is to estimate the derivative. The approximation improves as the spacing between the two points decreases.

The central di erence is the average of the forward and backward di erences. For uniform spacing

This average tends to cancel out the e ects of measurement error.

The second derivative of a function f(x) is the derivative of the first derivative of the function

f (xk) = df (x) dx

This function can be approximated using slopes of the first derivative. Using backward di erences and assuming uniform spacing ∆x

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diff Function

The diff function computes di erences between adjacent values in a vector, generating a new vector with one less value.

diff(x) For vector x, returns a vector of di erences between adjacent values in x: [x(2)-x(1) x(3)-x(2) ... x(n)-x(n-1)], where n is length(x). For a matrix x, returns the matrix of column di erences, [x(2:n,:) - x(1:n-1,:)].

Example:

>>x = [0,1,2,3,4,5];

>>y = [2,3,1,5,8,10];

>>dx = diff(x)

An approximate derivative f (x) is computed using diff by dividing a change in y = f(x) by a change in x.

>> df = diff(y)./diff(x) df =

Obviously, in this example, since samples of x are spaced by 1, dy and df are the same.

Note that the values of df are correct for both the forward-di erence and the backward-di erence approximation to the derivative, as explained above. The distinction between the two methods of approximating the derivative is determined by the values of x that correspond to the derivative dy. Since dy has length one less than that of x, it must be related to a vector xd that is the vector x truncated by one sample at either the beginning or the end. The backward-di erence approximation of the derivative is related to x truncated at the beginning:

>> xd = x(2:end) xd =

The forward-di erence approximation of the derivative is related to x truncated at the end:

>> xd = x(1:end-1) xd =

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The backward di erence and central di erence methods can be compared by considering a sinusoidal signal that is measured 51 times during one half-period. The measurements are in error by a Gaussian distributed error with standard deviation of 0.025. Figure 11.2 shows the measured data and the underlying sine curve. The following script implements the two methods. The results are shown in Figure 11.2. Clearly the central di erence method provides better results in this example.

%Comparison of numerical derivative algorithms x = [0:pi/50:pi];

n = length(x);

%Sine signal with Gaussian random error

yn = sin(x)+0.025*randn(1,n);

%Derivative of noiseless sine signal td = cos(x);

%Backward difference estimate noisy sine signal dynb = diff(yn)./diff(x);

subplot(2,1,1),plot(x(2:n),td(2:n),x(2:n),dynb,’o’),xlabel(’x’),...

ylabel(’Derivative’),axis([0 pi -2 2]),...

legend(’True derivative’,’Backward difference’)

%Central difference

dync = (yn(3:n)-yn(1:n-2))./(x(3:n)-x(1:n-2));

subplot(2,1,2),plot(x(2:n-1),td(2:n-1),x(2:n-1),dync,’o’),xlabel(’x’),...

ylabel(’Derivative’),axis([0 pi -2 2]),...

legend(’True derivative’,’Central difference’)

Figure 11.8: Measurements of a sine signal containing Gaussian distributed random errors

Example 11.2 Numerical di erentiation of a polynomial function

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x

Figure 11.9: Comparison of backward di erence and central di erence methods

Determine the linear acceleration of an object whose speed is defined by s(t) = t3 − 2t2 + 2 m/s, where t is in seconds, over the interval 0 to 5. Determine the specific acceleration at t = 2.5s.

Consider computing the derivative function at three di erent resolutions to determine the e ect of increasing the resolution. Begin with a time resolution of ∆t = 0.1 second, compute the speed at each of the time values and compute the acceleration by approximating the derivative of the speed with respect to time:

>>t = 0:0.1:5;

>>s = t.^3 - 2*t.^2 + 2;

>>ds = diff(s)./diff(t);

To determine the acceleration at t = 2.5s, it is first necessary to find the time index corresponding to this time. Relating t to find the index k and the time interval ∆t:

t = ∆t · (k − 1)

Thus

t

k = ∆t + 1

or

2.5

k = 0.1 + 1 = 26

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and

>> ds(26) ans =

9.3100

Thus, the acceleration is 9.31 meters/second2.

Repeating the computations for ∆t = 0.01 second:

>>t = 0:0.01:5;

>>s = t.^3 - 2*t.^2 + 2;

>>ds = diff(s)./diff(t);

The time index k for t = 2.5s:

k = 2.5 + 1 = 251 0.01

The computed acceleration:

>> ds(251) ans =

8.8051

Repeating again for a resolution ∆t = 0.001s, where we are interested in the acceleration at index

>>t = 0:0.001:5;

>>s = t.^3 - 2*t.^2 + 2;

>>ds = diff(s)./diff(t);

>>ds (2501)

ans = 8.7555

Note that the result decreases with each decrease in the time interval. To check the answer, analytically di erentiate s(t) to obtain:

s (t) = 3t2 − 4t

At t = 2.5s:

s (2.5) = 3(2.5)2 − 4 · 2.4 = 8.75m/s2

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