Tutorial 3 Arial2014

ME2 Computing- Tutorial 3: Numerical Differentiation/Integration

Summary • Use Matlab for numerical differentiation. • Use Matlab for numerical integration • Use Matlab to solve a differential equation

Introduction In engineering equations that govern the behaviour of a system are often in form of differential equations. As the name suggests they contain derivatives and can be difficult to solve analytically. There are very simple and effective techniques to approximately calculate a derivative numerically using a computer. These techniques can also be applied to solve differential equations where finding an analytical solution might prove to be difficult. Similarly numerical techniques can be used to integrate functions that can not easily be handled analytically. Thus the solution of equations can prove to be a very useful tool to approximate/describe the behaviour of an engineering system. In this tutorial some basic methods of numerical calculus will be explored. Then the numerical solution of a differential equation will be worked through and a simple engineering problem will be solved.

ME2 Computing 2014 1 Tutorial 3 (week 4)

Task A: Numerical differentiation

Consider an arbitrary function y = f(x) as shown in the above figure. The derivative of the function dy/dx is to be determined. The derivative is defined as the slope of the function at x, i.e.

xy

dxdy

x ∆∆

=→∆ 0

lim [ 1]

Assuming that ∆x is small several numerical approximations to the derivative are possible. The backward difference estimate at xi (as shown by the line A in the above figure) is defined by

1

1

−

−

−−

=ii

ii

backward xxyy

dxdy

[ 2]

(Pay attention to the first index i=1, for this value the backward difference is not defined as the index i-1=1-1=0 does not exist. Matlab array indices always start at 1! If it is not defined you will have to leave it out.) The forward difference estimate at xi (as shown by line B) is defined by

ii

ii

forward xxyy

dxdy

−−

=+

+

1

1 [ 3]

(Again pay attention to the indices, the last index i=n+1 is not defined as y only has n elements. N+1 does not exist and the forward difference is not defined.)


And there is also a central difference estimate at xi which uses the points to either side of xi and yi, i.e.

−−

+−−

=+

+

−

−

ii

ii

ii

ii

central xxyy

xxyy

dxdy

1

1

1

1

21

. [ 4]

For the following work save all your commands in an m-file so that you can execute them several times with slightly different settings.

1. Create a vector x with 20 linearly spaced numbers between 0 and 4π (use the linspace command).

2. From x create a new vector y = sin x.

3. Write a little script that calculates the backward difference

estimate of the derivative of y with respect to x.

4. Plot dy/dx and also the true derivative y = cos x. (Hint: use the diff function to calculate the difference between two consecutive vector elements. For more information on the diff function look it up in the Matlab help.)

5. Now also compute the forward and central difference estimates of

the derivative and plot all of the results on the same graph (use hold to display all graphs in the same plot, give the graphs different colours).

6. Find the maximum deviation from the true value (dy/dx = cos x)

for each scheme.

7. Now reduce ∆x using 100 elements between 0 and 4π (use linspace). What is the maximum deviation for each scheme now?

As expected the smaller the step ∆x becomes the more accurate the numerical evaluation of the derivative becomes.


An important aspect that governs the accuracy to which the derivative can be determined is the spread or error in the function/data that is to be numerically differentiated.

8. In order to explore this create a new vector containing elements y = (sin x) ± 0.01 random error. (Use the rand function to create the random error).

9. Again use your derivative scripts to evaluate the numerical

derivative of the new y function that contains errors. How large are the deviations of the different schemes now? Useful Matlab functions for this section: linspace - creates a vector with uniformly spaced elements diff - calculates the difference between adjacent elements in a

vector end - defines the end of a vector, useful in matrix manipulation length - returns the length of a vector rand - creates a random number

If you are unclear of how to use any of these functions either type “help function” (e.g. help mkdir) followed by enter into the Matlab command window or open the Matlab help browser and use the search tool to search for the documentation relevant to the desired function.

Answer Quiz Question Tutorial 3 Task A


Task B: Numerical Integration Integration calculates the area under a curve. Numerical integration is very useful when the analytical form of an integral is unknown or when it would be very difficult to find the solution otherwise. Consider a function y=f(x).

The easiest way to evaluate the integral of a function is by approximating each datapoint(x,y) by a bar of height y and width ∆x. Where ∆x= x2-x1. This method is very crude but gives an approximate answer. A better method is to use trapezoidal elements and use the trapezium rule for integration. Matlab has its own function to carry out integration using the trapezium rule, it is called by trapz(x,y). Where x and y are vectors of the function y=f(x). trapz(x,y) performs numerical integration given the 2 vectors x and y. In order for this to work x and y must be of the same length.

1. Integrate y=-(x-2)2+4 in the interval from 0 to 4 using the trapezium rule in matlab.

2. Compare your result to the analytical solution4

0

32

32

−=

xxI .

3. What is the maximum step size ∆x to avoid making the error larger

than 1%? (answer=0.4) Numerical integration using the trapezium method is most prone to error in regions where a function is changing rapidly (for example between x1 and x2 in the figure above). Using “adaptive” methods, i.e. by changing the step size locally where these fast changes occur, the accuracy of the result can be improved. trapz(x,y) is not adaptive. However the Matlab function quad is adaptive, but it can only be used on functions and not on vectors of data as for trapz.


quad(a,b) uses an adaptive form of Simpson’s rule (See your 1M maths notes for more information) to evaluate the integral of a function between two specified limits.

4. Use quad to integrate the function sqrt(x) from 0 to 1.

5. Now create your own function “myint” for y = -(x-2)2 + 4 and use quad to evaluate the integral from x = 0 to x = 4.

6. Compare the result with the one you found earlier.

The error tolerance to which quad evaluates the integral can be adjusted by an extra argument tol when calling the function. You can find out about this in the Matlab help.

Answer Quiz Question Tutorial 3 Task B

Useful Matlab functions for this section: linspace - creates a vector with uniformly spaced elements plot - plots two vectors trapz - numerical integral of a curve described by two data vectors quad - numerically integrates a function



Task C: Numerical solution of a differential equation (Euler method) Consider the simple Ordinary Differential Equation (ODE)

)()( tytFdtdy

= . [ 5]

Where F(t) is a known function and you want to determine y(t). While analytical solutions to this problem exist a numerical solution can also be found. The advantage of the numerical solution is that it is based on a simple difference equation and can be solved quickly even in cases when an analytical solution is not known. However there are usually drawbacks: it is only an approximation and if some parameters such as the step size are poorly selected, the numerical routine can lead to large errors. One of the simplest and crudest methods to solve ODE’s is the Euler method, which will be explored below: The derivative in equation [1] is expressed as a finite difference

ttytty

ttytty

dtdy

t ∆−∆+

≈∆

−∆+=

→∆

)()()()(lim0

[ 6]

Using this to re-write equation [1] results in

)()()()( tytFt

tytty=

∆−∆+

or

( )ttFtyttytFtytty ∆+=∆+=∆+ )(1)()()()()( . [ 7] This shows that we can evaluate any future value of y if we know an initial value of y (i.e. y(t)), the function F(t) and use a small time step ∆t (larger ∆t will cause larger errors but decrease the computation time). It is convenient to write the equation in terms of time steps rather than simply time, i.e.:

ttytFtyty kkkk ∆+=+ )()()()( 1 . [ 8]

1. Now use equation [8] to evaluate y(t) of equation [5] for F(t) = - 2, y(0) = 2, ∆t = 0.02 over the range t = 0 t = 3. (Hint: You will have to write a script that calculates a new y for each time step. The initial value y(0) gives you the starting point.)

2. Compare the answer to the analytical solution y=2e-2t.


3. In subtask 1 increase the time step size ∆t to 0.2 (and then to 0.4)

and recalculate y(t). Plot the result and the analytical solution (subtask 2) on the same graph and compare the two curves. Plot both curves in a different colour so that you can differentiate them.

4. Calculate the maximum deviation between the analytical solution

and the solution obtained using the Euler method for ∆t = 0.02, 0.2 and 0.4.

Answer Quiz Question Tutorial 3 Task C

Useful Matlab functions for this section: for… end - Loop that repeats the statements within the described block

for a predefined number of times plot - plots two vectors



Task D: RC circuit example

An RC circuit (resistor - capacitor), as shown above, has the following characteristic equation:

0=+ ydtdyRC . [ 9]

The value for R = 100kΩ and C = 1µF so that RC = 0.1, no voltage is applied and the initial voltage across the capacitor is y(0) = 2V so that the equation reduces to:

01.0 =+ ydtdy

. [ 10]

1. Write a little script that evaluates y as a function of t (from t=0 to t=3secs)

2. Find the the solution for y for ∆t=0.01 .

The power delivered by the circuit is the product of VI (voltage V and current I) and the energy stored in the capacitor is the integral of VI with respect to time.

3. Find the energy stored in the capacitor at t=0 by numerically integrating the power (VI). (Hint: V=RI, knowing V you can calculate I) The solution will always slightly change as ∆t is further reduced however the changes become less and less important. Usually a tolerance is set and if the solution changes by less than a predefined error tolerance say 0.01 in this case the result is close enough to the real solution.

4. Reduce the steps size from ∆t=0.01 (by dividing by 10) until the solution of the energy stored in the capacitor changes by less than 1%. The solution can then be believed to have converged to the real value.

Answer Quiz Question Tutorial 3 Task D

Useful Matlab functions for this section: plot - plots data trapz - numerical integral of a curve described by two data vectors



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Tutorial 3 Arial2014