AN ENGINEER’S GUIDE TO MATLAB 3rd Edition CHAPTER 4 PROGRAM FLOW CONTROL

Copyright © Edward B. Magrab 2009

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An Engineer’s Guide to MATLAB

AN ENGINEER’S GUIDE TO MATLAB

3rd Edition

CHAPTER 4

PROGRAM FLOW CONTROL


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Chapter 4 – Objective

Introduce various means of controlling the order in which a program’s expressions get evaluated and a set of relational and logical operators that are used to accomplish this control.


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Topics

• Introduction – Logical Operators

• Control of Program Flow

Branching – if Statement

Branching – switch Statement

Specified repetition – for Loop

Unspecified repetition – while Loop

Early Termination of a for or while Loop


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Program Control

• Achieved by four program flow control structures –

while, if, for, and switch

• Each time one of these statements appears, it must be followed at a later place within the program by an end statement.

• All expressions that appear between the control structure statement and the end statement are executed until all requirements of the structure are satisfied.

• Each of these control structure statements can appear as often as necessary within themselves or within other control structures.

When this occurs, they are called nested structures.


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• Control structures frequently rely on relational and logical operators to determine whether a condition has been met.

• When a condition has been met, the structure directs the program to a specific part of the program to execute one or more expressions.

• One can use the relational and logical operators to create a logical function whose output is 1 if the relational and logical operations are true and 0 if they are false.


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Relational and Logical Operators

Conditional

Mathematical symbol

MATLAB symbol

Relational operators equal not equal less than greater than less than or equal greater than or equal

Logical operators and or not

= < > AND OR NOT

== ~= < > <= >= & or && | or || ~


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When using control structures –

• Indent the statements following each control structure definition up to, but not including, the end statement.

Greatly improves the readability.

• When the structures are nested, the entire nested structure is indented, with each nested structure’s indentation preserved.

• When using MATLAB's editor, the indenting can be done automatically.


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Logical Operator –

Suppose that we want to create a function g(x) such that

g(x) = f(x) a x < b = 0 x < a and b x

The logical operator is formed by

y = (a<=x & x<b)

where a and b have been assigned numerical values prior to this statement and

(a<=x & x<b)

is the logical operator that has a value of 1 (true) when x a and x < b. Its value is 0 (false) for all other values of x.


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If we let

a = 1, b = 2f(x) = ex/2, x = [4, 1, 1, 4]

then a script using the logical operator is

a = -1; b = 2;x = [-4, -1, 1, 4];r = (a <= x)p = (x < b)logi = (r & p)gofx = exp(x/2).*logi

which, upon execution, yields


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r = 0 1 1 1p = 1 1 1 0logi = 0 1 1 0gofx = 0 0.6065 1.6487 0

where the intermediate expressions r, p, and logi were introduced to explicitly show that they are each a vector of logical results; ones (true) and zeros (false).

Notice that dot multiplication was employed because x and logi are each (1×4) vectors.


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In practice, the expressions r, p, logi and gofx are combined into one expression as shown below.

a = -1; b = 2;x = [-4, -1, 1, 4];gofx = exp(x/2).*((a<=x) & (x<b))


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This logical operator can be used to create the unit step function u(t), which is defined as

( ) = 1 0

= 0 < 0

u t t

t

If t varies by increments of 0.25 in the range 1 t 1, then the following script creates the unit step function

t = -1:0.25:1;UnitStep = (t>=0);disp(' t UnitStep')disp([t' UnitStep'])

t UnitStep -1.0000 0 -0.7500 0 -0.5000 0 -0.2500 0 0 1.0000 0.2500 1.0000 0.5000 1.0000 0.7500 1.0000 1.0000 1.0000


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Example –

Compare two vectors of equal length. Then

a = [4, 5, 6, 7, 8]; b = [4, 3, 2, 1, 8];d = (a == b)e = (a > b)

Its execution gives

d = 1 0 0 0 1e = 0 1 1 1 0


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Control of Program Flow

if Statement

The if statement is a conditional statement that branches to different parts of its structure depending on the satisfaction of certain conditional expressions.

The general form of the if statement is

if condition #1 expressions #1elseif condition #2 % (optional) expressions #2else % (optional) expressions #3end


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Example –

if j == 1 z = sin(x) ; if nnum <= 4 nr = 1 ; nc = 1; else nr = 1 ; nc = 2; endelse nr = 2; nc = 1; end

These statements executed only when j = 1 and nnum 4.

These statements executed only when j 1.

Executed only when j = 1.

This if statement encountered only when j = 1.

These statements executed only when j = 1 and nnum > 4.


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Note: When comparing a vector to a scalar, the condition is satisfied only when each element in the vector satisfies the condition.


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Switch Statement -

The switch structure is, essentially, an alternative to using a series of if-elseif-else-end structures.

The general form of the switch statement is

switch switch_expressioncase case_expression #1

statements #1case case_expression #2

statements #2case case_expression #n

statements #notherwise

statements #n+1end


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Example –

k = 3;switch k case 1 disp('Case 1') Executed only when k = 1

case {2, 3} Notice that cell is used

disp('Case 2 or 3' ) Executed only when k = 2, 3

case 9 disp('Case 9') Executed only when k = 9

otherwise disp('Otherwise') Executed only when k 1, 2, 3, or 9

end


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for Loop

A for loop repeats a series of statements a specific number of times. Its general form is

for variable = expression statementsend

where one or more of the statements can be a function of variable.


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Array Pre-allocation –

Consider a single for loop of the form

A = zeros(Nrow, 1); % Array pre-allocationfor r = 1:Nrow Statements A(r) = ...end

where Nrow is a positive integer that previously has been assigned a numerical value.

The addition of the array assignment statement

A = zeros(Nrow, 1);

ensures that the loop executes at maximum speed.


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For nested for loops, we have that

B = zeros(Nrow, Ncol); % Array pre-allocation for c = 1:Ncol % Column must be outer loop index Statements for r = 1:Nrow % Row must be inner loop index Statements B(r, c) = ... % Index order must be as shown endend

where Nrow and Ncol are positive integers that previously have been assigned numerical values.


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Example – Creation of a Sequentially-numbered Square Matrix

We shall generate an (N×N) matrix in which the elements of each row are such that

a11 = 1 and a1n = Na21 = N+1 and a2n = 2N…aN1 = (N 1)N +1 and aNN = N 2

The script is


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N = input('Enter a positive integer < 15: ');Matr = zeros(N, N);for r = 1:N Matr(r,1:N) = ((r-1)*N+1):r*N;enddisp(Matr)

Upon execution, we obtain

Enter a positive integer < 15: 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81


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Example – Dot Multiplication of Matrices

We shall perform the dot multiplication of two matrices A and B of the same order.

The script is equivalent to A.*B.

In our case, we will illustrate the procedure using A = magic(3) and B = A'.

However, in general, before the multiplication can be performed, one must ensure that the order of the matrices is equal.

The script is


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A = magic(3);B = A';[rA, cA] = size(A);[rB, cB] = size(B);if (rA~=rB)||(cA~=cB) error('Matrices must be the same size')endM = zeros(rA, cA);for c = 1:cA for r = 1:rA M(r, c) = A(r, c)*B(r, c); endenddisp(M)


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64 3 24 3 25 63 24 63 4


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Example – Analysis of the Frequency Spectrum of a Three Degree-of-Freedom System

Consider the following solution to a three degree-of-freedom system as a function of the forcing frequency .

12Y I A B

where

1

2

3

14 27 14 4

8 14 8 2.5

2.5 4 2.5 1

Y

Y Y B A

Y

We shall determine the maximum value of Yj, j = 1, 2, 3, when we take 1000 values of in the range 0 3.5.


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N = 1000;B = [14, 8, 2.5];A = [27, 14, 4; 14, 8, 2.5; 4, 2.5, 1];Om2 = linspace(0, 3.5, N).^2;sav = zeros(N, 3);for k = 1:N sav(k,:) = inv(eye(3)-Om2(k)*A)*B';endfor h = 1:3 [mx, ix] = max(sav(:,h)); disp(['Max of Y(' int2str(h) ') = ' num2str(mx, 6) ' at

Omega = ' num2str(sqrt(Om2(ix)), 4)])end


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Max of Y(1) = 1731.45 at Omega = 0.1682Max of Y(2) = 921.09 at Omega = 0.1682Max of Y(3) = 532.522 at Omega = 2.971


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Example - Total Interest of a Loan

Compute the total interest on a loan when the amount of the loan is L, its duration is m months, and its annual percentage interest Ia. The monthly payment pmon is determined from

-1- (1+ )mon m

iLp

i

where i = Ia/1200 is the monthly interest rate expressed as a decimal number.

As the loan is being paid off, a portion of the payment is used to pay the interest, and the remainder is applied to the unpaid loan amount.

The unpaid loan amount after each payment is called the balance.


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1,2,3n n-1

n mon n

n n-1 n

i = ib

P = p - i n = , ...,m

b = b - P

If b0 = L, then

where

in is the portion of pmon that goes towards the payment of the interest.

Pn is the portion of the payment that goes towards the reduction of the balance bn—that is, the amount required to pay off the loan.

The total interest paid at the end of the loan’s duration is

1

m

T jj

i i


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The script to compute iT is

loan = input('Enter loan amount: ');durat = input('Enter term of loan in months: ');int = input('Enter annual interest rate: ')/1200;ints = zeros(durat,1); % Pre-allocationprins = ints; % Pre-allocationbals = ints; % Pre-allocationpmon = (loan*int)/(1-(1+int)^(-durat));bals(1) = loan;for m = 2:durat+1 ints(m) = int*bals(m-1); prins(m) = pmon-ints(m); bals(m) = bals(m-1)-prins(m);enddisp(['Total interest = ', num2str(sum(ints),8)])

-1- (1+ )mon m

iLp

i

n n-1

n mon n

n n-1 n

i = ib

P = p - i

b = b - P


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Execution of the script gives

Enter loan amount: 100000Enter term of loan in months: 360Enter annual interest rate: 8Total interest = $164155.25

The first three lines were answered by the user, the last line is the result.


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Example - Specification of the Elements of an Array

We shall create an (n×n) matrix whose elements are either +1 or 1 such that

1 1 1

1 1 1( )

1 1 1M n n

The script is


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n = input('Enter the order of the square matrix: ');k = 1:n;M = zeros(n, n);OddRow = (-1).^(k-1);EvenRow = (-1).^k;for m = 1:2:n M(m,:) = OddRow; if m+1 <= n M(m+1,:) = EvenRow; endenddisp(M)

1 1 1

1 1 1

1 1 1M


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The execution of this script for n = 3 displays to the command window

Enter the order of the square matrix: 3

1 -1 1 -1 1 -1 1 -1 1

where the value 3 was entered by the user.


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Example - Sorting a Vector of Numerical Values in Ascending Order

We shall create a script that does the same thing sort.

H = [17, 12, 12, -6, 0, -14];LH = length(H);for k = 1:(LH-1) smin = H(k); for m = (k+1):LH if H(m) < smin smin = H(m); M = m; end end temp = H(k); H(k) = H(M); H(M) = temp;enddisp(H)

The execution of the script gives

-14 -6 0 12 12 17


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while Loop

The while loop repeats one or more statements an indefinite number of times, leaving the loop only when a specified condition has been satisfied.

Its general form is

while condition statementsend

where the expression defining condition is usually composed of one or more of the variables evaluated by statements.


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Example - Approximation to

The following expression converges to 1/ when and yo = 1/2 1 2ox

2 11 1 11 2 0,1,2,...n

n n n ny y x x n

where2

1 2

1 1

1 1

nn

n

xx

x

We shall show that the difference |1/ y4| is less

than 1015. The script is


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xo = 1/sqrt(2); yo = 1/2; n = 0;while abs(1/pi - yo) > 1e-15 xo = (1-sqrt(1-xo^2))/(1+sqrt(1-xo^2)); yo = yo*(1+xo)^2-2^(n+1)*xo; n = n+1;endfprintf(1, 'For n = %2.f, |1/pi-y_(n+1)| = %5.4e\n', n-1,

abs(1/pi - yo))

Execution of this script results in

For n = 3, |1/pi-y_(n+1)| = 4.9960e-016


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Example – Interval Halving and the Roots of Functions

Find a series of positive values of x (x1, x2, …) that make f(x) = 0, assuming that the sign of f(x) alternates as x increases

x

xstart + 5/2

/2

xstart + 11/4

f(x1) 0 f(x2) 0

f(x)

xstart + 2 xstart xstart +

f(x) > 0

f(x) < 0

/4

xstart + 3


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n = 5; a = pi; increment = 0.3; tolerance = 1e-6; xstart = 0.2; x = xstart; dx = increment;for m = 1:n s1 = sign(cos(a*x)); while dx/x > tolerance if s1 ~= sign(cos(a*(x+dx))) dx = dx/2; else x = x+dx; end end route(m) = x; dx = increment; x = 1.05*x;enddisp(route)

f(x) = cos(ax) 0

0.5000 1.5000 2.5000 3.5000 4.5000

x

xstart + 5/2

/2

xstart + 11/4

f(x1) 0 f(x2) 0

f(x)

xstart + 2 xstart xstart +

f(x) > 0

f(x) < 0

/4

xstart + 3


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Example – Convergence of a Series

Determine and display the number of terms that it takes for the series

2=1

1N

Nn

Sn

to converge to within 0.01% of its exact value, which is S = 2/6.


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series = 1; k = 2; exact = pi^2/6;while abs((series-exact)/exact) >= 1e-4 series = series+1/k^2; k = k+1;enddisp(['Number of terms = ' num2str(k-1)])

which, upon execution, displays to the command window

Number of terms = 6079

2=1

1N

Nn

Sn


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Early Termination of Either a for or while Loop

The break function is used to terminate either a for or while loop.

If the break function is within nested for or while loops, then it returns to the next higher level for or while loop.


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Example –

for j = 1:14 … b = 1 while b < 25 … if n < 0 break end

… end …end

When n < 0 the while loop is exited and the script continues from the next statement after this end statement

AN ENGINEER’S GUIDE TO MATLAB 3rd Edition CHAPTER 4 PROGRAM FLOW CONTROL

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