MATLAB NOTE

PLOT THE FUNCTION e-X/3SIN(X)
BETWEEN 0≤X≤4Π

DISPLAY FACILITIES

CONTD..

LINE SPECIFIERS IN THE plot() COMMAND
plot(x,y,‘line specifiers’)

Line Specifier Line Specifier Marker Specifier
Style
Color Type

Solid - red r plus sign +
dotted : green g circle o
dashed -- blue b asterisk *
dash-dot -. Cyan c point .
magenta m square s
yellow y diamond d
black k

» x = 1:2:50; Plots
» y = x.^2;
» plot(x,y) 2500
2000
1500
1000

500
0
0 5 10 15 20 25 30 35 40 45 50

Plots

» plot(x,y,'*-')

» xlabel('Values of x')

» ylabel('y') 2500

2000

1500

y

1000

500

0
0 5 10 15 20 25 30 35 40 45 50
Values of x

MULTIPLE GRAPHS

t=0:pi/100:2*pi; 1
y1=sin(t); 0.8
y2=sin(t+pi/2); 0.6
plot(t,y1,t,y2); 0.4
grid on 0.2

0
-0.2
-0.4
-0.6
-0.8

-1
01234567

MULTIPLE PLOTS

t=0:pi/100:2*pi;
y1=sin(t);
y2=sin(t+pi/2);
subplot(2,2,1)
plot(t,y1)
grid on
subplot(2,2,2)
plot(t,y2);
grid on

subplot(i,j,k)

• i is the number of rows of subplots in the plot
• j is the number of columns of subplots in the plot
• k is the position of the plot

INTERESTING FEATURE OF GENERATING

SINE CURVE

 x = 0:0.05:6; 1
 y = sin(pi*x); 0.8
 Y = (y >= 0).*y; 0.6
 plot(x,y,':',x,Y,'-') 0.4
0.2

0

-0.2

-0.4

-0.6

-0.8

-1
012 345 6

CONTINUED..

 x = 0:0.05:6; 1
 y = sin(pi*x); 0.8
 Y = (y >= 0).*y; 0.6
 plot(x,y,„.',x,Y,'-') 0.4
0.2

0

-0.2

-0.4

-0.6

-0.8

-1
012 345 6

OPERATORS (RELATIONAL, LOGICAL)

POLYNOMIALS

MATLAB FUNCTIONS FOR POLYNOMIALS

Contd..

Representing Polynomials:
x4 - 12x3 + 25x + 116
» P = [1 -12 0 25 116];

» roots(P)
ans =
11.7473

2.7028
-1.2251 + 1.4672i
-1.2251 - 1.4672i

» r = ans;
» PP = poly(r)
PP =

1.0000 -12.0000 -0.0000 25.0000 116.0000

Polynomial Multiplication

a = x3 + 2x2 + 3x + 4
b = 4x2 + 9x + 16

» a = [1 2 3 4];
» b = [4 9 16];
» c = conv(a,b)

c=

4 17 46 75 84 64

Evaluation of a
Polynomial

a = x3 + 2x2 + 3x + 4

» polyval(a, 2)

ans =

26

Symbolic Math

» syms x
» int('x^3')
ans =
1/4*x^4
» eval(int('x^3',0,2))
ans =

4
»

Solving Nonlinear Equations

nle.m
Function of the program:-

function f = nle(x)
f(1) = x(1) - 4*x(1)*x(1) - x(1)*x(2);
f(2) = 2*x(2) - x(2)*x(2) + 3*x(1)*x(2);

Program:-
x0 = [1 1]';
x = fsolve('nle', x0)

Solution:-
x=

0.2500
0.0000

TO FIND EIGEN VALUES AND EIGEN

VECTORS OF MATRICES

CONTINUED…

SOLVING LINEAR EQUATIONS

(TRY MANUALLY)

 Solve manually and tell me what is the answer..?
That is find out x=

y=
z=

SOLVING SET OF SIMULTANEOUS EQUATIONS

DIFFERENTATION

SOLVING DIFFERENTIAL EQUATIONS

PERFORMING INTEGRATION

NUMERICAL INTEGRATION

 Numerical integration of the integral f (x) dx is called
quadrature. MATLAB provides the following built-in functions
for numerical integration:

quad:
 It integrates the specified function over specified limits, based on

adaptive Simpson's rule.
 The general call syntax for both quad and quadl is as follows:
Syntax:-
integral = quad(„function‟, a, b)

dblquad: (It calculates double integration)
 MATLAB provides a function dblquad. The calling syntax for

dblquad is:
Syntax:-
I = dblquad(„function_xy‟, xmin, xmax, ymin, ymax)

FLOW CONTROL

CONTROL STRUCTURES

CONTROL STRUCTURES

CONTROL STRUCTURES

IF STATEMENT
(EXAMPLE)

 n = input(„Enter the upper limit: „);

 if n < 1

disp („Your answer is meaningless!‟)

 end

 x = 1:n; Jump to here if TRUE

 term = sqrt(x); Jump to here if FALSE
 y = sum(term)

EXAMPLE PROGRAM TO EXPLAIN IF LOOP

% Program to find whether roots are imaginary or not%
clc;
clear all;
a=input('enter value of a:');
b=input('enter value of b:');
c=input('enter value of c:');
discr = b*b - 4*a*c;
if discr < 0
disp('Warning: discriminant is negative, roots are imaginary');
end
Solution:-
Input:
enter value of a:1
enter value of b:2
enter value of c:3
Output:
Warning: discriminant is negative, roots are imaginary

EXAMPLE OF IF ELSE STATEMENT

Program:-
A = 2;

B = 3;
if A > B

'A is bigger'
elseif A < B

'B is bigger'
elseif A == B

'A equals B'
else

error('Something odd is happening')
end

Solution:-
ans =
B is bigger

IF STATEMENT EXAMPLE

Here are some examples based on the familiar quadratic formula.
1. discr = b*b - 4*a*c;
if discr < 0
disp('Warning: discriminant is negative, roots are imaginary');
end
2. discr = b*b - 4*a*c;
if discr < 0
disp('Warning: discriminant is negative, roots are imaginary');
else
disp('Roots are real, but may be repeated')
end
3. discr = b*b - 4*a*c;
if discr < 0
disp('Warning: discriminant is negative, roots are imaginary');
elseif discr == 0
disp('Discriminant is zero, roots are repeated')
else
disp('Roots are real')
end

EXAMPLE OF FOR LOOP

Problem: Draw graphs of sin(nπ x) on the interval −1 ≤ x ≤ 1 for

n = 1,2,....,8.We could do this by giving 8 separate plot
commands but it is much easier to use a loop.

Program:-

x=-1:0.05:1;
for n=1:8
subplot(4,2,n);
plot(x, sin(n*pi*x));
end

EXAMPLE OF FOR & WHILE LOOP

1.) % example of for loop%
Program:-
for ii=1:5
x=ii*ii
End

Solution: 3.) %example of while loop%
1 4 9 16 25 x=1
while x <= 100
2.) %example of while loop% x = 3*x
Program:- end
x=1 Solution:-
while x <= 10 X= 1 3 9 27
x = 3*x
End

Solution:
x=1 x=3 x= 9 x=27

SWITCH STATEMENT EXAMPLE

Program:-
n=input(„enter the value of n: ‟);
switch(rem(n,3))

case 0
m = 'no remainder'

case 1
m = „the remainder is one'

case 2
m = „the remainder is two'

otherwise
error('not possible')

end
Solution:-
enter the value of n: 8
m =the remainder is two

SPRING/MASS/DAMPER SYSTEM EXAMPLE

88

PROBLEM SOLVED BY USING
MATLAB AND SIMULINK

INTRODUCTION

PROBLEM

PROBLEM

FUNCTION OF THE PROGRAM

function f=programone (t,z)
m=3;
c=8;
k=100;
dzdt=[z(2); -(c/m)*z(2)-(k/m)*z(1)];

MAIN BODY OF THE PROGRAM

% For a single degree of freedom system in free vibration
clc;
clear all;
%Enter initial conditions
z0=[5;15];
%Enter time span for solution
tspan=[0 10];
%Call solver
[t,z]=ode45(„programone',tspan,x0);
%Set up plot
plot(t,z(:,1));

displacement in mmRESULT

single degree of freedom spring mass damper system behaviour
6

displacement
5

4

3

2

1

0

-1

-2

-3
0 5 10 15
time in seconds

DEMONSTRATION OF THE CONCEPT
WITH SIMPLE EXAMPLE

DEMONSTRATION OF THE CONCEPT
WITH SIMPLE EXAMPLE

DEMONSTRATION OF THE CONCEPT
WITH SIMPLE EXAMPLE

DEMONSTRATION OF THE CONCEPT
WITH SIMPLE EXAMPLEPROBLEM
(OSCILLATOR)

MODELLING PENDULUM WITH DAMPING