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Bisection Method

2009-07-25T23:19:00.000-07:00

Theory For BISECTION METHOD:

In mathematics, the bisection method is a root-finding algorithm which repeatedly divides an

interval in half and then selects the subinterval in which a root exists. It is a very simple and

robust method, but it is also rather slow.Suppose we want to solve the equation
f(x)=0,where f is a continuous function.

The bisection method starts with two points xl and xu such that f(xl) and f(xu) have opposite

signs. The intermediate value theorem says that f must have at least one root in the interval

[x_l,x_u]. The method now divides the interval in two by computing

xr = (x_l+x_u) / 2.

There are now two possibilities: either f(x_l) and f(x_u) have opposite

signs, or f(x_r) and f(x_l) have opposite signs. The bisection algorithm is then

applied recursively to the sub-interval where the sign change occurs.

To determine in which sub-interval the root lies:

If f(x_l).f(x_r)<0,then we set x_u=x_r

If f(x_l).f(x_r)>0, then we set x_l=x_r

If f(x_l).f(x_r)=0, then we set x_t=x_r

According to the condition we have to iterate again and again till then whenever |%error| or

|e_a|=e_s. Then we have to find out the %error. The rule for determine %error

is

%error or e_a = (x_rnew-x_rold)/

x_rnew*100%

MATLAB CODE FOR BI-SECTION METHOD OF GIVEN EUATION:

FUNCYION OF RLC:

function f_value=RLC(R)

t=0.05;

q0=5e-4;

q=5e-6;

L=5;

C=1e-4;

f_value=(exp(-R*t/(2*L)))*cos(sqrt((1/(L*C))-(R/(2*L))^2)*t)-q/q0;

MAIN PROGRAM:

clc;

clear all

tic

Rl=200;

Ru=400;

es=.0001;

i=0;

imax=25;

Rr=(Rl+Ru)/2;

fl=RLC(Rl)

while(1)

Rrold=Rr

Rr=(Rl+Ru)/2

fr=RLC(Rr)

if Rr~=0

ea=abs((Rr-Rrold)/Rr)*100

end

test=fl*fr

if test<0 ru="Rr">0

Rl=Rr;

fl=fr

else

ea=0;

end

if(ea=imax

break

end

fp=fopen('iea.txt','a+');

[err,msg]=ferror(fp,'clear');

if err~=0

disp('An error occurred while appending data to the file')

disp(msg)

end

fprintf(fp,'%0.0f\t%8.5e\n',i,ea);

err=fclose(fp);

if err~=0

disp('Could not close the file')

end

i=i+1;

end

disp('The value of resistance is:')

disp(Rr)

disp('The value of i is:')

disp(i)

toc

OUTPUT

fl =

-0.1631

Rrold =

300

Rr =

300

fr =

-0.0295

ea =

0

test =

0.0048

fl =

-0.0295

Rrold =

300

Rr =

350

fr =

0.0209

ea =

14.2857

test =

-6.1704e-004

Ru =

350

Rrold =

350

Rr =

325

fr =

-0.0032

ea =

7.6923

test =

9.3070e-005

fl =

-0.0032

Rrold =

325

Rr =

337.5000

fr =

0.0091

ea =

3.7037

test =

-2.8865e-005

Rr =

328.1513

fr =

-1.0717e-007

ea =

1.1625e-004

test =

5.2177e-014

fl =

-1.0717e-007

Ru =

337.5000

Rrold =

337.5000

Rr =

331.2500

fr =

0.0031

ea =

1.8868

test =

-9.6751e-006

Ru =

331.2500

Rrold =

331.2500

Rr =

328.1250

fr =

-2.6307e-005

ea = 0.9524

test =

8.2989e-008

fl =

-2.6307e-005

Rrold =

328.1250

Rr =

329.6875

fr =

0.0015

fl =

-2.0056e-006

Rrold =

328.1494

Rr =

328.1616

fr =

1.0144e-005

ea =

0.0037

test =

-2.0345e-011

Ru =

328.1616

Rrold =

328.1616

Rrold =

328.1513

Rr =

328.1515

fr =

8.2673e-008

ea =

5.8124e-005

test =

-8.8602e-015

Rr =

328.155

fr =

4.0694e-006

ea =

0.0019

test =

-8.1616e-012

Ru =

328.1555

Rrold =

328.1555

Rr =

328.1525

fr =

1.0319e-006

ea =

9.2998e-004

test =

-2.0696e-012

Ru =

328.1525

Rrold =

328.1525

Rr =

328.1509

fr =

-4.8686e-004

ea =

4.6499e-004

test =

9.7646e-013

fl =

-4.8686e-007

Rrold =

328.1509

Rr =

328.1517

fr =

2.7252e-007

ea =

2.3250e-004

test =

-1.3268e-013

Ru =

328.1517

Rrold =

328.151

Ru =

328.1515

The value of resistance is:

328.1515

The value of i is:

19

elapsed_time =

0.1410

>>

Disadvantages:

• It is relatively slow when compared with other root finding methods we will study,
especially when the function f (x) has several continuous derivatives about the root

x_r.

• The algorithm has no check to see whether the e is too small for the computer
arithmetic being used. [This is easily fixed by reference to the machine epsilon of the
computer arithmetic.]

• When we do this method by software (matlab) then need more time of execution for this method

because this method takes more number of iteration (i).

• It cannot be used to

find roots when the function is tangent is the axis and does not pass through the axis.
For example: f(x) = x₂

Discussion:

In this experiment we learn the MATLAB code of bi-section method for finding root by numerical

method. We also clarify the basic theory of bi-section method. In MATLAB code we use a user define

function named RLC so we learn the rule to define a ‘user defined function’ in MATLAB. Finally we

can say that we properly follow our teacher instruction that’s why we do not face any problem to

solve our problem.

Finding root of an equation using Newton-Rapshon Method

2009-07-23T05:33:00.000-07:00

Newton-Raphson Method, the most widely used of all root-locating formulas. If the function, f(x)
and an initial guess, xi is known then we can easily draw a tangent at the point [x_i, f(x_i)]. The

point where the tangent crosses the x-axis usually represents an improved estimate of the root.

The formula is derived geometrically from the straight line equation as the first derivative of the
function f(x) represents the slop of the tangent at that particular point.

A simple algorithm for the Newton-Raphson Method is given below:

1. Choose an initial guess that will be close to the true root.

2. An estimate of the root(ε_s) is determined by
x_i+1=x_i - f(x_i)/f’(x_i);
3. Evaluate the approximate percentage of errors(ε_a) by using the following formula:
ε_a= ((x_i+1- x_i)/ (x_i+1))*100%
If ε_a < ε_s, specified error tolerance, then stop the iteration; otherwise go to step 2. The process
is repeated until this condition is met.

In this experiment we will see how MATLAB can assist us in finding the roots of various functions using the numerical techniques Newton-Raphson method.

An experimental Circuit for the Newton raphson Method:

The circuit in Figure:1 charges the capacitor from the DC source when the switch is in its rest position.At t=0 sec,the is thrown at "2" position and the capacitor starts to discharge the RLC series circuit.

Figure1:An RLC circuit to charge and dischage a capacitor

The function is given as follows:
f(R)=e^-Rt/2Lcos[√ (1/LC+ (R/2L)²t]-q/q0
In this section,we have to calculate the resistance in MATLAB using Newton-Raphson Method
MATLAB Files:

Name this function file as the RLC.m

function f_value=RLC(R);
t=0.05;
q0=5e-4;
q=5e-6;
L=5;
C=1e-4;
f_value=(exp(-R*t/(2*L)))*cos(sqrt((1/(L*C))-(R/(2*L))^2)*t)-q/q0;

Code derivative function of the given equation:

function f_valueD=RLC_D(R);
f_valueD=(R*t)/((4*L^2)*(sqrt(1/(L*C)-(R/(2*L))^2)))*(exp(-R*t/(2*L)))*sin(sqrt((1/(L*C))-(R/(2*L))^2)*t)-(t/(2*L))*(exp(-R*t/(2*L)))*cos(sqrt((1/(L*C))-(R/(2*L))^2)*t);

%code for Newton-Raphson(Main File):

clc;
clear all
tic
Ri=200;
es=0.0001;
j=0;
jmax=25;
while(1)
fi=RLC(Ri)
fid=RLC_D(Ri)
Ril=Ri-fi/fid
if Ril~=0
ea=abs((Ril-Ri)/Ril)*100
end
if ea=jmax
break
end
fp=fopen('iea.txt', 'a+');
[err, msg]=ferror(fp, 'clear');
if err~=0
disp('An error occurred while appending data to the file')'
disp(msg)
end
fprintf(fp, '%0.0f\t %8.5e\n',j,ea);
err=fclose(fp);
if err~=0
disp('could not close the file')
end
Ri=Ril
j=j+1
end
disp('The value of resistance is: ')
disp(Ril)
disp('The value of j is: ')
disp(j)
toc

Output:

fi =
-0.1631
fid =
0.0016
Ril =
301.8218
ea =
33.7357
Ri =
301.8218
j =
1
fi =
-0.0275
fid =
0.0011
Ril =
326.9259
ea =
7.6788
Ri =
326.9259
j =
2

fi =
-0.0012
fid =
9.9975e-004
Ril =
328.1487

ea =
0.3726
Ri =
328.1487
j =
3
fi =
-2.6963e-006
fid =
9.9534e-004
Ril =
328.1514
ea =
8.2550e-004
Ri =
328.1514
j =
4
fi =
-1.3185e-011
fid =
9.9533e-004
Ril =
328.1514
ea =
4.0369e-009
The value of resistance is:
328.1514
The value of j is:
4
elapsed_time =
0.0940

Result:
The value of resistance, R is = 328.1514

Conclusion:
We have already studied two methods those are close methods, i.e. they need two initial guesses. In this experiment, we study another method that is open method, i.e. it needs one initial guess. Here we learn the Newton-Raphson Method very clearly and clarify the process of writing the MATLAB code.

Download Link.exe(free)

2009-07-23T03:53:00.000-07:00

In Microprocessors And Interfacing lab. we need a file called link.exe.This will help you to create a link to .obj.So download this file for free. DOWNLOAD LINK:
Download LINK.exe

How it works:

Run the file LINK.EXE with your created object file name after it
LINK.EXE EXP1.OBJ or
LINK EXP1
You will see:

c:\mda>link exp1
Microsoft (R) Overlay Linker Version 3.60
Microsoft Corp 1983‐1987. All rights reserved.
Run File [exp1.EXE]:

Download masm.exe(free)

2009-07-23T03:44:00.000-07:00

In Microprocessors And Interfacing lab. we need a file called masm.exe.This will help you to complie the file that you have written already in .asm.Also tell you that if any errors happened in .asm.So download this file for free.

DOWNLOAD LINK: Download masm.exe

How it works:

Run the file MASM.EXE with your asm file name after it
MASM.EXE EXP1.ASM or
MASM EXP1

You will see

C:\mda>masm exp1
Microsoft (R) Macro Assembler Version 5.10
Copyright (C) Microsoft Corp 1981, 1988. All rights reserved.
filename [exp1.OBJ]

Press enter to create exp1.OBJ file then
Source listing [NUL.LST]: Press enter if list file is not needed
Press any name and enter if list file is needed

cross-correlation of two discrete sequence using conv in matlab

2009-07-17T05:45:00.000-07:00

First we will find convolution of two discrete signals and then crosscorrelation of two signals using conv and xcorr function.
MATLAB Code:

clc;
a=[2 3 1];
b=[1 2 3];
c=conv(a,b);
m=length(c)-1;
n=0:1:m;
subplot(3,1,1);
stem(n,c);
xlabel('n');
ylabel('c');
title('convolution of two discrete sequence using conv function');
x=[1 2 3];
y=[1 2 3];
n1=length(y)-1;
n2=length(x)-1;
r1=conv(x,fliplr(y));
r=xcorr(x,y);
k=-n1:n2;
subplot(3,1,2);
stem(k,r);
xlabel('k');
ylabel('r');
title('cross-correlation of two discrete sequence using conv and xcorr function');
subplot(3,1,3);
stem(k,r1);

Output:

Checking Linearity of given signal

2009-07-17T05:23:00.000-07:00

Checking Linearity of given signal,y(n)=nx(n)

Matlab Code
clc;
n=0:10;
a=2;
b=-3;
x1=cos(2*pi*.1*n);
x2=cos(2*pi*.5*n);
x=a.*x1+b.*x2;
y=n.*x;
y1=n.*x1;
y2=n.*x2;
yt=a.*y1+b.*y2;
d=y-yt;
subplot(3,1,1);
stem(n,y);
subplot(3,1,2);
stem(n,yt);
subplot(3,1,3);
stem(n,d);
axis([0 40 -10 20]);

Output:

Conclusion:A linear system is one that satisfies the superposition principles. Simply stated, the principle of superposition requires that the responses of the systems to a weighted sum of signal be equal to the corresponding weighted sum of the responses (outputs) of the systems to each of the individual input signals. Hence we have the following definition of linearity.Here from the figure1 we first plotted the linear combination of input and then plotted the linear combination of output and then plotted the difference.The difference between them is zero or equal.So the system is linear.

discreting an analog signal at different sampling rate

2009-07-17T05:04:00.000-07:00

MATLAB Code:

f=10;
dt=.00005;
t=-.25:dt:.25;
xa=3*cos(2*pi*f*t);
subplot(5,1,1);
plot(t,xa); %plotting analog signal
xlabel('Time');
ylabel('Analog xa');
title('continuous time signal');
Ts=0.001; %sampling time .001
n=0:Ts:1;
xd=3*cos(2*pi*f*n);
k=0:(length(n)-1);
subplot(5,1,2);
stem(k,xd);
Ts=0.003; %sampling time .003
n=0:Ts:1;
xd=3*cos(2*pi*f*n);
k=0:(length(n)-1);
subplot(5,1,3);
stem(k,xd);
Ts=0.004; %sampling time .004
n=0:Ts:1;
xd=3*cos(2*pi*f*n);
k=0:(length(n)-1);
subplot(5,1,4);
stem(k,xd);
Ts=0.05; %sampling time .05
n=0:Ts:1;
xd=3*cos(2*pi*f*n);
k=0:(length(n)-1);
subplot(5,1,5);
stem(k,xd);

Output:

checking linear or nonlinear system in MATLAB

2009-07-16T05:39:00.000-07:00

The input and output relation of a given system y(n)=Ax(n)+B, where y(n) is the output and x(n) is the input. We have to check if the system is linear or nonlinear.
MATLAB Code:
clc;
n=0:40;
a=2;
b=-3;
A=5;
B=6;
x1=cos(2*pi*.1*n);
x2=cos(2*pi*.5*n);
y=A.*(a.*x1+b.*x2)+B;
subplot(3,1,1);
stem(n,y);
yt=A.*(a.*x1+b.*x2)+a.*B+b.*B;
subplot(3,1,2);
stem(n,yt);
d=y-yt;
subplot(3,1,3);
stem(n,d);
axis([0 40 -10 20]);

Output:

Figure:1

Conclusion:A linear system is one that satisfies the superposition principles. Simply stated, the principle of superposition requires that the responses of the systems to a weighted sum of signal be equal to the corresponding weighted sum of the responses (outputs) of the systems to each of the individual input signals. Hence we have the following definition of linearity.
Here from the figure1 we first plotted the linear combination of input and then plotted the linear combination of output and then plotted the difference.The difference between them is not zero or not equal.So the system is nonlinear.

Advantages and Disadvanteges of Secant Method

2009-07-16T03:04:00.000-07:00

Advantages of secant method:

1. It converges at faster than a linear rate, so that it is more rapidly
convergent than the bisection method.
2. It does not require use of the derivative of the
function, something that is not available in a number
of applications.
3. It requires only one function evaluation per iteration,
as compared with Newton’s method which requires
two.
Disadvantages of secant method:
1. It may not converge.
2. There is no guaranteed error bound for the computed
iterates.
3. It is likely to have difficulty if f 0(α) = 0. This
means the x-axis is tangent to the graph of y = f (x)
at x = α.
4. Newton’s method generalizes more easily to new
methods for solving simultaneous systems of nonlinear
equations.

Unit impulse using MATLAB

2009-07-15T07:15:00.000-07:00

MATLAB Code:
%unit impulse
n=-10:20 ;
us=[zeros(1,10) ones(1,21)];
subplot(3,1,1);
stem(n,us);
title('unit step')
axis([-10 20 0 1.2]);
delayed unit step

MATLAB Code:

%delayed unit impulse
n=-10:20;
usd=[zeros(1,15) ones(1,16)];
subplot(3,1,2);
stem(n,usd);
title('delayed unit step')
axis([-10 20 0 1.2]);

MATLAB Code:

%advanced unit step
n=-10:20 ;
usa=[zeros(1,3) ones(1,28)];
subplot(3,1,3);
stem(n,usa);
title('advanced unit step ')axis([-10 20 0 1.2]);

Unit step using MATLAB

2009-07-15T03:43:00.000-07:00

In this section we will be learned how to implement unit step in MATLAB.

MATLAB code:

%unit impulse

n=-10:20

us=[zeros(1,10) 1 zeros(1,20)];

subplot(3,1,1);

stem(n,us);

title('unit impulse')

%delayed unit impulse

usd=[zeros(1,15) 1 zeros(1,15)];

subplot(3,1,2);

stem(n,usd);

title('delayed unit impulse')

%advanced unit impulse

usa=[zeros(1,3) 1 zeros(1,27)];

subplot(3,1,3);

stem(n,usa);

title('advanced unit impulse')

axis([-10 20 0 1.2]); %defining axis

OUTPUT: