Chapter 1 Homework The largest numbers that could be added with the function

1.37 Write a user-defined MATLAB function that adds two integers in binary form according to the algo-

rithm of Problem 1.30. Name the function aplusb = addbin(a,b), where the input arguments a and b

are the numbers to be added in binary form (vectors with 1s and 0s that represent the binary numbers), and

the output argument aplusb is the result in binary form (a vector with 1s and 0s that represent the binary

number). The largest numbers that could be added with the function should be binary numbers with 15 1s.

Use the function in the Command Window to add 100111011 and 1100110.

Solution

function aplusb=addbin(a,b)

n=length(a); m=length(b);

maxi=max(m,n);

if m>n

end

if bb(i)==1

flag_b=1;

When the function is executed in the command window, the following is obtained:

>> a=[1 0 0 1 1 1 0 1 1];

>> b=[1 1 0 0 1 1 0];

>> aplusb=addbin(a,b)

aa =

1.38 Write a user-defined MATLAB function that adds two integers in decimal form. Name the function

c = adddec(a,b), where the input arguments a and b are the numbers to be added in decimal form, and

the output argument c is the result in decimal form. The addition is done in the following way: First, the

numbers a and b are converted to binary form by using the user-defined function intTObina written in

Problem 1.35. Then, the converted two numbers are added using the user-defined function addbin writ-

ten in Problem 1.37. Finally, the result is converted to decimal form by using the user-defined function

binaTOint written in Problem 1.34. The three user-defined functions intTObina, addbin, and

binaTOint can be written as subfunctions inside adddec. The numbers that are being added with the

function cannot be larger than 65,535 each. If a larger number is entered, an error message is displayed.

Use the function in the Command Window to add 60,000 and 12,087.

Solution

If the functions addbin and binaTOint are used as they are in the previous problems, the following

program will still work, but the error messages can be eliminated by replacing the criteria with:

%**************************************************************************

end

if (flag == 1)

break

end

for i=1:length(pint)

b(20-pint(i))=1;

end

b(20-pint(i))=1;

end

%**************************************************************************

function aplusb=addbin(a,b)

end

flag_a=0; flag_b=0; aplusb(1:maxi+1)=0; answer(1:maxi+1)=0;

carry=0;

aa(1:maxi)=0; bb(1:maxi)=0;

if large==n

aa(1:maxi)=a(1:n); bb(maxi-m+1:maxi)=b(1:m);

else

end

aplusb=ans;

%**************************************************************************

function d=binaTOint(b)

sum=0;

n=length(b);

When the function is executed in the command window, the following is obtained:

>> c=adddec(60000,12087)

c =

72087

1.39 Write a user-defined MATLAB function that implements the IDDD-643 standard of problem 1.13.

Name the function n = iddd643(num), where the input argument num is any real number, and the out-

put argument n is a 16-element-long vector with 1s and 0s. Use chopping if necessary. Use iddd643 for

determining n for the following numbers:

(a) 81. (b) 256.1875.

Solution

function n = iddd643(num)

IDDD643(1:16)=0;

base10 = num;

if base10 > 0.0

sign = 0;

else

%**********************************************************************

function b=intTObina(d)

large=0; b(1:20)=0;

for i=1:20

large=large+(2^(i-1));

end

if(d>large) disp(‘ERROR: The number entered is too large for this function to

handle’)

return

%**************************************************************************

function b=deciTObina(d)

large=0; b(1:30)=0;

if d > 0.0

sign = 1;

else

sign = 0;

end

for i=1:15

large=large+(2^(i-1));

end

%Convert digits to the left of the decimal point:

if(integer~=0)

flag = 0; icount = 1;

for j = 15:-1:1

temp = integer/(2^j);

if (temp >= 1)

pint(icount) = j; icount = icount + 1; integer = integer -(2^j);

if (integer == 1)

flag = 1; pint0 = 0;

end

for i=1:length(pint)

b(15-pint(i))=1;

end

When the function is executed in the command window, the following is obtained:

>> n = iddd643(81)

n =

1.40 Write a user-defined MATLAB function that determines the binary floating point representation of a

number written in decimal form. Name the function b = decTObinfloat(num), where the input argu-

ment num is a real number, and the output argument b is a two-element vector in which the first element is

the mantissa and the second element is the value of the exponent. Use the function to determine the binary

floating point representation of 55.6, 2143.75, and 0.00843.

Solution

function b = decTObinfloat(num)

format long

if num > 0.0

temp = dec*(2^j);

if (temp >= 1)

break

end

Max_expo = -j;

sci_bin = dec/(2^Max_expo);

mantissa = sci_bin – 1;

b(1)=mantissa; b(2)=Max_expo;

end

When the function is executed in the command window, the following is obtained:

>> b = decTObinfloat(55.6)

1.41 The value of can be calculated with the series:

(1.1)

Write a MATLAB program in a script file that calculates the value of by using n terms of the series and

calculates the corresponding true relative error. (For the true value of , use the predefined MATLAB vari-

able pi.) Use the program to calculate and the true relative error for:

(a) . (b) . (c) .

Solution

clear; clc;

n=input(‘Please enter the number of terms of the series desired:\n’);

total=0;

When the function is executed in the command window, the following is obtained:

(a)

Please enter the number of terms of the series desired:

10

For n= 10, the calculated value of pi is 3.04184

π

π41–()

n1– 1

2n1–

————––

n1=

41 1

3

—–

–1

5

—–1

7

—–

–1

9

—–1

11

—––

–…++ +





==

π

n10=

n20=

n40=

For n= 20, the calculated value of pi is 3.09162

The true relative error is 1.59056e-02 or 1.591 percent

>>

As can be seen, the convergence of this series is extremely slow, and many more terms are required before

accuracy can be obtained to three decimal places:

Please enter the number of terms of the series desired:

2000

1.42 The Taylor’s series expansion for is:

where x is in radians. Write a user-defined function that determines using Taylor’s series expansion.

For function name and arguments, use y=sinTaylor(x), where the input argument x is the angle in

degrees and the output argument y is the value for . Inside the user-defined function, use a loop for

adding the terms of the Taylor’s series. If is the nth term in the series, then the sum of the n terms is

. In each pass, calculate the estimated error E given by . Stop adding terms

when . Use sinTaylor for calculating:

(a) . (b) .

Compare the values calculated using sinTaylor with the values obtained by using MATLAB’s built-in

sind function.

Solution

function y=sinTaylor(x)

format long

xsin

xsin xx3

3!

—-–

–x5

5!

—-–x7

7!

—-–

–…++ 1–()

n

2n1+()!

—————–—-– x2n1+()

n0=

∞



==

xsin

an

Sn

SnSn1– an

+=

ESnSn1–

–

Sn1–

—————–—–

=

E0.000001≤

65°sin

195°sin

When the function is executed in the command window, the following is obtained:

(a)

>> y=sinTaylor(65)

y =

0.906307786213760