computer organization and architecture lab manual

COMPUTER ORGANIZATION & ARCHITECTURE LAB

1

LAB@1

Full Marks: 25 Pass Marks: 10

Marks Distribution

Mark Distribution Criteria Mark Remark

Attendance 8

Lab report 8 Viva 8

Discipline 1

Total 25

Lab Report Format

Title

Background Theory

MATLAB tools used in your lab program

Algorithm&Flowchart

Code

Output : Screen shots/handwritten output

Discussion & Conclusion

Integer representation (Fixed point representation):

Only have 0 & 1 to represent everything

Positive numbers stored in binary; e.g. 41=00101001

No minus sign and No period

Sign-Magnitude

Left most bit (MSB) is sign bit

0 means positive and 1 means negative e.g. +18 = 00010010 and -18 = 10010010

Problems: Need to consider both sign and magnitude in arithmetic&Two

representations of zero (+0 and -0)

Two’s compliment

2’s compliment representation uses the most significant bit (MSB) as sign bit

making it easy to rest whether the integer is negative or positive. It differs from

the use of sign magnitude representation in the way the other bits are interpreted.

For negation take the Boolean complement of each bit of corresponding positive

number and then add one to the resulting bit pattern viewed as unsigned integer.


2

+3 = 00000011 and -3 = 11111101

Benefits:

One representation of zero

Arithmetic works easily

Negating is fairly easy

3 = 00000011

Boolean complement gives 11111100 then add 1 to LSB11111101

Addition and Subtraction

Normal binary addition

Monitor sign bit for overflow

Take twos compliment of subtrahend and add to minuend

i.e. a - b = a + (-b) ,So we only need addition and complement circuits

For example

The 2’s complement of 0111 is 1001 and is obtained by leaving first 1 unchanged, and then

replacing 1’s by 0’s and 0’s by 1’s in the other three most significant bits. Let us see an example of subtraction using two binary numbers X = 10(1010) and Y = 9 (1001) and perform X-Y and Y-X.

X = 1010

2’s complement of Y = +0111

-----------

Sum = 10001

Discard end carry 24 = -10000

----------

Answer: X-Y = 0001

Y = 1001

2’s complement of X = +0110

-----------

Sum = 1111

There is no end carry

Answer is negative 0001 = 2’s compliment of 1111 and for addition of 2’s complement use

same algorithm.


3

Hardware for Addition and Subtraction

Figure: Hardware implementation of Addition/Subtraction

To implement the two arithmetic operations with hardware, it is first necessary that the two

numbers to be stored in registers. Let A and B be two registers that hold the magnitude of the

numbers and As and Bs be two flip-flops that hold the corresponding signs. The result of the

operation may be transferred to a third register.

The data path and hardware elements needed to accomplish addition and subtraction is shown in

Fig. 5.1. The center element is binary adder, which is presented two numbers for addition and

produces a sum and an overflow indication. The binary adder treats from two registers A and B.

the result may be stored in one of these registers or in the third register. The overflow indication

is stored in a 1-bit overflow flag. For subtraction, the subtrahend (B register) is passed through a

2’s complement unit so that its 2’s complement is presented to the adder (a-b=a + (-b)).


4

Logical operation:

The fundamental logic gate family

The seven fundamental logic gates are:

1. NOT (Inverter)

2. AND

3. NAND

4. OR

5. NOR

6. XOR

7. XNOR


5

Main program to add two unsigned binary number [4 bit binary adder]

%******* Main Program Code*******%

display('Enter First Binary Sequence')

for i=1:4

a(i)=input('');

end

display('Enter Second Binary Sequence')

for i=1:4

b(i)=input('');

end

carry=0;

for i=4:-1:1

c(i)=xors(xors(a(i),b(i)),carry);

carry=ors(ands(a(i),b(i)),ands(xors(a(i),b(i)),carry));

end

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

fprintf('%d',a)

fprintf('\n')

fprintf('%d',b)

fprintf('+\n')

display('=====================')

fprintf('%d%d',carry,c)

fprintf('\n')

Function for AND operation

function [res] = ands(a,b)

if(a==1) && (b==1)

res=1;

else

res=0;

end

end

Function for XOR operation

function [res]= xors(a,b)

if (a~=b)

res=1;

else

res=0;

end

end

Function for OR operation

function [res] = ors(a,b)

if (a==0) && (b==0)

res=0;

else

res=1;

end

end


6

Lab@2

%%***** Matlab code for subtraction *******%%

display('Enter First Number')

for i=1:4

a(i)=input('');

end

display ('Enter Second Number')

for i=1:4

b(i)=input('');

end

%%2's complement of Second Sequence

for i=1:4

b(i)=xor(1,b(i));

end

c=1;% for 2's complement we add 1 so initialize carry as 1

for i=4:-1:1

[s (i) c] = adder ( a(i) , b(i) , c);

end

display('result')

%% Carry decide the result is in negative or positive. ifCayy is 0 then

%% result is negative else result is positive

if (c==0)

for (i=1:4)

s(i) = xor(1,s(i));

end

car=1;

for (i=4:-1:1)

[s(i) , car] = adder(s(i),0,car);

end

%display('Negative')

fprintf('-')


7

fprintf('%d',s)

else

%display('Positive')

fprintf('+')

fprintf('%d',s)

end

% Plotting input and output

subplot(3,1,1)

stem(a,'R');

xlabel('First sequence');

subplot(3,1,2);

stem(b,'G');

xlabel('Second Sequence');

subplot(3,1,3);

stem(s,'B');

xlabel('Result(a-b)');

Function for Addition Operation

function [res,car] = adder(a,b,c)

res= xor(xor(a,b),c);

car= or(and(a,b),and(xor(a,b),c));

end


8

Lab@3

Multiplication Algorithm

Complex

Work out partial product for each digit

Take care with place value (column)

Add partial products

Example:

1011 Multiplicand (11 decimal)

X 1101 Multiplier (13 decimal)

1011 Partial products

0000 Note: if multiplier bit is 1 copy multiplicand (place value)

1011 otherwise zero

1011

10001111 Product (143 Dec)

Note: need double length result

Hardware Implementation:


9

The multiplier and multiplicand bits are loaded into two registers Q and M. A third register A is

initially set to zero. C is the 1 bit register which holds the carry bit resulting from addition. Now,

the control logic reads the bits of the multiplier one at a time. If Q0 is 1, the multiplicand is added

to the register A and is stored back in register A with C bit used for carry. Then all the bits of

CAQ are shifted to the right 1 bit so that C bit goes to An-1, A0 goes to Qn-1 and Q0 is lost. If Q0 is

0, no addition is performed just do the shift. The process is repeated for each bit of the original

multiplier. The resulting 2n bit product is contained in the QA register.

Flowchart for unsigned binary multiplication


10

%% Matlab code for Multiplication of two unsigned 4-bit binary number

clearall;

closeall;

R=[0 0 0 0 0 0 0 0];

display('Enter First Number');

for i=5:8

A(i)=input('');

end

display ('Enter Second Number');

for i=5:8

B(i)=input('');

end

display('***==============================***');

fprintf('Multiplicant=');

fprintf('%d',A);

fprintf('\n');

fprintf('Multiplier=');

fprintf('%d',B);

fprintf('\n');

fprintf('***************************');

subplot(3,1,1);

stem(A,'G');

xlabel('Multiplicant');

if(B(8)==1)

R=A;% Assign the given number First partial product

end

for i=7:-1:5

if(B(i)==1)

c=0;

for j=8:-1:1

[A(j) c]=adds(A(j),A(j),c);


11

end

c=0;

for j=8:-1:1

[ R(j) c]=adds(R(j),A(j),c);% shifting left

end

else

c=0;

for j=8:-1:1

[A(j) c]=adds(A(j),A(j),c);

end

end

end

%% Result Plotting

fprintf('\n');

fprintf('Multiplication result=');

fprintf('%d',R);

fprintf('\n');

display('***==============================***');

subplot(3,1,2);

stem(B,'R');

xlabel('Multiplier');

subplot(3,1,3);

stem(R);

xlabel('Result after multiplication');


12

Lab@4

Division of Unsigned Binary Integers

Hardware implementation of division algorithm

Restoring Division:

This method is called restoring division because the partial remainder is restored by adding the

divisor to the negative difference. Here division process requires controlled subtract-restore

operations. Whether the next operation is a subtraction or restoration, is controlled by the

result of the current operation.

Shift left

Divisor

Add/Subtract

An An-1 .....………. An

0 Mn-1 ……..…... Mn

Qn-1 ……..…… … Q0

N+1 Bit

Adder

Control

unit


13

Flowchart of Restoring Division algorithm

Algorithm

Step1: Initialize A 0, dividend (Q) & divisor (M) registers and counter 0

Step2: Shift left A, Q on binary position.

Step3: If MSB of A is 1, set Q (0) 0 and add M back to A (restore A), else set Q (0) 1

Step4: counter counter + 1; if couner ≠ n-1 where n is the number of bits in the dividend,

repeat process from step2 else stop the process. Finally remainder will be in A and

Quotient will be in Q.


14

%% Matlab Code for restoring Division algorithm [two 4-bit binary number %%

A = [ 0 0 0 0 ];

B = [0 0 0 0];

disp('Enter Dividend:')

for i=1:4

Q(i)=input('');

end

disp('Enter Divisor:')

for j=1:4

B(j)=input('');

end

%% Display Input Data

display('Dividend:');

Q

display('Divisor:');

B

%% code of 2's complement of Divisor

c=1;

for i=4:-1:1

[nB(i) c]=adds(xor(B(i),1),0,c);

end

%% Main code start

for cnt=1:4

% Shift operation (left Shift A, Q)

[A Q]=shifts(A,Q);

C=0;

% Subtraction operation (A=A-B)

for i=4:-1:1

[A(i) C]=adds(A(i),nB(i),C);

end

% Check MSB of A


15

if(A(1)==0)

Q(4)=1; % Set LSB of Q

else

Q(4)=0; % Set LSB of Q

c=0;

% Addition Operation (A=A+B)

for i=4:-1:1

[A(i) C] = adds(A(i),B(i),C);

end

end

end

%% Result Display

disp('Quotient: ')

Q

disp('Remainder: ')

A

%% Plotting Result

subplot(2,2,1);

stem(Q, 'Y');

xlabel('Dividend');

subplot(2,2,2);

stem(B,'B');

xlabel('Divisor');

subplot(2,2,3);

stem(Q,'R');

xlabel('Quotient');

subplot(2,2,4);

stem(A,'G');

xlabel('Remainder');

computer organization and architecture lab manual

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