Computer Organization Question set 3

Chapter 3 - arithmetic

1.  Data representation –                                                                                 

a) sign magnitude

     b) 1‘s complement

     c) 2’s complement

   Advantages, disadvantages, comparison

      2. r‘s complement, r-1’s complement

      3.Floating point number representation (8 bit,32 bit,64 bit)

      4.Special cases of IEEE 754 format.

      5.1’s complement arithmetic,2’s complement arithmetic

      6.Overflow and underflow detection

      7.Algorithm or flowchart – a) fixed point addition-subtraction

                                                    b) add –shift   multiplication

                                                    c) Booth’s multiplication

                                                   d) restore and non-restore division  

                                                   e) floating point addition, subtraction,                                                                        multiplication, division

     8.Common bus system

     9.Tristate buffer

    10.parallel adder, cla, adder-subtractor, carry save multiplier

    11. Bit sliced ALU

    12. 4bit ALU design

Computer Organization Question set 2

Chapter 2 – instruction cycle

1. Instruction cycle

2. Fetch, decode, execute

3. MAR, MBR, PC, SP, IR, DR, Flag register

4. PSW

Computer Organization Question set 1

Chapter 1 - Introduction

1. Generation of computer, advantages, disadvantages

2. IAS computer structure, block diagram.

3. Von-Neumann concept.

4. Bottleneck of Von-Neumann.

5. Harvard architecture

6.Comparison of Harvard and Von Neumann.

Memory

Block diagram of ROM
Short note - PROM,EPROM,EEPROM
SRAM,DRAM
Comparison SRAM,DRAM
Read and write operation of Static ram cell
Read and write operation of Dynamic ram cell
Combinational circuit using ROM
Binary to ASCII using ROM
Full Adder using ROM
Block Diagram of PLA
Combinational circuit using PLA
Comparison ROM,PLA
Block Diagram of PAL
Combinational circuit using PAL
Comparison ROM,PLA,PAL

Sequential circuit using Mealy model and Moore model

Explain Moore type of sequential machine.
Explain Mealy type of sequential machine.
Compare Mealy model and Moore model
State transition,state diagram
Transition table
Rules of conversion
Design serial adder (ignoring carry)
Output=1 when even number of 1's accepted,otherwise output=0
Output=1 when 3k numbers of 1's accepted,otherwise output=0, k>=0
Output=1 when 3k numbers of 1's accepted,otherwise output=0, k>0
Output=1 when 3k+1 numbers of 1's accepted,otherwise output=0, k>=0
Output=1 when even number of 1's and even numbers of 0's accepted,otherwise output=0
Output=1 when even number of 1's and odd numbers of 0's accepted,otherwise output=0
Output=1 when odd number of 1's and even numbers of 0's accepted,otherwise output=0
Output=1 when odd number of 1's and odd numbers of 0's accepted,otherwise output=0
Output=1 when 1011 sequence accepted,otherwise output=0
Sequence generator

Register

Defination of register
Comparison of register and counter
Serial in serial out register
Serial in parallel out register
Parallel in serial out register
Parallel in parallel out register
Register with parallel load (SR,D,JK)
Shift register (Left - SR,D,JK)
Shift register (Right - SR,D,JK)
Bidirectional shift register
Universal shift register

COUNTER

Synnchronous counter - up- 2 bit,3 bit,4bit
Synnchronous counter - down- 2 bit,3 bit,4bit
Synnchronous counter - up- decimal,octal
Synnchronous counter - down- decimal,octal
Synnchronous counter - up/down 4 bit
Synnchronous counter - up- mod 3,mod 4,mod 6,mod 9,mod 10,mod 12
Synnchronous counter - down- mod 3,mod 4,mod 6,mod 9,mod 10,mod 12
Asynnchronous counter - up- 2 bit,3 bit,4bit
Asynnchronous counter - down- 2 bit,3 bit,4bit
Asynnchronous counter - up- decimal,octal
Asynnchronous counter - down- decimal,octal
Asynnchronous counter - up/down 4 bit
Asynnchronous counter - up- mod 3,mod 4,mod 6,mod 9,mod 10,mod 12
Asynnchronous counter - down- mod 3,mod 4,mod 6,mod 9,mod 10,mod 12
Ring Counter
Johnson Counter
Self stopping counter

Flip-Flop

Flip-Flop

1. Sequential circuit defination, comparison with combnational circuit .
2. What do you mean by memory elements in sequential cir5cuit?
3. Latch - nand based,nor based, compare with flip-flop.
4. Flip-flop- truth table,circuit diagram, state diagram, characteristic equation.
5. Flip-flop - Edge-triggered , level-triggered.
6. Propagation Delay.
7. Race around condition .
8. Master slave flip-flop.
9.Flip flop conversion ( one flip-flop to another flipflop).
10. frequency divider .

C Program code of Bisection method

 C Program code of Bisection method ( by Shrayashi Majumder)

         1.   Find the root of the equation(x³ - 4x-9=0) using bisection method.

Code:

#include<stdio.h>

#include<math.h>

double f(double x)

{

return x*x*x-4*x-9;

}

void main()

{

int i=0;

double mid,a=0.0,b=1.0,x=0.0,s=1.0,s1=1.0;

while(1)

{

s=f(a)*f(b);

s1=f(-a)*f(-b);

if(s<0.0 || s1<0.0)

        break;

else

{

        a=b;

        b++;

}

}

if(s1<0.0)

{

a=-a;

b=-b;

}

printf("Range %f to %f\n \n",a,b);

printf("\n          a        b        f(a)       f(b)    mid");

while(1)

{

mid=(a+b)/2;

printf("\nStep%d:  %f %f %f %f %f",i++,a,b,f(a),f(b),mid);

if(f(mid)==0.0)

        break;

if(fabs(a-b)<0.0001)

        break;

        if(f(a)*f(mid)<0.0)

                b=mid;

        else

                a=mid;

}

printf("\nx=%f",mid);

}

Output:

Range 2.000000 to 3.000000

          a        b        f(a)       f(b)    mid

Step0:  2.000000 3.000000 -9.000000 6.000000 2.500000

Step1:  2.500000 3.000000 -3.375000 6.000000 2.750000

Step2:  2.500000 2.750000 -3.375000 0.796875 2.625000

Step3:  2.625000 2.750000 -1.412109 0.796875 2.687500

Step4:  2.687500 2.750000 -0.339111 0.796875 2.718750

Step5:  2.687500 2.718750 -0.339111 0.220917 2.703125

Step6:  2.703125 2.718750 -0.061077 0.220917 2.710938

Step7:  2.703125 2.710938 -0.061077 0.079423 2.707031

Step8:  2.703125 2.707031 -0.061077 0.009049 2.705078

Step9:  2.705078 2.707031 -0.026045 0.009049 2.706055

Step10:  2.706055 2.707031 -0.008506 0.009049 2.706543

Step11:  2.706055 2.706543 -0.008506 0.000270 2.706299

Step12:  2.706299 2.706543 -0.004118 0.000270 2.706421

Step13:  2.706421 2.706543 -0.001924 0.000270 2.706482

Step14:  2.706482 2.706543 -0.000827 0.000270 2.706512

x=2.706512

2.  Find the root of equation using bisection method.

Code:

#include<stdio.h>

#include<math.h>

double f(float c3, float c2 , float c1 , float cn , double x)

{

double r;

r=c3*x*x*x+c2*x*x+c1*x+cn;

return r;

}

void main()

{

int i=0;

float c3,c2,c1,cn;

double mid,omid=100.0,a=0.0,b=1.0,s=1.0,s1=1.0,tol;

printf("Enter coefficients:");

scanf("%f%f%f%f",&c3,&c2,&c1,&cn);

printf("Enter Tolerance:");

scanf("%lf",&tol);

while(1)

{

s=f(c3,c2,c1,cn,a)*f(c3,c2,c1,cn,b);

s1=f(c3,c2,c1,cn,-a)*f(c3,c2,c1,cn,-b);

if(s<0.0 || s1<0.0)

        break;

else

{

        a=b;

        b++;

}

}

if(s1<0.0)

{

        a=-a;

        b=-b;

}

printf("Range %f to %f",a,b);

printf("\n    \ta        b        f(a)       f(b)    mid");

while(1)

{

        mid=(a+b)/2;

        printf("\nStep %d: %f %f %f %f %f",i++,a,b,f(c3,c2,c1,cn,a),f(c3,c2,c1,cn,b),mid);

        if(f(c3,c2,c1,cn,mid)==0.0)

                break;

        if(fabs(a-b)<tol)

                break;

        if(f(c3,c2,c1,cn,a)*f(c3,c2,c1,cn,mid)<0.0)

                b=mid;

        else

                a=mid;

        omid=mid;

}

printf("\nx=%f",mid);

}

Output:

Enter coefficients:1 0 -4 -9

Enter Tolerance:0.0001

Range 2.000000 to 3.000000

        a        b        f(a)       f(b)    mid

Step 0: 2.000000 3.000000 -9.000000 6.000000 2.500000

Step 1: 2.500000 3.000000 -3.375000 6.000000 2.750000

Step 2: 2.500000 2.750000 -3.375000 0.796875 2.625000

Step 3: 2.625000 2.750000 -1.412109 0.796875 2.687500

Step 4: 2.687500 2.750000 -0.339111 0.796875 2.718750

Step 5: 2.687500 2.718750 -0.339111 0.220917 2.703125

Step 6: 2.703125 2.718750 -0.061077 0.220917 2.710938

Step 7: 2.703125 2.710938 -0.061077 0.079423 2.707031

Step 8: 2.703125 2.707031 -0.061077 0.009049 2.705078

Step 9: 2.705078 2.707031 -0.026045 0.009049 2.706055

Step 10: 2.706055 2.707031 -0.008506 0.009049 2.706543

Step 11: 2.706055 2.706543 -0.008506 0.000270 2.706299

Step 12: 2.706299 2.706543 -0.004118 0.000270 2.706421

Step 13: 2.706421 2.706543 -0.001924 0.000270 2.706482

Step 14: 2.706482 2.706543 -0.000827 0.000270 2.706512

x=2.706512

8085 Microprocessor Assignment

8085 assignments :

Download Link