8085 Microprocessor sample question set THEORY

Gaonkar Chapter 1 - CLICK HERE

Gaonkar Chapter  2 - CLICK HERE

Gaonkar Chapter  3 - CLICK HERE

Gaonkar Chapter  4 - CLICK HERE

GaonkarChapter  5 - CLICK HERE

Gaonkar Chapter  6 - CLICK HERE

Gaonkar Chapter  7 - CLICK HERE

Gaonkar Chapter  8 - CLICK HERE

Gaonkar Chapter  9 - CLICK HERE

Gaonkar Chapter  10 - CLICK HERE

Operating System Question Set

Operating System Question Set 
GALVIN

INTRODUCTION: -  CLICK HERE

PROCESS: CLICK HERE

THREAD: CLICK HERE

Process synchronisation: CLICK HERE

Memory management. CLICK HERE

File system: CLICK HERE

Disk Scheduling: CLICK HERE

Deadlock:  CLICK HERE

C Language QUESTION Theory Set :

set1:  CLICK

set2:  CLICK

set3: CLICK

Basic Electronics Set 2

1. Operation of a pnp transistor,  npn transistor
2. Basic circuit for using a pnp transistor as an amplifier
3. Why this is called bipolar transistor?
4.Why this is called current controlled device?
5. Emitter follower.
6. Common-base, common-emitter and common-collector amplifiers
7. Relationship between α and β.
8. Load line analysis,Q-point
9. Describe the operation of a transistor amplifier in CE configuration.
10.Differentiate between FET and BJT transistors.
11. What is the use of biasing? Draw the DC equivalent model.
12. Thermal Runaway, thermal resistance
13.  Draw a BJT fixed bias circuit and derive the expression for the stability factor ‘S’.
14.Advantages and  disadvantages of fixed bias circuit
15. what is the condition for thermal stability?
16. Explain thermal instability. What are the factors affecting the stability factor?

Basic Electronics Set 1

1. Conductor , insulator , semiconductor - definition, example.
2. Valence band, conduction band, band gap
3.  Intrinsic semiconductor, extrinsic semiconductor, doped semiconductors,doping
4. Hole,doping,Recombination
5. Diffusion and drift, drift current
6. filter,rectifier
7. Characteristics graph of forward biasing,reverse biasing
8.avalanche break down,Zener break down
9. LED
10.Depletion region
11.Diode current equation
12. Half wave rectifier
13.full wave rectifier-center trap,bridge rectifier(with filter and with out filter)
13. ripple factor,PIV,Load current,rectifier efficiency
14. Characteristics of Zener diode
15. Diode as a voltage regulator

Computer Organization Question set 7

Chapter 7 – I/O Organization

1. I/O controller or I/O interface

2. I/O driver

3. I/O port

4. Synchronous data transfer and asynchronous data transfer

5. Handshaking mode, strobe control mode (source initiated, destination initiated)

6. Programmed I/O

7. Interrupt I/O

8. DMA –DMA controller, handshaking mode, cycle stealing mode

9. Software interrupt, Hardware interrupt

10. Internal interrupt, External interrupt

11.Vector interrupts non-vectored interrupt
12. ISR

13. Bus Arbitration – daisy chaining, polling, independent request

14. Input output processor

15. Memory mapped I/O,I/O mapped I/O

16. Serial communication, parallel communication, advantages, disadvantages

17. Short note – SCSI,PCI,USB

Computer Organization Question set 6

Chapter 6 – Control unit

1. Definition control unit

2. Hardwired control unit, micro programmed control unit advantages, disadvantages, comparison

3. State table method, sequence counter, delay element

4. Control memory

5. Wilkes design

6. Horizontal micro instruction, vertical micro instruction, comparison

7. Encoding of micro instruction

8. Parallelism in micro instruction 

Computer Organization Question set 5

                                 Chapter 5 – Instruction set

1. Instruction set

2. Instruction set completeness

3. Instruction format

4. Accumulator organization

5. Stack organization

6. Three address instruction, two address instruction, one address instruction, Zero address instruction

7. Addressing mode

8. CISC/RISC advantages, disadvantages, comparison

9. Instruction pipeline definition only

Computer Organization Question set 4

Chapter 4 – Memory

1. Defination – seek time, capacity, Rotational delay

2. Short note – memory hierarchy

3. SRAM – read and write technique

4. DRAM – Read and Write technique

5. SRAM,DRAM – Comparison

6. RAM CHIP, ROM CHIP example.

7. LARGE memory using small chip, range of memory address map

8. Memory -1D,2D,2.5D,3D

9. PROM, EPROM, EEPROM,FLASH MEMORY – Comparison

10. Magnetic Disc- track, sector, cylinder, inter-sector gap, surface, platter

11.  Short note- optical drive, associative memory

12. Cache memory- cache hit, cache miss, locality of reference, spatial locality, temporal locality, direct mapping, associative mapping, se-associative mapping, write through, write back, two level cache

13. Virtual memory

14. Page, frame

15. Page replacement technique.

16. Desructive and nondestructive memory

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 :

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          C++ this function

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