Name– study of Boolean equations using the given ICs.
Aim – to study the manipulation of given equation into an equivalent logic circuit.
Apparatus – required ICs, circuit board, power supply +5V DC, LED, connecting wires, soldering iron, cutter etc.
Circuit diagram –
Draw the logic diagram of given equation (using basic gates only)
on left page of practical record book
with pin numbers of gates and output equation.


Pro ....etc

[:=Read Full Message Here=:]

i need this pdf file can you please make it available for me. thanx ....etc

[:=Read Full Message Here=:]

Procedure:-
1. Rewrite the given difference equation to have only the output terms on the LHS and input terms to be on RHS
2. Create a Matrix of Y coefficients, a
3. Create a Matrix of X coefficients, b
4. Generate the input sequence x(n)
5. Find the output of the system for the input sequence x(n)
6. Plot the input and the output

MATLAB program for difference equation without initial condition
clear
close all
clc

% THE INPUT SEQUENCE IS X(N) = U(N)
% THE DIFFERENCE EQUATION IS Y(N) – 0.9Y(N-1)= X(N)
% N ....etc

[:=Read Full Message Here=:]

Variable Voltage And Variable Frequency drives

Presented By:
P.NATARAJU S.E.V.S.RAGHU VARMA
3rd year E.E.E. 3rd year E.E.E.

Abstract:-
Our varying industrial needs demand more precise control of the outputs of our basic electrical prime movers i.e, THE MOTORS. Basically dc motors are easy to control compared to their ac counter parts, but they have their own limitations with increase in capacity. Conversely ac motors in particular squirrel cag ....etc

[:=Read Full Message Here=:]

PRESENTED BY:
Dr J Maiti


An Application of Structural Equation Modeling in Evaluating Accident/Injury Occurrences in Underground Coal Mines
Introduction
 The purpose of this presentation is to show how Structural Equation Modeling (SEM) can be applied to real life problem solving.
Stages in SEM
• Stage 1: Developing a Theoretically Based
Model
• Stage 2: Constructing a Path Diagram of Causal Relationships
– Elements of a Path Diagram
– Examples of Path Diagrams ....etc

[:=Read Full Message Here=:]

seminar on quadratic equation for 10 class
....etc

[:=Read Full Message Here=:]

//QuadEqn.java
package p;
interface Quadratic
{
void process();
}
public class QuadEqn implements Quadratic
{
double a,b,c,d;
public QuadEqn(double x,double y,double z)
{
a=x;
b=y;
c=z;
}
public void process()
{

d=b*b-(4*a*c);
if(d==0)
{
double r=-b/2*a;
System.out.println(Roots are equal.Root=\t+r);
}
else
if(d>0)
{
double sq=Math.sqrt(d);
double r1=(-b+sq)/2*a;
double r2=(-b-sq)/2*a;
System.out.println(Root1=\t+r1+\nRoot2=\t+r2);
}
else
System.out.pri ....etc

[:=Read Full Message Here=:]

Submitted by:Isha



Introduction

Laplace's equation
Laplace's equation is a partial differential equation. Partial differential equations (PDE) are a type of differential equation, i.e., a relation involving an unknown function (or functions) of several independent variables and their partial derivatives with respect to those variables. Partial differential equations are used to formulate, and thus aid the solution of, problems involving functions of several variable ....etc

[:=Read Full Message Here=:]

Plzz give a project on linear equation in two variables for class 10 ....etc

[:=Read Full Message Here=:]

clear
close all
clc
% SOLVE THE GIVEN DIFFERENCE EQUATION Y(N) - 1.5Y(N-1) + 0.5Y(N-2) = X(N), N >= 0
% X(N) = (1/4)^n * U(N) SUBJECT TO Y(-1) = 4 AND Y(-2) = 10
% CALCULATE AND PLOT THE OUTPUT OF THE SYSTEM
% FIND THE TRANSFER FUNCTION OF THE GIVEN DIFFERENCE EQUATION H(Z) = Y(Z)/X(Z)
% NUMERATOR POLYNOMIAL IS a AND DENOMINATOR POLYNOMIAL IS b
% FOR THE ABOVE EXAMPLE b = AND a =

b = ;
a = ;
% FIND THE INITIAL CONDITIONS FOR THE INPUT X SUCH THAT THE GIVEN INITIAL CONDITIONS
% FOR Y ARE OBTAIN ....etc

[:=Read Full Message Here=:]