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

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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

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The Radar Equation


Introduction


Before target information can be extracted from an echo signal, that signal must be of sufficient magnitude to overcome the effects of interference.
The radar equation is used to predict echo power and interfering power to assist in making the determination of whether or not this condition is met. Use of the radar equation accomplishes the following:

Assists in the design of radar systems to meet the detection specific ....etc

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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

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Hi, I am Apurva Dubey request you to provide me job cost sheet in blance sheet of Amul Company for 3 consecutive years. For the financial year 2014-2015, 2015-2016, 2016-2017. I am living at Vadodara and I am studying at Parul Univesity right now and I need this detail in order to do cost analysis for my project which is a part of MBA curriculum.

Regards,
Apurva Dubey
[email protected] ....etc

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If the secondary coil is attached to a load that allows current to flow, electrical power is transmitted from the primary circuit to the secondary circuit.Ideally,the transformer is perfectly efficient; all the incoming energy is transformed from the primary circuit to the magnetic field and into the secondary circuit. If this condition is met, the incoming electric power must equal the outgoing power.
Pincoming = IPVP = Poutgoing = ISVS

giving the ideal transformer equation

....etc

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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

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cost approximate cost sheet of amul milk product pdf ....etc

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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

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//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

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