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Title: The Radar Equation
Page Link: The Radar Equation -
Posted By: seminar addict
Created at: Thursday 26th of January 2012 02:01:19 PM
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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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Title: pneumatic conveyor design equation ppt
Page Link: pneumatic conveyor design equation ppt -
Posted By:
Created at: Monday 14th of January 2013 09:01:54 PM
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sir,
i need a ppt sugesting some useful equations which would help me to find the power ratings of the motor which can be used to drive a roots type blower for pneumatic conveyor applications.I have to transport TiO2 powder to a horizontal distance of 10 m,then vertically up through 25 m and dump it in a tank there.The pipeline used is nearly 6inch in diameter. and the discharge requirements is 7200kg/hr.
Please help me...
Thanking You,
PRABHU VIJAYAN ....etc

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Title: An Application of Structural Equation Modeling in Evaluating AccidentInjury Occurren
Page Link: An Application of Structural Equation Modeling in Evaluating AccidentInjury Occurren -
Posted By: seminar class
Created at: Wednesday 16th of March 2011 02:23:50 PM
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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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Title: SOLUTION OF DIFFERENCE EQUATION WITH INITIAL CONDITIONS
Page Link: SOLUTION OF DIFFERENCE EQUATION WITH INITIAL CONDITIONS -
Posted By: seminar class
Created at: Friday 06th of May 2011 06:10:52 PM
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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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Title: PROGRAM TO FIND ROOTS OF QUADRATIC EQUATION BY USING PACKAGES AND INTERFACE
Page Link: PROGRAM TO FIND ROOTS OF QUADRATIC EQUATION BY USING PACKAGES AND INTERFACE -
Posted By: project topics
Created at: Wednesday 06th of April 2011 03:48:35 PM
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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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Title: Study the manipulation of given equation into an equivalent logic circuit
Page Link: Study the manipulation of given equation into an equivalent logic circuit -
Posted By: seminar class
Created at: Friday 13th of May 2011 06:37:09 PM
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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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Title: SOLUTION OF DIFFERENCE EQUATION WITHOUT INITIAL CONDITIONS
Page Link: SOLUTION OF DIFFERENCE EQUATION WITHOUT INITIAL CONDITIONS -
Posted By: seminar class
Created at: Friday 06th of May 2011 06:05:41 PM
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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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Title: point by point solution of swing equation
Page Link: point by point solution of swing equation -
Posted By:
Created at: Monday 15th of October 2012 12:40:39 AM
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please give me materal related to slides on swing equation

thanks and regards
poocho ....etc

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Title: Laplaces equation
Page Link: Laplaces equation -
Posted By: seminar surveyer
Created at: Thursday 23rd of December 2010 07:31:04 PM
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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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Title: IDEAL POWER EQUATION
Page Link: IDEAL POWER EQUATION -
Posted By: smart paper boy
Created at: Wednesday 03rd of August 2011 12:00:44 PM
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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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