Logic :  A simple logic lies here. We declare another matrix B with order (N, M). The first matrix is traced through the first entry, till the end. The member is copied to another matrix with interchanging the column and rows count. Finally, the matrix B will be the transpose of the entered matrix A.

Please find the
other matrix  related programs from the index.

#include<stdio.h>
void main()
{
int A[5][5], B[5][5], i, j, m, n;
clrscr();
printf(“\n\n\t ENTER A ORDER OF THE MATRIX M,N…: “);
scanf(“%d,%d”,&m,&n);
printf(“\n\n\t ENTER THE ELEMENTS OF THE MATRIX..:\n\n”);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(20+j*4,12+i*2);
scanf(“%d”,&A[i][j]);
}
printf(“\n”);
}
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
B[j][i] = A[i][j];
printf(“\n\t THE TRANSPOSE OF A MATRIX IS…:\n\n\t\t “);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
printf(” %d”,B[i][j]);
printf(“\n\n\t\t “);
}
getch();
}
Download exe and source code here.

The post Program to find the transpose of a Matrix first appeared on Student Projects.

Logic :  The secondary diagonal elements will have a constant relation in their row and column column count. As the row count increases, the column count decreases, to maintain the sum of the two to ( M+1) where M is the order. By keeping this in mind, we can trace the matrix till the end using the help of a for loop.  The flag ‘sum’ is updated in each of the iterations.
Finally, we’ll get the sum of the diagonal elements after the end of the while loop.

You are always welcome with your suggestion and doubts into our discussion forum.

#include<stdio.h>
void main()
{
int A[5][5],i,j,m,n,sum = 0;
clrscr();
printf(“\n\n\t ENTER A ORDER OF THE MATRIX M,N…: “);
scanf(“%d,%d”,&m,&n);
printf(“\n\n\t ENTER THE ELEMENTS OF THE MATRIX..:\n\n”);
if(m == n)
{
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(20+j*4,12+i*2);
scanf(“%d”,&A[i][j]);
}
printf(“\n”);
}
for(i=1;i<=m;i++)
sum = sum + A[i][n+1-i];
printf(“\n\t THE SUM OF SECONDARY DIAGONAL OF A MATRIX IS…: %d”, sum);
}
else
{
printf(“\n\t THE ORDER OF THE MATRIX IS NOT CORRECT”);
printf(“\n\n\t HELP : ‘M’ SHOULD BE EQUAL TO ‘N'”);
}
getch();
}
Download exe and source code here.

The post To find the sum of secondary diagonal of a Matrix first appeared on Student Projects.

Logic :  The diagonal elements, are the elements having equal column and row value throughout. By keeping this in mind, we can trace the matrix till the end using the help of a for loop.  The flag ‘sum’ is updated in each of the iterations.
Finally, we’ll get the sum of the diagonal elements after the end of the while loop.

You are always welcome with your suggestion and doubts into our discussion forum.

#include<stdio.h>
void main()
{
int A[5][5],i,j,m,n,sum = 0;
clrscr();
printf(“\n\n\t ENTER A ORDER OF THE MATRIX M,N…: “);
scanf(“%d,%d”,&m,&n);
printf(“\n\n\t ENTER THE ELEMENTS OF THE MATRIX..:\n\n”);
if(m == n)
{
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(20+j*4,12+i*2);
scanf(“%d”,&A[i][j]);
}
printf(“\n”);
}
for(i=1;i<=m;i++)
sum = sum + A[i][i];
printf(“\n\t THE SUM OF PRIMARY DIAGONAL OF A MATRIX IS…: %d”, sum);
}
else
{
printf(“\n\t THE ORDER OF THE MATRIX IS NOT CORRECT”);
printf(“\n\n\t HELP : ‘M’ SHOULD BE EQUAL TO ‘N'”);
}
getch();
}
Download exe and source code here.

The post To find the sum of primary diagonal of a Matrix first appeared on Student Projects.

Logic : To multiply two matrices, the condition should be satisfy that the column count of the  first matrix A should equal the column count of the second matrix B. (The program will check for the validity of this condition i.e. if, N = P. If fails, outputs an error message.) The first row of A is multiplied with with the first column of B. i.e. the first member of the first row of A is multiplied with the first member of first column of B, stored in a variable sum, second with the second and the product is added up, similarly, the procedure grows till the end, and sum is updated in each multiply, and the sum will be the first member of the product matrix C. The procedure grows till the end. Finally we will get the product matrix C of order (M, Q).

The procedure can be modified slightly with the same algorithm, to add,  subtract,  transpose or others like, to trace the diagonals etc.

#include<stdio.h>
void main()
{
int A[5][5],B[5][5],C[5][5],i,j,m,n,p,q,k;
clrscr();
printf(“\n\n\t ENTER A ORDER OF THE FIRST MATRIX M,N…: “);
scanf(“%d,%d”,&m,&n);
printf(“\n\n\t ENTER A ORDER OF THE SECOND MATRIX P,Q…: “);
scanf(“%d,%d”,&p,&q);
if(n == p)
{
printf(“\n\n\t ENTER THE ELEMENTS OF THE FIRST MATRIX..:\n\n”);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(25+j*4,14+i*2);
scanf(“%d”,&A[i][j]);
}
printf(“\n”);
}
printf(“\n\t ENTER THE ELEMENTS OF THE SECOND MATRIX..:\n\n”);
for(i=1;i<=p;i++)
{
for(j=1;j<=q;j++)
{
gotoxy(25+j*4,20+m+i*2);
scanf(“%d”,&B[i][j]);
}
printf(“\n”);
}
for(i=1;i<=m;i++)
for(j=1;j<=q;j++)
{
C[i][j] = 0;
for(k=1;k<=n;k++)
C[i][j] = C[i][j] + (A[i][k] * B[k][j]);
}
printf(“\n\t THE PRODUCT OF TWO MATRICES IS…:\n\n\t\t\t “);
for(i=1;i<=m;i++)
{
for(j=1;j<=q;j++)
printf(” %d”,C[i][j]);
printf(“\n\n\t\t\t “);
}
}
else
{
printf(“\n\t THE ORDERS OF TWO MATRICES ARE NOT CORRECT”);
printf(“\n\n\t HELP : ‘N’ SHOULD BE EQUAL TO ‘P'”);
}
getch();
}
Download exe and source code here.

The post To multiply two Matrices first appeared on Student Projects.

Logic : The logic applied here is very simple, that what we do in the paper. The first entry of the first matrix  ( A[1,1] ) is subtracted by the first entry of the second Matrix  (B[1,1]) and stored as the first member of the resultant matrix (C[1,1]). The procedure is continued till the last number of the matrix.

The procedure can be modified slightly with the same algorithm, to add, multiply, or transpose etc.

#include<stdio.h>
void main()
{
int A[5][5], B[5][5], C[5][5], i, j, m, n;
clrscr();
printf(“\n\n\t ENTER A ORDER OF THE MATRICES M,N…: “);
scanf(“%d,%d”, &m, &n);
printf(“\n\n\t ENTER THE ELEMENTS OF THE FIRST MATRIX..:\n\n”);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(25+j*4,12+i*2);
scanf(“%d”,&A[i][j]);
}
printf(“\n”);
}
printf(“\n\t ENTER THE ELEMENTS OF THE SECOND MATRIX..:\n\n”);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(25+j*4,18+m+i*2);
scanf(“%d”,&B[i][j]);
}
printf(“\n”);
}
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
C[i][j] = A[i][j] – B[i][j];
printf(“\n\t THE DIFFERENCE OF TWO MATRICES IS…:\n\n\t\t\t “);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
printf(” %d”,C[i][j]);
printf(“\n\n\t\t\t “);
}
getch();
}
Download exe and source code here.

The post To find the difference of two Matrices first appeared on Student Projects.

Logic :  The logic applied here is very simple, that what we do in the paper. The first entry of the first matrix  ( A[1,1] ) is added up with the first entry of the second Matrix  (B[1,1]) and stored as the first member of the resultant matrix (C[1,1]). The procedure is continued till the last number of the matrices.

The procedure can be modified slightly with the same algorithm, to subtract, multiply, or transpose etc.

#include<stdio.h>
void main()
{
int A[5][5], B[5][5], C[5][5], i, j, m, n;
clrscr();
printf(“\n\n\t ENTER A ORDER OF THE MATRICES M,N…: “);
scanf(“%d,%d”, &m, &n);
printf(“\n\n\t ENTER THE ELEMENTS OF THE FIRST MATRIX..:\n\n”);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(25+j*4,12+i*2);
scanf(“%d”,&A[i][j]);
}
printf(“\n”);
}
printf(“\n\t ENTER THE ELEMENTS OF THE SECOND MATRIX..:\n\n”);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
gotoxy(25+j*4,18+m+i*2);
scanf(“%d”,&B[i][j]);
}
printf(“\n”);
}
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
C[i][j] = A[i][j] + B[i][j];
printf(“\n\t THE SUM OF TWO MATRICES IS…:\n\n\t\t\t “);
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
printf(” %d”,C[i][j]);
printf(“\n\n\t\t\t “);
}
getch();
}
Download exe and source code here.

The post To find the sum of two Matrices first appeared on Student Projects.

The idea is whenever any engine observes a red signal on its track it will start decreasing its speed gradually and stops automatically at some distance from the signal pole. After then when it gets green signal the driver can manually start the train and go on. In the mean time when train has not stopped yet and a red signal becomes green then it crosses the signal pole with low speed and then driver can slowly increase the speed.

So now before the driver observes the red signal the engine itself observes it and automatically starts decreasing speed and then stops. The driver can feel relax in driving because he doesn’t have to take care about red signal. Even if he forgets to take any action on red signal then also we can avoid accidents by the implementation of this idea.

General description:

What we have to do is we have to attach a transmitter with signal pole which will start transmitting signals only when the red light is on. If there is green light no transmission. The engine has a receiver which catches these transmitted signals and takes desire actions.

Both the transmitter and receiver are of RF type with minimum range of 2 Km. so that train can get enough time to decrease its speed and stop before the signal pole with minimum swapping distance of 100-200 mt.

Here in our project we have used IR transmitter and receiver instead of RF for demo purpose. But same idea can be easily implemented with RF also with a little more cost.

Lets first discuss the demonstration model.

Demonstration Model:

The train engine runs on 24V DC motor so that we can easily vary its speed by varying applied voltage. The switching voltage is applied in step of 18 V, 15 V, 12 V and 9 V (min speed). The 230 VAC is step-down to 24 VAC by 12-0-12, 2 Ampere step down transformer. As shown in figure this 24 VAC line runs parallel with track at the top of the train. Movable tapping are taken from this line and fed to the internal circuit of engine. These tapping slides as the train runs on the track and give continuous supply to circuit. The IR sensor is placed at the top of the engine, senses the signals transmitted by IR transmitter attached to signal pole. Train track is straight and 20 ft long. Signal pole is placed at the end of track and train starts from farther end.

The post Intelligent Train Engines first appeared on Student Projects.

Some photographs of this model:

Project Description:

Keyboard section:

There are six switches in this section. They are

1. Turn left.
2. Turn right.
3. Stop.
4. About turn.
5. Park left.
6. Park right.

Circuit diagram of keyboard is shown bellow.

Car section:

There are many sub sections in this section. They are

Motor:

We are using a 5V dc motor to drive the vehicle. The speed of the vehicle and its strength is controlled by the proper use of pulley. The rear wheel of the vehicle is connected to this motor through a pulley. This motor is meant for moving the vehicle both in forward and backward direction. Microcontroller (8051) controls the forward and backward movement of the vehicle in the following manner:

Here in the above circuit, T1, T2, T3, T4 are the NPN power transistor (2N3055). A0, A1, A2, A3 are the signals coming from the micro controller. With the specific combination of A0, A1, A2, A3 we can change the direction of rotation of motor as follows:

Case I: When   A0=high; A3=high; & A1=low; A2=low
The motor rotates in clockwise direction

Case II: When   A0=low; A3=low; & A1=high; A2=high
The motor rotates in anti-clockwise direction

Case III: When   A0=low; A3=low; A1=low; A2=low
The motor stops the rotation.

The post Robotic car using 8951 Microcontroller first appeared on Student Projects.