Random File Access in C

WAP to print the first character of each word from the string contained in a file “details.txt”. The file “details.txt” has “Deep Raj Bhujel” string in it and your program should print “D R B” on the console screen. [Random File Access: fseek(), ftell(), rewind()]

#include<stdio.h>

#include<stdlib.h>

int main()

{

    FILE *fp;

    char ch;

    //long pos;

    fp = fopen("details.txt","r");

    if(fp == NULL)

    {

        puts("Error opening file");

        exit(1);

    }

    //printf("%ld ",ftell(fp));

    ch = fgetc(fp);

    printf("%c ",ch);

    //printf("%ld ",ftell(fp));

    fseek(fp,4,SEEK_CUR);

    ch = fgetc(fp);

    printf("%c ",ch);

    //printf("%ld ",ftell(fp));

    fseek(fp,3,SEEK_CUR);

    ch = fgetc(fp);

    printf("%c ",ch);

    //printf("%ld ",ftell(fp));

    //rewind(fp);

    //printf("%ld ",ftell(fp));

    return 0;

}

Caesar Cipher in C

 #include<stdio.h>

int main(){

char inputtext[]="Deep Raj Bhujel", ciphertext[100], decipheredtext[100];

int i,shift=3;

for(i=0;inputtext[i]!='\0';i++){

if(isupper(inputtext[i]))

ciphertext[i]=(inputtext[i]-65+shift)%26 + 65;

else if(islower(inputtext[i]))

ciphertext[i]=(inputtext[i]-97+shift)%26 + 97;

else

ciphertext[i]=inputtext[i];

}

ciphertext[i]='\0';

printf("The encrypted string is %s\n\n",ciphertext);

for(i=0;ciphertext[i]!='\0';i++){

if(isupper(ciphertext[i]))

decipheredtext[i]=(ciphertext[i]-65-shift)%26 + 65;

else if(islower(ciphertext[i]))

decipheredtext[i]=(ciphertext[i]-97-shift)%26 + 97;

else

decipheredtext[i]=ciphertext[i];

}

decipheredtext[i]='\0';

printf("The decrypted string is %s",decipheredtext);

return 0;

}

Implementation of Preprocessor in C Programming

Q. Write macro definitions with arguments for calculation of area and perimeter of a triangle, a square and a circle. Store these macro definitions in a file called "areaperi.h". Include this file in your program, and call the macro definitions for calculating area and perimeter for different triangles, squares and circles.

Step 1: Create a header file (.h) declaring the prototypes of required functions.

float area1(int, int, int); // for triangle

int area2(int); // for square

float area3(float); // for circle

float perimeter1(int, int, int); // for triangle

int perimeter2(int); // for square

float perimeter3(float); // for circle

Step 2: Create a C programming file (.c) defining those functions.

#include<stdio.h>

#include<math.h>

#define PI 3.1416

float perimeter1(int a, int b, int c) // for triangle

{

    return a+b+c;

}

int perimeter2(int a) // for square

{

    return 4*a;

}

float perimeter3(float r) // for circle

{

    return 2*PI*r;

}

float area1(int a, int b, int c) // for triangle

{

    float s = perimeter1(a, b, c)/2;

    return sqrt(s*(s-a)*(s-b)*(s-c));

}

int area2(int a) // for square

{

    return pow(a,2);

}

float area3(float r) // for circle

{

    return PI*r*r;

}

Step 3: Write your C program (.c) including the areaperi.h file as #include "areaperi.h".

Standard AM, FM and PM Equations

Standard Mathematical Equations for AM, FM and PM

1. Amplitude Modulation (AM)

• : Carrier amplitude, : Carrier frequency
• :  Message signal amplitude, : Message signal frequency
• : Modulation index
• : Carrier Power, : Sideband Power, : Total Power
• : Efficiency
• : Bandwidth of the AM signal

Amplitude Modulation

Q. The equation of amplitude modulated wave is given by s(t) = 40[1+0.8cos(2π ×10^3t)]cos(4π×10^5t). Find the carrier power, the total sideband power and bandwidth of the signal. The value of resistor given is 30Ω.

Solution: Here,

We are given:

$$s(t)=40[1+0.8\cos (2\pi \cdot {10}^{3}t)]\cos (4\pi \cdot {10}^{5}t)$$

And:

• Load resistance $R = 30\, \Omega$
  

Step 1: Identify the standard AM form

Standard AM wave:

$$s(t)=A_c[1+\mu \cos (2\pi f_{m}t)]\cos (2\pi f_{c}t)$$

From the given equation:

• $A_c = 40$

• $\mu =0.8$

• $f_m={10}^3=1\text{kHz}$

• $f_c=\frac{4\pi \cdot {10}^5}{2\pi }=2\times {10}^5=200\text{kHz}$

a) Carrier Power $P_c$

$$P_c=\frac{A_c^2}{2R}=\frac{{40}^2}{2\cdot 30}=\frac{1600}{60}=\boxed{{\displaystyle 26.67\text{W}}}$$