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;

}

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 $P_c$

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

Frequency Modulation

Q. A single tone FM is represented by the voltage equation as v(t)=12cos(5x10^8t+5sin1250t). Determine following:

a) Carrier frequency

b) Modulating frequency

c) Modulation index

d) Maximum frequency deviation

Solution: Here,

We are given a single-tone FM signal:

$$v(t)=12\cos (5\times {10}^{8}t+5\sin (1250t))$$

This equation is in the general form of an FM signal:

$$v(t)=A_c\cos (2\pi f_{c}t+\beta \sin (2\pi f_{m}t))$$

Where:

• $f_c$ = carrier frequency

• $f_m$ = modulating frequency

• $\beta$ = modulation index

• $\Delta f = \beta f_m$ = frequency deviation

Step-by-step Extraction from Given Equation

We compare:

$$v(t)=12\cos (5\times {10}^{8}t+5\sin (1250t))$$

with:

$$v(t)=A_c\cos ({\omega }_{c}t+\beta \sin ({\omega }_{m}t))$$

From this, we get:

• $\omega_c = 5 \times 10^8 \Rightarrow f_c = \frac{\omega_c}{2\pi} = \frac{5 \times 10^8}{2\pi} \approx 79.58 \times 10^6 = 79.58 \, \text{MHz}$

• $\omega_m = 1250 \Rightarrow f_m = \frac{1250}{2\pi} \approx 199 \, \text{Hz}$
  

• $\beta = 5$

Amplitude Modulation

Q. An audio frequency signal 10sin(1000πt) is used for a single tone amplitude modulation with a carrier of 50sin(2π×10^(5)t). Calculate:

(i) Modulation index

(ii) Bandwidth requirement

(iii) Total power delivered if load = 60Ω.

Solution: Here,

We are given:

• Message signal: $m(t)=10\sin (1000\pi t)$

• Carrier signal: $c(t)=50\sin (2\pi \times {10}^{5}t)$

• Load Resistance: $R = 60\, \Omega$
  

Let’s solve the parts step-by-step.

(i) Modulation Index $\mu$

The modulation index is defined as:

$$\mu =\frac{A_m}{A_{\mathrm{c}}}$$

Where:

• $A_m =$Amplitude of message = 10

• $A_c=\; Amplitude\; of\; carrier\; =\; 50$

$$\mu =\frac{10}{50}=0.2$$

So, Modulation Index = 0.2

Linear Block Code in Error Detection and Correction

Design a code word of a C(8, 4) block code with any suitable generation matrix.

Line Spectrum of Signal

Q. Plot the line spectrums of x(t) =12 + 6sin (140πt +30°) - 9cos(80πt-70°).

Solution:

Let’s Analyze and Plot the Line Spectrum of

x(t) = 12 + 6sin(140πt+30⁰) − 9cos(80πt−70⁰)

Step 1: Convert all terms to cosine form

Use the identity: sin(θ) = cos(θ−90⁰)

So,

6sin(140πt+30⁰) = 6cos(140πt−60⁰)

−9cos(80πt−70⁰) = 9cos(80πt+110⁰)

Thus, x(t) = 12 + 6cos(140πt−60⁰) + 9cos(80πt+110⁰)

Step 2: Frequencies, Amplitudes, Phases

Term	Frequency (Hz)	Amplitude	Phase
12 (DC)	0	12	0°
6cos(140πt−600)	f=70	6	-60°
9cos(80πt+1100)	f=40	9	+110°

 We split each cosine into two spectral lines at ±f with half the amplitude.