Communication Systems

$ \begin{aligned} & H_t=\frac{1}{20000} R=\frac{R}{20000} \\ & H_r=\frac{R}{80000} \end{aligned} $

Maximum distance

$ \begin{aligned} d & =\sqrt{2 R H_t}+\sqrt{2 R H_r} \\ & =\sqrt{2 R \times \frac{R}{20000}}+\sqrt{2 R \times \frac{R}{80000}} \\ & =\frac{R}{100}+\frac{R}{200}=\frac{6.4 \times 10^6}{100}+\frac{6.4 \times 10^6}{200} \\ & =6.4 \times 10^4+3.2 \times 10^4 \\ & =9.6 \times 10^4 \mathrm{~m}=96 \times 10^3 \mathrm{~m}=96 \mathrm{~km} \end{aligned} $

$ \begin{aligned} f_m & =5 \mathrm{kHz} \\ f_c & =900 \mathrm{kHz} \end{aligned} $

Upper side band

$ \begin{aligned} & =f_c+f_m=900+5 \\ & =905 \mathrm{kHz} \end{aligned} $

Lower side band

$ \begin{aligned} & =f_c-f_m=900-5 \\ & =895 \mathrm{kHz} \end{aligned} $

$ \begin{aligned} &\text { Modulation index is given by }\\ &\begin{aligned} \mu & =\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }} \\ & =\frac{\frac{A_{\max }}{A_{\min }}-1}{\frac{A_{\max }}{A_{\min }}+1} \\ & =\frac{\frac{7}{3}-1}{\frac{7}{3}+1}=\frac{\frac{4}{3}}{\frac{10}{3}}=0.4 \end{aligned} \end{aligned} $

The ionosphere reflects high frequency (HF) radio waves, which correspond to the range:

$ 3\ \text{MHz} \text{ to } 30\ \text{MHz} $

This property allows long-distance radio communication (sky wave propagation).

✅ Correct answer:

Option B: $ 3 - 30\ \mathrm{MHz} $

$ \begin{aligned} & A_{\max }=25 \mathrm{~V} \\ & A_{\min }=5 \mathrm{~V} \end{aligned} $

$\therefore$ Modulation index

$ \begin{aligned} \mu & =\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }} \\ & =\frac{25-5}{25+5}=\frac{20}{30}=\frac{2}{3} \end{aligned} $

Step 1: Write down the given data

$ f = 1000 \, \text{kHz} = 1 \times 10^6 \, \text{Hz} $

Speed of electromagnetic wave in free space,

$ c = 3 \times 10^8 \, \text{m/s} $

Step 2: Find the wavelength

$ \lambda = \frac{c}{f} = \frac{3 \times 10^8}{1 \times 10^6} = 300 \, \text{m} $

Step 3: Relation between antenna length and wavelength

For efficient radiation, the minimum antenna length (usually a quarter-wave antenna) is:

$ L = \frac{\lambda}{4} $

Step 4: Calculate $ L $

$ L = \frac{300}{4} = 75 \, \text{m} $

✅ Answer: Option C — 75 m

$ \text { } \begin{aligned} & A_{\max }=A_m+A_c \\ & =\mu A_c+A_c=(\mu+1) A_c \\ \Rightarrow A_c & =\frac{A_{\max }}{\mu+1}=\frac{14}{0.4+1}=\frac{14}{1.4}=10 \mathrm{~V} \end{aligned} $

Given:

• Maximum amplitude $ A_{\max} = 30 \text{ mV} $

• Minimum amplitude $ A_{\min} = 5 \text{ mV} $

Formula for modulation index $ m $:

$ m = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}} $

Substitute the values:

$ m = \frac{30 - 5}{30 + 5} = \frac{25}{35} = \frac{5}{7} $

✅ Answer: Option C — $ \frac{5}{7} $

The range of line of sight communication between two antennas is given by

$ x=\sqrt{2 R h_T}+\sqrt{2 R h_R} $

Here, $h_T+h_R+h$

Let $\quad h_T=H$, then

$ \begin{aligned} & & h_R & =h-H \\ & & \therefore x & =\sqrt{2 R}[\sqrt{H}+\sqrt{h-H}] \\ & & \frac{d x}{d H}=\sqrt{2 R} & {\left[\frac{1}{2 \sqrt{H}}+\frac{1}{2 \sqrt{h-H}}(-1)\right]=0 } \\ & \Rightarrow & \frac{1}{2 \sqrt{H}} & =\frac{1}{2 \sqrt{h-H}} \\ & & H & =h-H \\ & \Rightarrow & H & =\frac{h}{2} \end{aligned} $

The correct answer is Option C: Pulse band modulation.

Here's a brief explanation of why:

Pulse duration modulation (often called Pulse Width Modulation, PWM) varies the width or duration of the pulse.

Pulse amplitude modulation (PAM) varies the amplitude of the pulse.

Pulse position modulation (PPM) varies the position of the pulse within a time frame.

Pulse band modulation is not a standard or recognized method in the classification of pulse modulation techniques.

Therefore, Option C is the one that does not belong.

Height of receiving antenna=80m$=h_t$

Maximum distance between

transmitting and receiving antenna in

line of slight mode $=d_m=57.6 \mathrm{~km}$ $=57600 \mathrm{~m}$

Radius of earth $=6.4 \times 10^6 \mathrm{~m}$

Let height of transmitting antenna $a=h_t$

We know that, maximum line of slight distance between two antennas

$ \begin{array}{ll}\, d_m=\sqrt{2 R h_T}+\sqrt{2 R h_T} \\ \Rightarrow \quad 57600=\sqrt{2 \times 6.4 \times 10^6 \times h_T} +\sqrt{2 \times 6.4 \times 10^6 \times 80} \\ =\sqrt{12.8 \times 10^6 \times h_T}+\sqrt{1024 \times 10^6} \\ =\sqrt{12.8 \times 10^6 \times h_T}+32000 \\ \Rightarrow \sqrt{12.8 \times 10^6 \times h_T}=25600=256 \times 10^{2} \\ \Rightarrow 128 \times 10^6 \times h_T=65536 \times 10^4 \\ \Rightarrow h_T=\frac{65536 \times 10^4}{12.8 \times 10^6} \\ \quad=\frac{5120}{10^2}=51.20 \\ h_T=51.20 \mathrm{~m} \end{array} $

Given:

Signal frequency, $ v_s = 10 \, \text{kHz} $

Carrier wave frequency, $ v_c = 3.61 \, \text{MHz} = 3610 \, \text{kHz} $

We can calculate the side band frequencies using these formulas:

Upper Side Band Frequency:

The formula for the upper side band frequency is:

$ v_u = v_c + v_s $

Substitute the given values:

$ v_u = 3610 + 10 $

$ v_u = 3620 \, \text{kHz} $

Lower Side Band Frequency:

The formula for the lower side band frequency is:

$ v_l = v_c - v_s $

Substitute the given values:

$ v_l = 3610 - 10 $

$ v_l = 3600 \, \text{kHz} $

Therefore, the upper side band frequency is 3620 kHz, and the lower side band frequency is 3600 kHz.

To determine the modulation index in amplitude modulation:

Given:

Carrier amplitude, $ A_c = 10 \, \text{V} $

Side band amplitude, $ A_s = 2 \, \text{V} $

The relationship between $ A_c $ and $ A_s $ is defined by the formula:

$ A_s = \frac{A_c \times m}{2} $

Here, $ m $ represents the modulation index. Rearranging the formula to solve for $ m $, we have:

$ m = \frac{2 A_s}{A_c} $

Substituting the given values:

$ m = \frac{2 \times 2}{10} = 0.4 $

Thus, the modulation index is 0.4.

The effective modulation index, when a carrier is modulated by two sine waves, can be calculated using the formula:

$ \mu = \sqrt{\mu_1^2 + \mu_2^2} $

Where:

$\mu_1 = 0.3$

$\mu_2 = 0.4$

To find the total modulation index:

$ \mu = \sqrt{(0.3)^2 + (0.4)^2} $

Calculating inside the square root:

$ = \sqrt{0.09 + 0.16} $

$ = \sqrt{0.25} $

$ = 0.5 $

Therefore, the total modulation index is 0.5.

Given:

Maximum voltage of an AM wave: $ V_{\max} $

Minimum voltage of an AM wave: $ V_{\min} $

Amplitude of the modulating signal is calculated as:

$ A_m = \frac{V_{\max} - V_{\min}}{2} $

Amplitude of the carrier signal is calculated as:

$ A_C = \frac{V_{\max} + V_{\min}}{2} $

The modulation index or modulation factor $ m $ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal:

$ m = \frac{A_m}{A_C} $

Substituting the expressions for $ A_m $ and $ A_C $, the modulation index becomes:

$ m = \frac{\frac{V_{\max} - V_{\min}}{2}}{\frac{V_{\max} + V_{\min}}{2}} \Rightarrow m = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}} $