Communication Systems

Step 1: Formula for Radio Horizon

The distance to the radio horizon, $ d $, can be found using the formula:
$ d = \sqrt{2Rh} $
Here, $ R $ is the radius of the Earth and $ h $ is the height of the antenna.

Step 2: Plug in the Values

Let $ R = 6.4 \times 10^6 $ meters (which is 6400 km) and $ h = 39.2 $ meters.

Step 3: Calculate Inside the Square Root

Multiply them:
$ 2 \times 6.4 \times 10^6 \times 39.2 $

Step 4: Take the Square Root

Find the square root of the product:
$ d = \sqrt{2 \times 6.4 \times 10^6 \times 39.2} $

When calculated:
$ d = 22.4 \times 10^3 $ meters

This is the same as 22.4 kilometers.

So, the radio horizon is 22.4 km away.

The correct answer is D) thermosphere.

Here's why:

1. Ionosphere: The reflection of radio waves is primarily the function of the ionosphere. The ionosphere is not a distinct layer in the same way as the troposphere or stratosphere, but rather a region within the upper atmosphere (primarily within the thermosphere and extending into the mesosphere) where gases are ionized by solar radiation.

2. Layers of the Ionosphere: The ionosphere consists of several ionized layers: D, E, F1, and F2.

   • D-layer: This is the lowest layer. It absorbs high-frequency (HF) waves during the day and largely disappears at night.

   • E-layer: This layer reflects some HF waves during the day but weakens considerably at night.

   • F-layer (F1 and F2 merging into F-layer at night): These are the highest and most intensely ionized layers. During the day, they are distinct F1 and F2 layers. At night, the D and E layers largely dissipate due to the absence of solar radiation, and the F1 and F2 layers merge into a single, strong F-layer. This strong F-layer at night is highly efficient at reflecting high-frequency (HF) radio waves back to Earth, enabling long-distance radio communication (skywave propagation).

3. "Particularly at night": This phrase is key. At night, the lower D and E layers fade, reducing absorption and allowing HF waves to reach the higher F-layer of the ionosphere (within the thermosphere), where they are reflected much more effectively.

Therefore, the thermosphere, which contains the F-layer of the ionosphere, is responsible for efficiently reflecting high-frequency waves, particularly at night.

$ \begin{aligned} & \text { Range, } d=\sqrt{2 R h} \\ & =\sqrt{2 \times 6400 \times 10^3 \times 980} \\ & =112 \times 10^3 \mathrm{~m}=112 \mathrm{~km} \end{aligned} $

The correct answer is:

✅ Option C: 750 MHz

Explanation:

Coaxial cable is a high-frequency transmission medium commonly used for cable television, internet, and long-distance communication.

• It typically offers a bandwidth range from a few kilohertz up to several hundred megahertz.

• The approximate frequency bandwidth of a standard coaxial cable used in communications is up to 750 MHz.

• Actual usable bandwidth depends on cable type, length, and quality (e.g., RG-6, RG-59, etc.).

Hence, Option C (750 MHz) is correct.

The correct answer is:

✅ Option B: $ 300 \, \text{Hz} - 3100 \, \text{Hz} $

Explanation:

Human speech contains frequency components roughly between 100 Hz and 8 kHz, but for commercial telephony, only the most important frequencies for intelligibility are transmitted.

• The telephone system limits the transmitted audio bandwidth to approximately 300 Hz – 3400 Hz, though sometimes it’s cited as 300 Hz – 3100 Hz, depending on standards and equipment.

• This limited range saves bandwidth while still allowing speech to sound natural and be easily understood.

So, the frequency range adequate for commercial telephonic communication is approximately 300 Hz to 3100 Hz (or 3400 Hz).

$ \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} $

Maximum distance between transmitting and receiving antennas is given by

$ D=\sqrt{2 h_{t} R}+\sqrt{2 h_{r} R} $

where, $ h_{t}= $ height of transmitting antenna

$ h_{r}= $ height of receiving antenna

$ R= $ Radius of earth

Given, $ h_{t}^{\prime}=2 h_{t} $

$ h_{r}^{\prime}=2 h_{r} \Rightarrow D^{\prime}=? $

Putting value in Eq. (i), we get

$ \begin{array}{l} \mathrm{g} \text { value in Eq. }(1), \text { we get } \\\\ D^{\prime}=\sqrt{2 \cdot 2 h_{t} \cdot R}+\sqrt{2 \cdot 2 h_{T} \cdot R} \\\\ D^{\prime}=\sqrt{2} \sqrt{2 h_{t} R}+\sqrt{2} \sqrt{2 h_{T} R} \\\\ D^{\prime}=\sqrt{2}\left(\sqrt{2 h_{t} R}+\sqrt{2 h_{T} R}\right) \end{array} $

From Eq. (i), we get

$ D^{\prime}=\sqrt{2} D $

Given the parameters for an amplitude modulated wave:

Maximum amplitude, $ V_{\text{max}} = 10 \, \text{V} $

Minimum amplitude, $ V_{\text{min}} = 2 \, \text{V} $

The formula to calculate the modulation index is as follows:

$ \mu = \frac{V_{\text{max}} - V_{\text{min}}}{V_{\text{max}} + V_{\text{min}}} $

Substituting the given values into the formula:

$ \mu = \frac{10 - 2}{10 + 2} = \frac{8}{12} $

Simplifying the fraction:

$ \mu = \frac{2}{3} $

Therefore, the modulation index is $\frac{2}{3}$.

The process of the loss of strength of a signal while propagating through a medium is called attenuation. It refers to the reduction in signal strength when signal energy is absorbed, reflected, refracted, or otherwise diminished as it travels through a medium. Attenuation can be influenced by factors such as the distance of transmission, the medium through which the signal passes, and the signal frequency.

In contrast:

Damping typically refers to the reduction of oscillations in a system, which might relate to signal but isn't specifically about transmission through a medium.

Amplification is the process of increasing the power or strength of a signal.

Modulation involves varying a carrier signal in order to encode information.

Thus, Option B is correct: attenuation.

Amplitude of each side band $=\frac{V_m}{2}$ $=\frac{12}{2}=6 \mathrm{~V}$

Given, Height of the transmitting antenna, $H_T=33.8 \mathrm{~m}$

$ H_T=0.0338 \mathrm{~km} $

Height of the receiving antenna,

$ H_R=64.8 \mathrm{~m} \Rightarrow H_R=0.0648 \mathrm{~km} $

Radius of earth, $R=6400 \mathrm{~km}$

As we know, $d_{\max }=\sqrt{2 \times R \times H}$

For both antennas,

$ \begin{aligned} d_{\max }= & \sqrt{2 \times 6400 \times 0.0338} \\ & +\sqrt{2 \times 6400 \times 0.0648} \\ d_{\max }= & \sqrt{12800 \times 0.0338} \\ & +\sqrt{12800 \times 0.0648} \\ d_{\max }= & 49.6 \mathrm{~km} \end{aligned} $

The maximum distance between the antennas for satisfactory communication in line of sight mode is 49.6 km .

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}} $

To determine the minimum height of the receiving antenna required to receive the signal in line of sight at a distance of 4 km from the transmitting antenna, we need to consider the curvature of the Earth. Let's denote the height of the receiving antenna as $h$ and the distance between the antennas as $d = 4 \mathrm{~km}$.

We can use the Pythagorean theorem to relate the Earth's radius $R$, the height of the receiving antenna $h$, and the distance between the antennas $d$:

$(R + h)^2 = R^2 + d^2$

Now, we can plug in the given values for $R$ and $d$:

$(6400 + h)^2 = 6400^2 + 4^2$

Expanding the equation and subtracting $6400^2$ from both sides gives:

$2 \times 6400 \times h + h^2 = 16$

Since $h$ is much smaller than the Earth's radius, we can ignore the $h^2$ term:

$2 \times 6400 \times h \approx 16$

Now, we can solve for $h$:

$h \approx \frac{16}{2 \times 6400} = \frac{1}{800} \mathrm{~km}$

Converting to meters and multiplying by $10^2$ to match the desired format:

$x = h \times 10^3 \times 10^2 = \frac{1}{800} \times 10^5 = 125$

So, the value of $x$ is 125

The bandwidth of an amplitude-modulated wave is given by $2B$, where $B$ is the bandwidth of the modulating signal.

In this case, the modulating signal is a message signal of frequency $3~\mathrm{kHz}$, so the bandwidth of the amplitude modulated wave is $2(3~\mathrm{kHz})=6~\mathrm{kHz}$.

Therefore, the correct answer is $6~\mathrm{kHz}$.

The range (or distance to the horizon) for line-of-sight communication is given by the formula:

$d = \sqrt{2hR}$,

where:

• $d$ is the range,
• $h$ is the height of the antenna, and
• $R$ is the radius of the Earth.

Given that $h = 98 \, \text{m} = 98 \times 10^{-3} \, \text{km}$ and $R = 6400 \, \text{km}$, we can substitute these values into the formula to find $d$:

$d = \sqrt{2 \times 98 \times 10^{-3} \times 6400} \approx 35.42 \, \text{km}$.

The surface area $A$ covered by the transmitting antenna is approximately a circle with radius equal to the range $d$. The area of a circle is given by the formula $A = \pi r^2$, where $r$ is the radius. Therefore, we can substitute $d$ into this formula to find $A$:

$A = \pi d^2 = \pi (35.42)^2 \approx 3940 \, \text{km}^{2}$.

Therefore, the surface area covered by the transmitting antenna is approximately $3940 \, \text{km}^{2}$.

In the case of Amplitude Modulation, the modulation index (µ) can also be represented as:

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

where:

• $A_{\max}$ is the maximum amplitude of the modulated wave,
• $A_{\min}$ is the minimum amplitude of the modulated wave.

Given that the modulation index is 60% or 0.6 and the minimum amplitude $A_{\min}$ is 3 V, we can rearrange this formula to solve for the maximum amplitude $A_{\max}$:

$ 0.6 = \frac{A_{\max} - 3}{A_{\max} + 3} $

Solving this equation gives:

$ A_{\max} = 12 \, \text{V} $

So, the correct answer is 12 V.

The range of transmission (R) of a TV tower (or any radio tower) depends on the height of the tower (h). This relationship is given by the formula:

$ R = \sqrt{2hR_E} $

where $R_E$ is the radius of the Earth. This formula comes from the geometry of the situation, considering the curvature of the Earth.

Now, if the height of the tower is increased, this will increase the range of transmission. Specifically, because the height appears under the square root in the formula, the range of transmission is proportional to the square root of the height. So, if the height is multiplied by some factor, the range will be multiplied by the square root of that factor.

In this case, the height of the tower is increased by 21%, which means the new height is $h_2 = 1.21h_1$. If we substitute this into the formula for the range, we get:

$ R_2 = \sqrt{2h_2R_E} = \sqrt{2(1.21h_1)R_E} = \sqrt{1.21} \sqrt{2h_1R_E} = 1.1R_1 $

This shows that the new range is 1.1 times the original range, or in other words, the range has increased by 10%.

For transmitting a signal, antenna should have a size comparable to the wavelength of the signal i.e., at least ${1 \over 4}$ in dimension, so that the antenna properly senses the time variation of the signal.

The amplitude of the carrier wave is varied in accordance with the message signal is termed as amplitude modulation.

So, statement I is correct but statement II is incorrect.

Theoretical attenuation $\to$ (IV)

Transducer $\to$ (III)

Demodulation $\to$ (II)

Repeater $\to$ (I)

Amplitude of each side band = $\frac{A_{\mathrm{message}}}{2}$

$\mathrm{A_{carrier}+A_{message}=120}$ ............ (1)

$\mathrm{A_{carrier}-A_{message}=80}$ .............. (2)

From (1) and (2)

$\mathrm{A_{message}=20~V}$

$\therefore$ Amplitude of each side band = 10 V

The modulation index (m) of an amplitude-modulated wave is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. It is a measure of the degree of modulation of the carrier wave and is given by the following equation:

$A_{m}=$ Amplitude of modulating wave $=1$ volt

$A_{c}=$ Amplitude of carrier wave $=2$ volt

Modulation index $=\frac{A_{m}}{A_{c}}=\frac{1}{2}=\frac{1}{2}$