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).

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 .

Digital signal uses binary system, that means $(0,1)$. This leads to discrete wave produce from digital signal.

They are in the form of rectangular waveform. Digital signal cannot provide a continuous set of value because it is not using Hexadecimal system, which gives continuous values.

Hence, the correct option is (c).

Given,

Amplitude modulated wave is

$ C_m(t)=10[1+0.6 \sin (1250 t)] \sin \left(10^8 t\right) $

Compare this wave equation with standard wave equation for amplitude modulation.

$ C_m(t)=A_c\left[1+\frac{A_m}{A_c} \sin \left(\omega_m t\right)\right] \sin \left(\omega_c t\right) $

where $A_c=$ Amplitude of carrier wave.

$A_m=$ Amplitude of modulating wave.

Modulation index is given by

$ M=A_m / A_c \Rightarrow M=0.6 $

Hence, option (d) is correct option.

Carrier frequency

$ \begin{aligned} f_c & =20 \mathrm{GHz} \\ & =20 \times 10^9 \mathrm{~Hz} \\ & =2 \times 10^{10} \mathrm{~Hz} \end{aligned} $

Utilised carrier frequency,

$ \begin{aligned} f_c^{\prime} & =20 \% \text { of } f_C \\ & =\frac{20}{100} \times 2 \times 10^{10} \\ & =4 \times 10^9 \mathrm{~Hz} \end{aligned} $

Bandwidth of one channel.

$ \begin{aligned} f_m & =5 \mathrm{kHz} \\ & =5 \times 10^3 \mathrm{~Hz} \end{aligned} $

No. of telephonic channel

$ \begin{aligned} n & =\frac{f_c^{\prime}}{f_m}=\frac{4 \times 10^9}{5 \times 10^3} \\ & =8 \times 10^5 \mathrm{~Hz} \end{aligned} $

The loss of strength of a signal white propagating through a medium is known as attenation.

Modulation is the process in which characteristics of carrier wave is changed according to modulating signal.

Demodulation is the process of separating the original message signal from the modulated signal.

Noise is unwanted signal or unpleasant signal.

Modulation : It is a process in which the message signal is mixed with carrier signal and then sent it. The need for modulation is, Original (Message) signal distorts at very less distance and cannot be sent over large distance.

So to overcome this issue signal is modulated in carrier signal and then sent over large distance.

The given problem involves calculating the sideband frequencies produced when a message signal modulates a carrier wave.

Carrier wave frequency: $ 6 \text{ MHz} = 6000 \text{ kHz} $

Message signal frequency: $ 10 \text{ kHz} $

Let $ C $ denote the carrier wave frequency and $ M $ represent the message signal frequency. The sideband frequencies are calculated as follows:

Upper Sideband Frequency:

$ C + M = 6000 \text{ kHz} + 10 \text{ kHz} = 6010 \text{ kHz} $

Lower Sideband Frequency:

$ C - M = 6000 \text{ kHz} - 10 \text{ kHz} = 5990 \text{ kHz} $

Thus, the final sideband frequencies are $ 6010 \text{ kHz} $ and $ 5990 \text{ kHz} $.

Given modulation index, $\mu=0.5$

Maximum amplitude, $A_{\text {max }}=10 \mathrm{~V}$

Minimum amplitude, $A_{\text {min }}=$ ?

We know that,

$ \begin{aligned} & \mu=\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }} \Rightarrow 0.5=\frac{10-A_{\min }}{10+A_{\min }} \\ \Rightarrow & 5+0.5 A_{\min }=10-A_{\min } \\ \Rightarrow & 1.5 A_{\min }=5 \\ \Rightarrow & A_{\min }=\frac{5}{1.5}=\frac{1}{0.3}=\frac{10}{3}=3.33 \mathrm{~V} \end{aligned} $

Peak voltage of carrier wave,

$ A_c=10 \mathrm{~V} $

Modulation index,

$ \mu=80 \%=0.8 $

If $A_m$ be the peak voltage of modulating signal, then

$ \begin{aligned} \mu & =\frac{A_m}{A_c} \\ \Rightarrow \quad A_m & =\mu A_c \\ & =0.8 \times 10 \\ & =8 \mathrm{~V} \end{aligned} $

The power radiated from a linear antenna is directly proportional to square of frequency. Thus, power radiated from the linear antenna increases with increasing frequency.

The range of frequency bands used for standard AM broadcast is $(540-1600) \mathrm{kHz}$.

Given, frequency of message signal,

$ f_m=15 \mathrm{kHz}=15 \times 10^3 \mathrm{~Hz} $

Let $f_c$ be the carrier frequency, then upper side band frequency $=f_c+f_m$

$ \begin{array}{ll} \Rightarrow & 1515=f_c+f_m \\ \Rightarrow & f_c+f_m=1515\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

Similarly, lower side band frequency

$ f_c-f_m=1485 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Adding Eqs. (i) and (ii), we have

$ \begin{aligned} 2 f_c & =3000 \\ \Rightarrow \quad f_c & =1500 \mathrm{kHz} \\ & =1.5 \times 10^6 \mathrm{~Hz} \\ & =1.6 \mathrm{MHz} \end{aligned} $

Given, height of transmission antenna,

$ h=40 \mathrm{~m}, $

Radius of earth,

$ \begin{aligned} R & =6400 \mathrm{~km} \\ & =6.4 \times 10^6 \mathrm{~m} \end{aligned} $

The maximum distance up to which signal can be broadcasted,

$ \begin{aligned} d_{\max } & =\sqrt{2 R h} \\ \therefore \text { Service area, } A & =\pi d_{\max }^2 \\ = & \pi \times 2 R h \\ = & \pi \times 2 \times 6.4 \times 10^6 \times 40 \\ = & 512 \pi \times 10^6 \mathrm{~m}^2 \end{aligned} $

The electromagnetic waves of frequency 6 GHz is in the range of microwaves.

So, they are used in satellite communication.

Peak voltage (amplitude) of message signal,

$ A_m=5 \mathrm{~V} $

Amplitude of carrier wave, $A_c=20 \mathrm{~V}$

∵ Modulation index, $m_f=\frac{A_m}{A_c}=\frac{5}{20}=\frac{1}{4}=0.25$

Television signals lies in region of 40 MHz to 900 MHz , so these are not reflected by ionosphere and can be communicated by using a satellite only. It is because, ionosphere reflects electromagnetic wave of frequencies in the range of 3 MHz to 30 MHz . Hence, Assertion is false but Reason is true.

$ \begin{aligned} &\text { Modulation index, }\\ &\mu=\frac{\text { Amplitude of message signal }}{\text { Amplitude of carrier wave }}=\frac{15 \mathrm{~V}}{30 \mathrm{~V}}=0.5 \end{aligned} $