Communication Questions in English

(N/A) An analog signal is a continuous signal that represents physical measurements. It varies smoothly with time and can take any value within a given range. Examples include sound waves,light intensity,and temperature variations. Unlike digital signals,which are discrete and represented by binary values ($0$ and $1$),analog signals are continuous in both time and amplitude.

(A) The standard expression for an amplitude modulated wave is given by $v(t) = A_c(1 + \mu \cos \omega_m t) \sin \omega_c t$,where $A_c$ is the carrier amplitude and $\mu$ is the modulation index.
Comparing this with the given expression $v_m = 5(1 + 0.6 \cos 6280 t) \sin (211 \times 10^4 t)$:
Carrier amplitude $A_c = 5 \; V$.
Modulation index $\mu = 0.6$.
The amplitude of the modulated wave varies between $A_{max} = A_c(1 + \mu)$ and $A_{min} = A_c(1 - \mu)$.
$A_{max} = 5(1 + 0.6) = 5(1.6) = 8 \; V$.
$A_{min} = 5(1 - 0.6) = 5(0.4) = 2 \; V$.
Thus,the minimum and maximum amplitudes are $2 \; V$ and $8 \; V$ respectively.

(A) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(V_{m})$ to the amplitude of the carrier wave $(V_{c})$:
$\mu = \frac{V_{m}}{V_{c}}$
Given that $\mu = \frac{1}{2}$ and $V_{m} = 2$,we substitute these values into the formula:
$\frac{1}{2} = \frac{2}{V_{c}}$
By cross-multiplying,we get:
$V_{c} = 2 \times 2$
$V_{c} = 4$

(D) For an antenna to be effective,its length $L$ should be a fraction of the wavelength $\lambda$ of the signal it transmits.
Typically,the minimum length of an antenna is given by $L = \frac{\lambda}{4}$.
Given the length of the antenna $L = 25\, m$.
Substituting the value into the formula: $25 = \frac{\lambda}{4}$.
Solving for $\lambda$: $\lambda = 25 \times 4 = 100\, m$.
Thus,the wavelength of the signal is $100\, m$.

(A) The distance to the horizon $d$ for an antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the earth.
The service area $A$ covered by the antenna is the area of a circle with radius $d$,given by $A = \pi d^{2}$.
Substituting the expression for $d$,we get $A = \pi (\sqrt{2Rh})^{2} = 2\pi Rh$.
Given $R = 6400\, km = 6400 \times 10^{3}\, m$ and $h = 30\, m$.
Converting $R$ to $km$ to match the required unit for area: $R = 6400\, km$ and $h = 30\, m = 0.03\, km$.
$A = 3.14 \times 2 \times 6400 \times 0.03$.
$A = 3.14 \times 384 = 1205.76\, km^{2}$.
Rounding off to the nearest integer,we get $A \simeq 1206\, km^{2}$.

(B) For line-of-sight communication between two antennas of height $h$,the maximum distance $D$ is given by the formula $D = \sqrt{2Rh_t} + \sqrt{2Rh_r}$.
Since the towers are identical,$h_t = h_r = h$,so $D = 2\sqrt{2Rh}$.
Given $D = 45 \, km$ and $R = 6400 \, km$.
Squaring both sides,we get $D^2 = 4(2Rh) = 8Rh$.
Therefore,$h = \frac{D^2}{8R}$.
Substituting the values: $h = \frac{45^2}{8 \times 6400} \, km = \frac{2025}{51200} \, km$.
$h \approx 0.03955 \, km = 39.55 \, m$.

(D) The bandwidth of an amplitude modulated signal is given by $BW = 2f_m$,where $f_m$ is the frequency of the message signal.
Given the message signal $m(t) = 5 \sin(1.57 \times 10^8 t)$,the angular frequency is $\omega_m = 1.57 \times 10^8 \text{ rad/s}$.
Since $\omega_m = 2\pi f_m$,we have $f_m = \frac{\omega_m}{2\pi}$.
Substituting the values,$f_m = \frac{1.57 \times 10^8}{2 \times 3.14} = \frac{1.57 \times 10^8}{6.28} = 0.25 \times 10^8 \text{ Hz} = 25 \times 10^6 \text{ Hz} = 25 \text{ MHz}$.
Therefore,the bandwidth $BW = 2f_m = 2 \times 25 \text{ MHz} = 50 \text{ MHz}$.

(B) The range of a $TV$ transmission tower is given by $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth and $h$ is the height of the antenna.
Case $(i)$: When the receiving antenna is at ground level,the range is $d_1 = \sqrt{2Rh}$,where $h = 20\, m$.
Case $(ii)$: When the receiving antenna is at a height $h' = 5\, m$,the total range is $d_2 = \sqrt{2Rh} + \sqrt{2Rh'}$.
The increase in range is $\Delta d = d_2 - d_1 = \sqrt{2Rh'}$.
The percentage increase $n\%$ is given by $\frac{\Delta d}{d_1} \times 100\%$.
$n = \frac{\sqrt{2Rh'}}{\sqrt{2Rh}} \times 100 = \sqrt{\frac{h'}{h}} \times 100$.
Substituting the values $h = 20\, m$ and $h' = 5\, m$:
$n = \sqrt{\frac{5}{20}} \times 100 = \sqrt{\frac{1}{4}} \times 100 = 0.5 \times 100 = 50\%$.
Therefore,the value of $n$ is $50$.

(D) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_{m})$ to the amplitude of the carrier signal $(A_{c})$.
Given,$A_{m} = 20 \text{ V}$ and $A_{c} = 80 \text{ V}$.
$\mu = \frac{A_{m}}{A_{c}} = \frac{20}{80} = 0.25$.
Percent modulation is given by $\mu \times 100$.
Percent modulation $= 0.25 \times 100 = 25 \%$.

(C) Given:
Modulating frequency $f_m = 2 \, kHz$.
Carrier frequency $f_c = 1 \, MHz = 1000 \, kHz$.
For Amplitude Modulation $(AM)$,the bandwidth is given by $BW = 2f_m$.
$BW = 2 \times 2 \, kHz = 4 \, kHz$.
Thus,Statement-$I$ is true.
The sideband frequencies are given by $f_c + f_m$ and $f_c - f_m$.
Upper Sideband $(USB)$ $= 1000 \, kHz + 2 \, kHz = 1002 \, kHz$.
Lower Sideband $(LSB)$ $= 1000 \, kHz - 2 \, kHz = 998 \, kHz$.
Thus,Statement-$II$ is true.
Therefore,both statements are correct.

(D) The total attenuation in the cable is given by: $\text{Total Attenuation} = (\text{Attenuation per km}) \times (\text{Length}) = -5 \, dB/km \times 20 \, km = -100 \, dB$.
The formula for gain in $dB$ is: $\text{Gain} = 10 \log_{10}(\frac{P_0}{P_i})$.
Here,$\text{Gain} = -100 \, dB$,$P_i = 0.1 \, kW = 100 \, W = 10^2 \, W$.
Substituting the values: $-100 = 10 \log_{10}(\frac{P_0}{10^2})$.
$-10 = \log_{10}(\frac{P_0}{10^2})$.
$10^{-10} = \frac{P_0}{10^2}$.
$P_0 = 10^{-10} \times 10^2 = 10^{-8} \, W$.
Comparing $P_0 = 10^{-8} \, W$ with $10^{-x} \, W$,we get $x = 8$.

(D) When a message signal is amplitude modulated with a carrier signal,the resulting modulated wave contains the carrier frequency $f_{c}$ and the sideband frequencies $(f_{c} + f_{m})$ and $(f_{c} - f_{m})$.
However,the carrier signal is the primary component that is radiated through the antenna to carry the information.
The wavelength $\lambda$ of the radiated signal is determined by the carrier frequency $f_{c}$ as $\lambda = \frac{c}{f_{c}}$,where $c$ is the speed of light in air.

(B) The bandwidth required for a single $AM$ broadcast station is given by twice the maximum modulating frequency.
Bandwidth per station $= 2 \times f_m = 2 \times 5\, kHz = 10\, kHz$.
To find the number of stations that can be accommodated in a total bandwidth of $90\, kHz$,we divide the total bandwidth by the bandwidth per station.
Number of stations $= \frac{\text{Total Bandwidth}}{\text{Bandwidth per station}} = \frac{90\, kHz}{10\, kHz} = 9$.
Therefore,$9$ stations can be accommodated.

(B) The standard equation for an amplitude modulated wave is given by $C_{m}(t) = A_{c}(1 + \mu \cos \omega_{m} t) \sin \omega_{c} t$.
Comparing this with the given equation $C_{m}(t) = 10(1 + 0.2 \cos 12560 t) \sin (111 \times 10^{4} t)$, we identify the angular modulating frequency $\omega_{m} = 12560 \ rad/s$.
We know that $\omega_{m} = 2 \pi f_{m}$, where $f_{m}$ is the modulating frequency.
Therefore, $f_{m} = \frac{\omega_{m}}{2 \pi} = \frac{12560}{2 \times 3.14} = \frac{12560}{6.28} = 2000 \ Hz$.
Converting this to $kHz$, we get $f_{m} = 2 \ kHz$.

(D) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_1$ and a receiving antenna of height $h_2$ is given by the formula:
$d = \sqrt{2Rh_1} + \sqrt{2Rh_2}$
Given:
$h_1 = 50\, m = 50 \times 10^{-3}\, km$
$h_2 = 80\, m = 80 \times 10^{-3}\, km$
$R = 6400\, km$
Substituting the values:
$d = \sqrt{2 \times 6400 \times 50 \times 10^{-3}} + \sqrt{2 \times 6400 \times 80 \times 10^{-3}}$
$d = \sqrt{640000 \times 10^{-3}} + \sqrt{1024000 \times 10^{-3}}$
$d = \sqrt{640} + \sqrt{1024}$
$d = 25.30 + 32 = 57.30\, km$
Rounding to the nearest provided option,the range is $57.28\, km$.

(A) The maximum distance $d$ for line-of-sight communication is given by the formula: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
Here,$R$ is the radius of the Earth $\approx 6400\, km = 6.4 \times 10^6\, m$.
$h_T = 320\, m$ and $h_R = 2000\, m$.
Substituting the values:
$d = \sqrt{2 \times 6.4 \times 10^6 \times 320} + \sqrt{2 \times 6.4 \times 10^6 \times 2000}$
$d = \sqrt{4096 \times 10^6} + \sqrt{25600 \times 10^6}$
$d = 64000 + 160000 = 224000\, m$.
Converting to kilometers: $d = 224\, km$.

(B) For an antenna to radiate signals effectively,the height of the antenna $h$ must be comparable to the wavelength $\lambda$ of the signal.
Typically,for effective radiation,the antenna height $h$ is related to the wavelength by $h \approx \lambda / 4$.
Given the height of the antenna $h = 400 \; m$.
Using the relation $h = \lambda / 4$,we get $\lambda = 4h$.
$\lambda = 4 \times 400 \; m = 1600 \; m$.
However,in the context of effective transmission range $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth $(6400 \; km)$,the range is given as $44 \; km$.
$44 \times 10^3 = \sqrt{2 \times 6.4 \times 10^6 \times 400}$.
$44 \times 10^3 = \sqrt{5120 \times 10^6} \approx 71.5 \times 10^3 \; m = 71.5 \; km$.
Since the question asks for the wavelength that can be radiated effectively,and considering the standard condition for antenna design where $\lambda$ should be of the order of the antenna dimensions,the most appropriate value provided in the options that satisfies the physical constraint $\lambda > h$ is $605 \; m$ (as it is the only value significantly larger than $400 \; m$).

(C) The range of $LOS$ communication is given by $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$,where $h_T$ and $h_R$ are the heights of the transmitting and receiving antennas respectively.
Given $h_T + h_R = 160 \, m$. Let $h_T = x$,then $h_R = 160 - x$.
$d = \sqrt{2R}(\sqrt{h_T} + \sqrt{h_R}) = \sqrt{2R}(\sqrt{x} + \sqrt{160 - x})$.
To maximize $d$,we differentiate with respect to $x$ and set it to zero: $\frac{d}{dx}(\sqrt{x} + \sqrt{160 - x}) = 0$.
$\frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{160 - x}} = 0 \implies \sqrt{x} = \sqrt{160 - x} \implies x = 80 \, m$.
Thus,$h_T = 80 \, m$ and $h_R = 80 \, m$.
Converting heights to km: $h_T = h_R = 0.08 \, km$.
$d_{max} = \sqrt{2 \times 6400 \times 0.08} + \sqrt{2 \times 6400 \times 0.08} = 2 \times \sqrt{12800 \times 0.08} = 2 \times \sqrt{1024} = 2 \times 32 = 64 \, km$.

(D) The bandwidth required for a single $A.M.$ station is twice the maximum frequency of the modulating audio signal.
$\text{Bandwidth per station} = 2 \times f_m = 2 \times 6 \, \text{kHz} = 12 \, \text{kHz}$.
The total available bandwidth is $6 \, \text{MHz} = 6000 \, \text{kHz}$.
The number of stations $N$ that can be broadcasted simultaneously is given by the ratio of the total bandwidth to the bandwidth per station:
$N = \frac{\text{Total Bandwidth}}{\text{Bandwidth per station}} = \frac{6000 \, \text{kHz}}{12 \, \text{kHz}} = 500$.
Therefore,$500$ stations can be broadcasted simultaneously.

(C) The maximum voltage of an amplitude-modulated wave is given by $V_{\max} = A_{c} + A_{m}$,where $A_{c}$ is the amplitude of the carrier wave and $A_{m}$ is the peak voltage of the modulating signal.
Given,$A_{c} = 160 \text{ V}$ and $V_{\max} = 200 \text{ V}$.
Substituting these values into the equation:
$200 = 160 + A_{m}$
$A_{m} = 200 - 160$
$A_{m} = 40 \text{ V}$.
Alternatively,using the minimum voltage formula: $V_{\min} = A_{c} - A_{m}$.
$120 = 160 - A_{m}$
$A_{m} = 160 - 120 = 40 \text{ V}$.
Thus,the peak voltage of the modulating signal is $40 \text{ V}$.

(A) The radius covered by the antenna is given by $r = \sqrt{2 R_e H_T}$.
Given $r = 150 \, km = 1.5 \times 10^5 \, m$ and $R_e = 6.5 \times 10^6 \, m$.
Squaring both sides: $r^2 = 2 R_e H_T$.
$H_T = \frac{r^2}{2 R_e} = \frac{(1.5 \times 10^5)^2}{2 \times 6.5 \times 10^6} = \frac{2.25 \times 10^{10}}{13 \times 10^6} \approx 1730.76 \, m \approx 1731 \, m$.
The population covered is the area covered multiplied by the population density.
Area $= \pi r^2 = 3.14 \times (150 \, km)^2 = 3.14 \times 22500 \, km^2 = 70650 \, km^2$.
Population $= \text{Area} \times \text{Density} = 70650 \, km^2 \times 2000 \, / km^2 = 141300000 = 1413 \times 10^5$.

(A) The modulation index $\mu$ is defined as the ratio of the peak voltage of the message signal $(A_m)$ to the peak voltage of the carrier wave $(A_c)$.
Given:
$A_m = 20 \, V$
$A_c = 20 \, V$
Using the formula:
$\mu = \frac{A_m}{A_c} = \frac{20}{20} = 1$
Therefore,the modulation index is $1$.

(B) Given: Carrier amplitude $A_C = 15 \, V$,Modulating signal amplitude $A_m = 5 \, V$.
The modulation index $\mu$ is defined as $\mu = \frac{A_m}{A_C} = \frac{5}{15} = \frac{1}{3}$.
The amplitude of both the upper sideband $(USB)$ and lower sideband $(LSB)$ in an $A.M.$ wave is given by $\frac{\mu A_C}{2}$.
Amplitude of $USB$ = $\frac{\mu A_C}{2} = \frac{(1/3) \times 15}{2} = \frac{5}{2} = 2.5 \, V$.
Amplitude of $LSB$ = $\frac{\mu A_C}{2} = \frac{(1/3) \times 15}{2} = \frac{5}{2} = 2.5 \, V$.
According to the problem,the amplitudes are $\frac{a}{10} \, V$ and $\frac{b}{10} \, V$.
So,$\frac{a}{10} = 2.5 \Rightarrow a = 25$ and $\frac{b}{10} = 2.5 \Rightarrow b = 25$.
Therefore,the value of $\frac{a}{b} = \frac{25}{25} = 1$.

(B) The maximum amplitude of an amplitude modulated wave is given by $A_{\max} = A_c + A_m = 12\, V$.
The minimum amplitude of an amplitude modulated wave is given by $A_{\min} = A_c - A_m = 3\, V$.
Adding these two equations: $2A_c = 15 \Rightarrow A_c = 7.5\, V$.
Subtracting the second from the first: $2A_m = 9 \Rightarrow A_m = 4.5\, V$.
The modulation index $\mu$ is defined as $\mu = \frac{A_m}{A_c} = \frac{4.5}{7.5} = 0.6$.
Given that the modulation index is $0.6\, x$,we have $0.6 = 0.6\, x$,which implies $x = 1$.

(B) The speed of light in a dielectric medium is given by $v = \frac{c}{\sqrt{\mu_{r} \varepsilon_{r}}}$.
Given $\mu_{r} = 1$ and $\varepsilon_{r} = 6.25$,the speed $v = \frac{3 \times 10^{8}}{\sqrt{6.25}} = \frac{3 \times 10^{8}}{2.5} = 1.2 \times 10^{8} \, m/s$.
For an antenna to radiate effectively,its length $L$ should be at least $\frac{\lambda}{4}$.
Thus,$\lambda = 4L = 4 \times 5.0 \times 10^{-3} \, m = 20 \times 10^{-3} \, m = 0.02 \, m$.
The frequency $f$ is given by $f = \frac{v}{\lambda} = \frac{1.2 \times 10^{8}}{0.02} = 60 \times 10^{8} \, Hz = 6 \, GHz$.

(C) The carrier frequency is $f_{c} = 3.5\, GHz = 3.5 \times 10^{9}\, Hz$.
The wavelength $\lambda$ of the carrier signal is given by $\lambda = \frac{c}{f_{c}}$,where $c = 3 \times 10^{8}\, m/s$ is the speed of light.
$\lambda = \frac{3 \times 10^{8}}{3.5 \times 10^{9}} = \frac{3}{35}\, m \approx 0.0857\, m = 85.7\, mm$.
The minimum length of the antenna required for efficient transmission is $\frac{\lambda}{4}$.
Minimum length $= \frac{85.7\, mm}{4} = 21.425\, mm \approx 21.4\, mm$.

(B) The correct matching is as follows:
$A.$ Facsimile is a method used to transmit a $Static\, Document\, Image$.
$B.$ Guided media channel refers to physical paths like $Optical\, Fiber$ through which signals are transmitted.
$C.$ Frequency Modulation is commonly used in $Local\, Broadcast\, Radio$.
$D.$ Digital Signal is represented by a $Rectangular\, wave$.
Therefore,the correct sequence is $A-I, B-IV, C-II, D-III$.

(B) The frequency range of optical fibre communication is approximately $1 \,THz$ to $1000 \,THz$. Since $100 \,THz$ falls within this range,optical fibre is the most efficient medium for transmitting such high-frequency signals. Other options like coaxial cables and twisted pairs are limited to much lower frequencies (typically in the $MHz$ to $GHz$ range).

(D) Given carrier wave: $y(t) = 40 \sin(10^7 \pi t)$,so $A_c = 40$ and $\omega_c = 10^7 \pi$.
Given modulating wave: $x(t) = 20 \sin(1000 \pi t)$,so $A_m = 20$ and $\omega_m = 10^3 \pi$.
The modulated wave equation is given by $E = A_c(1 + \mu \sin \omega_m t) \sin \omega_c t$,where $\mu = \frac{A_m}{A_c} = \frac{20}{40} = 0.5$.
Expanding the equation: $E = A_c \sin \omega_c t + \frac{\mu A_c}{2} \cos(\omega_c - \omega_m)t - \frac{\mu A_c}{2} \cos(\omega_c + \omega_m)t$.
The frequency components are $\omega_c$,$(\omega_c - \omega_m)$,and $(\omega_c + \omega_m)$.
The minimum frequency component is $(\omega_c - \omega_m)$.
The amplitude of this component is $\frac{\mu A_c}{2} = \frac{A_m}{2} = \frac{20}{2} = 10$.

(C) In amplitude modulation,the amplitude of a high-frequency carrier wave is varied in accordance with the instantaneous amplitude of the message signal (also known as the information signal).

(D) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula:
$d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$
Given:
$h_T = 25 \; m$
$h_R = 49 \; m$
$R = 64 \times 10^{5} \; m$
Substituting the values:
$d = \sqrt{2 \times 64 \times 10^{5} \times 25} + \sqrt{2 \times 64 \times 10^{5} \times 49}$
$d = \sqrt{2 \times 64 \times 10^{5}} \times (\sqrt{25} + \sqrt{49})$
$d = \sqrt{128 \times 10^{5}} \times (5 + 7)$
$d = \sqrt{1280 \times 10^{4}} \times 12$
$d = \sqrt{256 \times 5 \times 10^{4}} \times 12$
$d = (16 \times 10^{2} \sqrt{5}) \times 12$
$d = 192 \sqrt{5} \times 10^{2} \; m$
Comparing this with $K \sqrt{5} \times 10^{2} \; m$,we get $K = 192$.

(C) Statement $(a)$ is true: The antenna size must be at least $\lambda/4$ to radiate efficiently. For low frequencies,$\lambda = c/f$ is very large,making the antenna size impractical.
Statement $(b)$ is false: The power radiated by an antenna of length $l$ is proportional to $(l/\lambda)^2$. For low frequencies (large $\lambda$),the radiated power is extremely low.
Statement $(c)$ is true: Multiplexing is necessary to distinguish signals from different sources.
Statement $(d)$ is true: Modulation (superimposing low frequency signals onto high frequency carrier waves) allows for efficient transmission over long distances.
Thus,statements $(a), (c),$ and $(d)$ are correct.

(C) The standard equation for an amplitude modulated wave is $V_{AM} = A_c [1 + \mu \cos(\omega_m t)] \cos(\omega_c t)$.
Comparing this with the given equation $V_{AM} = 10[1 + 0.4 \cos(2 \pi \times 10^4 t)] \cos(2 \pi \times 10^7 t)$,we identify the modulating frequency $f_m$.
The term inside the cosine function for the modulating signal is $2 \pi f_m t = 2 \pi \times 10^4 t$,which gives $f_m = 10^4 \text{ Hz} = 10 \text{ kHz}$.
The bandwidth of an amplitude modulated wave is given by $BW = 2 f_m$.
Therefore,$BW = 2 \times 10 \text{ kHz} = 20 \text{ kHz}$.

(C) The frequency ranges for various signals are as follows:
$1$. Television signals require a bandwidth of approximately $6\,MHz$ for transmission.
$2$. Radio signals (specifically $AM$ broadcast) typically use a bandwidth of $02\,MHz$ (or $200\,KHz$ per channel,but in this context,$02\,MHz$ is the standard match).
$3$. High-quality music requires a wider frequency range,typically up to $20\,KHz$.
$4$. Human speech is generally transmitted within a frequency range of $03\,KHz$.
Matching these:
$A$ (Television signal) $\rightarrow$ $IV$ $(06\,MHz)$
$B$ (Radio signal) $\rightarrow$ $III$ $(02\,MHz)$
$C$ (High Quality Music) $\rightarrow$ $II$ $(20\,KHz)$
$D$ (Human speech) $\rightarrow$ $I$ $(03\,KHz)$
Therefore,the correct matching is $A-IV, B-III, C-II, D-I$.

(C) The range $d$ of a $TV$ transmission tower of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Let the initial height be $h_1 = 125\, m$ and the initial range be $d_1 = \sqrt{2Rh_1}$.
We want to double the range,so the new range $d_2 = 2d_1$.
Substituting the formula,we get $\sqrt{2Rh_2} = 2\sqrt{2Rh_1}$.
Squaring both sides,we get $2Rh_2 = 4(2Rh_1)$,which simplifies to $h_2 = 4h_1$.
Substituting the value of $h_1$,we get $h_2 = 4 \times 125\, m = 500\, m$.
The increase in height required is $\Delta h = h_2 - h_1 = 500\, m - 125\, m = 375\, m$.

(B) The frequency of the optical source is given by $f = \frac{c}{\lambda} = \frac{3 \times 10^{8} \,m/s}{1000 \times 10^{-9} \,m} = 3 \times 10^{14} \,Hz$.
The available channel bandwidth is $2 \%$ of this frequency: $\text{Bandwidth} = \frac{2}{100} \times 3 \times 10^{14} \,Hz = 6 \times 10^{12} \,Hz$.
The bandwidth required for one audio signal channel is $8 \,kHz = 8 \times 10^{3} \,Hz$.
The number of channels that can be accommodated is given by the ratio of the total available bandwidth to the bandwidth per channel:
$\text{Number of channels} = \frac{6 \times 10^{12} \,Hz}{8 \times 10^{3} \,Hz} = 0.75 \times 10^{9} = 75 \times 10^{7}$.

(D) The modulation index $\mu$ is defined as the ratio of the difference of maximum and minimum amplitudes to the sum of maximum and minimum amplitudes.
$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$
Given $V_{\max} = 6\,V$ and $V_{\min} = 2\,V$.
Substituting the values:
$\mu = \frac{6 - 2}{6 + 2} = \frac{4}{8} = 0.5$
To express this as a percentage:
$\text{Modulation index} = 0.5 \times 100\% = 50\%$.

(A) The coverage distance $d$ of a $TV$ tower of height $h$ is given by $d = \sqrt{2Rh}$.
Given the radius of the Earth $R = 6400 \, km$.
The area covered by the tower is $A = \pi d^2 = \pi (2Rh)$.
Given the population $P = 6.03 \times 10^5$ and population density $\rho = 100 \, \text{people/km}^2$.
The area $A$ is also given by $A = \frac{P}{\rho} = \frac{6.03 \times 10^5}{100} = 6030 \, \text{km}^2$.
Equating the two expressions for area: $6030 = \pi \times 2 \times 6400 \times h$ (where $h$ is in $km$).
$h = \frac{6030}{2 \times \pi \times 6400} \, km$.
$h = \frac{6030}{40212} \approx 0.150 \, km$.
Converting to meters: $h = 0.150 \times 1000 = 150 \, m$.

(B) Given:
$V_{\max} = 60\,V$
$V_{\min} = 20\,V$
The modulation index $\mu$ is given by the formula:
$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$
Substituting the values:
$\mu = \frac{60 - 20}{60 + 20} = \frac{40}{80} = 0.5$
The percentage modulation index is $\mu \times 100\% = 0.5 \times 100\% = 50\%$.

(B) Given: Frequency $f_{1} = 6\,MHz = 6 \times 10^{6}\,Hz$ and $f_{2} = 10\,MHz = 10 \times 10^{6}\,Hz$.
Speed of light $c = 3 \times 10^{8}\,m/s$.
The wavelength $\lambda$ is given by $\lambda = \frac{c}{f}$.
For $f_{1} = 6 \times 10^{6}\,Hz$,$\lambda_{1} = \frac{3 \times 10^{8}}{6 \times 10^{6}} = 50\,m$.
For $f_{2} = 10 \times 10^{6}\,Hz$,$\lambda_{2} = \frac{3 \times 10^{8}}{10 \times 10^{6}} = 30\,m$.
The wavelength bandwidth is the difference between the two wavelengths: $\Delta\lambda = \lambda_{1} - \lambda_{2} = 50\,m - 30\,m = 20\,m$.

(A) From the given figure,the amplitude of the modulating signal $A_m$ is the peak value of the square wave,which is $1 \text{ V}$.
The carrier wave is given by $C(t) = 5 \sin(8\pi t) \text{ V}$. Comparing this with the standard carrier wave equation $C(t) = A_C \sin(\omega_c t)$,we get the amplitude of the carrier wave $A_C = 5 \text{ V}$.
The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave:
$\mu = \frac{A_m}{A_C}$
Substituting the values:
$\mu = \frac{1}{5} = 0.2$
Thus,the modulation index is $0.2$.

(D) Let $d$ be the coverage range of the tower of height $h$.
The range $d$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
Thus,$d \propto \sqrt{h}$.
Let the initial height be $h_1 = 100\,m$ and the initial range be $d_1$.
Let the new height be $h_2$ and the new range be $d_2 = 3d_1$.
Since $d \propto \sqrt{h}$,we have:
$\frac{d_2}{d_1} = \sqrt{\frac{h_2}{h_1}}$
Substituting the values:
$3 = \sqrt{\frac{h_2}{100}}$
Squaring both sides:
$9 = \frac{h_2}{100}$
$h_2 = 900\,m$.
Therefore,the height of the tower should be increased to $900\,m$.

(D) Given:
Modulating frequency $f_m = 20\,kHz$.
Deviation ratio $\beta = 10$.
The deviation ratio is defined as $\beta = \frac{\Delta f}{f_m}$,where $\Delta f$ is the frequency deviation.
Therefore,$\Delta f = \beta \times f_m = 10 \times 20\,kHz = 200\,kHz$.
According to Carson's rule,the bandwidth $BW$ required for an $FM$ signal is given by:
$BW = 2(\Delta f + f_m)$
$BW = 2(200\,kHz + 20\,kHz)$
$BW = 2(220\,kHz) = 440\,kHz$.

(A) The modulation index is defined as $\mu = \frac{A_m}{A_c}$,where $A_m$ is the amplitude of the message signal and $A_c$ is the amplitude of the carrier wave.
To avoid distortion in the amplitude modulated wave,the modulation index must satisfy the condition $\mu \leq 1$.
If $\mu > 1$,over-modulation occurs,which leads to the distortion of the signal and interference between the carrier frequency and the message frequency.

(A) The modulation index $m$ is defined as $m = \frac{A_m}{A_c}$,where $A_m$ is the amplitude of the modulating signal and $A_c$ is the amplitude of the carrier wave.
Given the variation in amplitude is $2A_m = 8\,V$,we find $A_m = 4\,V$.
The maximum amplitude of the $AM$ wave is given by $A_{max} = A_c + A_m = 9\,V$.
Substituting $A_m = 4\,V$,we get $A_c + 4 = 9$,which implies $A_c = 5\,V$.
Therefore,the modulation index is $m = \frac{4}{5} = 0.8$.

(D) The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal $(A_m)$ to the amplitude of the carrier wave $(A_c)$.
From the given figure,the amplitude of the modulating square wave signal is $A_m = 1 \text{ V}$.
The carrier wave is given by $c(t) = 2 \sin(8 \pi t)$,which is in the form $c(t) = A_c \sin(\omega_c t)$. Thus,the amplitude of the carrier wave is $A_c = 2 \text{ V}$.
The modulation index is calculated as:
$\mu = \frac{A_m}{A_c} = \frac{1}{2} = 0.5$.

(A) The standard frequency ranges for communication systems are as follows:
$1$. $AM$ Broadcast: $540-1600\,kHz$
$2$. $FM$ Broadcast: $88-108\,MHz$
$3$. Television: $54-890\,MHz$
$4$. Satellite Communication: $3.7-4.2\,GHz$
Matching these with the given lists:
$A$ ($AM$ Broadcast) matches with $II$ $(540-1600\,kHz)$.
$B$ ($FM$ Broadcast) matches with $I$ $(88-108\,MHz)$.
$C$ (Television) matches with $IV$ $(54-890\,MHz)$.
$D$ (Satellite Communication) matches with $III$ $(3.7-4.2\,GHz)$.
Therefore,the correct matching is $A-II, B-I, C-IV, D-III$.

(C) Given:
Message signal frequency $f_m = 5\,kHz$.
Carrier wave frequency $f_c = 2\,MHz = 2000\,kHz$.
In amplitude modulation,the bandwidth is defined as the difference between the upper sideband frequency and the lower sideband frequency.
Bandwidth $= (f_c + f_m) - (f_c - f_m) = 2f_m$.
Substituting the values:
Bandwidth $= 2 \times 5\,kHz = 10\,kHz$.

(A) The atmospheric layers and their approximate altitudes are as follows:
$1$. Troposphere: Extends up to approximately $10 \ km$ from the Earth's surface $(A-III)$.
$2$. $D$-Part of Ionosphere: Located at approximately $65-75 \ km$ altitude $(D-I)$.
$3$. $E$-Part of Ionosphere: Located at approximately $100 \ km$ altitude $(B-IV)$.
$4$. $F_2$-Part of Ionosphere: Located at approximately $300 \ km$ altitude $(C-II)$.
Thus,the correct matching is $A-III, B-IV, C-II, D-I$.

(D) The maximum line of sight distance $(d_M)$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula: $d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
Given: $h_T = 80\,m$,$h_R = 80\,m$,and $R = 6.4 \times 10^6\,m$.
Substituting the values: $d_M = \sqrt{2 \times 6.4 \times 10^6 \times 80} + \sqrt{2 \times 6.4 \times 10^6 \times 80}$.
$d_M = 2 \times \sqrt{2 \times 6.4 \times 10^6 \times 80} = 2 \times \sqrt{1024 \times 10^6} = 2 \times 32 \times 10^3\,m$.
$d_M = 64 \times 10^3\,m = 64\,km$.