Communication Questions in English

(D) For an amplitude modulated wave,the maximum voltage is given by $V_{max} = A_c + A_m$ and the minimum voltage is given by $V_{min} = A_c - A_m$,where $A_c$ is the carrier amplitude and $A_m$ is the message signal amplitude.
Adding these two equations: $V_{max} + V_{min} = 2A_c \implies A_c = \frac{V_{max} + V_{min}}{2}$.
Subtracting these two equations: $V_{max} - V_{min} = 2A_m \implies A_m = \frac{V_{max} - V_{min}}{2}$.
The modulation factor $m$ is defined as the ratio of the message amplitude to the carrier amplitude: $m = \frac{A_m}{A_c}$.
Substituting the values: $m = \frac{(V_{max} - V_{min})/2}{(V_{max} + V_{min})/2} = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$.

(B) The characteristics of an optical fibre are as follows:
$(i)$ It works on the principle of total internal reflection.
$(ii)$ It consists of a core made of glass, silica, or plastic with a refractive index $n_1$, surrounded by a cladding with a refractive index $n_2$, where $n_1 > n_2$. The refractive index of the core can be uniform (step-index) or vary gradually (graded-index).
$(iii)$ Optical fibres exhibit extremely low transmission loss.
$(iv)$ Optical fibres are immune to electromagnetic interference from external electric and magnetic fields because they transmit light signals rather than electrical signals.
Therefore, the statement that optical fibres are subjected to electromagnetic interference is $NOT$ true.

(C) From the given waveform,the maximum voltage $V_{\max} = 10 \ V$ and the minimum voltage $V_{\min} = 2 \ V$.
The modulation index (or modulation factor) $\mu$ is given by the formula:
$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$
Substituting the values:
$\mu = \frac{10 - 2}{10 + 2} = \frac{8}{12} = \frac{2}{3}$
Thus,the modulation factor is $\frac{2}{3}$.

(B) For an $AM$ wave,the maximum voltage is given by $V_{max} = V_c + V_m$,where $V_c$ is the carrier amplitude and $V_m$ is the message signal amplitude.
The minimum voltage is given by $V_{min} = V_c - V_m$.
Adding these two equations: $V_{max} + V_{min} = 2V_c \implies V_c = \frac{V_{max} + V_{min}}{2}$.
Subtracting these two equations: $V_{max} - V_{min} = 2V_m \implies V_m = \frac{V_{max} - V_{min}}{2}$.
The modulation index $\mu$ is defined as $\mu = \frac{V_m}{V_c}$.
Substituting the values: $\mu = \frac{(V_{max} - V_{min})/2}{(V_{max} + V_{min})/2} = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$.

(B) The radio horizon distance $d_T$ for a transmitting antenna of height $h_T$ is given by the formula $d_T = \sqrt{2Rh_T}$, where $R$ is the radius of the Earth. Option $B$ states $d_T = \sqrt{Rh_T}$, which is incorrect because the factor of $2$ is missing. Therefore, statement $B$ is the $WRONG$ statement.

(B) The attenuation in decibels $(dB)$ is given by the formula: $10 \log_{10} (I/I_0) = -\alpha_{dB} x$,where $\alpha_{dB}$ is the attenuation constant in $dB/km$.
Given that the intensity falls by $50$ percent,the final intensity $I = 0.5 I_0$,which means $I/I_0 = 1/2$.
The distance $x = 50 \ km$.
Substituting these values into the formula:
$10 \log_{10} (1/2) = -\alpha_{dB} \times 50$
$10 (\log_{10} 1 - \log_{10} 2) = -50 \alpha_{dB}$
Since $\log_{10} 1 = 0$ and $\log_{10} 2 \approx 0.3010$:
$10 (0 - 0.3010) = -50 \alpha_{dB}$
$-3.010 = -50 \alpha_{dB}$
$\alpha_{dB} = 3.010 / 50 = 0.0602 \ dB/km$.

(B) The modulation index $\mu$ is defined as the ratio of the peak voltage of the modulating signal $(A_m)$ to the peak voltage of the carrier wave $(A_c)$.
$\mu = \frac{A_m}{A_c}$
Given,$\mu = 75 \% = 0.75$ and $A_c = 12 \, V$.
Substituting these values into the formula:
$0.75 = \frac{A_m}{12}$
$A_m = 0.75 \times 12$
$A_m = 9 \, V$.
Therefore,the peak voltage of the modulating signal should be $9 \, V$.

(A) Given:
Carrier frequency $f_{c} = 1.5 \, MHz = 1500 \, kHz$
Modulating frequency $f_{m} = 10 \, kHz$
The lower side-band $(LSB)$ frequency is given by:
$f_{LSB} = f_{c} - f_{m} = 1500 \, kHz - 10 \, kHz = 1490 \, kHz$
The upper side-band $(USB)$ frequency is given by:
$f_{USB} = f_{c} + f_{m} = 1500 \, kHz + 10 \, kHz = 1510 \, kHz$
Thus,the lower and upper side-band frequencies are $1490 \, kHz$ and $1510 \, kHz$ respectively.

(C) The modulation index $(m_a)$ is defined as the ratio of the amplitude of the modulating signal $(E_m)$ to the amplitude of the carrier signal $(E_c)$.
Given:
Amplitude of modulating signal,$E_m = 15 \text{ V}$.
Amplitude of carrier signal,$E_c = 60 \text{ V}$.
The depth of modulation in percentage is given by:
$\text{Depth of modulation} = \frac{E_m}{E_c} \times 100$
$= \frac{15}{60} \times 100$
$= \frac{1}{4} \times 100 = 25\%$.

(D) Statement-$1$ is true: Sky waves (frequencies between $2 \text{ MHz}$ to $30 \text{ MHz}$) are used for long-distance radio communication by reflecting off the ionosphere. They are less stable than ground waves because they depend on the fluctuating properties of the ionosphere.
Statement-$2$ is true: The ionosphere is a region of the upper atmosphere containing ions and free electrons. Its density and height vary significantly due to solar radiation, changing from hour to hour, day to day, and season to season.
Conclusion: Since the stability of sky waves depends directly on the properties of the ionosphere, Statement-$2$ provides the correct explanation for why sky waves are less stable (as mentioned in Statement-$1$). Therefore, both statements are true and Statement-$2$ is the correct explanation of Statement-$1$.

(C) The carrier wave is represented as $e_c(t) = E_c \sin(\omega_c t)$,where $E_c$ is the peak amplitude of the carrier wave.
Power is defined as the mean square voltage divided by resistance: $P_c = \frac{V_{rms}^2}{R}$.
Since the root mean square $(RMS)$ value of a sinusoidal wave is $E_{rms} = \frac{E_c}{\sqrt{2}}$,the power is calculated as:
$P_c = \frac{(E_c / \sqrt{2})^2}{R} = \frac{E_c^2}{2R}$.

(B) The bandwidth required for a single amplitude modulated $(AM)$ station is twice the highest modulating frequency $(f_m)$.
Given,$f_m = 15\, kHz$.
Therefore,the bandwidth per channel = $2 \times f_m = 2 \times 15\, kHz = 30\, kHz$.
The total available bandwidth is $300\, kHz$.
The number of stations that can be accommodated = $\frac{\text{Total Bandwidth}}{\text{Bandwidth per channel}} = \frac{300\, kHz}{30\, kHz} = 10$.

(C) The resonant frequency of the tank circuit is given by $f_c = \frac{1}{2\pi\sqrt{LC}}$.
Given: $L = 49 \times 10^{-6}\,H$ and $C = 2.5 \times 10^{-9}\,F$.
$f_c = \frac{1}{2\pi\sqrt{49 \times 10^{-6} \times 2.5 \times 10^{-9}}} = \frac{1}{2\pi\sqrt{122.5 \times 10^{-15}}} = \frac{1}{2\pi\sqrt{1.225 \times 10^{-13}}} \approx \frac{1}{2\pi \times 3.5 \times 10^{-7}} = \frac{10^7}{7\pi} \approx 454.7\,kHz$.
For an audio signal frequency $f_m = 12\,kHz$,the sideband frequencies are $f_c - f_m$ and $f_c + f_m$.
Lower sideband $= 454.7 - 12 = 442.7\,kHz$.
Upper sideband $= 454.7 + 12 = 466.7\,kHz$.
The range is approximately $442\,kHz - 466\,kHz$.

(A) Given: Modulation index $m = 80\% = 0.8$.
Carrier wave peak voltage $E_c = 14\,V$.
We know that the modulation index $m$ is defined as the ratio of the peak voltage of the modulating signal $(E_m)$ to the peak voltage of the carrier wave $(E_c)$.
Formula: $m = \frac{E_m}{E_c}$.
Rearranging the formula to find $E_m$: $E_m = m \times E_c$.
Substituting the values: $E_m = 0.8 \times 14\,V = 11.2\,V$.
Therefore,the peak voltage of the modulating signal is $11.2\,V$.

(D) The modulation index $m$ is given by the ratio of the peak voltage of the modulating signal $V_m$ to the peak voltage of the carrier wave $V_c$.
$m = \frac{V_m}{V_c} = \frac{5 \, V}{25 \, V} = 0.2$.
Given,the frequency of the carrier wave $f_c = 1.2 \, MHz = 1200 \, kHz$.
The frequency of the modulating signal $f_m = 20 \, kHz$.
The sideband frequencies are given by $f_c \pm f_m$.
Lower sideband frequency $= f_c - f_m = 1200 \, kHz - 20 \, kHz = 1180 \, kHz$.
Upper sideband frequency $= f_c + f_m = 1200 \, kHz + 20 \, kHz = 1220 \, kHz$.
Thus,the modulation index is $0.2$ and the sidebands are at $1180 \, kHz$ and $1220 \, kHz$.

(A) For efficient transmission and reception of signals,the length $l$ of a linear antenna should be comparable to the wavelength $\lambda$ of the signal. Typically,$l = \frac{\lambda}{4}$ or $l = \frac{\lambda}{2}$ are standard lengths. Among the given options,the length must be a fraction of $\lambda$.
The power radiated $P_{eff}$ by a linear antenna is proportional to the square of the ratio of the antenna length to the wavelength. Specifically,$P_{eff} \propto \left(\frac{l}{\lambda}\right)^2$. Since $l$ is proportional to $\lambda$,we have $P_{eff} \propto \left(\frac{\lambda}{\lambda}\right)^2$ is not the case here; rather,for a fixed antenna length,the power radiated is inversely proportional to the square of the wavelength,i.e.,$P_{eff} \propto \frac{1}{\lambda^2}$.
Thus,$P_{eff} = K \left(\frac{1}{\lambda}\right)^2$. Comparing this with the options,option $A$ provides the correct relationship for power,and the length $\lambda$ is a standard order of magnitude for antenna design.

(C) The bandwidth of an $AM$ signal is defined as the difference between the maximum and minimum frequencies in the signal.
For the human speech signal,the frequency range is $200\,Hz$ to $2700\,Hz.$ The bandwidth is $BW_1 = 2700\,Hz - 200\,Hz = 2500\,Hz.$
When both signals are sent together,the total frequency range spans from the lowest frequency of the speech signal $(200\,Hz)$ to the highest frequency of the music signal $(15200\,Hz.)$ The total bandwidth is $BW_{total} = 15200\,Hz - 200\,Hz = 15000\,Hz.$
The ratio of the total bandwidth to the speech signal bandwidth is $\frac{BW_{total}}{BW_1} = \frac{15000\,Hz}{2500\,Hz} = 6.$

(B) The standard equation for an $AM$ wave is $C_m(t) = A_c \sin(\omega_c t) + \frac{\mu A_c}{2} \cos((\omega_c - \omega_m)t) - \frac{\mu A_c}{2} \cos((\omega_c + \omega_m)t)$.
Comparing this with the given equation $C_m(t) = 30 \sin(300\pi t) + 10 \cos(200\pi t) - 10 \cos(400\pi t)$:
$1$. Carrier frequency: $\omega_c = 300\pi \Rightarrow 2\pi f_c = 300\pi \Rightarrow f_c = 150 \text{ Hz}$.
$2$. Sideband frequencies: $\omega_c - \omega_m = 200\pi$ and $\omega_c + \omega_m = 400\pi$.
Subtracting the two: $2\omega_m = 200\pi \Rightarrow \omega_m = 100\pi \Rightarrow 2\pi f_m = 100\pi \Rightarrow f_m = 50 \text{ Hz}$.
$3$. Modulation index: $\frac{\mu A_c}{2} = 10$. Since $A_c = 30$,we have $\frac{\mu(30)}{2} = 10 \Rightarrow 15\mu = 10 \Rightarrow \mu = 10/15 = 2/3$.

(B) Given: Height of transmitting antenna $h_T = 50\, m$ and height of receiving antenna $h_R = 32\, m$.
Radius of the Earth $R \approx 6.4 \times 10^6\, m$.
The maximum line-of-sight distance $d_M$ is given by the formula: $d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
Substituting the values:
$d_M = \sqrt{2 \times 6.4 \times 10^6 \times 50} + \sqrt{2 \times 6.4 \times 10^6 \times 32}$.
$d_M = \sqrt{640 \times 10^6} + \sqrt{409.6 \times 10^6}$.
$d_M = (25.298 \times 10^3) + (20.238 \times 10^3) = 45.536 \times 10^3\, m$.
Converting to kilometers: $d_M \approx 45.5\, km$.

(B) Sky wave propagation utilizes the ionosphere to reflect radio waves back to Earth. This mode of propagation is effective for frequencies in the range of $5\,MHz$ to $25\,MHz$. Frequencies lower than this are absorbed by the ionosphere,while frequencies higher than this penetrate the ionosphere and escape into space.

(B) Sky wave propagation is a mode of radio wave propagation in which radio waves are reflected or refracted back to the Earth from the ionosphere.
This phenomenon is primarily used for long-distance communication.
The ionosphere acts as a reflecting medium for radio waves within a specific frequency range.
According to standard communication physics,the frequency range suitable for sky wave propagation is typically between $2\, MHz$ and $30\, MHz$.
Comparing this with the given options,the range $8 - 25\, MHz$ falls within this specified limit.
Therefore,option $B$ is the correct answer.

(C) The standard equation for an amplitude modulated wave is given by $C_m(t) = A_c(1 + \mu \sin \omega_m t) \sin \omega_c t$,where $\mu$ is the modulation index.
Given the carrier wave $C(t) = A \sin \omega_c t$ and the modulating signal $m(t) = A \sin \omega_m t$,the amplitude of the carrier is $A_c = A$ and the amplitude of the modulating signal is $A_m = A$.
The modulation index $\mu$ is defined as $\mu = \frac{A_m}{A_c} = \frac{A}{A} = 1$.
Substituting $\mu = 1$ and $A_c = A$ into the general equation,we get $C_m(t) = A(1 + 1 \cdot \sin \omega_m t) \sin \omega_c t = A(1 + \sin \omega_m t) \sin \omega_c t$.
Thus,the modulated signal is $C_m(t) = A(1 + \sin \omega_m t) \sin \omega_c t$ and the modulation index is $1$.

(C) The critical frequency $(f_c)$ is defined as the highest frequency that is reflected back by the ionosphere when incident vertically. Waves with frequencies higher than the critical frequency penetrate the ionosphere and are not reflected. Statement $C$ claims that waves with frequencies higher than $30 \, MHz$ cannot penetrate the ionosphere, which is incorrect because frequencies above the critical frequency (typically $3-30 \, MHz$ for the ionosphere) easily penetrate it.

(B) In communication systems,noise primarily affects the amplitude of the signal. Frequency Modulation $(FM)$ is generally more noise-tolerant than Amplitude Modulation $(AM)$. Among the given options,$Short-wave$ signals (often associated with specific propagation characteristics and modulation techniques) are historically recognized for better performance in long-distance communication compared to $Long-wave$ or $Medium-wave$ signals,which are more susceptible to atmospheric noise and interference.

(A) Broadcasting antennas are generally designed to be vertical type. This is because vertical antennas radiate electromagnetic waves uniformly in all directions in the horizontal plane,which is ideal for broadcasting to a wide area.

(B) The given expression for the amplitude modulated wave is $\vec E = \hat i E_c (1 + \frac{E_m}{E_c} \cos \omega_m t) \cos \omega_c t$.
Expanding this expression,we get $\vec E = \hat i [E_c \cos \omega_c t + E_m \cos \omega_m t \cos \omega_c t]$.
Using the trigonometric identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we can write:
$\vec E = \hat i [E_c \cos \omega_c t + \frac{E_m}{2} (\cos(\omega_c + \omega_m)t + \cos(\omega_c - \omega_m)t)]$.
This expression shows that the modulated wave consists of three distinct frequency components:
$1$. The carrier frequency: $\omega_c$
$2$. The upper sideband frequency: $\omega_c + \omega_m$
$3$. The lower sideband frequency: $\omega_c - \omega_m$
Therefore,the frequencies present are $\omega_c, \omega_c + \omega_m$,and $\omega_c - \omega_m$.

(C) The frequency of the source is given by $f = \frac{c}{\lambda}$.
Substituting the values,$f = \frac{3 \times 10^8\,m/s}{800 \times 10^{-9}\,m} = \frac{3 \times 10^8}{8 \times 10^{-7}} = 0.375 \times 10^{15}\,Hz = 3.75 \times 10^{14}\,Hz$.
The available signal bandwidth is $1\%$ of the source frequency,so $\text{Bandwidth} = 0.01 \times f = 0.01 \times 3.75 \times 10^{14}\,Hz = 3.75 \times 10^{12}\,Hz$.
The number of channels $n$ that can be accommodated for $TV$ signals of bandwidth $6\,MHz$ $(6 \times 10^6\,Hz)$ is given by $n = \frac{\text{Total Bandwidth}}{\text{Bandwidth per channel}}$.
$n = \frac{3.75 \times 10^{12}}{6 \times 10^6} = 0.625 \times 10^6 = 6.25 \times 10^5$.

(A) The maximum distance $d$ for $LOS$ communication is given by the formula: $d = \sqrt{2R h_T} + \sqrt{2R h_R}$.
Given:
Height of transmitter tower $h_T = 140\, m$
Height of receiving antenna $h_R = 40\, m$
Radius of Earth $R = 6.4 \times 10^6\, m$
Substituting the values:
$d = \sqrt{2 \times 6.4 \times 10^6 \times 140} + \sqrt{2 \times 6.4 \times 10^6 \times 40}$
$d = \sqrt{1792 \times 10^6} + \sqrt{512 \times 10^6}$
$d = (42.33 \times 10^3) + (22.63 \times 10^3)$
$d = 64.96 \times 10^3\, m \approx 65\, km$.

(D) Given,modulation frequency $(f_m)$ = $250\, kHz$.
It is given that $f_m$ is $10\%$ of the carrier frequency $(f_c)$.
So,$250\, kHz = 0.10 \times f_c$.
Therefore,$f_c = 2500\, kHz$.
To avoid interference,the minimum frequency separation between two $AM$ stations must be $2 \times f_m = 2 \times 250\, kHz = 500\, kHz$.
Thus,the next available broadcast frequency should be $f_c + 500\, kHz = 2500\, kHz + 500\, kHz = 3000\, kHz$ or $f_c - 500\, kHz = 2000\, kHz$.
Comparing with the given options,$2000\, kHz$ is the correct choice.

(C) $1$. From the graph,the amplitude of the signal varies between $8 \, V$ and $10 \, V$. This can be expressed as $A(t) = 9 + 1 \sin(\omega_m t) \, V$.
$2$. The time period of the modulating signal (envelope) is $T_m = 100 \, \mu s = 100 \times 10^{-6} \, s = 10^{-4} \, s$. The angular frequency is $\omega_m = \frac{2\pi}{T_m} = \frac{2\pi}{10^{-4}} = 2\pi \times 10^4 \, rad/s$.
$3$. The time period of the carrier wave is $T_c = 8 \, \mu s = 8 \times 10^{-6} \, s$. The angular frequency is $\omega_c = \frac{2\pi}{T_c} = \frac{2\pi}{8 \times 10^{-6}} = 0.25 \times 10^6 \pi = 2.5\pi \times 10^5 \, rad/s$.
$4$. The general form of an amplitude modulated signal is $V(t) = (A_c + A_m \sin(\omega_m t)) \sin(\omega_c t)$.
$5$. Substituting the values,we get $V(t) = (9 + 1 \sin(2\pi \times 10^4 t)) \sin(2.5\pi \times 10^5 t) \, V$.

(C) The maximum amplitude of the modulated wave is given by $A_{max} = A_c + A_m = 160\, V$.
The minimum amplitude of the modulated wave is given by $A_{min} = A_c - A_m = 40\, V$.
Adding these two equations: $(A_c + A_m) + (A_c - A_m) = 160 + 40$,which gives $2A_c = 200$,so $A_c = 100\, V$.
Subtracting the second equation from the first: $(A_c + A_m) - (A_c - A_m) = 160 - 40$,which gives $2A_m = 120$,so $A_m = 60\, V$.
The modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave:
$\mu = \frac{A_m}{A_c} = \frac{60}{100} = 0.6$.

(C) The covering range $d$ of a $TV$ transmission tower of height $h$ is given by the formula $d = \sqrt{2hR}$,where $R$ is the radius of the Earth.
From this relation,we can see that $d \propto \sqrt{h}$.
Let the initial range be $d_1 = \sqrt{2h_1R}$ and the new range be $d_2 = \sqrt{2h_2R}$.
We want the new range to be double the initial range,so $d_2 = 2d_1$.
Substituting the expressions,we get $\sqrt{2h_2R} = 2 \sqrt{2h_1R}$.
Squaring both sides,we get $2h_2R = 4(2h_1R)$.
This simplifies to $h_2 = 4h_1$.
Therefore,the height of the tower must be multiplied by $4$ to double the covering range.

(A) In modern optical fiber communication systems,the signal attenuation is minimized at specific infrared wavelengths.
Specifically,the wavelength of carrier waves used in these networks is typically close to $1500 \, nm$ to ensure low loss and high efficiency during transmission.

(A) The range $d$ for line of sight communication is given by the formula: $d = \sqrt{2Rh_T} + \sqrt{2Rh_R}$.
Given: $d = 50 \times 10^3\, m$,$R = 6.4 \times 10^6\, m$,$h_R = 70\, m$.
Substituting the values:
$50 \times 10^3 = \sqrt{2 \times 6.4 \times 10^6 \times h_T} + \sqrt{2 \times 6.4 \times 10^6 \times 70}$.
$50000 = \sqrt{12.8 \times 10^6 \times h_T} + \sqrt{896 \times 10^6}$.
$50000 = \sqrt{12.8 \times 10^6} \times \sqrt{h_T} + 29933$.
$50000 - 29933 = 3577.7 \times \sqrt{h_T}$.
$20067 = 3577.7 \times \sqrt{h_T}$.
$\sqrt{h_T} \approx 5.61$.
$h_T \approx 31.47\, m \approx 32\, m$.

(B) The standard expression for an amplitude modulated $(AM)$ wave is given by:
$v_{AM} = (v_0 + A_m \sin \omega_m t) \sin \omega_c t$
Given the message signal is $A \cos \omega t$ and the carrier wave is $v_0 \sin \omega_0 t$,the modulated signal is:
$v_{AM} = (v_0 + A \cos \omega t) \sin \omega_0 t$
$v_{AM} = v_0 \sin \omega_0 t + A \cos \omega t \sin \omega_0 t$
Using the trigonometric identity $2 \sin A \cos B = \sin(A + B) + \sin(A - B)$,we can write:
$A \cos \omega t \sin \omega_0 t = \frac{A}{2} [2 \sin \omega_0 t \cos \omega t] = \frac{A}{2} [\sin(\omega_0 + \omega)t + \sin(\omega_0 - \omega)t]$
Substituting this back into the equation:
$v_{AM} = v_0 \sin \omega_0 t + \frac{A}{2} \sin(\omega_0 + \omega)t + \frac{A}{2} \sin(\omega_0 - \omega)t$
This matches option $B$.

(C) The physical size of an antenna,such as a dipole antenna,is typically related to the wavelength $\lambda$ of the signal being transmitted or received.
Since the wavelength $\lambda$ is given by $\lambda = c/f$,where $c$ is the speed of light and $f$ is the carrier frequency,the length of the antenna $L$ is proportional to $\lambda$ (e.g.,$L = \lambda/2$ or $L = \lambda/4$).
Therefore,$L \propto 1/f$.
This implies that the physical size of the transmitter and receiver antenna is inversely proportional to the carrier frequency.

(B) $1$. $Optical\, Fiber\, Communication - Infrared\, light$: Optical fiber communication uses light signals, typically in the infrared region $(850\, nm, 1300\, nm, 1550\, nm)$, to minimize attenuation and scattering in glass fibers.
$2$. $Radar - Radio\, waves$: Radar (Radio Detection and Ranging) systems use radio waves to detect the position, velocity, and other characteristics of remote objects.
$3$. $Sonar - Ultrasound$: Sonar (Sound Navigation and Ranging) uses ultrasonic sound waves to detect objects underwater or in robotics, as these waves can propagate through media effectively.
$4$. $Mobile\, Phones - Microwaves$: Mobile phones operate using microwave frequencies (typically in the $GHz$ range) for wireless communication between the handset and the base station.
Therefore, the correct matching is $A-Q, B-S, C-P, D-R$.

(A) 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 $(A_c)$:
$\mu = \frac{A_m}{A_c} = \frac{2}{4} = 0.5$.
The carrier frequency $f_c$ is obtained from $2 \pi f_c = 20000 \pi$,which gives $f_c = 10000 \text{ Hz} = 10 \text{ kHz}$.
The modulating frequency $f_m$ is obtained from $2 \pi f_m = 2000 \pi$,which gives $f_m = 1000 \text{ Hz} = 1 \text{ kHz}$.
The lower sideband frequency $(LSB)$ is given by $f_c - f_m = 10 \text{ kHz} - 1 \text{ kHz} = 9 \text{ kHz}$.
Therefore,the modulation index is $0.5$ and the lower sideband frequency is $9 \text{ kHz}$.

(D) The electromagnetic spectrum is divided into various frequency bands for communication purposes.
$UHF$ stands for Ultra High Frequency.
The frequency range for the $UHF$ band is defined as $300 \, MHz$ to $3000 \, MHz$ (or $0.3 \, GHz$ to $3 \, GHz$).
Therefore,the correct option is $D$.

(D) The standard equation for an amplitude-modulated wave is given by $e = A_c(1 + \mu \sin(\omega_m t)) \sin(\omega_c t)$, where $\mu$ is the modulation index.
Comparing the given equation $e = 10(1 + 6 \sin(50t)) \sin(10^8t)$ with the standard form:
$A_c = 10$
$\mu = 6$
$\omega_m = 50 \text{ rad/s}$
$\omega_c = 10^8 \text{ rad/s}$
Thus, the modulation index $\mu$ is $6$.

(A) For an antenna to be efficient for transmission,its length $L$ should be related to the wavelength $\lambda$ of the signal. The most common condition for a resonant antenna is that its length should be a quarter of the wavelength,i.e.,$L = \frac{\lambda}{4}$.
Given the antenna length $L = 100\,m$.
Substituting the value: $100 = \frac{\lambda}{4}$.
Therefore,$\lambda = 100 \times 4 = 400\,m$.

(B) basic communication system follows this logical flow:
$1$. Information source: The origin of the message.
$2$. Transmitter: Converts the information into a signal suitable for transmission.
$3$. Channel: The medium through which the signal travels.
$4$. Receiver: Extracts the signal from the channel.
$5$. User of information: The final destination or recipient of the message.
Therefore, the sequence is $(B) \to (A) \to (D) \to (E) \to (C)$, which corresponds to $BADEC$.

(A) For an amplitude-modulated wave,the bandwidth required for one channel is twice the maximum frequency of the modulating signal.
Bandwidth $(BW)$ per channel $= 2 \times f_{max} = 2 \times 5\, kHz = 10\, kHz$.
The total available bandwidth is $150\, kHz$.
The number of stations that can be accommodated is given by the ratio of the total bandwidth to the bandwidth per channel.
Number of stations $= \frac{\text{Total Bandwidth}}{\text{Bandwidth per channel}} = \frac{150\, kHz}{10\, kHz} = 15$.

(A) The range $r$ of an antenna is given by the formula $r = \sqrt{2hR}$,where $h$ is the height of the antenna and $R$ is the radius of the Earth.
Initially,the range is $r = \sqrt{2hR}$.
If the height is doubled,the new height becomes $h' = 2h$.
The new range $r'$ is given by $r' = \sqrt{2h'R}$.
Substituting $h' = 2h$ into the equation,we get $r' = \sqrt{2(2h)R} = \sqrt{2} \cdot \sqrt{2hR}$.
Since $r = \sqrt{2hR}$,we have $r' = \sqrt{2} r$.

(D) Ground wave propagation is a method of radio wave propagation that uses the area between the surface of the Earth and the ionosphere for waveguide.
Ground waves are typically used for low-frequency and medium-frequency communication.
However,the specific range mentioned in the options refers to the wavelength range used for ground-based communication systems,which is typically between $10^{-3}\, m$ and $0.1\, m$ for specific microwave applications or line-of-sight communication.

(B) For $100\%$ modulation,the modulation index $m_a = 1$.
The total power radiated in an $AM$ wave is given by $P_t = P_c (1 + \frac{m_a^2}{2})$,where $P_c$ is the carrier power.
The useful power is the power contained in the sidebands,given by $P_{sb} = P_c \frac{m_a^2}{2}$.
The ratio of useful power to total power is $\frac{P_{sb}}{P_t} = \frac{P_c \frac{m_a^2}{2}}{P_c (1 + \frac{m_a^2}{2})} = \frac{m_a^2}{2 + m_a^2}$.
Substituting $m_a = 1$:
$\frac{P_{sb}}{P_t} = \frac{1^2}{2 + 1^2} = \frac{1}{3}$.
Thus,the useful part of the total power radiated is $1/3$ of the total power.

(C) Sky wave propagation relies on the reflection of radio waves by the ionosphere. The ionosphere can only reflect electromagnetic waves with frequencies up to approximately $30\, MHz$. For frequencies greater than $30\, MHz$,the waves penetrate the ionosphere and escape into space instead of being reflected back to Earth. Therefore,sky wave propagation is not possible for frequencies greater than $30\, MHz$.

(D) Audible frequency waves have very low frequencies and cannot be transmitted over long distances directly because they require antennas of impractical sizes and suffer from high attenuation.
To transmit these signals,they are superimposed (modulated) onto high-frequency carrier waves.
Radio-frequency waves are ideal for this purpose because they have short wavelengths (on the order of $100\, m$) and high energy,allowing them to travel long distances through space.
Therefore,e.m. waves of audible frequency are superimposed with radio-frequency waves.

(N/A) Ground waves: These are radio waves that travel along the surface of the Earth. They are typically used for low-frequency and medium-frequency communication. Because they follow the curvature of the Earth,they can travel beyond the horizon,but they are heavily attenuated by the Earth's surface.
$(b)$ Trade winds: These are prevailing patterns of easterly surface winds found in the tropics,within the lower portion of the Earth's atmosphere,in the lower section of the troposphere near the Earth's equator. They blow from the subtropical high-pressure belts towards the equatorial low-pressure belt.

(B) transducer is an electronic device that converts one form of energy into another.
Common examples include microphones (which convert sound energy into electrical energy) and loudspeakers (which convert electrical energy into sound energy).
They are essential components in communication systems and sensors.