Communication Questions in English

(D) In optical communication,the modulation of a light wave involves varying one of its parameters to encode information. The most common method used in optical fiber communication is Intensity Modulation $(IM)$,where the intensity of the light wave is varied in accordance with the modulating signal. Therefore,the characteristic that is modulated is the intensity of the light wave.

(A) The modulation index $\mu$ is defined as $\mu = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}$.
For $100\%$ modulation,the modulation index $\mu = 1$.
Substituting this into the formula: $1 = \frac{E_{max} - E_{min}}{E_{max} + E_{min}}$.
This implies $E_{max} + E_{min} = E_{max} - E_{min}$.
Subtracting $E_{max}$ from both sides,we get $E_{min} = -E_{min}$,which means $2E_{min} = 0$.
Therefore,$E_{min} = 0$.

(A) In satellite communication systems,the signal transmitted from the Earth station to the satellite is called the $up\,\text{link}$. This typically uses a higher frequency,$6\,\text{GHz}$,to overcome atmospheric attenuation.
The signal transmitted from the satellite back to the Earth station is called the $down\,\text{link}$. This typically uses a lower frequency,$4\,\text{GHz}$,to reduce power requirements on the satellite.
Therefore,the $6\,\text{GHz}$ frequency is used for the $up\,\text{link}$ and the $4\,\text{GHz}$ frequency is used for the $down\,\text{link}$.

(B) The audible frequency range for human beings is typically considered to be from $20 \, Hz$ to $20 \, kHz$.
Therefore,the bandwidth of an audio signal is $20 \, kHz - 20 \, Hz \approx 20 \, kHz$.
This range is standard for audio communication systems.

(C) Signal attenuation in an optical fiber refers to the loss of optical power as light travels through the fiber.
The primary mechanisms responsible for this attenuation are:
$1$. Absorption: This occurs due to the presence of impurities in the glass (like $OH^-$ ions) and the intrinsic properties of the material,which convert light energy into heat.
$2$. Scattering: This is primarily caused by Rayleigh scattering,which arises from microscopic variations in the density and refractive index of the glass material during the manufacturing process.
Therefore,both absorption and scattering are the main causes of signal attenuation in optical fibers.

(C) The frequency range for ground wave propagation is typically up to $2 \text{ MHz}$.
For sky wave propagation, the frequency range is generally between $3 \text{ MHz}$ and $30 \text{ MHz}$.
Since the given frequency is $10 \text{ MHz}$, which falls within the $3 \text{ MHz}$ to $30 \text{ MHz}$ range, the mode of propagation is sky wave propagation.
Therefore, the correct option is $C$.

(B) In frequency modulation $(FM)$,the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal (message signal).
While the frequency of the carrier wave changes,its amplitude remains constant.
Therefore,the correct statement is that the frequency of the carrier wave varies according to the amplitude of the modulating signal.

(D) Radio waves can propagate from one place to another using different modes depending on their frequency range:
$1$. Ground wave propagation: Used for low frequencies (up to $2 \text{ MHz}$), where waves follow the curvature of the Earth.
$2$. Sky wave propagation: Used for medium and high frequencies ($3 \text{ MHz}$ to $30 \text{ MHz}$), where waves are reflected by the ionosphere.
$3$. Space wave propagation: Used for very high frequencies (above $30 \text{ MHz}$), where waves travel in a straight line (line-of-sight) between transmitting and receiving antennas.
Since all three methods are standard modes of radio wave propagation, the correct answer is $D$.

(C) Coaxial cables are widely used in communication systems for transmitting high-frequency signals.
Due to their construction,they offer a much higher bandwidth compared to twisted pair cables.
The typical bandwidth of a coaxial cable system is approximately $250 \ MHz$ to $750 \ MHz$.
Among the given options,$250 \ MHz$ is the standard value associated with coaxial cable bandwidth in physics textbooks.

(C) The modulation index $\beta$ for Frequency Modulation $(FM)$ is defined as $\beta = \frac{\Delta f}{f_m}$,where $\Delta f$ is the frequency deviation and $f_m$ is the modulation frequency.
From this formula,it is clear that the modulation index $\beta$ is inversely proportional to the modulation frequency $f_m$.
Therefore,$\beta \propto \frac{1}{f_m}$.

(C) Optical fiber communication offers significant advantages over traditional metallic cables like two-wire lines or coaxial cables.
$1$. $Large$ $bandwidth$: Optical fibers use light waves (high frequency), which allow for a much higher data transmission capacity compared to electrical signals in copper wires.
$2$. $Low$ $transmission$ $loss$: The attenuation (signal loss) in optical fibers is extremely low, allowing signals to travel over long distances without the need for frequent repeaters.
Therefore, the correct combination is large bandwidth and low transmission loss.

(C) The $UHF$ (Ultra High Frequency) range lies between $300 \ MHz$ and $3000 \ MHz$.
These frequencies are not reflected by the ionosphere (sky waves) and are not suitable for ground wave propagation due to high absorption by the Earth's surface.
Therefore,$UHF$ waves are transmitted via space wave propagation (line-of-sight communication),which involves direct transmission between the transmitting and receiving antennas.

(C) In radio broadcasting,modulation is used to transmit sound waves over long distances.
Generally,both Amplitude Modulation $(AM)$ and Frequency Modulation $(FM)$ are used in radio broadcasting.
However,in the context of standard radio broadcasting,$AM$ (Amplitude Modulation) is historically and technically the primary method used for long-range transmission.
Therefore,the correct option is $C$.

(B) Modulation is primarily used for three reasons:
$1$. To reduce the size of the antenna: An antenna size should be of the order of $\lambda/4$. For low-frequency audio signals,$\lambda$ is very large,requiring an impractically large antenna. Modulation shifts the signal to a higher frequency,reducing $\lambda$ and thus the antenna size.
$2$. To reduce the fractional bandwidth: The ratio of signal bandwidth to carrier frequency is reduced,which makes the design of the antenna and the tuning circuit easier.
$3$. To allow for multiplexing: By modulating different signals to different carrier frequencies,multiple signals can be transmitted simultaneously over the same channel.
Option $B$ is incorrect because modulation does not reduce the time interval (propagation delay) between the transmitter and receiver; this is determined by the speed of the signal (usually the speed of light) and the distance between them.

(B) Space wave propagation involves the transmission of electromagnetic waves directly from the transmitting antenna to the receiving antenna,typically through line-of-sight communication.
This mode of propagation is effective for frequencies above $30 \ MHz$.
Specifically,it is used for $VHF$ $(30-300 \ MHz)$,$UHF$ $(300-3000 \ MHz)$,and $SHF$ $(3-30 \ GHz)$ ranges.
Among the given options,$VHF$,$UHF$,and $SHF$ all utilize space wave propagation. However,in standard physics curriculum contexts,$VHF$ is the most commonly cited range for this mode of propagation in introductory problems. Given the options,$VHF$ is the primary range associated with this phenomenon.

(B) signal is a time-varying quantity that carries information.
Lightning flashes (option $A$) and radiations from the Sun (option $C$) are natural phenomena.
Fluorescent tube light flashes (option $B$) are generated by human-made devices using electrical circuits,making them man-made signals.
Therefore,the correct option is $B$.

(B) Computers process and store information in the form of binary digits ($0$ and $1$).
Therefore,the data exchange between two computers is carried out using digital signals.
Digital signals represent data as a sequence of discrete values,which is the standard method for computer communication.

(D) The characteristic impedance $(Z_0)$ of a coaxial cable is determined by its physical dimensions and the dielectric constant of the insulating material between the inner and outer conductors.
Standard coaxial cables used in telecommunications and radio frequency applications are designed to have a characteristic impedance typically ranging from $50$ $\Omega$ to $75$ $\Omega$.
Specifically,$50$ $\Omega$ is the standard for data transmission and radio frequency equipment,while $75$ $\Omega$ is the standard for video and television signals.
Therefore,the range $50$ $\Omega$ to $70$ $\Omega$ is the most accurate choice among the given options.

(B) The efficiency of an antenna in radiating electromagnetic waves depends on its size relative to the wavelength of the signal. For effective transmission,the antenna length $L$ should be comparable to the wavelength $\lambda$ of the signal (e.g.,$L = \lambda/4$). Since $\lambda = c/f$,where $c$ is the speed of light and $f$ is the frequency,a higher frequency $f$ results in a smaller wavelength $\lambda$. Smaller wavelengths allow for smaller,more practical antenna sizes. Therefore,high-frequency waves can be transmitted more efficiently.

(C) The process of superimposing a low-frequency information signal (baseband signal) onto a high-frequency carrier wave is known as modulation.
This process is essential for efficient transmission of signals over long distances,as high-frequency waves can travel further and require smaller antenna sizes.
Therefore,the correct option is $C$.

(B) tuning fork produces a continuous sound wave that varies smoothly with time.
Since the amplitude and frequency of the sound wave change continuously,it represents an analog signal.
Digital signals,on the other hand,are discrete and represent data as a sequence of values (usually $0$ and $1$).
Therefore,the sound produced by a tuning fork is an analog signal.

(B) The refractive index $(\mu)$ of the ionosphere depends on the electron density $(N)$ and the frequency $(f)$ of the radio wave. The relationship is given by $\mu = \sqrt{1 - \frac{81N}{f^2}}$.
As we move from the lower layer to the upper layer of the ionosphere, the electron density $(N)$ increases.
Since $\mu = \sqrt{1 - \frac{81N}{f^2}}$, an increase in $N$ leads to a decrease in the value of $\mu$.
Therefore, the refractive index of the ionosphere decreases as we move from the lower layer to the upper layer.

(C) In standard television transmission systems,the audio signal is modulated using Frequency Modulation $(FM)$ to ensure high-quality sound and reduce noise interference. The video signal is modulated using Amplitude Modulation $(AM)$ (specifically Vestigial Sideband Modulation,which is a form of $AM$) to efficiently transmit the large bandwidth required for visual information. Therefore,the audio signal is frequency modulated and the video signal is amplitude modulated.

(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 $(A_c)$.
Mathematically, $\mu = \frac{A_m}{A_c}$.
Given that the amplitude of the modulated signal is equal to the amplitude of the carrier wave, we have $A_m = A_c$.
Therefore, $\mu = \frac{A_m}{A_m} = 1$.
To express this as a percentage, we multiply by $100$.
Thus, the modulation index is $1 \times 100 = 100\%$.

(B) In a communication system,the receiver section often receives unwanted noise or signals along with the desired signal. $A$ filter circuit is specifically designed to allow certain frequencies to pass while attenuating or reducing the magnitude of unwanted signals or noise. Therefore,a filter circuit is used to reduce the magnitude of unwanted signals in the receiver section.

(B) Sky wave propagation involves the reflection of radio waves by the ionosphere back to the Earth's surface.
This mode of propagation is effective for frequencies in the range of approximately $2 \text{ MHz}$ to $30 \text{ MHz}$.
Frequencies below $2 \text{ MHz}$ are absorbed by the ionosphere,while frequencies above $30 \text{ MHz}$ (like those used in space wave propagation) penetrate the ionosphere and escape into space.
Therefore,the range $2 \text{ MHz}$ to $20 \text{ MHz}$ falls within the suitable band for sky wave propagation.

(D) In amplitude modulation,the carrier wave frequency $(f_c)$ must be significantly higher than the modulating signal frequency $(f_m)$ to ensure efficient transmission and to avoid overlapping of sidebands.
Given,$f_m = 3 \ kHz = 3000 \ Hz$.
Comparing the given options,$30 \ Hz$,$300 \ Hz$,and $3000 \ Hz$ are either less than or equal to the signal frequency.
$3 \ MHz = 3000 \ kHz$ is much greater than $3 \ kHz$.
Therefore,the carrier frequency must be $3 \ MHz$.

(C) The process of superimposing a low-frequency message signal (information) onto a high-frequency carrier wave is called $Modulation$. This is done to make the signal suitable for long-distance transmission,as high-frequency waves can travel much further and require smaller antenna sizes. Therefore,the correct option is $C$.

(C) coaxial cable is a type of transmission line that consists of an inner conductor surrounded by a tubular insulating layer,surrounded by a tubular conducting shield.
Due to its construction,it offers a much higher bandwidth compared to twisted pair cables.
The typical bandwidth of a coaxial cable is approximately $750 \ MHz$.
Therefore,the correct option is $C$.

(C) In broadcasting,$AM$ (Amplitude Modulation) is widely used primarily because it allows for the design of very simple and inexpensive receivers. The detection of an $AM$ signal can be achieved using a simple envelope detector (like a diode and an $RC$ circuit),which significantly reduces the cost and complexity of the receiver compared to $FM$ or other complex modulation schemes. Therefore,option $C$ is the correct reason.

(B) $Transducer$ is an electronic device that converts energy from one form to another. Common examples include microphones (converting sound energy to electrical energy) and loudspeakers (converting electrical energy to sound energy). Therefore,the correct option is $B$.

(D) For ground wave propagation,the electric field vector of the electromagnetic wave must be perpendicular to the surface of the Earth to avoid short-circuiting the electric field. Since the electric field is vertical,the antenna must be a vertical antenna. Therefore,transmitting antennas are generally of the vertical type.

(D) Audio signals have very low frequencies (typically $20 \ Hz$ to $20 \ kHz$).
The length of the antenna $(L)$ must be of the order of the wavelength $(\lambda)$ of the signal,i.e.,$L \approx \lambda / 4$.
The wavelength is given by $\lambda = c / f$,where $c$ is the speed of light $(3 \times 10^8 \ m/s)$.
If we take a frequency of $15 \ kHz$,then $\lambda = (3 \times 10^8) / (15 \times 10^3) = 20,000 \ m = 20 \ km$.
Therefore,the required antenna length would be $L = 20,000 / 4 = 5,000 \ m = 5 \ km$.
Constructing an antenna of such a large length is not practically feasible,hence audio signals cannot be transmitted directly.

(B) The maximum range of a radar is given by the radar range equation:
$R_{max} = \left( \frac{P_t A^2 S}{4\pi \lambda^2 P_{min}} \right)^{1/4}$
Where:
$P_t$ = Maximum transmitted power
$A$ = Effective area of the receiving antenna
$S$ = Radar cross-section of the target
$\lambda$ = Wavelength of the radar signal
$P_{min}$ = Minimum detectable power of the receiver
From the equation,it is clear that $R_{max} \propto (P_t)^{1/4}$.
Therefore,the maximum range is proportional to the fourth root of the maximum transmitted power.

(A) microphone is a transducer that converts sound energy (mechanical waves) into electrical energy. The electrical signal produced by a microphone varies continuously with time in accordance with the sound waves. Such a signal,which varies continuously with time,is known as an analog signal. Therefore,the output of a microphone is an analog signal.

(D) Radio waves can be transmitted from one place to another using different modes of propagation depending on their frequency range.
$1$. Ground wave propagation is used for low-frequency signals (up to $2 \text{ MHz}$).
$2$. Sky wave propagation is used for medium and high-frequency signals (between $3 \text{ MHz}$ and $30 \text{ MHz}$) by reflecting them off the ionosphere.
$3$. Space wave propagation is used for very high-frequency $(VHF)$ and ultra-high-frequency $(UHF)$ signals (above $30 \text{ MHz}$) using line-of-sight communication.
Since all three methods are used for radio wave transmission, the correct answer is $D$.

(B) The bandwidth required for a video signal to transmit moving pictures is approximately $4.2 \text{ MHz}$. This is a standard value used in television broadcasting to ensure high-quality image transmission.

(C) Sound waves are low-frequency waves (typically $20 \ Hz$ to $20 \ kHz$).
For efficient transmission of electromagnetic waves,the antenna length should be at least $\lambda/4$,where $\lambda$ is the wavelength.
For a frequency of $15 \ kHz$,the wavelength $\lambda = c/f = (3 \times 10^8 \ m/s) / (15 \times 10^3 \ Hz) = 20 \ km$.
The required antenna height would be $\lambda/4 = 5 \ km$,which is practically impossible to construct.
Furthermore,low-frequency waves are heavily absorbed by the atmosphere.
Therefore,sound waves are modulated onto high-frequency carrier waves for transmission.

(A) In traditional radio and television broadcasting,information is transmitted using analog signals. An analog signal is a continuous wave that changes over a time period. While modern systems are transitioning to digital,the fundamental definition for standard radio and television broadcasting in the context of physics curricula refers to analog signals.

(D) In amplitude modulation, the carrier frequency $(f_c)$ must be significantly higher than the modulating (audio) frequency $(f_m)$ to ensure efficient transmission and signal processing.
Typically, the carrier frequency is chosen to be at least $100$ times the modulating frequency.
Given $f_m = 500 \, \text{Hz}$.
Among the given options, $50,000 \, \text{Hz}$ $(50 \, \text{kHz})$ is the only value that satisfies the condition $f_c \gg f_m$.
Therefore, the correct carrier frequency is $50,000 \, \text{cycles/sec}$.

(D) For an antenna of length $l$ radiating a signal of wavelength $\lambda$,the power radiated $P$ is given by the relation $P \propto (l/\lambda)^2$.
Since the wavelength $\lambda$ is constant for a given frequency,the radiated power $P$ is directly proportional to the square of the length of the antenna,i.e.,$P \propto l^2$.
Thus,the correct option is $D$.

(D) In amplitude modulation $(AM)$,the amplitude of the high-frequency carrier wave is varied in accordance with the instantaneous value of the low-frequency modulating signal (or message signal).
Therefore,the amplitude of the carrier wave changes in proportion to the instantaneous value of the modulating wave.

(A) The formula for Maximum Usable Frequency $(MUF)$ is given by: $\text{MUF} = f_c \sec(i)$,where $f_c$ is the critical frequency and $i$ is the angle of incidence.
Given: $f_c = 50 \text{ MHz}$ and $i = 74^{\circ}$.
$\text{MUF} = 50 \times \sec(74^{\circ})$.
Since $\cos(74^{\circ}) \approx 0.2756$,then $\sec(74^{\circ}) = \frac{1}{\cos(74^{\circ})} \approx \frac{1}{0.2756} \approx 3.628$.
$\text{MUF} = 50 \times 3.628 = 181.4 \text{ MHz}$.
Rounding to the nearest provided option,the correct value is $181.5 \text{ MHz}$.

(D) Optical fibers are widely used in modern communication systems due to their extremely high bandwidth capacity.
They operate in the frequency range of light,which is very high compared to radio waves or microwaves.
The bandwidth of an optical fiber is typically in the range of $100 \ GHz$ or even higher,allowing for the transmission of massive amounts of data.
Therefore,the correct option is $D$.

(D) basic communication system consists of three essential elements: the transmitter,the transmission channel,and the receiver.
Noise is an external,unwanted signal that interferes with the communication process,but it is not considered a functional component of the system itself.

(C) The range $d$ of a transmitting antenna of height $h$ is given by the formula: $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth.
From this relation,we can see that $d \propto \sqrt{h}$.
If the range is to be doubled,let the new range be $d' = 2d$ and the new height be $h'$.
Then,$d' = \sqrt{2Rh'} \implies 2d = \sqrt{2Rh'}$.
Squaring both sides,we get $4d^2 = 2Rh'$.
Since $d^2 = 2Rh$,we substitute this into the equation: $4(2Rh) = 2Rh'$.
$8Rh = 2Rh' \implies h' = 4h$.
Therefore,the height of the transmitting antenna must be quadrupled.

(A) Commercial $FM$ (Frequency Modulation) radio broadcasting operates in the Very High Frequency $(VHF)$ band.
According to standard communication physics, the frequency range allocated for commercial $FM$ radio is $88 \, \text{MHz}$ to $108 \, \text{MHz}$.
Therefore, the correct option is $A$.

(A) $Twisted\, pair\, wire$ consists of two insulated copper wires twisted together to reduce electromagnetic interference.
It is commonly used for telephone lines and local area networks $(LAN)$.
The typical bandwidth of a $Twisted\, pair\, wire$ is limited,and it is generally used for transmitting signals in the frequency range of $0$ to $5\, MHz$.

(A) Optical communication typically involves the transmission of light signals through optical fibers. An optical fiber acts as a physical path that guides the light from the transmitter to the receiver. Since the signal is confined within a physical structure (the fiber),it is classified as a guided medium.

(D) Optical fibers operate on the principle of total internal reflection.
They are made of glass or plastic and are immune to electromagnetic interference because they transmit light signals rather than electrical signals.
Therefore,the statement that optical fibers can be affected by external electromagnetic interference is false.
All other statements regarding low transmission loss,homogeneous core with cladding,and varying refractive indices are correct characteristics of optical fibers.