371 + Interview Questions in COMMUNICATION-SYSTEMS IN GENERAL AWARENESS Page 3 InterviewSolution

101.	Distinguish between "Point to Point" and "Broadcast" modes of communication.
Answer» Point to point mode of communication is a communication over a link between a single transmitter and receiver. Example of such a system is a telephone. Broad case mode of communication is a communication in which large number of receivers are linked to a single transmitter. Examples of such a system are radio, television.

Discussion

102.	What is meant by data retrival?
Answer» Data retrieval is recovering the original data form modulated wave.

Discussion

103.	The frequency of amplitude modulated wave is equal toA. Frequency of carrier wave `(f_c)`B. Frequency of modulating signal `(f_m)`C. `(f_c+f_m)/2`D. `(f_c-f_m)/2`
Answer» Correct Answer - A

Discussion

104.	What are radiowaves?
Answer» The electromagnetic waves of frequency ranging from a few kilo hertz to about few hundred mega hertz (i.e., wavelength 0.3 m and above) are called radio waves.

Discussion

105.	Name the two basic modes of communication system.
Answer» Two basic modes of communication are (i) Ground wave propagation (ii) Sky wave propagation.

Discussion

106.	What does a modem represent?
Answer» A modem is a device that modulates and demodulates.

Discussion

107.	An audio signal of amplitude 0.1 V is used is used in amplitude modulation of a carrier wave of amplitude 0.2 V. Calculate the modulation index.
Answer» Correct Answer - 0.5 Here, `A_m = 0.1 V, A_c = 0.2 V, mu = ? ` `mu = A_(m)/A_(c) = 0.1/0.2 = 0.5` .

Discussion

108.	What is the maximum distance, the radiowaves can cover in a single reflection from ionosphere?
Answer» Correct Answer - 4000 km.

Discussion

109.	What is fax?
Answer» Fax is the electronic reproduction of a document at a distance.

Discussion

110.	An audio signal of 0.1 V is used in amplitude modulation of a career wave of amplitude 0.3. The modulation index isA. 3B. `1/3`C. `0.03`D. None of these
Answer» Correct Answer - B

Discussion

111.	In space communication, the sound waves can be sent from one place to anotherA. through spaceB. through wiresC. by superimposing it on undamped electromagnetic wavesD. by superimposing it on damped electromagnetic waves.
Answer» Correct Answer - C

Discussion

112.	What should be the ratio of the critical frequencies for reflection or radiowaves from E and `F_1` layers of the ionosphere. If their electrons densities are `2xx10^11//m^3` and `3xx10^11//m^3` respectively ?A. `sqrt(1/2)`B. `sqrt(2/3)`C. `sqrt(3/2)`D. `sqrt(2/5)`
Answer» Correct Answer - B Critical Frequency `v_c=9 (N_"max")^(1//2)` `therefore v_c prop (N_"max")^(1//2)` `therefore v_E/v_(F_1) =((2xx10^11)^(1//2))/((3xx10^11)^(1//2)) = (2/3)^(1//2) = sqrt2/3`

Discussion

113.	Which one of the following statement is wrong ?A. Radio horizon of the transmitting antenna for space wave is `d_T=sqrt(2 Rh)` , where R is the radius of the earth and h is the height of the transmitting antenna .B. Within the skip distance neither the ground waves nor the sky waves are receivedC. Radiowaves in the frequency range 30 MHz to 60 MHz are called sky wavesD. For the propagation of ground waves horizontal aerials are used
Answer» Correct Answer - C For sky waves the frequencies are between 3 MHz and 30 MHz is a correct statement ( c) is wrong.

Discussion

114.	Write short note on FAX System.
Answer» FAX System: FAX service provides automatic communication through printed words between subscribers in cities and towns provided with Electronic Telephone Exchanges. FAX .subscribers establish their own connections.

Discussion

115.	The frequency of a wave propagating in D-region having refractive index of 0.5 isA. `420` kc/s .B. 300 kc/s.C. `207.33` kc/s.D. none of these
Answer» Correct Answer - C Electron density of D-region is `10^(9) m^(-3)` ` mu = sqrt(1-(81.45N)/v^(2))` ` 0.5 = sqrt(1-(81.45xx10^(9))/v^(2))` ` :. v = 207 .33 kc//s`

Discussion

116.	In space communcation , the information can be passed from on place to another at a distance of 100 km inA. `1 s `B. `0.5` sC. `0.03` sD. none of these
Answer» Correct Answer - D `t = ("distance")/("velocity")` ` = (100 xx 10^(3))/(3 xx 10^(8))` ` = 0.33 " "10^(-3)` sec

Discussion

117.	Radiowaves of constant amplitude can be generated withA. FilterB. RectifierC. FETD. Oscillator
Answer» Correct Answer - D

Discussion

118.	Which of the following statement is wrong?A. Ground wave propagation can be sustained at frequncies 500 kHz to 1500 kHzB. Satellite communication is useful for the frquencies above 30 MHzC. Sky wave propagation is useful in the range of 30 to 40 MHzD. space wave propagation takes place through tropospheric space
Answer» Sky wave propagation is possible for the radiowaves of frequency range from 2 MHz to 30 MHz and not in the range 30 to 40 MHz.

Discussion

119.	The T.V. transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received, (taking the radius of earth to be `6.4xx10^(6)m)` isA. 100 kmB. 60 kmC. 55 kmD. 50 km
Answer» Correct Answer - C

Discussion

120.	What will be the area of reception of the signals transmitted by the TV antenna of length 300 m and radius of earth is 6400 km ?A. `12058 " km"^(2)`B. `300 "km"^(2)`C. `500 " km"^(2)`D. `1258" km"^(2)`
Answer» Correct Answer - A Area converted ` = pi d^(2)` ` = 2 xx 3.14 xx 6400 xx 10^(3) xx 300` ` = 1258 km^(2)`

Discussion

121.	The height of a T. V. tower is 75 m. If we want to double its coverage range, then the height of the T. V. tower should beA. 150 mB. 300 mC. 200 mD. 500 m
Answer» Correct Answer - B `d=sqrt(2Rh) therefore d prop sqrth` `d_2/d_1=sqrt(h_2/h_1)` `therefore h_2/h_1=(d_2/d_1)^2=4` `therefore h_2`=75 x 4 = 300 m

Discussion

122.	The T.V. transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received, (taking the radius of earth to be `6.4xx10^(6)m)` isA. 100 kmB. 60 kmC. 55 kmD. 50 km.
Answer» `d=sqrt(2hR)=sqrt(2xx240xx64xx10^(6))` `=55.426xx10^(3)m=55km`

Discussion

123.	A TV tower has a height of 500 m at a given place. Calculate the coverage range, if the radius of the earth is 6400 km.
Answer» Here, ` h = 500 m , R = 6400 xx 1000m` Coverage range, ` d = sqrt(2hR)` `= sqrt (2xx 500 xx 6400 xx 1000)` ` 8 xx 10^4 m = 80 km`.

Discussion

124.	ATV tower has a height of 100m. How much population is covered by TV broadcast. If the average population density around the tower is `1000 km^(-2)` ? (radius of earth =`6.4xx10^(6)m`)
Answer» `h=100m, R=6.37xx10^(6)`, Average population density `=1000km^(-2)=1000(10^(3))^(-2)m^(-2)=10^(-3)m^(-2)` Distance up to which the transmissionn could be viewed `d=sqrt(2hR)` Total area over which transmisiion could be viewed `pi^(2)=2pihR` Population covered `=10^(-3)xx2pihR=10^(-3)xx2xx3.14xx100xx6.37xx10^(6)=40` lakh

Discussion

125.	ATV tower has a height of 100m. How much population is covered by TV broadcast. If the average population density around the tower is `1000 km^(-2)` ? (radius of earth =`6.4xx10^(6)m`)A. `10^(3)`B. `10^(6)`C. `4xx10^(6)`D. `4xx10^(9)`
Answer» Here, h=100, population density `p=1000km^(-2)` `=(1000)/((1000)^(2)m^(2))=10^(-3)m^(-2)` Radius of the area covered by T.V. Broadcast `d=sqrt(2Rh)` Population covered=Area of rangexxpopulation density `pi d^(2)xxp=pi(2Rh)xxp` `=3.14xx[2xx(6.4xx10^(6))xx100]xx10^(-3)` `=40.12xx10^(5)cong4xx10^(6)`

Discussion

126.	What is the value of frequency at which an e.m. wave must be propagated for the D - region to have a refractive index of 0.49 ? Electron density for D-layer is `10^(9) m^(-3)`.
Answer» Here, `mu = 0.49 , N = 10^9 m^(-3), v = ?` As `mu_r = mu/mu_0 = sqrt(1-(81.45N)/v^2)` `:. 0.49 = sqrt(1-((81.45 xx 10^9))/v^2)` or `0.24 = 1- (81.45 xx 10^9)/v^2` or `v = ((81.45 xx 10^9)/(1-0.24))^(1//2) = 3.27 xx 10^5 Hz` .

Discussion

127.	A T. V. tower has a height of 100 m. How much population is covered by the T.V. broadcast if the average population density around the tower is 1500 `km^(-2)` ? (Radius of earth ` = 6.37 xx 10^(6)`m. )
Answer» Here, ` h = 100 m, R = 6.37 xx 10^(6)m , ` population density, `rho = 1500 km^(-2) = 1500 xx 10^(-6)m^(-2)` Population covered `= rho xx pi d^2 = rho pi 2 h R` ` = 1500 xx 10^(-6) xx 22/7 xx 2 xx 100 xx 6.37 xx 10^6` ` = 6 xx 10^6` .

Discussion

128.	A TV transmitting antenna is 125 m tall. How much service area this transmitting antenna cover, if the receiving antenna is at the ground level? Radius of earth `= 6400 km`.
Answer» Here, `h = 125m`, `R = 6400 km = 6.4 xx 10^6 m` Area covered `= pid^2 = pi xx 2 h R [:. d = sqrt (2hR)]` `= 3.14 xx 2 xx 125 xx 6.4 xx 10^6` `5024 xx 10^6m^2 = 5024 km^2`

Discussion

129.	Why is the communication using line of sight mode limited to frequencies above 40 MHz?
Answer» The electromagnetic waves having frequencies above 40 MHz can not be reflected by ionosphere. Therefore, their propagation through sky wave is not possible. Such high frequency waves are heavily absorbed while progressing over the surface of earth. Therefore, their ground wave propagation is also not possible. The only way by whch such waves can propagate is the line of sight mode.

Discussion

130.	Long distance radio broadcasts use short wave bands. Explain why?
Answer» Because ionosphere easily reflects the waves in this band.

Discussion

131.	Demodulation is the reverse process of modulation, which is performed in a receiver to recover the original modulating signals. At the receiving end, the signals are generally weak due to attenuation in the channel. The receiver selects only the desired signal, rejecting the unwanted (noise) signals and amplifies the same. Read the above passage and answer the following questions: (i) What are the main functions of a demodulator? (ii) Give any two examples of demodulators. (iii) What does the phenomemom of demodulation imply in day to day life?
Answer» (i) The main functions of a demodulator are: (a) selecting the desired signal and rejecting the unwanted signals. (b) demodulating the desired signal and amplifying it. (c) displaying the original modulating signal in a desired manner. (ii) All receivers like television and radio are demodulators. (iii) In day to day life, demodulation is so common. An aeroplane, for example carries us to our destination town. we get off the plane and meet our loved ones. The carrier is to be left behind. In a demodulator, carrier wave is by passed and only the original modulating signal is reproduced.

Discussion

132.	Why is the amplitude of modulating signal kept less than the amplitude of carrier wave ?
Answer» If amplitude of modulating signal `(A_m)` exceeds the amplitude of carrier wave `(A_c)`, the carrier is over modulated `(mu_a gt1)`. This causes distortion during reception.

Discussion

133.	The question has statement - 1 and statement - 2 Of the four choices given after the statements , choose the one that best describes the two statements statement - 1 : Sky wave signals are used for long distance radio communication . These signals are in generel , less stable then ground wave signals statement - 2 : The state of inosphere varies from to hour day and season to season .A. Statement-1 is true, statement-2 is false,B. Statement-1 is true, statement-2 is true, statement-2 is the correct explanation of statement-1C. statement-1 is true, statement-2 is not the correct explanation of statement-1D. statement-1 is false, statement -2 is true
Answer» Correct Answer - D

Discussion

134.	The question has statement - 1 and statement - 2 Of the four choices given after the statements , choose the one that best describes the two statements statement - 1 : Sky wave signals are used for long distance radio communication . These signals are in generel , less stable then ground wave signals statement - 2 : The state of inosphere varies from to hour day and season to season .A. Statement-1 is true: Statement-2 is true, and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Sttement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is false.D. Statement-1 is false, but Statement-2 is true.
Answer» Since the ionospheric properties change with time and season hence sky wave signals are in general less stable than ground wave signals. Thus statement-2 is correct and is a correct explanation of statement-1.

Discussion

135.	Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?A. The side band frequencies are 1506 KHz and 1494 KHzB. The bandwidth required for amplitude modulation is 6 MHzC. The bandwidth required for amplitude modulation is 3 MHzD. The side band frequencies are 1503 KHz and 1497 KHz
Answer» Correct Answer - D Frequency (fc) of carrier signal = 1.5 MHz `therefore f_c = 1.5 xx10^6` Hz = 1500 KHz Frequency of modulating signal (fm) = 3 KHz `therefore` Upper side band fequency (USB) = 1500 + 3 = 1503 KHz and Lower side band frequency (LSB) = 1500- 3 = 1497 KHz and bandwidth =`2f_m=2xx3`= 6 KHz Thus (d) is the correct choice .

Discussion

136.	Rakesh and Rajesh are 8th class students. They are fond of watching cricket match, particularly when it is played between Australia and India. They observed that most of the players, when they are in the field, apply a cream on their face. They did not know its reason. One day they asked this question to their teacher. The teacher thought it to be a good question and explained the reason for applying this cream to the whole class. Based on this paragraph, answer the following questions: (i) In your opinion, what explanation did the teacher offer to the students in the class? (ii) Why is small ozone layer on top of the stratosphere considered crucial for human survival? (iii) Write any two values displayed by Rakesh and Rajesh and their class teacher?
Answer» (i) The sun lotion used by players on their face absorbs a part of ultraviolet radiation coming from sun and thus causing less damage to the skin of players using it. (ii) The small ozone layer present on the top of the stratosphere absorbs a part of ultraviolet radiations emitted by sun and going towards earth. As a result of it, a small portion of ultraviolet radiation of sun can reach the earth. Due to it, there is no harm to human beings. (iii) Rakesh and Rajesh shown curiosity and keeness for knowing the scientific facts. Teachers showed maturity and enjoyed while explaining the scientific facts to all the students.

Discussion

137.	In amplitude modulationA. only the amplitude is changed but frequency remains sameB. both the amplitude and frequency shcange equallyC. both the amplitude and frequency change unequallyD. none of the above
Answer» Correct Answer - A

Discussion

138.	Modulation factor determines-A. only the strength of the transmitted signalB. only the quality of the transmitted signalC. both the strength and quality of the signalD. none of the above
Answer» Correct Answer - C

Discussion

139.	A bandwidth of 6MHz is available for A.M. transmission . The maximum frequency of the audio signals used for modulating the carrier wave is 5 KHz. How many stations can be broadcasted simultaneously within this band without interferring with each other ?A. 400B. 500C. 600D. 700
Answer» Correct Answer - C Bandwidth for each station = 2 x maximum frequency of audio signal = 2 x 5 KHz = 10 KHz = `10xx10^3` Hz = `10^4` Hz `therefore` Available bandwidth =6 MHz = `6xx10^6` Hz `therefore` No. of stations that can be broadcast within the bandwidth of 6 MHz =`(6xx10^6)/10^4`=600

Discussion

140.	A band width of 5 MHz is available for AM transmission. If the maximum audio signal frequency used for modulating the carrier is not to exceed 5 kHz, how many stations can be broad cast within this band simultaneously without interfering with eachother?
Answer» Correct Answer - 500 Band width required for each station `= 2 xx max`. freq. of audio signal `= 2 xx 5 kHz = 10 kHz` Max. number of stations , which can be broadcast within band width of `5 kHz`

Discussion

141.	An antenna is a deviceA. converts electromagnetic energy into radio frequency signalB. converts radio frequency signal into electromagnetic energyC. converts guided electromagnetic waves into free space electromagnetic waves and vice versaD. none of these
Answer» Correct Answer - C

Discussion

142.	For sky wave propagation of a 18 MHz signal, what should be the minimum electron density in ionosphere ?A. `4xx10^(12)m^(-3)`B. `2xx10^(12)m^(-3)`C. `4xx10^(14)m^(-3)`D. `8xx10^(14)m^(-3)`
Answer» `v=9(N)^(1//2)` or `N=(v^(2))/(9^(2))=`((18xx10^(6))^(2))/(81)=4xx10^(12)m^(-3)`

Discussion

143.	The process of changing some characteristic of a carrier wave in accordance with the intensity of the signal is called.A. amplificationB. rectificationC. modulationD. oscillation
Answer» Correct Answer - C

Discussion

144.	For sky wave propagation of a ` 15 MHz` signal what should be the minimum electron density in ionosphere ?
Answer» Correct Answer - `2.78 xx 10^(912) m^(-3)` Here , ` v_(c ) = 15 MHz = 15 xx 10^(6) Hz , N_(max) = ?` As ` v_( c ) = 9 (N_(max))^(1//2)` or `N_(max) = ((v_(c ))/( 9))^(2) = (v_( c)^(2))/( 81) = ((15 xx 10^(6))^(2))/(81)` ` = 2.78 xx 10^(12) m^(-3)`

Discussion

145.	For sky wave propagation of a `10 MHz` signal, what should be the minimum electron density in ionosphere?A. `1-2xx10^(8)m^(-3)`B. `1-2xx10^(12)m^(-3)`C. `1-2xx10^(14)m^(-3)`D. `1-2xx10^(16)m^(-3)`
Answer» As, `v_(c)=9(N_("max"))^(1//2)` or `N_("max")=((v_(c))/(9))^(2)=(v_(c)^(2))/(81) =((10xx10^(6))^(2))/(81)` `=1.2xx10^(12)m^(-3)`

Discussion

146.	Which of the following is/are correct with respect to frequency modulationA. the amplitude of the modulated wave varies as the frequency of carrier waveB. the frequency of the modulated wave varies as the amplitude of the modulating waveC. modulation index of frequency modulation is `mu=("change in frequency")/("original frequency")`D. frequency modulation vary the carrier frequency within some small range about its original value.
Answer» In frequency modulation, the frequency of the modulated wave varies as the amplitude of the modulating wave, modulation index is `("Change in frequencuy")/("Original frequency"), i.e., (Deltaf)/(f)`. In frequency modulation, frequency of carrier wave is varied within small ragne about original frequency.

Discussion

147.	Audio signal cannot be transmitted becauseA. the signal has more noiseB. the signal cannot be amplified for distance communicationC. the transmitting antenna length is very small to designD. the required length of the transmitting antenna is very large and impracticable
Answer» Correct Answer - D For direct transmission of audio signal the length of the antenna will be very large and impracticable

Discussion

148.	Audio signal cannot be transmitted becauseA. the signal has more noiseB. the signal cannot be amplified for distance communicationC. the transmittig antenna length is very small to designD. the transmitting antenna length is very large and impracticable
Answer» Correct Answer - D

Discussion

149.	An audio signal of frequency `v_a` is to be transmitted as an electormagnetic wave (i) directly as such (ii) through its use as modulating signal on a carrier wave of frequency `v_c`. State the ratio of the size of antenna that can properly sense the time variation of the signal.
Answer» As size of antenna `prop (1)/("Frequency")` `:.` Required ratio ` = v_(c ) : v_( a )`

Discussion

150.	An audio signal of 15 kHz frequency cannot be transmitted over long distance without modulation because.A. the size of the required antenna would be at least `5 km` which is not convenientB. the audio signal can not be transmitted through sky wavesC. the size of the required antenna would be at least `20 km` , which is not convenientD. effective power transmitted would be very low , if the size of the antenna is less than ` 5 km`
Answer» Correct Answer - A::B::D Here , `v_(m) = 15 kHz = 15 xx 10^(3) Hz` `lambda_(m) = (c )/( v_(m)) = ( 3 xx 10^(8))/( 15 xx 10^(3)) = (1)/(5) xx 10^(5) m` Size of the antenna required , ` l = ( lambda)/( 4) = (1)/(4) ((1)/(5) xx 10^(5)) = 5 xx 10^(3) m = 5 km` The audio signals are of low frequency waves . They cannot be transmitted through sky waves as they are absorbed by atmosphere. If the size of the antenna is less than ` 5 km` , the effective power transmission would be very low because of deviation from resonance wavelength of wave and antenna length.

Discussion

Explore topic-wise InterviewSolutions in .

Distinguish between "Point to Point" and "Broadcast" modes of communication.

What is meant by data retrival?

The frequency of amplitude modulated wave is equal toA. Frequency of carrier wave `(f_c)`B. Frequency of modulating signal `(f_m)`C. `(f_c+f_m)/2`D. `(f_c-f_m)/2`

What are radiowaves?

Name the two basic modes of communication system.

What does a modem represent?

An audio signal of amplitude 0.1 V is used is used in amplitude modulation of a carrier wave of amplitude 0.2 V. Calculate the modulation index.

What is the maximum distance, the radiowaves can cover in a single reflection from ionosphere?

What is fax?

An audio signal of 0.1 V is used in amplitude modulation of a career wave of amplitude 0.3. The modulation index isA. 3B. `1/3`C. `0.03`D. None of these

In space communication, the sound waves can be sent from one place to anotherA. through spaceB. through wiresC. by superimposing it on undamped electromagnetic wavesD. by superimposing it on damped electromagnetic waves.

What should be the ratio of the critical frequencies for reflection or radiowaves from E and `F_1` layers of the ionosphere. If their electrons densities are `2xx10^11//m^3` and `3xx10^11//m^3` respectively ?A. `sqrt(1/2)`B. `sqrt(2/3)`C. `sqrt(3/2)`D. `sqrt(2/5)`

Write short note on FAX System.

The frequency of a wave propagating in D-region having refractive index of 0.5 isA. `420` kc/s .B. 300 kc/s.C. `207.33` kc/s.D. none of these

In space communcation , the information can be passed from on place to another at a distance of 100 km inA. `1 s `B. `0.5` sC. `0.03` sD. none of these

Radiowaves of constant amplitude can be generated withA. FilterB. RectifierC. FETD. Oscillator

The T.V. transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received, (taking the radius of earth to be `6.4xx10^(6)m)` isA. 100 kmB. 60 kmC. 55 kmD. 50 km

What will be the area of reception of the signals transmitted by the TV antenna of length 300 m and radius of earth is 6400 km ?A. `12058 " km"^(2)`B. `300 "km"^(2)`C. `500 " km"^(2)`D. `1258" km"^(2)`

The height of a T. V. tower is 75 m. If we want to double its coverage range, then the height of the T. V. tower should beA. 150 mB. 300 mC. 200 mD. 500 m

The T.V. transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received, (taking the radius of earth to be `6.4xx10^(6)m)` isA. 100 kmB. 60 kmC. 55 kmD. 50 km.

A TV tower has a height of 500 m at a given place. Calculate the coverage range, if the radius of the earth is 6400 km.

ATV tower has a height of 100m. How much population is covered by TV broadcast. If the average population density around the tower is `1000 km^(-2)` ? (radius of earth =`6.4xx10^(6)m`)

ATV tower has a height of 100m. How much population is covered by TV broadcast. If the average population density around the tower is `1000 km^(-2)` ? (radius of earth =`6.4xx10^(6)m`)A. `10^(3)`B. `10^(6)`C. `4xx10^(6)`D. `4xx10^(9)`

What is the value of frequency at which an e.m. wave must be propagated for the D - region to have a refractive index of 0.49 ? Electron density for D-layer is `10^(9) m^(-3)`.

A T. V. tower has a height of 100 m. How much population is covered by the T.V. broadcast if the average population density around the tower is 1500 `km^(-2)` ? (Radius of earth ` = 6.37 xx 10^(6)`m. )

A TV transmitting antenna is 125 m tall. How much service area this transmitting antenna cover, if the receiving antenna is at the ground level? Radius of earth `= 6400 km`.

Why is the communication using line of sight mode limited to frequencies above 40 MHz?

Long distance radio broadcasts use short wave bands. Explain why?

Why is the amplitude of modulating signal kept less than the amplitude of carrier wave ?

In amplitude modulationA. only the amplitude is changed but frequency remains sameB. both the amplitude and frequency shcange equallyC. both the amplitude and frequency change unequallyD. none of the above

Modulation factor determines-A. only the strength of the transmitted signalB. only the quality of the transmitted signalC. both the strength and quality of the signalD. none of the above

A bandwidth of 6MHz is available for A.M. transmission . The maximum frequency of the audio signals used for modulating the carrier wave is 5 KHz. How many stations can be broadcasted simultaneously within this band without interferring with each other ?A. 400B. 500C. 600D. 700

A band width of 5 MHz is available for AM transmission. If the maximum audio signal frequency used for modulating the carrier is not to exceed 5 kHz, how many stations can be broad cast within this band simultaneously without interfering with eachother?

An antenna is a deviceA. converts electromagnetic energy into radio frequency signalB. converts radio frequency signal into electromagnetic energyC. converts guided electromagnetic waves into free space electromagnetic waves and vice versaD. none of these

For sky wave propagation of a 18 MHz signal, what should be the minimum electron density in ionosphere ?A. `4xx10^(12)m^(-3)`B. `2xx10^(12)m^(-3)`C. `4xx10^(14)m^(-3)`D. `8xx10^(14)m^(-3)`

The process of changing some characteristic of a carrier wave in accordance with the intensity of the signal is called.A. amplificationB. rectificationC. modulationD. oscillation

For sky wave propagation of a ` 15 MHz` signal what should be the minimum electron density in ionosphere ?

For sky wave propagation of a `10 MHz` signal, what should be the minimum electron density in ionosphere?A. `1-2xx10^(8)m^(-3)`B. `1-2xx10^(12)m^(-3)`C. `1-2xx10^(14)m^(-3)`D. `1-2xx10^(16)m^(-3)`

Audio signal cannot be transmitted becauseA. the signal has more noiseB. the signal cannot be amplified for distance communicationC. the transmitting antenna length is very small to designD. the required length of the transmitting antenna is very large and impracticable

Audio signal cannot be transmitted becauseA. the signal has more noiseB. the signal cannot be amplified for distance communicationC. the transmittig antenna length is very small to designD. the transmitting antenna length is very large and impracticable

An audio signal of frequency `v_a` is to be transmitted as an electormagnetic wave (i) directly as such (ii) through its use as modulating signal on a carrier wave of frequency `v_c`. State the ratio of the size of antenna that can properly sense the time variation of the signal.