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Communication System : Previous Year's Questions


Q.1      What are the three basic units of a communication system?

[Delhi 2008C]

Sol.          The three basic units of communication are transmitter, medium /channel and receives.


Q.2      Draw a block diagram showing the important component in a communication system. What is the function of a transducer?

[Foreign 2011]

Sol. c9


Q.3      What is the meaning of the term attenuation in communication system?

[All India 2008C]

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


Q.4      What is the function of a transmitter in a communication system?

[Foreign 2011]

Sol.         A transmitter processes the incoming message signal, makes it suitable for transmission.


Q.5      What does the term transducer mean in an electronic communication system?

[Delhi 2009C]

Sol.

In electronic communication system a transduces is a device that converts signals (emw) to electrical form or vice-versa.


Q.6       Name the type of communication in which the signal is a discrete and binary coded version of the message or information.

[Delhi 2000]

Sol.          Binary coded signals are in the form of pulse.


Q.7       “A communication satellite is essentially a repeater station in space.” Justify this statement by analyzing the function of a repeater.

[All India 2009 C]

Sol.

A repeater is a combination of a receives and a transmitter. It pickes up signals, amplifies and retransmits it. A satellite also receives signal, amplifies at and retransmit it to ground station. Thus the given statement is justified.


Q.8     What is the range of frequencies used for TV transmission? What is common between these waves and light waves?

[Delhi 2010]

Sol.

Range of frequencies for T.V. transmission are 54 MHz to 890 MHz The common feature between these waves and light waves is that both travel at the same speed and both are electromagnetic waves.


Q.9      Name the mode of propagation of radio waves which travel in a straight line from the transmitting antenna to the receiving antenna.

[All India 2008C]

Sol.          Space Waves


Q.10     Name any two types of transmission media that are commonly used for transmission of signals. Write the range of frequencies of signals for which these transmission media are used.

[All India 2010C]

Sol.

Sky waves and ground waves.
Sky waves – Range few MHz up to 30 to 40 MHz
Ground waves - < few MHz


Q.11      What is the ground wave communication? On what factors does the maximum range of propagation in this mode depend?

[All India 2011]

Sol.

When the waves propagates near to the surface, the waves glide over the surface of the earth, they are called ground waves. The maximum range of coverage depends on the transmitted power and frequency.


Q.12     Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz upto 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode? Why is there an upper limit to frequency of waves used in this mode?

[All India 2010]

Sol.       Sky waves. c8

The ionosphere acts as a reflector for a certain range of frequencies (3 to 30 MHz). Waves above 30 MHz penetrate the ionosphere and escape.


Q.13     (a) Why is communication using line of sight mode limited to frequencies above 40 MHz?
        
(b) A transmitting antenna at the top of a tower has a height 32 in and the height of the receiving antenna is 50 m. What is the maximum distance of them for satisfactory communication in line of
sight mode?

[Delhi 2010]

Sol.

(a) Line of sight mode is limited above 40 MHz, as these waves cannot propagate as sky waves.

(b) Maximum distance {d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}}
 = \sqrt {2 \times 64 \times {{10}^5} \times 32} + \sqrt {2 \times 64 \times {{10}^5} \times 50} \,m
 = 64\, \times {10^2} \times \sqrt {10} + 8 \times {10^3} \times \sqrt {10} \,m
 = 144 \times {10^2} \times \sqrt {10} \,m
 = 45.5\,km


Q.14     Distinguish between sky wave and space wave propagation. Explain brief, with the help of suitable diagram, how space waves are used for line of sight communication?

[Delhi 2009C]

Sol.

Sky Waves : Waves that propagate by reflection from ionosphere. Waves within the frequency range 3 MHz to 30 MHz can be transmitted in this way. Higher frequencies cannot be transmitted.

Space Waves :- A space wave travels in a straight line from the transmitting antennas to the receiving antennas. Frequencies above 40 MHz can be transmitted in this way. The range is limited by the curvature of the the earth.

9

Maximum range {d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_r}}
(where R is the radius of earth, {h_T} the height of transmitted antenna and {h_R} the height of the receiving antenna.)


Q.15     Distinguish between sinusoidal and pulse shaped signals.

[All India 2009C]

Sol.

When the signal is in the form of continuous variation of amplitude and can be writes in the sinusoidal form, it is called a sinusoidal signal. When the signal is in the form of discrete variations or pulse, the it is called a pulse signal.

c6


Q.16     What is the length of a dipole antenna to transmit signals of frequency 200 MHz?

[Delhi 2000C]

Sol.

The length of the antenna should be atleast {\lambda \over 4}.
For a carrier frequency 5 \times {10^8}\,Hz
\lambda = {C \over \upsilon } = {{3 \times {{10}^8}} \over {5 \times {{10}^8}}} = 0.6\,m
Therefore Length of antenna is  = {{0.6} \over 4} = 0.15\,m


Q.17     Why is frequency modulation preferred over amplitude modulation for transmission of music?

[Delhi 2007]

Sol.

In frequency modulation, naturally occurring noise is reduced, whereas is amplitude modulation, due to attenuation the signal is distorted and causes noise or disturbance in the signal.


Q.18      What is space wave communication? Write the range of frequencies suitable for space wave communication.

[All India 2011]

Sol.

When waves travel direct from the transmitter to the receiver. It is called space wave communications. The range suitable is above 40 MHz.


Q.19      Write two factors justifying the need of modulating a signal. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ?

[All India 2010]

Sol.

(a) Size of Antenna : For an electromagnetic waves of 20 KHz, the size of antenna is 15 km. This is not possible to construct and operate. So there is a need to translate law frequency into high frequency before transmissions.

(b) Effective power radiated : Power radiate \alpha \,{1 \over {{\lambda ^2}}}. It means power radiated increases with decreasing wavelength or increasing frequency. Therefore high frequency carrier waves are needed.


Q.20     Give reasons for the following :
            
(a) For ground wave transmission, size of antenna (1) should be comparable to wavelength (l) of signal i,e.,= l/ 4.
            
(b) Audio signals, converted into an electromagnetic wave, are not directly transmitted.
             (c) The amplitude of a modulating signal is kept less than the amplitude of carrier wave.

[Delhi 2011C]

Sol.

(a) For high efficiency of signal radiation, the antennas should have a size of atleast {\lambda \over 4}.
(b)
Audio signal if converted to electromagnetic signal will require antenna of atleast size 15 km, which is impractical and signals of different transmitter would mix up.

(c ) The amplitude of the modulating signal is kept less than the carrier waves so that no distortion occurs in the modulated wave.


Q.21    In standard AM broadcast, what mode of propagation is used for transmitting a signal? Why is this mode of propagation limited to frequencies upto a few MHz?

[Foreign 2010]

Sol.

In standard AM broadcast , ground wave is used as mode of propagation. This mode is limited up to a few MHz, because the attenuation of surface waves increases very rapidly with increase in frequency.


Q.22   A transmission antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is 49 m. What is the maximum distance between them, for satisfactory communication in the line of sight mode? (Radius of earth = 6400 km)

[Delhi 2008]

Sol.

{d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}}
 = \sqrt {2 \times 6400 \times {{10}^3} \times 36} + \sqrt {2 \times 6400 \times {{10}^3} \times 49}
 = 1040\sqrt {20}
 = 4648.8\,m

 = 4.65\,km


Q.23     Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram, the mode of propagation used in the frequency range from a few MHz to 40 MHz.

[Delhi 2012]

Sol.

The three modes of propagation are.
(i) Ground wave
(ii) Space wave
(iii) Sky wave
Mode of propagations used in the frequency range of few MHz to 40 MHz is sky wave.
Sky waves: 8

The ionosphere acts as a reflector for a certain range of frequencies (3 to 30 MHz). Waves above 30 MHz penetrate the ionosphere and escape.


Q.24    Draw a schematic diagram showing the (a) Ground wave (b) Sky wave (c) Space wave propagation modes for electromagnetic waves.

[All India 2011C]

Sol.7  


 

Q.25    Consider an optical communication system operating at A. = 800 nm. Suppose, only 1 % of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting. -
           
(a) audio-signals requiring a bandwidth of 8 kHz?
            (b) video TV signals requiring an approximate band-width of 4.5'MHz? Support your answer with suitable calculations.

[All India 2006]

Sol.

Wavelength of carrier wave 800 nm.
Therefore frequency  = {C \over \lambda } = {{3 \times {{10}^8}} \over {800 \times {{10}^{ - 9}}}} = {3 \over 8} \times {10^{15}}
If 1% is a variable, the 1% of above.
\upsilon ' = {3 \over 8} \times {10^{15}} \times {1 \over {100}} = {3 \over 8} \times {10^{13}}.\,Hz = 3.75 \times {10^{12}}\,Hz

(a) Band width = 8000 Hz.
No. of channels = {{\upsilon '} \over {Band\,width}}
 = {{3.75 \times {{10}^{12}}} \over {8000}}  = 4.7 \times {10^8}

(b) Band width = 4.5 MHz
No. of channels = {{\upsilon '} \over {Band\,width}} = {{3.75 \times {{10}^{12}}} \over {4.5 \times {{10}^6}}} = 8.33 \times {10^5}


Q.26    Why do we need a higher Bandwidth for transmission of music compared to that for commercial telephonic communication ?

[Delhi 2009]

Sol.

Higher bandwidth is required for transmission of music because of the high frequencies produced by the musical instruments of the range 20 Hz – 20 KHz. Speech signals range from 300 Hz - 3100 Hz.


Q.27    Distinguish between frequency modulation and amplitude modulation. Why is an FM signal less susceptible to noise than an AM signal?

[All India 2006,2007D]

Sol. c4

In frequency modulation the frequency of the carrier wave is modulated. In amplitude modulation, amplitude of the carrier wave is modulated. Due to attenuation, amplitude decrease affecting the amplitude modulation. Thus AM is more susceptible to noise than FM.


Q.28    Write briefly any two factors which demonstrate the need for modulating a signal. Draw, a suitable diagram to show amplitude modulation using a sinusoidal signal as the modulating signal.

[All India 2011]

Sol.

(a) Size of Antenna : For an electromagnetic waves of 20 KHz, the size of antenna is 15 km. This is not possible to construct and operate. So there is a need to translate law frequency into high frequency before transmissions.

(b) Effective power radiated : Power radiate \alpha \,{1 \over {{\lambda ^2}}}. It means power radiated increases with decreasing wavelength or increasing frequency. Therefore high frequency carrier waves are needed.

5


Q.29     State the two main reasons explaining the need of modulation for transmission of audio signals.

c13
The diagrams, given above show a carrier wave c(t), that is to be (amplitude) modulated by a modulating signal m(t). Draw the general shape of the resulting AM wave. Define its modulation index.

[All India 2010C]

Sol.

(a) Size of Antenna : For an electromagnetic waves of 20 KHz, the size of antenna is 15 km. This is not possible to construct and operate. So there is a need to translate law frequency into high frequency before transmissions.

(b) Effective power radiated : Power radiate \alpha \,{1 \over {{\lambda ^2}}}. It means power radiated increases with decreasing wavelength or increasing frequency. Therefore high frequency carrier waves are needed.


Q.30    What is modulation? Explain the need of modulating a low frequency information signal. With the help of diagrams, differentiae be r twegn PAM and PDM.

[All India 2006C,2007,2007D,2009D,2010F]

Sol.

(a) When a signal is superimposed on a high frequency wave, which acts as a carrier by change its amplitude, frequency or phase it is known as modulation.

(b)
Low frequency waves cannot be transmitted as such due to the following reasons. 1. Size of antenna : Required would be very large and hence is impractical. 2. Effective power radiated increases with frequency. So low frequency need to be converted to high frequency for good transmission. 3. The low frequency signals cannot be separated into different bands of signals, and will become mixed up.


Q.31    Define the term Modulation. Name three different types of modulation used for a message signal using a sinusoidal continuous carrier wave. Explain, the meaning of any one of these.

[Delhi 2000,2006C]

Sol.

Modulation –  When a signal is superimposed on a high frequency wave, which acts as a carrier by change its amplitude, frequency or phase it is known as modulation.
The three different modulation are : Amplitude modulation ,Frequency modulation ,Phase modulation
Amplitude modulation means that the amplitude of the carrier wave is modulated as per the amplitude of the signal to be transmitted.

5

 


Q.32    For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is 2 V. Calculate the modulation index. Why is modulation index generally kept less than one?

[Foreign 2011]

Sol.

Maximum amplitude = {A_C} + {A_m} = 10
Minimum amplitude = {A_C} - {A_m} = 2
We get , {A_C} = 6 {A_m} = 4

Modulation indices = {{{A_m}} \over {{A_C}}} = {4 \over 6} = {2 \over 3}
Modulation index is kept less than 1 to avoid distortions.


Q.33    (a) Define modulation index.
               (b) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?

[2011 D,2012F]

Sol.

(a) The ratio of amplitude of modulating signal to the amplitude of carrier signal is called modulation index. i.e., m = {{{A_m}} \over {{A_C}}}
(b) {{A_m}} is kept less than less than {{A_C}} to avoid distortion of modulating signal.


Q.34    Draw a block diagram of a simple amplitude modulation. Explain briefly how amplitude modulation is achieved?

[All India 2008,2008C,2008D]

Sol. 5

The modulating wave is superimposed on the carrier wave in such a way that the amplitude of carrier wave changes according to the modulating signal.


Q.35    Define modulation index. Give its physical significance. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is 2 V. Determine the modulation index m.

[Foriegn 2012]

Sol.

The ratio of amplitude of modulating signal to the amplitude is carrier signal is called modulation index. i.e., m = {{{A_m}} \over {{A_C}}}
Its physical significance is that it shows to what eastern the amplitude carrier wave has been changed when it is superimposed by the modulating signal.
Maximum amplitude = {A_C} + {A_m} = 10
Minimum amplitude = {A_C} - {A_m} = 2
We get , {A_C} = 6 , {A_m} = 4
Modulation indices = {{{A_m}} \over {{A_C}}} = {4 \over 6} = {2 \over 3}
Modulation index is kept less than 1 to avoid distortions.


 

Q.36    Draw a plot of the variation of amplitude versus \omega for an amplitude modulated wave. Define modulation index.State its importance for effective amplitude modulation.

[Delhi 2008]

Sol.6

The ratio of amplitude of modulating signal to the amplitude is carrier signal is called modulation index. i.e., m = {{{A_m}} \over {{A_C}}} Modulations index is important as it should not be greater than 1 to avoid distortion.


Q.37      Draw a block diagram of a detector for an amplitude modulated signal explaining briefly the function of each of its components.

[Delhi 2008C]

Sol. c1

Detections is the process of recovering the modulating signal from the modulated carrier wave. The signal is allowed to pass through a rectifier to produce the output. This output is passed through an RC circuit so that the high frequency wave is removed. Only the wave form is left which is the required signal.


Q.38     What is the range of frequencies used in satellite communication? What is common between these waves and light waves?

[Delhi 2010]

Sol.          3.7 GHz - 6.4 GHz.


Q.39    What should be the length of dipole antenna for a carrier wave of frequency 5 \times 10^8 Hz?

[All India 2007]

Sol.         Length of dipole antenna = {\lambda \over 4} = {1 \over 4} \times {C \over \upsilon } = {{3 \times 10^8 } \over {4 \times 5 \times 10^8 }} = 0.15\,m


Q.40     What is remote sensing. Give its two applications.

[All India 2007]

Sol.

Acquring knoweldge about an object or phenomenon without making physical contact is called remote sensing. Application : Mapping of earth is surface , archaelogical surveys.


Q.41    Consider an optical communication system operating at A. = 800 nm. Suppose, only 1 % of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting. - (a) audio-signals requiring a bandwidth of 8 kHz? (b) video TV signals requiring an approximate band-width of 4.5'MHz? Support your answer with suitable calculations.

[ All India 2006]

Sol.

Frequency of wavelength 800 nm. 0 = {C \over \lambda } = {{3 \times {{10}^8}} \over {800 \times 9}} = {3 \over 8} \times {10^{15}}Hz

(a)
No. of channels for audio signals of bandwith 8 KHz (when 1% is used).  = {3 \over 8} \times {{{{10}^{15}}} \over {8 \times 100}} \times {1 \over {100}}  = 4.6875 \times {10^9}\,Channels

(b) No. of channels for video signals of bardwith 4.5 MHz (when 1% is used)  = {3 \over 8} \times {{{{10}^{15}}} \over {4.5 \times {{10}^6}}} \times {1 \over {10}}  = 8.33 \times {10^6}\,\,Channels  


Q.42     Why do we need a higher bandwidth for transmission of music compared to that for commercial telephonic communication?

[Delhi2009]

Sol.

The range of musical instruments is from (20-20,000) Hz. So bandwith is (9,920) Hz , where Telephonic conversation range is 300 to 3100 Hz. So bandwidth is 2800 Hz.


Q.43     Carrier wave,c(t) = A_c \,Sin\,\omega _c t is amplitude modulated by a modulating signal, m(t) = A_m \,Sin\,\omega _m t . The maximum and minimum amplitudes, of the resulting AM wave, are found to be 16 V and 4 V respectively. Calculate the modulation index.

[Delhi 2010C]

Sol.

{A_C} + {A_m} = 16\,V
{A_C}{A_m} = 4\,V

Therefore {A_C} = 10\,\,;\,\,\,{A_m} = 6
Modulation index  = {{{A_m}} \over {{A_C}}} = {6 \over {10}} = {3 \over 5} = 0.6


Q.44    A message signal of frequency 10 kHz and peak voltage of 10 V is used to modulate frequency of 1 MHz and peak voltage of 20 V. Determine (a) modulation index (b) the side bands produced .

[All India 2009C]

Sol.

{A_C} = 20\,V {A_m} = 10\,V

(a) Modulation index  = {{{A_m}} \over {{A_C}}} = {{10\,V} \over {20\,V}} = {1 \over 2} = 0.5

(b) Side bands = 1 MHz  \pm 10 KHz
= 1000 KHz  \pm 10 KHz

Upper bands 1010 KHz
Lower bands 990 KHz


Q.45    Figure shows a block diagram of a transmitter identify the boxes X and Y and write their functions.

[Foreign 2012]

c11

Sol.

X – Amplitude modulator The message signal is modulated with the carrier wave.
Y – Power amplifier The modulated signal is provided with enough power by the amplifies transmission of signal.


Q.46    In the given block diagram of a receiver, identify the boxes labelled as X and Y and write their functions.

[All India-2012]

c12

Sol.

X – Intermediate frequency stage   Here the carrier wave frequency is charged to a lower frequency.
Y – Amplifier, It amplifies the signal before bearing sent to output



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