Basic block diagram of analog communication system


Communications refer to sending, receiving and processing of information by electrical means, that is it means exchanging information between transmitter and receiver.

In early 1840’s the type of communication used was Wire telegraphy later on the forms are as telephony, Radio communication (possible with the invention of triode tube, Satellite communications and fibre optics(with the invention of transistors and IC’s and semi-conductor devices), that means communications become more advanced with increasing emphasis on computer and other data communications.

A modern communication system is concerned with

before transmission:- 

  • sorting:- sorting for the right message.
  • Processing:- processing is to make that message more suitable for transmission.
  • storing:- storing that message before transmission.

then the actual transmission of that message takes place (processing and filtering  noise)

at the receiver:-

  • decoding:-decoding the original message.
  • storage:-storing a copy of that message.
  • interpretation:-and analyzing for the correctness of that message.

the different forms of modern communication systems includes Mobile communications,Computer communications, Radio telemetry etc.

to become familiar with communication systems one needs to know about amplifiers and oscillators that means fundamentals of electronic circuits must be known, with these concepts as a background the every day communication concepts like noise, modulation and information theory as well as various types of systems may be studied.

The most general form of Communication system ( one or two blocks may differ) is shown in the figure basic terminology used in Communication systems is message signal /information/data,channel,noise,modulation, encoding and decoding. Communication system is meant for communicating messages between Transmitter and Receiver (or) source & destination.


source or information source is the primary block in communication system which generates original message / actual message. 

i.e, selecting one message (actual message) from a group of messages itself is called as sorting data (or) information. Source generates message which may be in any form like words, code , symbols, sound signal, images, videos etc.among these the desired message has been selected and conveyed.

A transducer is one which converts one form of energy into electrical energy because the message from information source may not be always in electrical form, a transducer is used in between source and transmitter as a separate block sometimes (or) may be a part of Tx r.


Txr is meant for the following tasks

  • restriction of range of audio frequencies (i.e, limiting the bandwidth of the message signal).
  • Amplification.
  • Modulation. 

In general modulation is said to be the main function of the transmitter.


The medium that exists between transmitter and receiver is called as channel. The function of channel is to provide connection between transmitter  and receiver, two types of channels are  there wired/point to point  and wireless/broadcasting channels.

Point to point channels are generally wired channels(i.e, a physical medium exists) like Microwave links, optical fibre links etc. 

Microwave links:- these links are used in telephone transmission.In these type of links guided EM waves are used to transmit from Txr to Rxr.

optical fibre links:- used in low-loss high speed data transmission and uses optical fibers as the medium .

Broadcast channels:- the medium or channel is wireless here, in broadcasting a single transmitter can send information to many receivers simultaneously, satellite broadcasting system is one such system.

during the process of transmission and reception, the signal gets distorted due to noise in the channel, noise may interfere with the signal at any point but noise in the channel has greatest effect on the signal.


The main function of the receiver is to reproduce the message signal in electrical form from the distorted received signal. This reproduction process is called demodulation (or) detection , in general this demodulation may be assumed as the reverse process of modulation carried out in transmission. 

there are a great variety of receivers in communication systems, the type of receiver chosen depends on type of modulation, operating frequency ,its range  and type of destination required. Most common receiver is superheterodyne receiver .

                            crystal receiver with head phones
                                  Radio receiver

so many types of receivers are available from a very simple crystal receiver with headphones to radar receiver etc.

Destination:- It is the final stage of any communication system. it would be a loud speaker / a display device/simply a load etc depending up on the requirement of the system.

flooding (static)

This is another type of static algorithm.

the main concept of flooding is to sent every incoming packet on a line to every other outgoing line except the line it arrived on.

flooding generates a large no.of duplicate packets, sometimes infinite unless we may take certain measures.

the measures are as follows:-

  • one measure is use of hop count in the header of each packet and decrement this count at each hop when count reaches to zero discard the packet.
  • How to take this hop count is another problem. Generally it is set to the length of path from source to destination and in worst cases the full diameter of the subnet.

  • another way is avoid sending a packet more than once through a router this is possible by using sequence no.
  • i.e, a source router (which generates packets) can put a sequence no. to each packet and each router will maintain a list of sequence nos. and if sees a packet with same sequence no in the list that packet is discarded (not flooded).

another way of flooding is of use selective flooding.

i.e, with this the router wouldn’t send every incoming packet on every line instead the router will send packets in a particular direction only.

i.e, east bound packets are sent on east side routers and similarly on  west side by west side routers.

even flooding is cumbersome, it has some uses


  1. used in military applications.
  2. used in distributive data base applications in which to update all data bases concurrently.
  3. used in broadcast Routing.

flooding is used rather than any other algorithm since flooding chooses shorter path between two nodes where other algorithms may not.

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Reconstruction filter(Low Pass Filter)

Reconstruction filter (Low Pass Filter) Procedure to reconstruct actual signal from sampled signal:-

Low Pass Filter is used to recover original signal from it’s samples. This is also known as interpolation filter.

An LPF is that type of filter which passes only low frequencies up to cut-off frequency and rejects all other frequencies above cut-off frequency.

For an ideal LPF, there is a sharp change in the response at cut-off frequency as shown in the figure.

i.e, Amplitude response becomes suddenly zero at cut-off frequency which is not possible practically that means an ideal LPF is not physically realizable.

i.e, in place of an  ideal LPF a practical filter is used.

In case of a practical filter, the amplitude response decreases slowly to zero (this is one of the reason why we choose  f_{s}>2f_{m})

This means that there exists a transition band in case of practical Low Pass Filter in the reconstruction of original signal from its samples.

Signal Reconstruction (Interpolation function):-

The process of reconstructing a Continuous Time signal x(t) from it’s samples is known as interpolation.

Interpolation gives either approximate (or) exact reconstruction (or) recovery of CT signal.

One of the simplest interpolation procedures is known as zero-order hold.

Another procedure is linear interpolation. In linear interpolation the adjacent samples (or) sample points are connected by straight lines.

We may also use higher order interpolation formula for reconstructing the CT signal from its sample values.

If we use the above process (Higher order interpolation) the sample points are connected by higher order polynomials (or) other mathematical functions.

For a Band limited signal, if the sampling instants are sufficiently large then the signal may be reconstructed exactly by using a LPF.

In this case an exact interpolation can be carried out between sample points.

Mathematical analysis:-

A Band limited signal x(t) can be reconstructed completely from its samples, which has higher frequency component fm Hz.

If we pass the sampled signal through a LPF having cut-off frequency of  fm  Hz.

From sampling theorem  

g(t) = x(t).\delta _{T_{s}}(t).

g(t)=\frac{1}{T_{s}}\left \{ 1+2\cos \omega _{s}t+2\cos 2\omega _{s}t+2\cos 3\omega _{s}t+..... \right \}.

g(t)     has a multiplication factor  \frac{1}{T_{s}}. To reconstruct  x(t)  (or)  X(f) , the sampled signal must be passed through an ideal LPF of Band Width of  f_{m}  Hz and gain  T_{s} .

\left | H(\omega ) \right |=T_{s} \ for \ -\omega _{m}\leq \omega \leq \omega _{m}.

h(t) = \frac{1}{2\pi } \int_{-\omega _{m}}^{\omega _{m}}T_{s}e^{j\omega t}\ d\omega.

h(t) = 2f_{m}T_{s} \ sinc(2\pi f_{m}t).

If sampling is done at Nyquist rate , then Nyquist interval is  T_{s} = \frac{1}{2f_{m}}.

 therefore  h(t) = \ sinc(2\pi f_{m}t).

h(t) = 0.      at all Nyquist instants  t= \pm \frac{n}{2f_{m}}  , when    g(t)    is applied at the input to this filter the output will be  x(t)  .

Each sample in g(t)  results a sinc pulse having amplitude equal to the strength of sample. If we add all these sinc pulses that gives the original signal  x(t) .

g(t) = x(kT_{s})\delta (t-kT_{s}).

x(t) =\sum_{k} x(kT_{s})\ h (t-kT_{s}) .

x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m} (t-kT_{s})).

x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m}t-k\pi ) .

This is known as interpolation formula

It is assumed that the signal  x(t) is strictly band limited but in general an information signal may contain a wide range of frequencies and can not be strictly band limited this means that the maximum frequency in the signal can not be predictable.

then it is not possible to select suitable sampling frequency  fs  .

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Circuit Switched Networks

A Circuit Switched N/w consists of a set of switches connected by physical links.

A connection b/w ‘2’ stations is dedicated path made of one (or) more links. Each connection uses only one dedicated channel on each link.

i.e, each link is divided into n channels either by using TDM (or) FDM.

This circuit consists of 4 switches I, II, III and IV and Multiplexers with n=’3′ channels and one link.

In some circuits Multiplexing can be implicitly included in the switch fabric it self. In this circuit the end systems are connected to a switch for simplicity consider ‘2’ end systems A and M, connected to the switches I and III.

when A needs to communicate with M . A needs to request to a connection to M, which must be accepted by all switches and by M it self- which is called setup phase.

a channel circuit is reserved on each link and the combination of circuits forms a dedicated path. After establishing path data transfer can take place. The next phase is tear down.

i.e, after all data have been transferred. Generally circuit-switching takes place at the physical layer.

Before Communication (starting), the stations must make reservation for the resources like channels, switch buffers switch i/o ports switch processing time and are dedicated during the entire duration of data transfer until the tear down phase.

Data transferred is not packatized that is data is send as a continuous flow b/w source and destination.

there is end-to-end addressing in setup phase.

The 3 phases involved are:-

Circuit switched N/w’s requires ‘3’  setup phases

  1. Connection-setup.
  2. Data transfer.
  3. Tear down.

Setup Phase:-

A dedicated circuit is established before the ‘2’ communicating parties talk to each other.

i.e, creating  a dedicated channels b/w switches. To communicate A with M . initially a requesting process as follows

A to I, I to IV and IV to III, III to M and an acknowledgement in the reverse order after the reception of ‘ack’ a connection is established.

Data Transfer Phase:-

In this phase data transfer occurs b/w the ‘2’ devices.

Tear down phase:-

To disconnect , a signal is sent to each switch to release the resources by any one of station.

Efficiency of Circuit Switched Network:-

These are less efficient in terms of allocated resources. Since all the resources are allocated during the entire duration of the connection  and these resources are un available to other connections.

Delay in this type of N/w’s is due to establishment of connection , data transfer and disconnecting the circuit.

Switching at the physical layer in the traditional telephone N/w uses the circuit switching approach.

Broadcast Routing(dynamic)

In some applications hosts need to send messages to many (or) all other hosts like weather reports, stock market updates (or) live radio programs.

 i.e, sending a packet to all destinations simultaneously is called Broadcasting.

 Different methods of Broadcasting:-

  • first method is to send a packet to all destinations. This is a method wasteful of Band width and source needs to know the complete list of all destinations.

so this is least desirable one.

  • flooding is another way to broadcast a packet, the problem with flooding is that it generates too many packets and also consumes too much of Band width.
  • Third way is to use multi destination routing

In this technique each packet contains a list of destinations (or) a bit map for those destinations.

when a packet arrives at a router,  the router checks all the output lines it requires. The router generates a new copy of the packet for each output line after sufficient number of hops each packet will carry only one destination.

i.e, multi destination routing is like separately addressed packets (to B,C,D,E & D) must follow the same route one of them pays full fare and rest are free.

  • The fourth type of method is to use sink tree (or) spanning tree.

A spanning tree is a subset of subnet that includes all the routers but contains no loops.

if each router knows which of it’s lines belong to spinning tree then it broadcasts packet to all the lines except the one it arrived on.

This is efficient method in terms of Band width usage but problem is to maintain the knowledge of all the nodes of spanning tree at a routes.

  • Last method is to use Reverse path forwarding to approximate behavior of spanning tree.

Consider a subnet and it’s sink tree for router I as root node and how reverse path algorithm works in figure (C) 

on the first hop I sends packets to F, H, J & N. on the second hop eight packets are generated among them 5 are given to preferred paths indicated as circles (A,D,G,O,M)

of the 6 packets generated in third hop only 3 are given to preferred paths (C,E & K) the others are duplicates.

in the fourth hop to B and L after this broadcasting terminates.

advantages of reverse path forwarding:-

  • it is easy to implement.
  • it does not require routers to known about spanning trees.
  • it does not require any special mechanism to stop the process (as like flooding).

The principle is: if a packet arrives on a line if it is preferred one to reach the source it gets forwarded.

if it arrives on a line that is not preferred one that packet is discarded as a duplicate.



when a packet arrives at ‘L’ the preferred paths are N and P so it forwards the packets to both N and P and if a packet arrives at ‘K’, there the preferred path is M, and N is not preferred so it forwards the packet to M and discards to N.

This is reverse path forwarding.


aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

When a Continuous Time  band-limited signal is sampled at, then the successive cycles of the spectrum of the sampled signal overlap with each other as shown below


Some aliasing is produced in the signal this is due to under-sampling.

aliasing is the phenomenon in which a high-frequency component in the frequency spectrum of the signal takes as a low-frequency component in the spectrum of the sampled signal.

Because of aliasing, it is not possible to reconstruct x(t) from g(t) by low pass filtering.

The spectral components are in the overlapping regions and hence the signal is distorted.

Since any information signal contains a large no.of frequencies so the decision of sampling frequency always becomes a problem.

A signal is first passed through LPF  before sampling.

i.e, it is band limited by this LPF which is known as a pre-alias filter.

To avoid aliasing

  1. Pre-alias filter must be used to limit the bandwidth of the signal to f_{m}  Hz.
  2. Sampling frequency must be  f_{s}>2f_{m}.

Pre-alias filter means before sampling is passed through an LPF to make a perfect band-limited signal.


FSK Generator /BFSK generator

we know that the input to the FSK Generator is a binary sequence 1010…etc.

FSK generator uses two product modulators upper-modulator and lower-modulator with carriers



A level shifter is there in which the output of the level shifter is

when the input is a binary ‘1’ and ‘0’ volts for the input ‘0’ level shifter.


The working of the FSK generator is as follows when the input binary sequence is ‘1’

on the upper modulator 1 has been shifted to a voltage  so that the output of product modulator 1 is

and on the lower modulator input ‘1’ is passed through an inverter and if the output of the inverter is ‘0’ then the output of the level shifter will not change it remains at ‘0’ volt itself.

then the product modulator 2 output is 

then the overall output

similarly, when the input sequence is a binary ‘0’ 


Frequency Shift Keying (FSK)/BFSK

In a Binary FSK system, symbols 1 and 0 are distinguished from each other by transmitting one of two sinusoidal waves that differ in frequency by a fixed amount.


The frequency of the carrier signal shifted to two frequencies  for symbols ‘1’ and ‘0’ transmission.

The equation for FSK signal is

\[ S_{FSK}(t)= \sqrt{\frac{2E_{b}}{T_{b}}} \ \ cos (2\pi f_{c_{i}}t),\ 0\leq t\leq T_{b} , \ i=1,2 \]

\[ S_{FSK}(t)= 0 elsewhere \]



 is generally a high frequency.

 is a low frequency and vice-versa is also true.






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Analog Communications Quiz or Tutorials


  1. Find the modulation index if the amplitude of message signal is Two thirds of the amplitude of carrier signal ———————.
  2. The diagonal clipping in Amplitude Demodulation (using envelope detector) can be avoided if RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and ω is carrier frequency both in rad/sec) ( )
    a. RC<1/W             b. RC>1/W                  c. None of the above         d. RC>1/ω.
  3. Given AM signal is  XAM(t)= 10 (1+0.25 sin 2πfmt) cos 2πfct , then The average side band power given for the above AM signal is ( )
    a. 25W                    b. 12.5W               c.1.5625W                    d.3.125W.
  4. Given AM signal is XAM(t) = 100 (1+0.85 cos 2πfmt) ,then The total power required for the above AM signal is ( )
    a. 25W                   b. 12.5W               c.6.806KW                       d. None of the above.
  5. Consider the AM signal Ac cos Wct + 2 cos Wct cos Wmt   for the demodulation of the signal using envelope detector the minimum value of Ac should be ( ).
    a. 2                          b. 1                         c. 0.5                                    d. 0.
  6. Given AM signal is SAM(t) = 100 (1+0.3 cos 2πfmt +0.4 sin 2πfmt) cos 2πfct i. i.The total power required for the above AM signal is ( )
    a. 5.625W              b. 5.625KW             c.6.806KW                  d. None of the above.
    ii.Modulation index is ( )
    a. 0.53                     b. 0.5                         c. 0.2                               d. None of the above.
    iii.The carrier power is ( )
    a. 5KW                    b.6KW                       c. 7KW                           d.100KW.
    iv. Total current flowing through the transmitter if carrier current is 5A ( ).
    a. 5mA                      b.5.303A                 c. none of these          d. 25A.
  7. If the band width of message signal is 5KHz and the carrier frequency is 200KHz then upper sideband frequencies are( )
    a. 205KHz,190KHz                                                b.205KHz,-205KHz
    c. 205 KHz,-195 KHz                                           d. None of the above.
  8. If the highest frequency of message signal is 5KHz and the carrier frequency is 200KHz then lower sideband frequencies are( )
    a. 205KHz,190KHz                                                b.205KHz,-205KHz
    c. 195 KHz,-195 KHz                                            d. None of the above.
  9. If the bandwidth of message signal is 500Hz then the bandwidth of Amplitude Modulated signal is ( )
    a. 500Hz                    b. 1000Hz                    c.2KHz                     d. None of the above.
  10. If the message m(t)= 10cos⁡2000πt and carrier signal is                                      C(t) = 25 cos 200000 πt then draw the amplitude spectrum of AM signal————— .


  1. Angle modulation is a technique in which the ————— of the is varied with respect to instantaneous values of ————————— by keeping as constant. 2M.
  2. Write the expression for Angle Modulated signal –. 1M.
  3. An Angle Modulated signal is given as 𝑥(𝑡) = 100 cos(2𝜋𝑓𝑐 𝑡 + 4 sin(1000𝜋𝑡)) where 𝑓𝑐 = 10 𝑀𝐻𝑧. 6M.
    i. The Peak frequency deviation is ( )
    a. 2K                           b. 4000                   c. 4π                      d. 8π.
    ii. The Peak- phase deviation is ( ).
    a. 4                              b. 6                           c. 0                        d. None of the above.
    iii. The power of the Modulated signal is ( ).
    a. 10KW.                   b. 5 W                       c. 5 KW                d. 50W.
  4. The amount of change in carrier frequency produced by modulating signal is known as ( ). 1M.
    a. phase deviation                                                            b. amplitude deviation
    c. Frequency deviation                                                  d. none of the above.
  5. The total Transmitted power in FM is equal to the power of ( ) 1M.
    a. An AM signal.                                                            b. an unmodulated carrier
    c. Message signal                                                         d. all of the above.
  6. A 20 MHz carrier is frequency modulated by a sinusoidal signal with frequency 1KHz such that peak frequency deviation is 100KHz what will be the modulation index ( ) 2M.
    a. 100                           b.101                                c. 99                         d.200.
  7. The bandwidth for above FM system will be ( ) 2M.
    a. 101 KHz                 b. 202 KHz                     c. 99 KHz              d. 100 KHz.
  8. Which one of the following is an indirect method of generating FM ( ) 1M.
    a. Armstrong method                                                   b. Varactor diode modulator
    c. Reactance BJT modulator.                                     d. Reactance FET Modulator.
  9. In which of the Modulation system when the modulating frequency is doubled the modulation index reduces to half while modulating voltage remains constant ( ) 2M.                                                                                                        a. Phase                   b. Amplitude                   c. Frequency       d. None of the above.
  10. In FM, the frequency deviation is ( ) 2M.                                                                         a. Proportional to modulating frequency.                                                                         b. Proportional to amplitude of modulating signal.                                                     c. Constant.                                                                                                                                   d. Zero.
  11. In indirect method of FM generation FM is obtained from ( ) 1M.                         a. AM                    b. PM                 c. DSB                              d. FM
  12. Write Carson’s rule –. 1M.
  13. The Bandwidth of NBFM is given as –. 1M.
  14. A 25 MHz carrier is modulated by a 400Hz audio sine wave. The carrier voltage is 4V and the maximum deviation is 10 KHz. The modulation index will be( ) 2M.                                                                                                                                  a. 2.5                  b. 5                     c. 15                        d. 25
  15. For the above problem write the expression of FM wave will be———————————————————————————————–.1M.
  16. For the problem in 14 write the expression of PM wave———————————————————————————————————.1M.
  17. Standard FM broadcast stations uses a maximum bandwidth of ( ) 1M.              a. 150 KHz                   b.75KHz.            c. 200KHz         d. 15KHz.
  18. Which type of oscillators are preferred for carrier generation because of their good frequency stability ( ) 1M.                                                                                 a. LR               b.LC                       c. Crystal                         d. RC.
  19. The oscillator whose frequency is varied by an input voltage is called as ————————————————. 1M.
  20. Maximum deviation results at what point on modulating signal if the system is FM( ) 1M.                                                                                                                      a. Zero crossing of m(t)                                                                                                              b. Peak negative amplitude and peak positive amplitude of m (t).                        c. None of the above.                                                                                                                d. Both a and b.

(Radio receivers and Transmitters)

  1. Radio receivers are classified into how many types ( ). 2M.
    a. Three                   b. two                      c. four                             d. none of the above.
  2. The ability of a radio receiver to amplify weak signals is called as ( ). 2M.
    a. Fidelity             b. Selectivity            c. sensitivity                  d. all of the above.
  3. The phenomenon of Picking up of same short wave station at two nearby points on the receiver dial is known as ( ). 2M.
    a. Fidelity            b. sensitivity               c. Double spotting               d. selectivity.
  4. The ability of a receiver to reject unwanted signals is called ( ). 2M.
    a. Selectivity           b. Fidelity                  c. sensitivity                  d. Double spotting
  5. Standard broadcast AM receivers tuned in the frequency range of 540 KHz to 1640 KHz has an intermediate frequency of ( ). 2M.
    a. 455 KHz                 b.1MHz                      c. 20Hz                      d. 200Hz.
  6. Standard broadcast FM receivers tuned in the frequency range of 88MHz -108 MHz has an intermediate frequency of ( ). 2M.
    a. 455 KHz                    b.1MHz                   c. 20Hz                      d. 10.7MHz.
  7. Television receivers in the VHF band(54MHz-223MHz) and in the UHF band(470MHz-940MHz) use an IF between 26MHz and 46MHz with the two most popular values ( ). 3M.
    a. 36 MHz and 46 MHz                                                b. 455 KHz and 46 kHz.
    c. 36 KHz and 46 KHz.                                                 d. none of the above.
  8. In a broadcast FM receiver if the local oscillator is tuned to 98.7 MHz then the image frequency is ( ). 3M.
    a. 88MHz              b. 109.4MHz                   c. 96 MHz                d. none of the above.
  9. In a broadcast AM receiver if the signal is tuned to 530 KHz then Intermediate frequency, local oscillator frequency and image frequency are( ). 3M.
    a. 200 kHz, 730 KHz and 1000 kHz.
    b. 10.7MHz, 15.37MHz and 1000 KHz.
    c. 455 KHz, 985 KHz and 1440 KHz.
    d. None of the above.
  10. In communications, Audio frequency range is —————-. 2M.
  11. In communications, Radio frequency range is—————–. 2M.
  12. Draw the radio frequency spectrum with detailed values starting from Very Low Frequencies (VLF) to Extreme High Frequencies (EHF). 5M.
  13. Draw the block diagram of TRF receiver (only diagram). 5M.

Pulse Modulation Techniques

  1. In Pulse Position Modulation, the drawbacks are ( ) 2M.                 a.Synchronization is required between transmitter and receiver                       b. Large bandwidth is required as compared to PAM                                                 c. None of the above                                                                                                                 d. Both a and b.
  2. In PWM signal reception, the Schmitt trigger circuit is used ( ) 2M.                    a. To remove noise                                                                                                                    b. To produce ramp signal                                                                                                      c. For synchronization                                                                                                            d. None of the above.
  3. In pulse width modulation, ( ) 2M.                                                                                    a. Synchronization is not required between transmitter and receiver              b. Amplitude of the carrier pulse is varied                                                                      c. Instantaneous power at the transmitter is constant                                              d. None of the above.
  4. In different types of Pulse Width Modulation, ( ) 2M.                                                a. Leading edge of the pulse is kept constant                                                                b. Trailing edge of the pulse is kept constant                                                                c. Centre of the pulse is kept constant                                                                              d. All of the above.
  5. In Pulse time modulation (PTM), ( ) 2M.                                                                         a. Amplitude of the carrier is constant                                                                             b. Position or width of the carrier varies with modulating signal                           c. Pulse width modulation and pulse position modulation are the types of PTM                                                                                                                                                 d. All of the above.
  6. Drawback of using PAM method is ( ) 2M.                                                                      a. Bandwidth is very large as compared to modulating signal                                  b. Varying amplitude of carrier varies the peak power required for transmission                                                                                                                                  c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver                                                                                                                                          d. All of the above.
  7. Pulse time modulation (PTM) includes ( ) 2M                                                              a. Pulse width modulation                                                                                                        b. Pulse position modulation                                                                                                  c. Pulse amplitude modulation                                                                                              d. Both a and b.
  8. Calculate the Nyquist rate for sampling when a continuous time signal is given by  x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt ( ) 3M.                            a. 300Hz                                                                                                                                            b. 600Hz                                                                                                                                            c. 150Hz                                                                                                                                          d. 200Hz.
  9. Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt ( ) 3M.                                                              a. 100 Hz                                                                                                                                            b. 200 Hz                                                                                                                                          c. 400 Hz                                                                                                                                          d. 250 Hz.
  10. A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is ( ) 2M.                                                                                          a. fs > 2fm                                                                                                                                        b. fs < 2fm                                                                                                                                      c. fs = 2fm                                                                                                                                        d. fs ≥ 2fm.
  11. The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if ( ) 2M.                                                                                                    a. The sampled signal is passed through low pass filter                                              b. Filter has the cut off frequency wm                                                                              c. Both a and b                                                                                                                              d. None of the above.
  12. The frequency spectrum of x(t) is X(f) is given as follows 6M.

Draw the frequency spectrum of sampled signal by assuming suitable values for sampling frequency under the following conditions
i. Over sampling                      ii. Under sampling                         iii. fs = 2fm.

Analog Communication Lab viva questions


Amplitude Modulation:

  1. What is meant by Modulation? What is the need for modulation?
  2. What are different types of analog modulation techniques?
  3. What are the other names of message signal? What are the other names of carrier signal?
  4. Write the equation of AM signal and explain each parameter in that equation?
  5. Define Amplitude Modulation? Define modulation depth or modulation index?
  6. What is the range of Audio frequency signals? What is the range of Radio frequency signal?
  7. What are the applications of Amplitude modulation?
  8. How many generation methods are there to generate an AM wave? What are the methods of demodulation of an AM wave?
  9. Explain the operation of diode detector circuit?
  10. Write the formula for modulation index? Differentiate under, over and perfect modulation in AM?
  11. As the amplitude of message signal increases, modulation index increases or decreases?
  12. Define single tone modulation? In laboratory type of AM is single tone modulation or not?
  13. Draw the frequency spectrum of AM wave?
  14. If modulation index is 100%, calculate the ratio of total power to carrier power of an AM wave?
  15. If µ=1 in an AM wave what is the amount of power saving in an AM wave? What is the band width of an AM wave?
  16. Explain the operation of AM modulator? Explain the operation of 8038 circuit in AM modulator?
  17. Explain the procedure of Amplitude modulation? What is the significance of Emax and Emin points in AM wave?
  18. Plot message, carrier and AM signals?
  19. What is meant by envelope detector?
  20. The frequency of AM wave follows — (message signal frequency or carrier frequency)?
  21. The amplitude of AM wave at fc +fm is— and The amplitude of AM wave at fc -fm is—–
  22. In amplitude modulation the amplitude of ——— is changing with respect to ——
  23. Envelope of AM signal follows————– (message signal/ carrier signal)?
  24. What are the advantages and disadvantages of AM?
  25. How demodulated signal differs from original signal in AM?
  26. The two important distortions that can appear in the demodulated output of an envelope detector are————– and—————————- –.
  27. Differentiate high-level and low-level modulations in AM?
  28. What is trapezoidal rule?

Balanced Modulator:

  1. What are the disadvantages of AM?
  2. Most of the power in AM spectrum is carried by ————
  3. Define DSBSC modulation?
  4. How DSBSC is more efficient than AM in terms power saving, explain?
  5. What is meant by frequency response?
  6. Draw the magnitude response or amplitude spectrum of DSBSC signal?
  7. The signal generated by balanced modulator is———–
  8. Draw the wave form of DSBSC wave and AM wave, and differentiate those two waveforms?
  9. Give the equation of DSBSC signal?
  10. What are the generation methods of DSBSC?
  11. What are the demodulation methods of DSBSC?
  12. What is the bandwidth of DSBSC signal?
  13. Define Costas loop and it’s operation?
  14. Amount of power saving in DSBSC signal is————
  15. Coherent detection means?
  16. Give the practical applications of balanced modulator?
  17. Explain the operation of product modulator?
  18. Why the circuit is called balanced modulator?
  19. If the circuit is operating in balanced state, the modulation index value is——- –.
  20. Explain the working procedure of 1496 IC for the generation of DSBSC wave?
  21. As message signal amplitude increases, carrier suppression in dB’s ———
  22. Plot message, carrier and DSBSC waves and explain each wave clearly.
  23. How do you differentiate modulation by demodulation?
  24. Explain the significance of local oscillator frequency in modulators and in
  25. Differentiate synchronous and non synchronous detection techniques in analog modulators?
  26. The phase shift at zero crossings in DSBSC wave is——- –.
  27. What is Quadrature carrier multiplexing?
  28. How DSBSC is different from SSB?

Frequency Modulation:

  1. Define Frequency modulation? How it is different from phase modulation?
  2. Write equation of FM wave, explain each parameter in it?
  3. Draw the amplitude spectrum of FM wave?
  4. Give the Carson’s rule in FM?
  5. Define modulation index β, frequency deviation?
  6. Differentiate Narrow band FM with Wide band FM?
  7. Explain the FM operation using 8308IC?
  8. Draw message, carrier and FM waves and explain each wave clearly?
  9. Explain the methods for generation of FM and its demodulation?
  10. How FM wave is different from PM wave?
  11. Give the practical applications of FM?
  12. State advantages and disadvantages of FM?
  13. The range of speech signals is——— –.
  14. Type of Modulation used in radios is——- –.
  15. Type of modulation used for voice signals in T.V — and for video signals in V is—- –.
  16. Noise immunity is more in which analog modulation technique———– –.
  17. FM is more robust to noise compared to AM, why?
  18. Carson’s rule is for———- –.
  19. In commercial FM broadcasting, the audio frequency range handled is only up to—- –.
  20. The transmission band width required for commercial FM broadcasting is——– –.
  21. Define Hilbert transform?
  22. Explain capture effect in FM broadcasting?

Pre-emphasis and De-emphasis:

  1. Define pre-emphasis and De-emphasis processes in
  2. Why Pre-emphasis is used at Transmitter of FM and de-emphasis at FM receiver?
  3. Draw the pre-emphasis circuit and explain its working in detail?
  4. Draw de-emphasis circuit and explain its working in detail?
  5. Draw the frequency response characteristics of pre-emphasis and de-emphasis explain each one in detail?
  6. Calculate the cut-off frequencies of pre-emphasis and de-emphasis circuits practically
  7. Pre-emphasis circuit operation is similar to——— –.
  8. De-emphasis circuit operation is similar to——— –.
  9. What is the necessity of boosting up high frequencies in frequency modulation communication system?
  10. Define 3dB frequencies?

Sampling and reconstruction:

  1. Define sampling theorem? What is the need for sampling?
  2. What are the necessary and sufficient condition for sampling and reconstruction of a signal?
  3. Define Nyquist rate and Nyquist interval in sampling theorem?
  4. If message frequency is 2 KHz and sampling frequency is 2 KHz,4 KHz, 8 KHz and 16 KHz in each case the number of samples are—————————– –.
  5. What are different types of sampling techniques?
  6. What was the effect on sampled signal if fs < 2 fm ?
  7. Draw the amplitude spectrum of sampled signal if fs < 2 fm, fs =2 fm, fs > 2 fm.
  8. What is aliasing effect in sampling? How to avoid it?
  9. Why do we use pre-filtering in sampling?
  10. What do you mean by reconstruction of sampling theorem?
  11. What are the types of filters used in reconstruction?
  12. Define sample and hold process?
  13. Differentiate second order, fourth order and sixth order low pass filters in reconstruction process.
  14. Explain the sampling and reconstruction process in detail by using the trainer
  15. Define band pass sampling?
  16. How sampling is different from PAM?
  17. Define a continuous time signal or an analog signal. Give some examples of analog signals.
  18. Define a discrete time signal. Give some examples of discrete
  19. What is the difference between discrete and a digital signal?
  20. Define a digital signal? Give some
  21. What is the need for converting a continuous signal into a discrete
  22. Explain about zero-order hold circuit.
  23. How to convert an analog signal into digital signals?

Digital signal processors operates———— as inputs.As the number of samples increases, the reconstruction of original signal becomes?

Pulse Amplitude Modulation:

  1. What is the basic principle of PAM?
  2. Name some Pulse Modulation techniques?
  3. Define PAM?
  4. How PAM is different from AM?
  5. Can we produce a PAM signal using a sampling circuit?
  6. Differentiate PAM output with sampling output?
  7. Does PAM come under Analog modulation technique or Digital Modulation technique?
  8. What is the Bandwidth of PAM?
  9. Compare BW of PAM and AM?
  10. Draw waveforms of PAM. explain each one briefly.
  11. What are the advantages of PAM over AM?
  12. What are the advantages and Disadvantages of PAM?
  13. Explain the working procedure PAM kit?
  14. Can we use PAM technique in TDM?
  15. Differentiate uni-polar and bi-polar PAM.
  16. What do you mean by zero order holding? And draw the circuit diagram of zero-order hold circuits?
  17. What are the drawbacks of PAM?
  18. Explain the working of PAM demodulation circuit?
  19. Define Flat-Top sampling?
  20. Draw the circuit diagram of Flat-Top Sampled circuit?
  21. What was the roll off characteristics of sinc pulse?

    Pulse Width Modulation (PWM):

    1. Define PWM?
    2. Differentiate PWM, PAM and PPM?
    3. Name the applications of Mono-stable multivibrator?
    4. What is a Multivibrator?
    5. Differentiate Monostable, Bi stable and Astable Multivibrators?
    6. How a Monostable Multivibrator produces a PWM signal?
    7. What are the other names of PWM?
    8. Define Pulse Duration Modulation?
    9. What is Pulse Time Modulation?
    10. Draw PAM and PWM signals and each one in detail.
    11. Draw PWM signal with respect to message signal?
    12. In PWM —————- of Pulse carrier signal is changing with respect to message signal.
    13. Explain the operation of PWM circuit.
    14. 555 timer in Monostable mode produces————.
    15. 555 timer in Astable mode produces——————.
    16. What are the advantages of PWM over PAM?
    17. What is the difference between PWM and FM?
    18. Which type of noise is affecting the amplitude of PWM signal?
    19. Which system is more immune to noise (PWM or PAM)?
    20. What are the disadvantages of PWM?
    21. What are the applications of PWM?
    22. Band Width of PWM is—————-.
    23. Band width of PAM is—————-.

      Pulse Position Modulation (PPM):

      1. Define PPM?
      2. The information is conveyed by ————- of Pulses in PPM.
      3. In PWM information is conveyed by————— of pulses.
      4. In PAM information is conveyed by————— of pulses.
      5. What are the advantages and disadvantages of PPM?
      6. Compare PPM with Phase Modulation.
      7. PAM is similar to————————–.
      8. PWM is similar to————————–.
      9. PPM is similar to————————–.
      10. Differentiate Analog Modulation Techniques with Pulse Modulation Techniques.
      11. What are the applications of PPM?
      12. Draw PPM signal with respect to message signal.
      13. Draw PPM signal with respect to PWM signal.
      14. Explain the operation of PPM Modulator?

        Phase Locked Loop (PLL):

        1. What are the applications of PLL?
        2. Why this circuit is called Phase Locked Loop?
        3. What are the three components of PLL circuitry?
        4. Explain the operation of PLL by using a Block Diagram?
        5. Define free-running frequency?
        6. Define Lock range and Capture range of a PLL?
        7. What is meant by Frequency synthesizer?
        8. Why PLL is used in FM Receivers/
        9. How PLL is used in FSK demodulation circuits?
        10. What do you mean by Lock state in a PLL?
        11. What is meant by Pull in time in PLL?
        12. Phase Detector or Phase Comparator is used for ——————.
        13. Why VCO is used in feedback loop of PLL?
        14. What are the input and Output signals of a VCO in a PLL?
        15. What are the advantages and Disadvantages of PLL?
        16. Why Lock range is greater than Capture range in a PLL?

          Time Division Multiplexing (TDM):

          1. Define the concept of Time Division Multiplexing?
          2. Differentiate Multiplexing and Sampling?
          3. What are the different types of Multiplexing Techniques?
          4. Does TDM come under analog Multiplexing or Digital Multiplexing?
          5. Define a frame in a TDM?
          6. Why synchronization is required in TDM?
          7. Why multiplexing is required?
          8. What do you mean by inter-leaving gaps in TDM frame?
          9. If two signals of frequencies 2KHz and 4 KHz are multiplexed in time –domain then draw TDM signal
            1. Without inter-leaving gap.
            2. with inter-leaving gap of 10 ms.
          10. What are the advantages and Disadvantages of TDM?
          11. Differentiate TDM with FDM?
          12. What are the advantages of TDM over FDM?
          13. Explain the operation of TDM using trainer kit used.
          14. Synchronous TDM means?

            Additional Questions:

            1. What is a filter?
            2. Differentiate Active and Passive Filters?
            3. Why LPF is used in Demodulation Circuits?
            4. Why pre-filtering is required in sampling?
            5. Anti-aliasing is achieved by using———– in Sampling Circuits?
            6. Define Single-tone modulation in AM and FM?
            7. Why FM receivers are more immune to AM receivers?
            8. BW of Narrow band FM is ——————.
            9. BW of AM is ———————————-.
            10. How to calculate Image frequency of a Radio Receiver?
            11. Define Power Spectral Density.
            12. Define AWGN noise.
            13. Define SNR.
            14. Where do we use Hilbert Transform?
            15. Over modulation in AM means….
            16. What is µ value when AM wave is similar to DSBSC wave?

Digital Communications Slip Test Questions

Slip test-1

  1. In a PCM system, if the Quantization levels are increased from 2 to 8,                 i. Find the change in Signal to Quantization Noise ratio.                                ii.Find the change in Transmission Bandwidth.
  2.  i. Convert the following signal x(t) = 10 cos (200πt) to a discrete signal x[n] if sampling rate is 1000Hz.                                                                                       ii. Write the advantages of Digital Communication system over Analog Communication system.


  1. Draw the Basic Block Diagram of Digital Communication System and explain each block in detail.                                                                                                                             
  2. Deduce the expression for Signal to Quantization noise ratio in PCM system.                                                                  Slip test-2
  1. A PCM source transmits four samples (messages) with a rate 2B samples /second. The probabilities of occurrence of these 4 samples (messages) are equally likely Find out the information rate of the source.                                                           
  2. A source produces 26 symbols with equal probability what is the average information produced by this source?     


  1. State Mutual information. Prove any three properties of Mutual information.                 
  2. A source alphabet has 10 symbols with the given probabilities 0.02, 0.04, 0.17, 0.02, 0.16, 0.06, 0.03, 0.27, 0.20, 0.03 construct Shannon-Fano coding and calculate the efficiency.                                                                                                                             

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