Normal incidence on a perfect conductor

Normal incidence on a perfect conductor

whenever an EM Wave travelling in one medium impinges second medium the wave gets partially transmitted and partially reflected depending up on the type of the second medium.

Assume the first case in Normal incidence that is Normal incidence on a Perfect conductor.

i.e an EM wave propagating in free space strikes suddenly a conducting Boundary which means the other medium is a conductor.

The figure shows a plane Wave which is incident normally upon a boundary between free space and a perfect conductor.

assume the wave is propagating in positive z-axis and the boundary is z=0 plane.

The transmitted wave  since the electric field intensity inside a perfect conductor is zero.

The incident   and reflected   waves are in the medium 1  that is free space.

The energy transmitted is zero so the energy absorbed by the conductor is zero and entire wave is reflected to the same medium

now incident wave is 

  in free space  for medium 1

 ( a wave propagating in positive z-direction) and the reflected wave is  ( a wave propagating in positive z-direction).


by using tangential components  .

The resultant wave is   .


the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations


This is the wave equation which represents standing wave , which is the contribution of incident and reflected waves. as this wave is stationary it does not progress.

it has maximum amplitude at odd multiples of   and minimum amplitude at multiples of  .

Similarly The resultant Magnetic field is

The resultant wave is   .


the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations


this wave is  a stationary wave  it has minimum amplitude at odd multiples of   and maximum amplitude at multiples of  .


Nature of Magnetic materials

In order to find out the various types of materials in magnetic fields and their behaviour we use the knowledge of the action of magnetic field on a current loop with a simple model of an atom

Magnetic materials are classified on the basis of presence of magnetic dipole moments in the materials.

a charged particle with angular momentum always contributes to the permanent contributions to the angular moment of an atom

1. orbital magnetic dipole moment.

2. electron spin moment.

3. Nuclear spin magnetic moment.

Orbital Magnetic dipole Moment:-

The simple atomic model is one which assumes that there is a central positive nucleus surrounded by electrons in various circular orbits.

an electron in an orbit is analogous to a small current loop and as such experiences a torque in an external magnetic field, the torque tending to align the magnetic field produced by the orbiting electron with the external magnetic field.

Thus the resulting magnetic field at any point in the material would be greater than it would be at that point when the other moments were not considered.

so there are Quantum numbers which describes the orbital state of notion of electron in an atom there are n,l and ml

n-principal Quantum number, which determines the energy of an electron.

l-Orbital Quantum number which determines the angular momentum of orbit.

ml-magnetic Quantum number which determines the component of magnetic moment along the direction of an electric field.

electron spin Magnetic Moment:-

The angular momentum of an electron is called spin of the electron. as electron is a charged particle the spin of the electron produces magnetic dipole moment because electron is spinning about it’s own axis and thus generates a magnetic dipole moment.

\pm 9X 10^{-24} A-m^{2} is the value of electron spin when we consider an atom those electrons which are in shells which are not completely filled with contribute to a magnetic moment for the atom.

Nuclear spin Magnetic Moment:-

a third contribution of the moment of an atom is caused by nuclear spin this provides a negligible effect on the overall magnetic properties of material

That is the mass of the nucleus is much larger than an electron thus the dipole moments due to nuclear spin are very small.

so the total magnetic dipole moment of an atom is nothing but the summation of all the above mentioned .

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.

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  .

1 Star

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.                                                                                                                             

Digital Communications Quiz with solutions



  1. The band width  needed to transmit Television video plus audio signals of bandwidth 4.2 MHz using Binary PCM  quantization level of 512 is (  c     )         a. 2 MHz            b. 25.6MHz                        c.37.8MHz                    d.75.6MHz
  2. A signal m(t)has a bandwidth of 1.0 KHz and exhibits a maximum rate of change of 2.0 volts/sec. The signal is sampled at a sampling frequency of 20 KHz and quantized using delta modulator. The minimum step size to avoid slope overload is (  b    )                                                                                                                      a. 1.0mV        b. 0.1mV                c.10.0mV                      d.0.01mV
  3. For a 10 bit PCM system, the signal to Quantization noise ratio is 62 dB. If the number of bits are increased by 2, then the signal to Quantization noise ratio will be (  c      )                                                                                                                    a. Increased by 6 dB                                   b.  Decrease by 6 dB                                       c. Increased by 12 dB                                  d. Decrease by 12 dB
  4. Write the condition to eliminate slope overload error in a Delta modulation system—————————
  5. Write the condition to eliminate slope overload error in a Delta modulation system if the input is a single-tone signal—————————
  6. For which value of A , A-law has linear transfer characteristics——————–.
  7. For which value of µ , µ-law has linear transfer characteristics——————–.
  8. In a PCM system if the step size is 5V , then the Quantization noise in dB is   (  d    )                                                                                                                                             a. -5 dB.          b. -3.18dB       c. -10dB                      d. 3.18dB.
  9. Draw the characteristics of Mid-rise and Mid-tread type Quantizers ————–
  10. The maximum slope of the signal ———————-
  11. The sampling rate of the signal ————————–
  12. The step size in a 8- bit pcm system if the input signal to PCM is oscillating between [+4V,-4V] is—————————–



  1. Entropy is a measure of ——————————– 3M.  
  2. The capacity of a band – limited AWGN channel in terms of kbps if the average received signal power to noise power spectral density is 1000 and the bandwidth is approximately infinite is (  a    )   [Hint: Shannon’s bound Cinfinity =  S/No]                                                                                                                        a.  1.44        b. 1.08                                c. 0.72                           d. 0.36
  3. If Y= g(X) where g denotes a deterministic function, then the conditional entropy H(Y/X) is (  b ) 3M.                                                                                                           a. ≠ 1         b.  = 0                           c. = 1                           d. ≠ 0
  4. A Source generates three symbols with probability of 0.25, 0.25, and 0.50 at a rate of 3000 symbols per second. Assuming the symbols are generated independently from the source, the most efficient source encoder would have average bit rate of ( b ) 4M.                                                                                             a. 6000 bits/sec                                                         b. 4500 bits/sec                                 c. 3000 bits/sec                                                         d. 1500 bit/sec.
  5. A source generates four equi-probable symbols. If the source coding is adopted, the average length of the code for 100% efficiency is ( c )                       a. 6 bits / symbol                                                          b.   3 bits / symbol                           c.  2 bits / symbol                                                        d.  4 bits / symbol
  6. Draw the channel diagram of Binary Erasure channel and write its channel matrix ………………………..
  7. For a Binary Symmetric Channel the entropies of X & Y are if the conditional probability p=0.5, where X & Y are input & output random variables of BSC  (   d     )3M                                                                                                                          a. 1,0           b. 0,0               c. 0,1                           d. 1,1
  8. Match the following   (  d    )                                                                                                       a. Lossless channel                                           1.  only one non-zero element in each row                                                                                                                                        b. Deterministic channel                                2. Only one non-zero element in each row and each column.                                                                                                   c. Noiseless channel                                           3. Only one Non-zero element in each column.                                                                                                                                d. for a noiseless channel                                  4. H(Y/X)=0 and H(X/Y)= 0. a. none of these                                                  b.   b-3,c-4,a-1,d-2                                    c. c-4,d-3,a-1,b-2                                             d.  b-1,c-2,a-3,d-4.
  9. Given a Binary Symmetric Channel, the expression for Entropy is (pis conditional probability of error   ——————- 3M.  

Digital Communications Unit-3 Quiz with Solutions



  1. The number of Parity check bits in an (n, k) Linear Block codes are (    b      )      a.n                b. (n-k)                       c.  (n+k)                       d. k
  2. The Hamming Weight of the following code words 10011101 & 00111100 is      (   c   ) 2M.                                                                                                                                       a. None of these      b. 4, 5              c.5, 4                           d.3,4
  3. A cyclic code can be generated using———————— and A block can be generated using——————-.(   c   )                                  a.Generator matrix & Generator polynomial.                                                          b.Generator matrix & Generator matrix.                                                            c.Generator polynomial & Generator matrix.                                                            d.None of the above.
  4. The rate of a Block code is the ratio of(    c     )                                                                a.Message length to Block length.                 b.Block length to message length.  c.Message weight to Block length.                 d.None of the mentioned.
  5. The syndrome in LBC is calculated using , where Y represents received code word   (     a       ) 2M                                                                                                                      a. S= Y HT          b. S = YH                    c. S= YT H                  d. S= YT HT
  6.  A non-Zero value of Syndrome in a Block code represents ( b    ) 2M.                  a.No error during transmission.     b.An error occurred during transmission.   c.Both a and  b                                         d.None of the above.
  7. The transmitted code word(X) in an LBC can be obtained from received code word( Y) by using the equation (  a   ), where E represents error vector.              a. X= E + Y           b.  X = X.Y                  c. X= E.Y                    d. X= X/Y
  8. The parity check matrix of a (6,3) block code if Generator matrix is G  3M.             ——————-.
  9. In the above question find the Code words corresponding to message vectors [110] and [111]——————-. Ans: [110110], [111000].
  10. For the Q.8. Find the syndrome value when the received code word is 001111—-4MAns: syndrome value [0 0 1]
  11. For a (7,4) cyclic code , the generator polynomial is given as find the codeword  for the data 1100       Non systematic codeword is—————- .Ans:  1011100              Systematic codeword is——————–.Ans: 1011100.



    1. The modulation technique that provides minimum probability of error is (  b   ). a. ASK                  b. PSK                   c. DPSK                        d.FSK
    2. At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by (  c  )                                                                                                         a. 6 dB                b. 2 dB           c.  3 dB           d.  0 dB             e. None of these.
    3. If Eb, the energy per bit of a binary digital signal, is 10-5watt-sec and the one-sided power spectral density of the white noise, No= 10-6 W/Hz, then the output SNR of the matched filter is  (   d    )                                               a. 26 dB         b.  20 dB          c. 10 dB          d. 13 dB          e. None of these
    4. In which system, bit stream is portioned into even and odd stream (   c   )  a. BPSK                  b.  MSK            c. QPSK                     d.  FSK
    5. Optimum filter can be called as———when the input noise is white noise (  a   ) 2M.                                                                                                                        a. Matched filter   b.  High pass filter        c. Low pass filter      d. None of these
    6. The probability of error of ASK is ————————————.
    7. The probability of error of FSK is ————————————.
    8. Write the expression for QPSK modulation —————————————-2M.
    9. Draw the block diagram of DPSK system————————————.
    10. A pulse g(t) = A cos(πt/2T) for 0 ≤ t ≤ T is transmitted over an AWGN channel with two sided noise power spectral density No/2 Watts/Hz. The impulse response of the matched filter is (  a     )                                                           a. A sin (πt/2T)                                                            b.   A rec (πt/2T)                           c. A rec (π(T-t)/2T)                                                    d. A sin (π(T-t)/2T)           
    11. If the probability of error function of a modulation scheme is Pe = (½) erfc(x) thenthe same Pe interms of Q-function is ( b   )                                             a. Q ( 31/2 * x)                    b. Q (21/2 * x)                c. Q(x)             d. none of these.



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