Interview Questions and Answers on Digital Modulation Techniques

Q.1. What are the types of Digital Modulation Techniques ? Explain .


Ans. Basically, digital modulation techniques may be classified into coherent or non-coherent techniques, depending on whether the receiver is equipped with a phase-recovery circuit or not. The phase-recovery circuit ensures that the oscillator supplying the locally generated carrier wave to the receiver is synchronized to the oscillator supplying the carrier wave used to originally modulate the incoming data stream in the transmitter .

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(i) Coherent Digital Modulation Techniques

Coherent digital modulation techniques are those techniques which employ coherent detection. In coherent detection, the local carrier generated at the receiver is phase locked with the carrier at the transmitter. Thus, the detection is done by correlating received noisy signal and locally generated carrier. The coherent detection is a synchronous detection.

(ii) Non-coherent Digital Modulation Techniques

Non-coherent Digital Modulation Techniques are those techniques in which the detection process does not need receiver carrier to be phase locked with transmitter carrier.

The advantage of such type of system is that the system becomes simple. But the drawback of such system is that the error probability increases.

Q.2. State various continuous wave (CW)  modulation systems .

Ans. ASK, FSK, PSK, QPSK, MSK, M-ary PSK etc.

Q.3. Represent ASK mathematically .

Ans. VASK(t) = d(t) sin (2πfct)       where d(t) = 1 or 0 .

Q.4. State the bandwidth requirement of ASK system.

Ans. Maximum bandwidth Bmax = 2 fHz.

Q.5. In BASK the information is transmitted via ________ .

Ans. Amplitude switching of the carrier .

Q.6. In a CW modulation system, the input signal is _______ and transmitted signal is _________ .

Ans. Digital, Analog (sinusoidal) .

Q.7. In a BPSK system , the information is transmitted via ________ .

Ans. Phase change of the carrier.

Q.8. The phase difference between the two possible transmitted signals in BPSK is ____ .

Ans. 180° or π radians.

Q.9. Express the BPSK signal mathematically .


Q.10. What does √(2Ps) represent ?

Ans. √(2Ps) is equal to amplitude A of the carrier .

Q.11. The noise immunity of BPSK system is  _______ than that of the BASK system .

Ans. Better.

Q.12. What type of receiver is used for the BPSK detection ?

Ans. Synchronous detector .

Q.13. What is its speciality ?

Ans. The received BPSK signal is multiplied with the locally generated synchronous carrier.

Q.14. What is the advantage of synchronous detection ?

Ans. Error probability is reduced.

Q.15. What is the disadvantage of the BPSK receiver ?

Ans. The receiver is complicated, because it requires a synchronous carrier to be generated locally .

Q.16. What is the role of bit synchronizer in the BPSK receiver ?

Ans. The bit synchronizer will open and close the switched connected in the integrator circuit. It ensures that the integration will take place exactly over one bit duration .

Q.17. What is the maximum Bandwidth of BPSK system ?

Ans. B.W. = 2 fHz

Q.18. State the value of Euclidean distance for BPSK.

Ans. d = 2√Eb

Q.19. Bandwidth of BPSK is _______ than that of BFSK.

Ans. Lower

Q.20. What is merit of DPSK on BPSK ?

Ans. DPSK eliminates the ambiguity of whether the received data was inverted or not .

Q.21. What type of receiver is used for DPSK ?

Ans. We do not require a synchronous carrier to demodulate DPSK .

Q.22. How many phases are transmitted in DPSK ?

Ans. Two (same as BPSK ).

Q.23. What is the symbol duration in DPSK ?

Ans. Symbol duration T= 2 Tb

Q.24. What is the bandwidth of DPSK ?

Ans. BW = fb  which is half of the bandwidth of BPSK .

Q.25. State important disadvantages of DPSK .


  1. Noise in one bit interval causes error in two bit positions
  2. Error rate in DPSK is higher than that of BPSK
  3. The effect of noise is higher in DPSK than BPSK

Q.26. State true or false : Errors occur in pairs in DPSK.

Ans. True

Q.27. QPSK is a ________ .

(a)  Two level modulation

(b) Multilevel modulation

(c) None of the above

Ans. (b) Multilevel modulation

Q.28. The bit rate is ________ than baud rate in QPSK .

Ans. Double .

Q.29. Express QPSK mathematically .


Q.30. How many phases are transmitted in QPSK ?

Ans. Four

Q.31. What is the phase difference between the adjacent messages in QPSK ?

Ans. π/4 radians or 45°.

Q.32. In QPSK, bo(t) and be(t) ________.

Ans. can not change simultaneously .

Q.33. The offset between bo(t) and be(t) is ________ .

Ans. Tb

Q.34. The maximum phase change in OQPSK is ______ .

Ans. π/4

Q.35. The maximum phase change in QPSK is _______ .

Ans. π/2

Q.36. The Euclidean distance for QPSK is ____ .

Ans. d = 2 √Eb

Q.37. The error probability of QPSK is ______ .

(a) Better than BPSK

(b) Inferior to BPSK

(c) same as BPSK

Ans. (c) same as BPSK

Q.38. Bandwidth of QPSK is _____ as compared to that of BPSK .

Ans. Half

Q.39. What are the advantages of QPSK system ?


  1. BW is small
  2. Noise immunity is high
  3. Bit rate doubles for same baud rate

Q.40. What is the main advantage of M-ary PSK ?

Ans. The reduction in the channel bandwidth decreases with increase in number of messages (M) . This is the advantage .

Q.41. Compare error probability of M-ary PSK and QPSK .

Ans. Probability of error increases as M is increased .

Q.42. Represent M-ary PSK signal mathematically.


Q.43. State the expression for Euclidean distance for M-ary PSK .

Ans. d = 2√Esin (π/8)

Q.44. What is the bandwidth of M-ary PSK ?

Ans. BW = 2 fb/N , where N= Number of bits per message

Q.45. What is the relation between M and N ?

Ans. M = 2N

Q.46. What is the bandwidth of a 16-ary PSK system ?

Ans. BW = 2 fb/N = 2 fb/4 = fb/2 Hz

Q.47. What is QASK ?

Ans. QASK is a system in which amplitude and phase both of a transmitted signal are changed.

Q.48. Why is amplitude changed along with phase ?

Ans. Because it gives a better noise performance.

Q.49. The noise immunity of a 16 QAM is _____ than 16-ary PSK .

Ans. Better

Q.50. State the expression for Euclidean distance for 16-QAM system.

Ans. d = √(0.6Eb)

Q.51. Compare 16-QAM with QPSK on the basis of error rates .

Ans. 16-QAM has lower error rate as compared to that of QPSK.

Q.52. What is the bandwidth of QASK .

Ans. BW = 2fb/N

Q.53. State the expression for BFSK .


Q.54. How is a message transmitted in BFSK ?

Ans. The frequency of a sinusoidal carrier is switched between two values to represent 0 or 1 .

Q.55. What is the bandwidth of BFSK ?

Ans. BW = 4 f

Q.56. State true or false: BFSK signal consists of two BPSK terms.

Ans. True

Q.57. The bandwidth efficiency of BFSK IS _______ as compared to BPSK .

Ans. Worse

Q.58. What type of receiver is used for BFSK reception?

Ans. Asynchronous detector

Q.59. Error probability of BFSK is ________ as compared to BPSK .

Ans. Higher

Q.60. What is meant by orthogonal BFSK ?

Ans. In orthogonal BFSK, the two carriers u1(t) and u2(t) are orthogonal i.e., frequencies fand fare integer multiple of fb.

Q.61. What is the advantage of M-ary FSK over BFSK ?

Ans. Better noise immunity

Q.62. State the expression for bandwidth of an M-ary FSK .

Ans. BW = 2N+1 fb/N

Q.63. What are the disadvantages of M-ary FSK ?


  1. A number of correctly tuned filters are required
  2. Large bandwidth requirement

Q.64. What type of reception technique is used for M-ary FSK ?

Ans. Asynchronous demodulation using filters.

Q.65. What is the effect of abrupt phase change on amplitude in QPSK ?

Ans. Related with abrupt phase changes, there are always sudden amplitude variations. OQPSK is less affected than QPSK .

Q.66. What is the speciality of MSK in this contest ?

Ans. MSK exhibits  a phase continuity. So there are no amplitude variations in the transmitted signal.

Q.67. MSK is basically _________.

Ans. FSK

Q.68. What is the expression for Euclidean distance for MSK ?

Ans. d = √(4Eb)

Q.69. Compare error probability for MSK with QPSK .

Ans. Since, the expression for d is same as that for QPSK, the error probability of MSK is same as that of QPSK .

Q.70. What is the BW of MSK ?

Ans. BW = 1.5 f