## [MCQ’s]Principal of Communication [IT]

#### Module 01

1. What we use for impedance matching in RF amplifiers?
a) RC coupling
b) Transformer coupling
c) Direct coupling
d) RF coupling
Explanation: Impedance matching is the input impedance of an electrical load or the output impedance of corresponding electrical signal source to maximize power transfer or minimize signal reflection from the load. Transformer coupling is basically used for impedance matching in RF amplifiers. It is usually used with a small load for power amplification.

2. Neutralization cancels unwanted feedback by bypassing the feedback to the neutral or ground plane.
a) True
b) False
Explanation: Neutralization cancels unwanted feedback by adding feedback out of phase with the unwanted feedback. It is a method of filtering noise out of signals.

3. In a receiver, distortion can occur in ________
a) Mixer
b) Detector
c) IF amplifiers
d) Either mixer or detector or IF amplifiers
Explanation: Distortion in a receiver can occur in either mixer or in the detector. It can also occur in IF amplifiers. Distortion is the change in the shape of the waveform.

4. How we limit the response of a receiver to a weak signal?
a) by the AGC
b) by the noise that is generated in the receiver
c) by the dynamic range of the receiver
d) by the type of detector circuit being used
Explanation: The response of a receiver to weak signals is limited by the noise. This noise is generated in the receiver only. Noise is the unwanted signal that is present in the passband of the signal.

5. When aliasing will take place?
a) Sampling signals less than Nyquist Rate
b) Sampling signals more than Nyquist Rate
c) Sampling signals equal to Nyquist Rate
d) Sampling signals at a rate which is twice of Nyquist Rate
Explanation: Aliasing causes different signals to become indistinguishable when sampled. It happens when the sampling rate is less than Nyquist rate. To prevent aliasing, sampling signals should be at a rate which is twice of Nyquist Rate.

6. A resonant circuit is a simple form of bandpass filter.
a) True
b) False
Explanation: The quality factor Q, of a resonant circuit is a measure of quality of resonant circuit, thus indicating the performance of the resonant circuit. Bandwidth can be seen as the ratio of the resonant frequency to quality factor. A higher value for Q means a more narrow bandwidth, which we want in many applications.

7. Which statement is true for high level amplitude modulation?
a) all RF amplifiers are of class A
b) all RF amplifiers can be nonlinear
c) minimum RF power is required
d) maximum RF power is required
Explanation: In high level AM, modulation is done at high power of carrier and modulating signal therefore output power is high. Therefore, power amplifiers are not used to boost the carrier and modulating signal. Thus, all RF amplifiers are non-linear.

8. Which statement is true for low level amplitude modulation?
a) all RF amplifiers are of class A
b) all RF amplifiers can be linear
c) minimum RF power is required
d) maximum RF power is required
Explanation: For low level AM, modulation is done at low power of carrier and modulating signal, so power amplifiers are used to boost the carrier and modulating signal. Therefore, output power is low. Thus, the amplifiers used are linear.

9. Which two networks can be used for impedance matching?
a) pi network and T network
b) pi network and bridge network
c) bridge network and T network
d) pi network and omega network
Explanation: If a source with low impedance is connected with a load with high impedance then the power that can pass through the connection is limited by higher impedance. Generally the two networks, pi and T are used for impedance matching.

10. Which statement is true about frequency multipliers?
a) they are essentially balanced modulators
b) they are essentially class C amplifiers
c) they are essentially class AB amplifiers
d) they are essentially mixers
Explanation: Frequency multiplier multiplies the frequency of the input by an integer, such that the output is a multiple of the input. Frequency multipliers are generally class C amplifiers.

11. Which of the following is the right way of representation of equation that contains only the positive frequencies in a given x(t) signal?
a) X+(F)=4V(F)X(F)
b) X+(F)=V(F)X(F)
c) X+(F)=2V(F)X(F)
d) X+(F)=8V(F)X(F)
Explanation: In a real valued signal x(t), has a frequency content concentrated in a narrow band of frequencies in the vicinity of a frequency Fc. Such a signal which has only positive frequencies can be expressed as X+(F)=2V(F)X(F)
Where X+(F) is a Fourier transform of x(t) and V(F) is unit step function.

12. What is the equivalent time –domain expression of X+(F)=2V(F)X(F)?
a) F(+1)[2V(F)]*F(+1)[X(F)]
b) F(-1)[4V(F)]*F(-1)[X(F)]
c) F(-1)[V(F)]*F(-1)[X(F)]
d) F(-1)[2V(F)]*F(-1)[X(F)]
Explanation: Given Expression, X+(F)=2V(F)X(F).It can be calculated as follows
x+(t)=X+(F)ej2πFtdF
=F1[2V(F)]F1[X(F)]

13. In time-domain expression, x+(t)=F1[2V(F)]F1[X(F)]. The signal x+(t) is known as
a) Systematic signal
b) Analytic signal
c) Pre-envelope of x(t)
d) Both Analytic signal & Pre-envelope of x(t)
Explanation: From the given expression, x+(t)=F1[2V(F)]F1[X(F)].

14. In equation x+(t)=F1[2V(F)]F1[X(F)], if F1[2V(F)]=δ(t)+j/πt and F1[X(F)] = x(t). Then the value of ẋ(t) is?
a) 1πx(t)t+τdτ
b) 1πx(t)tτdτ
c) 1π2x(t)tτdτ
d) 1π4x(t)tτdτ
Explanation: x+(t)=[δ(t)+j/πt]x(t)
x+(t)=x(t)+[j/πt]x(t)
(t)=[j/πt]x(t)
=1πx(t)tτdτ Hence proved.

15. If the signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt, -∞ < t < ∞ when excited by the input signal x(t) then such a filter is called as __________
a) Analytic transformer
b) Hilbert transformer
c) Both Analytic & Hilbert transformer
d) None of the mentioned
Explanation: The signal ẋ(t) can be viewed as the output of the filter with impulse response h(t) = 1/πt,
-∞ < t < ∞ when excited by the input signal x(t) then such a filter is called as Hilbert transformer.

16. What is the frequency response of a Hilbert transform H(F)=?
a) ⎧⎩⎨j(F>0)0(F=0)j(F<0)
b) ⎧⎩⎨⎪⎪j0j(F<0)(F=0)(F>0)
c) ⎧⎩⎨⎪⎪j0j(F>0)(F=0)(F<0)
d) ⎧⎩⎨⎪⎪j0j(F>0)(F=0)(F<0)
Explanation: H(F) =h(t)ej2πFtdt
=1π1/te2πFtdt
=⎧⎩⎨⎪⎪j0j(F>0)(F=0)(F<0)
We Observe that │H (F)│=1 and the phase response ⊙(F) = -1/2π for F > 0 and ⊙(F) = 1/2π for F < 0.

17. What is the equivalent lowpass representation obtained by performing a frequency translation of X+(F) to Xl(F)= ?
a) X+(F+Fc)
b) X+(F-Fc)
c) X+(F*Fc)
d) X+(Fc-F)
Explanation: The analytic signal x+(t) is a bandpass signal. We obtain an equivalent lowpass representation by performing a frequency translation of X+(F).

18. What is the equivalent time domain relation of xl(t) i.e., lowpass signal?
a) xl(t)=[x(t)+j(t)]ej2πFct
b) x(t)+j ẋ(t) = xl(t)ej2πFct
c) xl(t)=[x(t)+j(t)]ej2πFct & x(t)+j ẋ(t) = xl(t)ej2πFct
d) None of the mentioned
Explanation: xl(t)=x+(t)ej2πFct
=[x(t)+j(t)]ej2πFct
Or equivalently, x(t)+j ẋ(t) =xl(t)ej2πFct.

19. If we substitute the equation xl(t)=uc(t)+jus(t) in equation x (t) + j ẋ (t) = xl(t) ej2πFct and equate real and imaginary parts on side, then what are the relations that we obtain?
a) x(t)=uc(t)cos2πFct+us(t)sin2πFct; ẋ(t)=us(t)cos2πFctuc(t)sin2πFct
b) x(t)=uc(t)cos2πFctus(t)sin2πFct; ẋ(t)=us(t)cos2πFct+uc(t)sin2πFct
c) x(t)=uc(t)cos2πFct+us(t)sin2πFct; ẋ(t)=us(t)cos2πFct+uc(t)sin2πFct
d) x(t)=uc(t)cos2πFctus(t)sin2πFct; ẋ(t)=us(t)cos2πFctuc(t)sin2πFct
Explanation: If we substitute the given equation in other, then we get the required result

20. In the relation, x(t) = uc(t)cos2πFctus(t)sin2πFct the low frequency components uc and us are called _____________ of the bandpass signal x(t).
c) Triplet components
d) None of the mentioned
Explanation: The low frequency signal components uc(t) and us(t) can be viewed as amplitude modulations impressed on the carrier components cos2πFct and sin2πFct, respectively. Since these carrier components are in phase quadrature, uc(t) and us(t) are called the Quadrature components of the bandpass signal x (t).

21. What is the other way of representation of bandpass signal x(t)?
a) x(t) = Re[xl(t)ej2πFct]
b) x(t) = Re[xl(t)ejπFct]
c) x(t) = Re[xl(t)ej4πFct]
d) x(t) = Re[xl(t)ej0πFct]
Explanation: The above signal is formed from quadrature components, x(t) = Re[xl(t)ej2πFct] where Re denotes the real part of complex valued quantity.

22. In the equation x(t) = Re[xl(t)ej2πFct], What is the lowpass signal xl (t) is usually called the ___ of the real signal x(t).
a) Mediature envelope
b) Complex envelope
c) Equivalent envelope
d) All of the mentioned
Explanation: In the equation x(t) = Re[xl(t)e(j2πFct)], Re denotes the real part of the complex valued quantity in the brackets following. The lowpass signal xl (t) is usually called the Complex envelope of the real signal x(t), and is basically the equivalent low pass signal.

23. If a possible representation of a band pass signal is obtained by expressing xl (t) as xl(t)=a(t)ejθ(t) then what are the equations of a(t) and θ(t)?
a) a(t) = u2c(t)+u2s(t)−−−−−−−−−−√ and θ(t)=tan1us(t)uc(t)
b) a(t) = u2c(t)u2s(t)−−−−−−−−−−√ and θ(t)=tan1us(t)uc(t)
c) a(t) = u2c(t)+u2s(t)−−−−−−−−−−√ and θ(t)=tan1uc(t)us(t)
d) a(t) = u2s(t)u2c(t)−−−−−−−−−−√ and θ(t)=tan1us(t)uc(t)
Explanation: A third possible representation of a band pass signal is obtained by expressing xl(t)=a(t)ejθ(t) where a(t) = u2c(t)+u2s(t)−−−−−−−−−−√ and θ(t)=tan1us(t)uc(t).

24. What is the possible representation of x(t) if xl(t)=a(t)e(jθ(t))?
a) x(t) = a(t) cos[2πFct – θ(t)]
b) x(t) = a(t) cos[2πFct + θ(t)]
c) x(t) = a(t) sin[2πFct + θ(t)]
d) x(t) = a(t) sin[2πFct – θ(t)]
Explanation: x(t) = Re[xl(t)ej2πFct]
= Re[a(t)ej[2πFct+θ(t)]]
a(t)cos[2πFct+θ(t)]
Hence proved.

25. In the equation x(t) = a(t)cos[2πFct+θ(t)], Which of the following relations between a(t) and x(t), θ(t) and x(t) are true?
a) a(t), θ(t) are called the Phases of x(t)
b) a(t) is the Phase of x(t), θ(t) is called the Envelope of x(t)
c) a(t) is the Envelope of x(t), θ(t) is called the Phase of x(t)
d) none of the mentioned
Explanation: In the equation x(t) = a(t) cos[2πFct+θ(t)], the signal a(t) is called the Envelope of x(t), and θ(t) is called the phase of x(t).

26. Notch is a ________
a) High pass filter
b) Low pass filter
c) Band stop filter
d) Band pass filter
Explanation: Notch filter is a band stop filter that allows most frequencies to pass through it, except frequencies in a specific range. It is just opposite of a band-pass filter. High pass filter allows higher frequencies to pass while Low pass filter allows lower frequencies to pass through it.

27. Sin wave is ________
a) Aperiodic Signal
b) Periodic Signal
c) Random Signal
d) Deterministic Signal
Explanation: Periodic signal is that which repeats itself after a regular interval. Sin wave is a periodic function since it’s value can be determined at any point of time, as it repeats itself at a regular interval. Aperiodic Signal does not repeat itself at regular interval of time. Random signals are the signals which have uncertain values at any time. While Deterministic signals are the signals which are constant over a period of time.

28. What is the role of channel in communication system?
a) acts as a medium to send message signals from transmitter to receiver
b) converts one form of signal to other
c) allows mixing of signals
d) helps to extract original signal from incoming signal
Explanation: Channel acts as a medium to transmit message signal from source transmitter to the destination receiver. Transducer converts a signal from one form of energy to other. Mixer allows mixing of signals while Demodulator helps to extract original message signal from incoming signal.

29. Sum of a periodic and aperiodic signal always be an aperiodic signal.
a) True
b) False
Explanation: Periodic signal is a signal which repeats itself after a regular interval. While Aperiodic Signal does not repeat itself at regular interval of time.
For example: Let f(x) = sin(x), be a periodic function with period 2π and g(x) = −sin(x) + sin(√2x), be an aperiodic function. Now the sum of both i.e. f(x) + g(x) = sin(√2x), which is a periodic function.
Therefore, the sum of a periodic and aperiodic signal can be periodic.

30. Noise is added to a signal ________
a) In the channel
b) At receiving antenna
c) At transmitting antenna
d) During regeneration of information
Explanation: Noise is an unwanted signal that gets mixed with the transmitted signal while passing through the channel. The noise interferes with the signal and provides distortion in received signal. The transmitting antenna transmits modulated message signal while the receiving antenna receives the transmitted signal. Regeneration of information refers to demodulating the received signal to produce the original message signal.

31. Agreement between communication devices are called ________
a) Transmission medium
b) Channel
c) Protocol
d) Modem
Explanation: Protocol is a set of rules that looks after data communication, by acting as an agreement between communication devices. Channel is the transmission medium or the path through which information travels. Modem is a device that modulates and demodulates data.

32. What is the advantage of superheterodyning?
a) High selectivity and sensitivity
b) Low Bandwidth
d) Low fidelity
Explanation: The main advantage of superheterodyning is that it provides high selectivity and sensitivity. It’s bandwidth remains same. It has high adjacent channel rejection and high fidelity.

33. Low frequency noise is ________
a) Flicker noise
b) Shot noise
c) Thermal noise
d) Partition Noise
Explanation: Flicker noise is a type of electronic noise which is generated due to fluctuations in the density of carrier. It’s also known as 1/f as it’s power spectral density increases with a decrease in frequency or increase in offset from a signal.

34. Relationship between amplitude and frequency is represented by ________
a) Time-domain plot
b) Phase-domain plot
c) Frequency-domain plot
d) Amplitude-domain plot
Explanation: Relationship between amplitude and frequency is represented by a frequency-domain plot. Also, it represents the relation between phase and frequency. While a time-domain plot shows how a signal varies over time.

35. A function f(x) is even, when?
a) f(x) = -f(x)
b) f(x) = f(-x)
c) f(x) = -f(x)f(-x)
d) f(x) = f(x)f(-x)
Explanation: Geometrically a function f(x) is even, if plot of the function is symmetric over y-axis. Algebraically, for any function f(x) to be even, f(x) = f(-x).
While for a function f(x) to be odd, f(x) = -f(-x).

36. Cellular concept replaces many low power transmitters to a single high power transmitter.
a) True
b) False
Explanation: Cellular concept is a system level idea that replaces a single high power transmitter to many low power transmitters. High power transmitters lead to large cell, and thus it was impossible to use the same frequencies throughout the systems. But, it is possible with low power transmitter.

37. Why neighbouring stations are assigned different group of channels in cellular system?
a) To minimize interference
b) To minimize area
c) To maximize throughput
d) To maximize capacity of each cell
Explanation: Neighbouring base stations are assigned different group of channels. It minimizes the interference between base stations and the users under their control.

38. What is a cell in cellular system?
a) A group of cells
b) A group of subscribers
c) A small geographical area
d) A large group of mobile systems
Explanation: Cell is a small geographic area in a cellular system. Each cellular base station within a cell is allocated a group of radio channels that could be used in another cell.

39. What is frequency reuse?
a) Process of selecting and allocating channels
b) Process of selection of mobile users
c) Process of selecting frequency of mobile equipment
d) Process of selection of number of cells
Explanation: Frequency reuse is the process of using the same radio frequencies on radio transmitter sites within a geographic area. They are separated by sufficient distance to cause minimal interference with each other.

40. Which of the following is a universally adopted shape of cell?
a) Square
b) Circle
c) Triangle
d) Hexagon
Explanation: Hexagonal cell shape is a simplistic model of radio coverage for each base station. It has been universally adopted since the hexagon permits easy and manageable analysis of a cellular system.

41. Actual radio coverage of a cell is called __________
a) Fingerprint
b) Footprint
c) Imprint
d) Matrix
Explanation: Actual radio coverage of a cell is known as the footprint. It is determined from field measurements or propagation prediction models. Although the real footprint is amorphous in nature, a regular cell shape is needed for systematic system design.

42. Why the shape of cell is not circle?
a) Omni directionality
b) Small area
c) Overlapping regions or gaps are left
d) Complex design
Explanation: Circle is the first natural choice to represent the coverage area of a base station. But while adopting this shape, adjacent cells cannot be overlaid upon a map without leaving gaps or creating overlapping regions.

43. What is the main reason to adopt hexagon shape in comparison to square and triangle?
a) Largest area
b) Simple design
c) Small area
d) Single directional
Explanation: For a given distance between the center of a polygon and its farthest perimeter points, the hexagon has the largest area. Thus, by using the hexagon geometry, the fewest number of cells can cover a geographic region.

44. Which type of antenna is used for center excited cells?
a) Dipole antenna
b) Grid antenna
c) Sectored antenna
d) Omnidirectional antenna
Explanation: For center excited cells, base station transmitters are used at the center of cell. To cover the whole cell, omnidirectional antenna is the best choice for base station transmitters.

45. Which type of antenna is used for edge excited cells?
a) Omnidirectional antenna
b) Grid antenna
c) Sectored directional antenna
d) Dipole antenna
Explanation: For edge excited cell, mostly base station transmitters are placed on three of the six cell vertices. To cover the assigned portion of a cell, sectored directional antenna is the best choice.

46. For a cellular system, if there are N cells and each cell is allocated k channel. What is the total number of available radio channels, S?
a) S=k*N
b) S=k/N
c) S=N/k
d) S=kN
Explanation: If there is a cellular system with total of S duplex channels. Each cell is allocated a group of k channels and there are total N cells in the system, S channels are divide among N cells into unique and disjoint channel groups. Therefore, total number of radio channel is the product of total number of cells in the system (N) and number of channel allocated to each cell (k).

47. What is a cluster in a cellular system?
a) Group of frequencies
b) Group of cells
c) Group of subscribers
d) Group of mobile systems
Explanation: Cluster is group of N cells. These cells use the complete set of frequency available for the cellular system at that location.

48. What is a frequency reuse factor for N number of cells in a system?
a) N
b) N2
c) 2*N
d) 1/N
Explanation: The frequency reuse factor is defined as 1 over the number of cells in the cluster of the system (N). It is given by 1/N since each cell within a cluster is only assigned 1/N of the total available channels in the system.

49. Capacity of a cellular system is directly proportional to __________
a) Number of cells
b) Number of times a cluster is replicated
c) Number of Base stations
d) Number of users
Explanation: The capacity of a cellular system is directly proportional to the number of times a cluster is replicated in a fixed area. If the cluster size N is reduced while the cell size is kept constant, more clusters are required to cover a given area, and hence more capacity is achieved.

50. A spectrum of 30 MHz is allocated to a cellular system which uses two 25 KHz simplex channels to provide full duplex voice channels. What is the number of channels available per cell for 4 cell reuse factor?
a) 150 channels
b) 600 channels
c) 50 channels
d) 85 channels
Explanation: Total bandwidth is 30 MHz. And the channel bandwidth is 50 KHz/duplex channel (25KHz*2). Therefore, total available channels are 600 channels (30,000/50). For 4 cell reuse factor, total number of channels available per cell will be 150 channels (600/4).

51. Which among the following is not used as an attribute for classifying spectrum sharing?
a) Architecture
b) Spectrum allocation behaviour
c) Spectrum access technique
d) Spectrum sensing behaviour
Explanation Based on the architecture, the spectrum sharing techniques are categorized as centralized and decentralized. Based on spectrum allocation behaviour, the techniques are categorized as cooperative and non-cooperative. Based on the spectrum access technique, the methods are categorized as overlay and underlay.

52. Which among the following statement is false about cooperative sensing?
a) Cooperative spectrum sharing is alternately known as collaborative spectrum sharing
b) Cooperative spectrum sharing studies the effect of communication of one node on other nodes
c) Cooperative spectrum sharing withholds the interference information
d) Cooperative spectrum sharing minimizes collision caused by users accessing the same portion of the spectrum
Explanation: Cooperative spectrum sharing studies the effect of communication of one node on other nodes. It also involves distributing the interference measurements among the users in the network. This information is utilized by spectrum allocation algorithms which provide spectrum access to users.

53. Which among the following is true non-cooperative spectrum sharing?
a) Non-cooperative spectrum sharing optimizes spectrum sharing algorithms
b) Non-cooperative spectrum sharing reduces effective spectrum utilization
c) Non-cooperative spectrum sharing distributes interference measurements to other users in the network
d) Non-cooperative spectrum sharing works only with spectrum broker
Example: Non-cooperative spectrum sharing only considers the communication of the one particular node. Due to the lack of information about other users, it exhibits reduced spectrum utilization. However due to minimal communication requirements, it is suitable for certain resource-constrained practical applications.

54. What is the parameter used to differentiate the overlay and underlay spectrum sharing technique?
c) Degree of interference exhibited towards primary user
d) Degree of interference exhibited towards secondary user
Explanation: The overlay and underlay spectrum sharing techniques are classified based on the amount of interference directed towards the primary users while operating in the licensed spectrum band. Overlay exhibits zero interference while the underlay technique exhibits considerable interference which is regarded as noise by primary users.

55. Which among the following method is employed in underlay spectrum sharing?
a) Time division multiple access
c) Frequency division multiple access
d) Time division duplexing
Explanation: Underlay spectrum sharing technique employs spread spectrum techniques while transmitting in the licensed portion of the spectrum band. Spread spectrum involves stretching a signal over a bandwidth which is wider than its original bandwidth. Therefore bandwidth utilization and interference towards the primary user is increased.

56. Which among the following is true about centralized spectrum sharing?
a) Centralized spectrum sharing does not rely on fixed infrastructure units
b) Centralized spectrum sharing holds each node responsible for spectrum allocation
c) Centralized spectrum sharing reduces the overhead of communication
d) Centralized spectrum sharing has a controlling entity for spectrum allocation and spectrum access
Explanation: Centralised spectrum sharing consists of a central entity for performing spectrum allocation and spectrum access. Each node in the network forwards information on the spectrum and the interference to the central entity. The central entity builds the spectrum allocation map using the collected information.

57. Which among the following has to be carried out after the spectrum is allocated for communication?
a) Listen for the arrival of the primary user
b) Handshake mechanism between transmitter and receiver
c) Estimate the duration of communication
d) Forward interference information to spectrum broker
Explanation: The selected portion of the spectrum must be informed to the receiver of data. This is accomplished by invoking a transmitter-receiver handshake protocol that is used for effective communication in xG networks.

58. Which among the following is not a step in spectrum sharing?
a) Spectrum sensing
b) Spectrum mobility
c) Spectrum access
d) Spectrum management
Explanation: Spectrum sharing process is integrated with spectrum sensing, spectrum allocation, spectrum access, handshake mechanism, and spectrum mobility. Thus the selection of best available spectrum, prevention of collision due to the overlapping portion of the spectrum, and seamless connection for communication regardless of the location of the user involves spectrum sharing.

59. Which among the following layer operation is similar to that of spectrum sharing?
a) Physical
c) Network
d) Transport
Explanation: The data link layer particularly the MAC portion is responsible for collision avoidance. Likewise, spectrum sharing is responsible for avoiding collision between licensed and unlicensed users of the licensed spectrum band as well as for avoiding collision among xG users in the unlicensed band.

60. Which among the following is a constraint observed in hardware constrained MAC protocol for cognitive radio networks?
a) Hidden terminal problem
b) Synchronization among xG users
c) Synchronization between xG user and primary user
Explanation: Sensing in a limited geographic area with a fixed bandwidth limit on secondary users is a challenge for hardware constrained MAC protocol. Thus hidden terminal problem is observed in the hardware constrained MAC protocol. The demand for synchronization among xG users is a problem of concern for cognitive MAC protocol.

61. Transmitter-receiver handshake does not involve any central entity such as the primary base-station.
a) True
b) False
Explanation: The transmitter-receiver handshake conveys the selected spectrum for communication to the receiver. This protocol is not limited between a transmitter and a receiver and may also include external entities such as a primary base-station especially in methods such as cooperative spectrum sharing.

62. Which among the following terms should replace the label ‘A’?

a) Centralised intra-network spectrum sharing
b) Centralised inter-network spectrum sharing
c) Distributed inter-network spectrum sharing
d) Distributed intra-network spectrum sharing
Explanation: xG networks are generally overlapped to maximize the opportunistic usage of a licensed spectrum band. An example of a centralized inter-network approach is the common spectrum coordination channel protocol designed to support the combined existence of IEEE 802.11b and IEEE 802.16a.

63. In amplitude modulation, which among the following is constant?
a) Amplitude
b) Frequency
c) Wave length
d) Time period
Explanation: In amplitude modulation, the carrier wave has constant frequency and the modulating wave information is conveyed by the amplitude of the carrier waves.

64. Modern phase techniques are capable of __________
a) Resolving modulation
b) Resolving amplitude
c) Resolving frequency
d) Resolving wave length
Explanation: Modern phase comparison techniques are able to possess a better resolving capacity than the remaining techniques. They can resolve better than 1/1000 part of a wavelength.

65. Lower frequency is not suitable in_________
a) Direct transmission
b) Distance calculation
c) Determination of wavelength
d) Determination of frequency
Explanation: The range of lower frequency is not suitable in case of direct transmission through the atmosphere because it may involve in atmospheric conditions like interference, reflection, fading and scattering. This may decrease the impact of frequency which may reduce the information being transmitted.

66. Which of the following represents the correct set of modulation classification?
a) Frequency, time period
b) Frequency, amplitude
c) Amplitude, wavelength
d) Wavelength, frequency
Explanation: The interference technique can be eradicated by modulation, which involves two classifications. They are amplitude and frequency modulations, which can be super imposed during phase comparison.

67. Which of the following indicates the correct set of frequency employed in measuring process?
a) 7*106 to 5*108 Hz
b) 7.5*106 to 4.5*108 Hz
c) 7.5*106 to 5.9*108 Hz
d) 7.5*106 to 5*108 Hz
Explanation: In general, the present situation needs a frequency range of approximately 7.5*106 to 5*108 Hz. This can be used in order to determine the distance between the points and also employed in EDM instruments.

68. Which of the following is constant in the case of frequency modulation?
a) Modulation
b) Wavelength
c) Amplitude
d) Frequency
Explanation: In frequency modulation, the carrier wave has constant amplitude and the modulating wave information is conveyed by the amplitude of the carrier waves.

69. Which can’t be done in high frequency zones?
a) Phase comparison
b) Super imposition of waves
c) Distance measurement
d) Wavelength measurement
Explanation: In high frequency zones, the phase comparison techniques cannot be applied. The high frequency may be determined as 5*108 Hz which may correspond to a wave length of 0.6 m.

70. Modulating wave can also be known as ______
a) Total wave
b) Measuring wave
c) Super wave
d) Incubation wave
Explanation: Modulation involves the overcoming of the problems raised due to the interference, scattering, etc. In this, the measuring wave is super imposed on a carrier wave of high frequency, so it is also known as measuring wave.

71. If 10mm is the accuracy considered, what will be the maximum value of λ for 1/1000 part?
a) 10000 m
b) 10 cm
c) 10 m
d) 10000 cm
Explanation: The maximum value of the wave length can be determined by multiplying assumed wave length with the accuracy considered, which means, λ = 10*1000 = 10 m.

72. Frequency modulation is equipped in all EDM instruments.
a) True
b) False
Explanation: In frequency modulation, the carrier wave has constant amplitude and frequency varies in proportion to the amplitude of the modulating wave. Frequency modulation is used in all EDM instruments, while amplitude modulation is done in visible light instruments and infrared instruments.

73. What is the full-form of UTP?
a) Unshielded Twisted-Pair copper wire
b) Unshielded Twisted-Pair Coaxial wire
c) Uniquely Twisted-Pair copper wire
d) Uniquely Twisted-Pair Coaxial wire
Explanation: UTP stands for Unshielded Twisted-Pair copper wire. Unshielded in UTP refers to the lack of metallic shielding around copper wires. The twisted pair helps to minimize electronic interference.

74. NIC stands for ________
a) Network Interface Card
b) Network Interface Cable
c) Network Interface Code
d) Network Internal Cord
Explanation: NIC stands for Network Interface Card. It is a computer circuit board that is installed in a computer so that it can be connected to a network.

75. Why CSMA/CD network has a maximum length for cables?
a) to increase the data rate
b) to decrease the data rate
c) to prevent packets from reaching all other nodes
d) to make sure that all other nodes hear a collision in progress
Explanation: CSMA/CD is a media access control used in early Ethernet technology for local area networking. Cables used in CSMA/CD has a maximum length when compared to others to make sure that all other nodes hear a collision in progress.

76. The Clapp oscillator is _________
a) an oscillator made using FET
b) a type of crystal controlled oscillators
c) modified form of Colpitts oscillator
d) modified form of Hartley oscillator
Explanation: Clapp oscillator is just an advanced form of Colpitts oscillator.

77. What is the total voltage, when two noise voltage V1 and V2 are combined?
a) V1×V1+V2×V2−−−−−−−−−−−−−−√
b) V1+V22
c) V1×V2−−−−−−√
d) V1 + V2
Explanation: When two noise voltage V1 and V2 are combined, the resulting total voltage VT comes out to be V1×V1+V2×V2−−−−−−−−−−−−−−√.

78. Shot noise is generated in ________
a) resistors
b) inductors
c) transistors and diodes
d) capacitors
Explanation: Shot noise originate from the discrete nature of electric charge. It is mostly generated in transistors and diodes.

79. Radians per second is equal to _________
a) 2πf
b) phase angle
c) phase deviation
d) phase swing
Explanation: Radians per second is equal to the value 2πf.

80. Repeaters are not required for fiber optic cable lengths upto ________
a) 10km
b) 100km
c) 5km
d) 12km
Explanation: Repeaters are devices which help in retransmitting a signal more powerfully. If the length of fiber optic cable is up to 100 km, there is no need for repeaters in that circuit.

81. Which of the following is not a name for phases present in a system of material in various conditions?
a) Phase diagram
b) Equilibrium diagram
c) Interstitial diagram
d) Constitutional diagram
Explanation: A phase diagram is a graphical representation of the phases present in the system of materials at various temperatures, pressures, and compositions. These diagrams show the constitution of alloys as a function of temperature under equilibrium conditions. They are otherwise also known as equilibrium or constitutional diagrams.

82. Which of the following cannot be obtained using a phase diagram?
a) Melting temperatures of various phases
b) Temperature range for solidification
c) Equilibrium solid solubility
d) Purity of materials
Explanation: A phase diagram is a graphical representation of the phases present in the system of materials at various temperatures, pressures, and compositions. It can be used to determine the melting temperature of various phases, the range of solidification, and the equilibrium solid solubility of one element in another.

83. A specific body of material or a series of alloys with the same compositions is/are known as _________
a) Component
b) System
c) Alloy
d) Solute
Explanation: A component is considered as a pure metal or compound of an alloy, whereas alloys are mixtures of two or more metals or non-metals. The system may be defined as either a specific body of material or a series of possible alloys consisting of the same components.

84. How many types of systems are applicable for phase diagrams?
a) One
b) Two
c) Three
d) Four
Explanation: System may be defined as either a specific body of material or a series of possible alloys consisting of the same components. A system having one component is called as a unary system. Similarly, two, three, and four component systems are called binary, ternary, and quaternary systems respectively.

85. The maximum concentration of solute that can be added is defined as ____________
a) Solution limit
b) Solubility limit
c) Concentration
d) Degrees of freedom
Explanation: Solubility limit is defined as the maximum concentration of solute that may be added without the formation of a new phase. An excess addition may result in the formation of another solid solution or compound. Degrees of freedom is defined as the number of independent variables that can be changed independently without changes in the phases of the system.

86. In this figure, what does O denote?

a) Melting point
b) Boiling point
c) Triple point
d) Vaporization point
Explanation: This graph illustrates the pressure-temperature diagram for a one-component phase diagram of the H2O system. The point O is marked as the triple point, which is the point at which three phases exist at the same time.

87. What is the triple point of water?
a) 0.0022oC
b) 0.0098oC
c) 0.022oC
d) 0.098oC
Explanation: The triple point is defined as the point at which three phases (liquid, solid, and vapor) exist at the same time. The triple point of water is 0.0098oC at pressure 4.58 mm of Hg.

88. How is Gibb’s phase rule defined?
a) C+P+1
b) C+P+2
c) C-P+2
d) C-P
Explanation: The number of phases present in an alloy depends on the number of elements of which it is composed. The Gibb’s phase rule is given by the equation F = C – P + 2. Here, F is the degrees of freedom, C is the number of components, and P is the number of phases.

89. What is the line QR in this fooling curve known as?

a) Latent heating line
b) Eutectic compound system
c) Binary solid
d) Horizontal thermal arrest
Explanation: Cooling curves are obtained by plotting the measured temperatures at equal intervals during the cooling period of a melt to a solid. This diagram shows the cooling curve for a pure metal or compound. PQ is a uniform curve, whereas QR is known as the horizontal thermal arrest.

90. Separation of single-phase solid regions from two-phase solid regions is done by _________
a) Solidus line
b) Liquidus line
c) Solvus line
d) Eutectic point
Explanation: The liquidus line separates liquid and liquid+solid phase regions, whereas solidus line separates solid and solid+liquid phase regions. A solvus line separates single-phase solid regions from two-phase solid regions.

91. The point at which two liquidus lines meet is known as __ ________
a) Eutectic point
b) Isothermal point
c) Solvus point
d) Peritectic point
Explanation: The liquidus line separates liquid and liquid+solid phase regions. Two liquidus lines meet at a point defined as the eutectic point. The corresponding temperature and composition are defined as eutectic temperature and eutectic composition.

92. Which reaction does this equation denote?
Liquid + Solid 1 → Solid 2
a) Eutectic
b) Peritectic
c) Eutectoid
d) Peritectoid
Explanation: In peritectic reactions, a solid and a liquid, due to the action of cooling, transform isothermally and reversibly into a solid with another composition. It is also known as an upside-down eutectic reaction.

93. Which reaction does this equation denote?
Solid 1 + Solid 2 →Solid 3
a) Eutectic
b) Peritectic
c) Eutectoid
d) Peritectoid
Explanation: In peritectoid reactions, two solid phases isothermally and reversibly transform into a solid with a third and different composition. It is also known as an upside-down eutectoid reaction. Such reactions are commonly found in Ni-Zn, Fe-Nb, Cu-Sn, and other systems.

94. In M-ary FSK, as M increases error
a) Increases
b) Decreases
c) Does not get effected
d) Cannot be determined
Explanation: In M-ary FSK as M increases error decreases.

95. In M-ary FSK as M tends to infinity, probability of error tends to
a) Infinity
b) Unity
c) Zero
d) None of the mentioned
Explanation: In M-ary FSK as M tends to infinity, probability of error becomes zero.

96. For non coherent reception of PSK _____ is used.
a) Differential encoding
b) Decoding
c) Differential encoding & Decoding
d) None of the mentioned
Explanation: For non coherent reception of PSK, differential encoding is used at the transmitter and decoding is used at the receiver.

97. Which modulation technique have the same bit and symbol error probability?
a) BPSK
b) DPSK
c) OOK
d) All of the mentioned
Explanation: BPSK, DPSK, OOK and non coherent FSK have same bit and symbol error probability.

98. An amplifier uses ______ to take input signal.
a) DC power
b) AC power
c) DC & AC power
d) None of the mentioned
Explanation: An amplifier uses DC power to take an input signal and increase its amplitude at the output.

99. Which has 50% maximum power efficiency?
a) Class A
b) Class B
c) Class AB
d) None of the mentioned
Explanation: Class A amplifiers have 50% maximum power efficiency.

100. Which generates high distortion?
a) Class A
b) Class B
c) Class C
d) Class AB
Explanation: Class C amplifiers generate high distortion and it is closer to switch than an amplifier.

101. Class B linear amplifiers have maximum power efficiency of
a) 50%
b) 75%
c) 78.5%
d) None of the mentioned
Explanation: Class B linear amplifiers have maximum power efficiency of 78.5%.

102. Which has the maximum power efficiency?
a) Class A
b) Class B
c) Class C
d) Class AB
Explanation: Class C has the maximum power efficiency when compared to the other linear amplifiers.

103. Free space in idealization which consists
a) Transmitter
d) None of the mentioned
Explanation: Free space is an idealization that consists of only transmitter and receiver.

#### Module 02

1. Which of the following is the Analysis equation of Fourier Transform?
a) F(ω)=f(t)ejωtdt
b) F(ω)=0f(t)ejωtdt
c) F(ω)=0f(t)ejωtdt
d) F(ω)=f(t)ejωtdt
Explanation: For converting time domain to frequency domain, we use analysis equation. The Analysis equation of Fourier Transform is F(ω)=f(t)ejωtdt.

2. Choose the correct synthesis equation.
a) f(t)=12πF(ω)ejωtdω
b) f(t)=12πF(ω)ejωtdω
c) f(t)=12π0F(ω)ejωtdω
d) f(t)=12π0F(ω)ejωtdω
Explanation: Synthesis equation converts from frequency domain to time domain. The synthesis equation of fourier transform is f(t)=12πF(ω)ejωtdω.

3. Find the fourier transform of an exponential signal f(t) = e-at u(t), a>0.
a) 1a+jω
b) 1ajω
c) 1a+jω
d) 1ajω
Explanation: Given f(t)= e-at u(t)
We know that u(t)={01t<0t>0
Fourier transform,
F(ω)=f(t)ejωtdt=eatu(t)ejωtdt=0e(a+jω)tdt
F(ω) = 1a+jω, a>0.

4. Find the fourier transform of the function f(t) = e-a|t|, a>0.
a) 2aa2ω2
b) 2aa2+ω2
c) 2aω2a2
d) aa2+ω2
Explanation: The given two-sided exponential function f(t) = e-a|t|, a>0 can be expressed as
f(t)={eateatt0t0
The Fourier transform is
F(ω)=f(t)ejωtdt=0f(t)ejωtdt+0f(t)ejωtdt
F(ω)=1a+jω+1ajω=2aa2+ω2.

5. Gate function is defined as ______________
a) G(t)={10|t|<τ2elsewhere
b) G(t)={10|t|>τ2elsewhere
c) G(t)={10|t|τ2elsewhere
d) G(t)={10|t|τ2elsewhere
Explanation: A gate function is a rectangular function defined as
G(t)=rect(tτ)={10|t|<τ2elsewhere
Where τ is pulse width.

6. Find the fourier transform of the gate function.
a) 1ωsin(ωτ2)
b) 1ωcos(ωτ2)
c) 2ωsin(ωτ2)
d) 2ωcos(ωτ2)
Explanation: Gate function is defined as
G(t)={10|t|<τ2elsewhere
The fourier transform is F(ω)=f(t)ejωtdt=τ/2τ/2ejωtdt=2ωsin(ωτ2).

7. Choose the wrong option.
a) G(t) = rect(tτ)
b) G(t) = u(t + τ2) – u(t-τ2)
c) G(ω) = τ sa(wτ2)
d) G(f) = τ sinc(f)
Explanation: Fourier transform of gate function, G(ω) = 2ωsin(wτ2)
Multiplying and dividing by τ we get
G(ω)=τsin(wτ2)wτ2=τsin(2πfτ2)2πfτ2=τsin(πτf)πτf=τsinc(τf).

8. Bandwidth of the gate function is __________
a) τ Hz
b) 1τ Hz
c) 2τ Hz
d) 2τ Hz
Explanation: The practical bandwidth of the gate function corresponds to the first zero crossing in the spectrum. Therefore, the bandwidth of the pulse or gate function is 2πτ or 1τ Hz.

9. Which of the following is not a fourier transform pair?
a) u(t)πδ(ω)+1jω
b) sgn(t)2jω
c) A2πδ(ω2)
d) G(t)sa(ωτ2)
Explanation: G(t)sa(ωτ2) is not a fourier transform pair.
G(t)τsa(ωτ2) (or) G(t)G(t)τsinc(τf).

10. Find the fourier transform of the unit step function.
a) πδ(ω) + 1ω
b) πδ(ω) + 1jω
c) πδ(ω) – 1jω
d) δ(ω) + 1jω
Explanation: We know that sgn(t) = 2u(t) – 1.
u(t) = 12[sgn(t)+1] Its Fourier transform is F[u(t)] = 12 F[sgn(t)] + 12 F[1]
As the Fourier transforms F[1] = 2πδ(ω) and [sgn(t)] = 2jω, hence
F[u(t)] = πδ(ω) + 1jω.

11. Fourier transform NMR spectrometer allows NMR transitions to be observed simultaneously.
a) True
b) False
Explanation: Fourier transform NMR spectrometer is a type of NMR spectrometer. It allows samples to be observed simultaneously instead of serially.

12. Fourier transform NMR spectrometer has which of the following characteristics?
a) Increased sensitivity, long time to obtain data
b) Decreased sensitivity, long time to obtain data
c) Increased sensitivity, reduced time to obtain data
d) Decreased sensitivity, reduced time to obtain data
Explanation: Fourier transform NMR spectrometer has increased sensitivity. It takes less time to obtain NMR data.

13. Which of the following cannot be done due to the multiplex advantage?
a) Repetitive signals can be summed
b) Repetitive signals can be averaged
c) Increases signal to noise ratio
d) Decreases signal to noise ratio
Explanation: The multiplexing advantage allows improvement of the signal to noise ratio. It also allows repetitive signals to be summed and averaged.

14. Two coils are necessary for Fourier transform NMR spectroscopy.
a) True
b) False
Explanation: Only one coil is necessary for Fourier transform NMR spectroscopy. The coil serves as both antenna for transmitting and receiving RF radiation.

15. Which of the following is the disadvantage of conventional mode of spectrometer or continuous-wave NMR spectrometer?
a) They are unstable
b) Maintenance is difficult
c) High operating cost
d) Excitation is inefficient
Explanation: The disadvantage of conventional mode of a spectrometer or continuous-wave NMR spectrometer is that the excitation is inefficient. Only a narrow band of frequencies are contributing to the signal at a time.

16. Which of the following must be done to improve sensitivity?
a) Frequency sweep mode must be preferred
b) Field sweep mode must be preferred
c) Single channel excitation must be preferred
d) Multichannel excitation must be preferred
Explanation: Sweep techniques have less sensitivity. To improve sensitivity, multichannel excitation must be preferred.

17. Which of the following is the disadvantage of multichannel excitation?
a) Low sensitivity
b) Low resolution
c) Small number of frequencies is present
d) Uneconomical
Explanation: Large number of frequencies is present in multichannel excitation. This is uneconomical since a receiver is required for each channel.

18. How can the need for an array of narrow-band filters and detectors be eliminated?
a) By using multi-channel excitation
b) By using a detector for each frequency
c) By reducing the number of detectors
d) By using Fourier transform
Explanation: Fourier transform plays the role of a multichannel receiver. Hence, the need for array of narrow-band filters can be eliminated

19. Fourier transform can be accomplished by using which of the following components?
a) Spin decoder
b) Detector
c) Spectrum analyser
d) Oscilloscope
Explanation: Fourier transform can be accomplished by using a spectrum analyser. Any complex waveform can be converted to frequencies using Fourier transform.

20. A 15.4 MHz crystal generates the _________ resonance frequency.
a) Hydrogen
b) Deuterium
c) Tritium
d) Helium
Explanation: A 15.4 MHz crystal generates the deuterium resonance frequency. This resonance signal is used to lock the magnetic signal to clock frequency.

21. If x(n)=xR(n)+jxI(n) is a complex sequence whose Fourier transform is given as X(ω)=XR(ω)+jXI(ω), then what is the value of XR(ω)?
a) n=0xR (n)cosωn-xI (n)sinωn
b) n=0xR (n)cosωn+xI (n)sinωn
c) n=xR (n)cosωn+xI (n)sinωn
d) n=xR (n)cosωn-xI (n)sinωn
Explanation: We know that X(ω)=n= x(n)e-jωn
By substituting e-jω = cosω – jsinω in the above equation and separating the real and imaginary parts we get
XR(ω)=n=xR (n)cosωn+xI (n)sinωn

22. If x(n)=xR(n)+jxI(n) is a complex sequence whose Fourier transform is given as X(ω)=XR(ω)+jXI(ω), then what is the value of xI(n)?
a) 12π2π0[XR(ω) sinωn+ XI(ω) cosωn] dω
b) 2π0[XR(ω) sinωn+ XI(ω) cosωn] dω
c) 12π2π0[XR(ω) sinωn – XI(ω) cosωn] dω
d) None of the mentioned
Explanation: We know that the inverse transform or the synthesis equation of a signal x(n) is given as
x(n)=12π2π0 X(ω)ejωn dω
By substituting e = cosω + jsinω in the above equation and separating the real and imaginary parts we get
xI(n)=12π2π0[XR(ω) sinωn+ XI(ω) cosωn] dω

23. If x(n) is a real sequence, then what is the value of XI(ω)?
a) n=x(n)sin(ωn)
b) –n=x(n)sin(ωn)
c) n=x(n)cos(ωn)
d) –n=x(n)cos(ωn)
Explanation: If the signal x(n) is real, then xI(n)=0
We know that,
XI(ω)=n=xR(n)sinωnxI(n)cosωn
Now substitute xI(n)=0 in the above equation=>xR(n)=x(n)
=> XI(ω)=-n=x(n)sin(ωn).

24. Which of the following relations are true if x(n) is real?
a) X(ω)=X(-ω)
b) X(ω)=-X(-ω)
c) X*(ω)=X(ω)
d) X*(ω)=X(-ω)
Explanation: We know that, if x(n) is a real sequence
XR(ω)=n= x(n)cosωn=>XR(-ω)= XR(ω)
XI(ω)=-n= x(n)sin⁡(ωn)=>XI(-ω)=-XI(ω)
If we combine the above two equations, we get
X*(ω)=X(-ω)

25. If x(n) is a real signal, then x(n)=1ππ0[XR(ω) cosωn- XI(ω) sinωn] dω.
a) True
b) False
Explanation: We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that, xR(n)=x(n)=12π2π0[XR(ω) cosωn- XI(ω) sinωn] dω
Since both XR(ω) cosωn and XI(ω) sinωn are even, x(n) is also even
=> x(n)=1ππ0[XR(ω) cosωn- XI(ω) sinωn] dω

26. If x(n) is a real and odd sequence, then what is the expression for x(n)?
a) 1ππ0[XI(ω) sinωn] dω
b) –1ππ0[XI(ω) sinωn] dω
c) 1ππ0[XI(ω) cosωn] dω
d) –1ππ0[XI(ω) cosωn] dω
Explanation: If x(n) is real and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently
XR(ω)=0
XI(ω)=2n=1x(n)sinωn
=>x(n)=-1ππ0[XI(ω) sinωn] dω

27. What is the value of XR(ω) given X(ω)=11aejω,|a|<1?
a) asinω12acosω+a2
b) 1+acosω12acosω+a2
c) 1acosω12acosω+a2
d) asinω12acosω+a2
Explanation: Given, X(ω)=11aejω, |a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)=1aejω(1aejω)(1aejω)=1acosωjasinω12acosω+a2
This expression can be subdivided into real and imaginary parts, thus we obtain
XR(ω)=1acosω12acosω+a2.

28. What is the value of XI(ω) given 11aejω, |a|<1?
a) asinω12acosω+a2
b) 1+acosω12acosω+a2
c) 1acosω12acosω+a2
d) asinω12acosω+a2
Explanation: Given, X(ω)=11aejω, |a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)=1aejω(1aejω)(1aejω)=1acosωjasinω12acosω+a2
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(ω)=asinω12acosω+a2.

29. What is the value of |X(ω)| given X(ω)=1/(1-ae-jω), |a|<1?
a) 112acosω+a2
b) 11+2acosω+a2
c) 112acosω+a2
d) 11+2acosω+a2
Explanation: For the given X(ω)=1/(1-ae-jω), |a|<1 we obtain
XI(ω)=(-asinω)/(1-2acosω+a2) and XR(ω)=(1-acosω)/(1-2acosω+a2)
We know that |X(ω)|=XR(ω)2+XI(ω)2−−−−−−−−−−−−−−√
Thus on calculating, we obtain
|X(ω)| = 112acosω+a2.

30. If x(n)=A, -M<n<M,; x(n)=0, elsewhere. Then what is the Fourier transform of the signal?
a) Asin(M12)ωsin(ω2)
b) A2sin(M+12)ωsin(ω2)
c) Asin(M+12)ωsin(ω2)
d) sin(M12)ωsin(ω2)
Explanation: Clearly, x(n)=x(-n). Thus the signal x(n) is real and even signal. So, we know that
X(ω)=XR(ω)=A(1+2n=1cosωn)
On simplifying the above equation, we obtain
X(ω)=Asin(M+12)ωsin(ω2).

31. What is the Fourier transform of the signal x(n)=a|n|, |a|<1?
a) 1+a212acosω+a2
b) 1a212acosω+a2
c) 2a12acosω+a2
d) None of the mentioned
Explanation: First we observe x(n) can be expressed as
x(n)=x1(n)+x2(n)
where x1(n)= an, n>0
=0, elsewhere
x2(n)=a-n, n<0 =0, elsewhere Now applying Fourier transform for the above two signals, we get X1(ω)=11aejω and X2(ω)=aejω1aejω
Now, X(ω)=X1(ω)+ X2(ω)=11aejω+aejω1aejω=1a212acosω+a2.

32. If X(ω) is the Fourier transform of the signal x(n), then what is the Fourier transform of the signal x(n-k)?
a) ejωk. X(-ω)
b) ejωk. X(ω)
c) e-jωk. X(-ω)
d) e-jωk. X(ω)
Explanation: Given
F{x(n)}= X(ω)=n=x(n)ejωn
=>F{x(n-k)}=n=x(nk)ejωn=ejωk.n=x(nk)ejω(nk)
=>F{x(n-k)}= e-jωk. X(ω)

33. What is the convolution of the sequences of x1(n)=x2(n)={1,1,1}?
a) {1,2,3,2,1}
b) {1,2,3,2,1}
c) {1,1,1,1,1}
d) {1,1,1,1,1}
Explanation: Given x1(n)=x2(n)={1,1,1}
By calculating the Fourier transform of the above two signals, we get
X1(ω)= X2(ω)=1+ e + e-jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X1(ω). X2(ω)=(1+2cosω)2=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}

34. What is the energy density spectrum of the signal x(n)=anu(n), |a|<1?
a) 11+2acosω+a2
b) 112acosω+a2
c) 112acosωa2
d) 11+2acosωa2
Explanation: Given x(n)= anu(n), |a|<1
The auto correlation of the above signal is
rxx(l)=11a2 a|l|, -∞< l <∞
According to Wiener-Khintchine Theorem,
Sxx(ω)=F{rxx(l)}=[11a2].F{a|l|} = 112acosω+a2

35. How do we represent a pairing of a periodic signal with its fourier series coefficients in case of continuous time fourier series?
a) x(t) ↔ Xn
b) x(t) ↔ Xn+1
c) x(t) ↔ X
d) x(n) ↔ Xn
Explanation: In case of continuous time fourier series, for simplicity, we represent a pairing of a periodic signal with its fourier series coefficients as,
x(t) ↔ Xn
here, x(t) is the signal and Xn is the fourier series coefficient.

36. What are the properties of continuous time fourier series?
a) Linearity, time shifting
b) Linearity, time shifting, frequency shifting
c) Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution
d) Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation
Explanation: Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation are some of the properties followed by continuous time fourier series. Integration and conjugation are also followed by continuous time fourier series.

37. Integration and conjugation are also followed by continuous time fourier series?
a) True
b) False
Explanation: Linearity, time shifting, frequency shifting, time reversal, time scaling, periodic convolution, multiplication, differentiation are some of the properties followed by continuous time fourier series. Integration and conjugation are also followed by continuous time fourier series.

38. If x(t) and y(t) are two periodic signals with coefficients Xn and Yn then the linearity is represented as?
a) ax(t) + by(t) = aXn + bYn
b) ax (t) + by(t) = Xn + bYn
c) ax(t) + by(t) = aXn + Yn
d) ax(t) + by(t) = Xn + Yn
Explanation: ax(t) + by(t) = aXn + bYn, x(t) and y(t) are two periodic signals with coefficients Xn and Yn.

39. How is time shifting represented in case of periodic signal?
a) If x(t) is shifted to t0, Xn is shifted to t0
b) x(t-t0), Yn = Xn e-njwt0
c) Xn = x(t-t0), Yn = Xn e-njwt0
d) Xn = x(-t0), Yn = Xn e-njwt0
Explanation: If x(t) and y(t) are two periodic signals with coefficients Xn and Yn, then if a signal is shifted to t0, then the property says,
Xn = x(t-t0), Yn = Xne-njwt0

40. What is the frequency shifting property of continuous time fourier series?
a) Multiplication in the time domain by a real sinusoid
b) Multiplication in the time domain by a complex sinusoid
c) Multiplication in the time domain by a sinusoid
d) Addition in the time domain by a complex sinusoid
Explanation: If x(t) and y(t) are two periodic signals with coefficients Xn and Yn,
Then y(t)= ejmwtx(t)↔Yn=Xn-m.
Hence, we can see that a frequency shift corresponds to multiplication in the time domain by complex sinusoid whose frequency is equal to the time shift.

41. What is the time reversal property of fourier series coefficients?
a) Time reversal of the corresponding sequence of fourier series
b) Time reversal of the last term of fourier series
c) Time reversal of the corresponding term of fourier series
d) Time reversal of the corresponding sequence
Explanation: x(t)↔ Xn
Y(t) = x(-t)↔Yn=X-n.
That is the time reversal property of fourier series coefficients is time reversal of the corresponding sequence of fourier series.

42. It does not depend whether the signal is odd or even, it is always reversal of the corresponding sequence of fourier series.
a) True
b) False
Explanation: It does depend whether the signal is odd or even.
If the signal is even, the reversal is positive and if the signal is odd, the reversal is negative.

43. Why does the signal change while time scaling?
a) Because the frequency changes
b) Time changes
c) Length changes
d) Both frequency and time changes
Explanation: x(t)↔Xn
Y(t) = x(at)↔Yn = Xn
Hence, the fourier coefficients have not changed but the representation has changed because of changes in fundamental frequency.

44. What is the period of the signal when it is time shifted?
a) Changes according to the situation
b) Different in different situation
c) Remains the same
d) Takes the shifted value
Explanation: The period of the periodic signal does not change even if it is time shifted.
If x(t) and y(t) are two periodic signals with coefficients Xn and Yn, then if a signal is shifted to t0, then the property says,
Xn = x(t-t0), Yn = Xne-njwt0.

45. How is the discrete time impulse function defined in terms of the step function?
a) d[n] = u[n+1] – u[n].
b) d[n] = u[n] – u[n-2].
c) d[n] = u[n] – u[n-1].
d) d[n] = u[n+1] – u[n-1].
Explanation: Using the definition of the Heaviside function, we can come to this conclusion.

46. What is the definition of the delta function in time space intuitively?
a) Defines that there is a point 1 at t=0, and zero everywhere else
b) Defines that there is a point 0 at t=0, and 1 everywhere else
c) Defines 1 for all t > 0, and 0 else
d) Defines an impulse of area 1 at t=0, zero everywhere else
Explanation: Arises from the definition of the delta function. There is a clear difference between just the functional value and the impulse area of the delta function.

47. Is it practically possible for us to provide a perfect impulse to a system?
a) Certainly possible
b) Impossible
c) Possible
d) None of the mentioned
Explanation: The spread of the impulse can never be restricted to a single point in time, and thus, we cannot achieve a perfect impulse.

48. The convolution of a discrete time system with a delta function gives
a) the square of the system
b) the system itself
c) the derivative of the system
d) the integral of the system
Explanation: The integral reduces to the the integral calculated at a single point, determined by the centre of the delta function.

49. Find the value of 2sgn(0)d[0] + d[1] + d[45], where sgn(x) is the signum function.
a) 2
b) -2
c) 1
d) 0
Explanation: sgn(0)=0, and d[n] = 0 for all n not equal to zero. Hence the sum reduces to zero.

50. Where h*x denotes h convolved with x, x[n]*d[n-90] reduces to
a) x[n-89].
b) x[n-91].
c) x[n=90].
d) x[n].
Explanation: The function gets shifted by the center of the delta function during convolution.

51. Where h*x denotes h convolved with x, find the value of d[n]*d[n-1].
a) d[n].
b) d[n-1].
c) d2[n].
d) d2[n-1].
Explanation: Using the corollary, if we take d[n] to be the ‘x’ function, it will be shifted by -1 when convolved with d[n-1], thus rendering d[n-1].

52. How is the continuous time impulse function defined in terms of the step function?
a) u(t) = d(d(t))/dt
b) u(t) = d(t)
c) d(t) = du/dt
d) d(t) = u2(t)
Explanation: Using the definition of the Heaviside function, we can come to this conclusion.

53. In which of the following useful signals, is the bilateral Laplace Transform different from the unilateral Laplace Transform?
a) d(t)
b) s(t)
c) u(t)
d) all of the mentioned
Explanation: The bilateral LT is different from the aspect that the integral is applied for the entire time axis, but the unilateral LT is applied only for the positive time axis. Hence, the u(t) [unit step function] differs in that aspect and hence can be used to differentiate the same.

54. What is the relation between the unit impulse function and the unit ramp function?
a) r = dd(t)/dt
b) d = dr/dt
c) d = d2(r)/dt2
d) r = d2(d)/dt2
Explanation: Now, d = du/dt and u = dr/dt. Hence, we obtain the above answer.

55. The term heterodyning refers to
a) Frequency conversion
b) Frequency mixing
c) Frequency conversion & mixing
d) None of the mentioned
Explanation: The method heterodyning means frequency conversion and mixing and this results in a spectral shift.

56. The causes for error performance degradation in communication systems are
a) Interference
b) Electrical noise
c) Effect of filtering
d) All of the mentioned
Explanation: The main causes of error performance degradation are interference electrical noise effect of filtering and also due to the surroundings.

57. Thermal noise in the communication system due to thermal electrons
a) Can be eliminated
b) Cannot be eliminated
c) Can be avoided upto some extent
d) None of the mentioned
Explanation: Thermal noise which cannot be eliminated is caused by the motion of thermal electrons causes degradation in system.

58. White noise has _______ power spectral density.
a) Constant
b) Variable
c) Constant & Variable
d) None of the mentioned
Explanation: The AWGN has constant power spectral density.

59. Which are called as hard decisions?
a) Estimates of message symbol with error correcting codes
b) Estimates of message symbol without error correcting codes
c) All of the mentioned
d) None of the mentioned
Explanation: If error correcting codes are not present, the detector output consists of estimates of the message symbol which is also called as hard decisions.

60. The filter which is used to recover the pulse with less ISI is called as
a) Matched filter
b) Correlator
c) Matched filter & Correlator
d) None of the mentioned
Explanation: The optimum filter used to recover the pulse with best possible signal to noise ratio and less or no ISI is called as correlator or matched filter.

61. The composite equalizing filter is the combination of
a) Receiving and equalizing filter
b) Transmitting and equalizing filter
c) Amplifier and equalizing filter
d) None of the mentioned
Explanation: The functions of both receiving and equalizing filter can be performed by only the equalizing filter alone. Thus equalizing filter is the combination of equalizing and receiving filter.

62. The sample from the demodulation process consists of sample which is _______ to energy of the received symbol and _____ to noise.
a) Directly and inversely proportional
b) Inversely and directly proportional
c) Both directly proportional
d) Both inversely proportional
Explanation: The output symbol of the sampler consists of sample which is directly proportional to the energy of the received signal and inversely proportional to the noise.

63. The average noise power of white noise is
a) 0
b) Infinity
c) 1
d) None of the mentioned
Explanation: White noise is a idealized process with two sided spectral density equal to constant N0/2 and frequencies varying from minus infinity to plus infinity. Thus the average noise power is infinity.

64. The channel may be affected by
a) Thermal noise
b) Interference from other signals
c) Thermal noise & Interference from other signals
d) None of the mentioned
Explanation: A channel can be modelled as a linear filter with additional noise. The noise comes from thermal noise source and also from interference from other signals.

65. Channels display multi-path due to
a) Scattering
b) Time delayed reflections
c) Diffraction
d) All of the mentioned
Explanation: Wireless wide-band channels display multi-path due to time delayed reflections, diffraction and also scattering.

#### Module 03

1. Carrier signal in modulation technique is _______ signal.
a) High frequency
b) Low frequency
c) High amplitude
d) Low amplitude
Explanation: Carrier signal in modulation technique is a high frequency signal. In amplitude modulation, the amplitude of a high frequency carrier signal is varied in accordance to the instantaneous amplitude of the modulating signal.

2. Modulation index of an AM signal is ratio of __________ to the _______
a) Peak carrier amplitude, Peak message signal amplitude
b) Peak message signal amplitude, Peak carrier amplitude
c) Carrier signal frequency, Message signal frequency
d) Message signal frequency, Carrier signal frequency
Explanation: The modulation index k of an AM signal is defined as the ratio of the peak message signal amplitude to the peak carrier amplitude. The modulation index is often expressed as a percentage. It is also called percentage modulation.

3. If the peak message signal amplitude is half the peak amplitude of the carrier signal, the signal is _____ modulated.
a) 100%
b) 2%
c) 50%
d) 70%
Explanation: The modulation is also expressed in percentage. It is also called percentage modulation. The signal is said to be 50% modulated if the peak message signal amplitude is half the peak amplitude of the carrier signal.

4. A percentage of modulation greater than ___________ will distort the message signal.
a) 10%
b) 25%
c) 50%
d) 100%
Explanation: A percentage of modulation greater than 100% will distort the message signal if detected by an envelope detector. In this case the lower excursion of the signal will drive the carrier amplitude below zero, making it negative (and hence changing its phase).

5. The RF bandwidth of AM is ____________ the maximum frequency contained in the modulating message signal.
a) Equal
b) Two times
c) Four times
d) Ten times
Explanation: The RF bandwidth of an AM signal is equal to BAM=2fm. It is double the maximum frequency contained in the modulating message signal. AM spectrum consists of an impulse at the carrier frequency and two sidebands which replicate the message spectrum.

6. Single sideband AM systems occupy same bandwidth as of conventional AM systems.
a) True
b) False
Explanation: Single sideband (SSB) AM systems transmit only one of the sidebands (either upper or lower) about the carrier. Hence, they occupy only half the bandwidth of conventional AM systems.

7. How is the performance of SSB AM systems in fading channels?
a) Poor
b) Best
c) Good
d) Average
Explanation: SSB systems have the advantage of being very bandwidth efficient. But their performance in fading channels is very poor. For proper detection, the frequency of the oscillator at the product detector mixer in the receiver must be same as that of the incoming carrier frequency.

8. Which of the following is a disadvantage of tone-in-band SSB system?
a) High bandwidth
c) Effects of multipath
d) Generation and reception of signal is complicated
Explanation: Tone-in-band SSB systems has the advantage of maintaining the low bandwidth property of the SSB signals, while at the same time providing good adjacent channel protection. The tone in band system employs feedforward automatic gain and frequency control to mitigate the effects of multipath induced fading.

9. FFSR in AM systems stands for ________
a) Feedforward signal regeneration
b) Feedbackward signal regeneration
c) Feedbackward system restoration
d) Feedforward system restoration
Explanation: FFSR stands for Feedforward signal regeneration. If the pilot tone and the information bearing signal undergo correlated fading, it is possible at the receiver to counteract the effects of fading through signal processing based on tracking of pilot tone. This process is called FFSR.

10. AM demodulation technique can be divided into _____ and _____ demodulation.
a) Direct, indirect
b) Slope detector, zero crossing
c) Coherent, noncoherent
Explanation: AM demodulation techniques may be broadly divide into two main categories. They are called coherent and noncoherent demodulation. They are differentiated by the knowledge of transmitted carrier frequency and phase at the receiver.

11. Non coherent detection requires the knowledge of transmitted carrier frequency and phase at the receiver.
a) True
b) False
Explanation: Non coherent detection does require the knowledge of phase information. However, coherent detection requires knowledge of the transmitted carrier frequency and phase at the receiver.

12. A product detector in AM systems is also called ___________
a) Envelope detector
b) Differentiator
c) Integrator
d) Phase detector
Explanation: A product detector is also called a phase detector. It forms a coherent demodulator for AM signals. It is a down converter circuit which converts the input bandpass signal to a baseband signal.

13. AM system use only product detector for demodulation. They never use envelope detectors.
a) True
b) False
Explanation: AM systems can use either product detector or envelope detector for demodulation. As a rule, envelope detectors are useful when input signal power is at least 10dB greater than noise power, whereas product detectors are able to process the AM signals with input signal to noise ratios well below 0 dB.

14. LCD uses ________
a) sematic crystals
b) twisted nematic crystals
c) nematic crystals
d) cholesteric crystals
Explanation: LCD uses liquid crystal display. It uses twisted nematic crystals which are a type of liquid crystal, consisting of a substance called the nematic. The nematic liquid crystal is placed between two plates of polarized glass.

15. Which of the following stage is present in FM receiver but not in AM receiver?
a) Amplitude limiter
b) Demodulator
c) AM amplifier
d) Mixer
Explanation: Amplitude Limiter circuit is used in FM receiver to remove the noise or any variation in amplitude present in the received signal. Thus, the output of the amplitude limiter has a constant amplitude. So it is only used in frequency modulation and not in amplitude modulation.

16. Function of duplexer in a RADAR is to permit the use of same antenna for transmission and reception.
a) True
b) False
Explanation: A duplexer is being an electronic unit, allows bi-directional communication over the same path. The transmitter and receiver can communicate simultaneously. In radar, the duplexer isolates the receiver from the transmitter while allowing them to share a common antenna.

17. Single Sideband Modulation (SSB) is generally reserved for point-to-point communication.
a) True
b) False
Explanation: A point-to-point communication refers to bidirectional communication between only one transmitter and one receiver. In SSB-SC modulation technique, the carrier is suppressed and only one of the two side-bands are transmitted. Thus, it reduces power consumption and lessens bandwidth. Thus, it is preferred for point-to-point communication.

18. For an AM transmitter, class C amplifier can be used after the modulation stage.
a) True
b) False
Explanation: In an AM transmitter, the required transmission power is obtained from class C amplifier, as it is a power amplifier, for low-level or high-level modulation. So it is not used after the modulation stage.

19. For which of the modulated system, the linear amplified modulated stage is used?
a) low level amplitude modulated system
b) high level amplitude modulated system
c) high level frequency modulated system
d) low level frequency modulated system
Explanation: In low-level modulation, the generation of amplitude modulated signal takes place at low power levels. The generated AM signal is then amplified using a chain of linear amplifiers, which are required to avoid waveform distortion. Thus, linear amplified modulated stage is used in low level amplitude modulated system.

20. When noise is passed through a narrow band filter, the output of filter should be?
a) triangular
b) square
c) parabolic
d) sinusoidal
Explanation: Narrow band filter is used to isolate a narrow band of frequencies from a wider bandwidth signal. It is a combination of band pass and band reject filter. When noise gets passed through it, the output of it should be sinusoidal.

21. A narrow band noise can exist in _________
a) AM only
b) PCM only
c) FM only
d) AM and FM both
Explanation: Narrow band filter is used to isolate a narrow band of frequencies from a wider bandwidth signal. It is a combination of band pass and band reject filter. So it can be used in both AM and FM to pass a band of frequencies or to attenuate a band of frequencies.

22. The upper and lower sideband frequencies for 5KHz amplitude modulation with a 30KHz carrier frequency will be?
a) 35KHz and 25KHz
b) 34KHz and 24KHz
c) 25KHz and 35KHz
d) 0.35KHz and 0.25KHz
Explanation: Upper sideband frequency will be (30 + 5) = 35 KHz and Lower sideband frequency will be (30 – 5) = 25 KHz.

23. Phase array radar can track many targets together.
a) True
b) False
Explanation: A phased array radar is an array of radiating elements, with each having a phase-shifter. The phase of the signal being emitted from the radiating element is changed to produce beams, thereby producing constructive or destructive interference for steering the beams in the required direction. Thus, it can track many targets together.

24. A duplex arrangement use separate frequencies for transmission.
a) True
b) False
Explanation: In duplex communication, two-way interaction is favourable simultaneously. Thus, a cordless telephone is duplex which uses separate frequencies for transmission in base and portable units.

25. VSB modulation is used in televisions because it avoids phase distortion at low frequencies.
a) True
b) False
Explanation: Vestigial Sideband Modulation (VSB) is a type of amplitude modulation in which the carrier and only one sideband is completely transmitted and the other sideband is partly transmitted. Thus, television production is done using VSB modulation as it reduces bandwidth to half.

26. A cordless telephone that uses separate frequencies for transmission in base and portable units is called _________
a) half duplex
b) duplex
c) simplex
d) one-way communication
Explanation: In duplex communication, two-way interaction is favourable simultaneously. Thus, a cordless telephone is duplex which uses separate frequencies for transmission in base and portable units.

27. Which polarization is used to reduce the depolarization effect on received waves?
a) Circular polarization
b) Linear polarization
c) Atomic polarization
d) Dipolar polarization
Explanation: In circular polarization at each point the electric field of electromagnetic wave has a constant magnitude but its direction changes as it rotates with time at a steady rate, in a plane perpendicular to the direction of propagation of wave. It is used to reduce depolarization effect on received waves.

28. Circular polarization involves critical alignment between transmitting and receiving antenna.
a) True
b) False
Explanation: In circular polarization at each point the electric field of the electromagnetic wave has a constant magnitude but its direction changes as it rotates with time at a steady rate, in a plane perpendicular to the direction of propagation of the wave. It is used to reduce depolarization effect on received waves. It does not involve alignment between transmitting and receiving antenna.

29. It is only the reflected color that decided the color of an object.
a) True
b) False
Explanation: Color of any object is decided by the reflected color for opaque object and wavelength transmitted through it for transparent object, while both reflector color and wavelength transmitted are considered for a translucent object.

30. What do you understand by the term “carrier” in modulation?
a) voltage to be transmitted
b) resultant wave
c) voltage for which amplitude, phase or frequency can be varied
d) voltage for which amplitude, phase or frequency remains constant
Explanation: Carrier wave is the wave with frequency higher than the message signal, whose certain characteristics like amplitude, phase or frequency are varied with respect to the instantaneous amplitude of the message signal. Thus forming the modulated wave which is the wave to be transmitted.

31. Carrier wave in modulation is a resultant wave.
a) True
b) False
Explanation: Carrier wave is the wave with frequency higher than the message signal, whose certain characteristics like amplitude, phase or frequency are varied with respect to the instantaneous amplitude of the message signal. Thus forming the modulated wave which is the wave to be transmitted.

32. For a low level AM system, amplifier modulated stage must have _________
a) harmonic devices
b) linear devices
c) non-linear devices
d) class A amplifiers
Explanation: In low-level modulation, the generation of amplitude modulated signal takes place at low power levels. The generated AM signal is then amplified using a chain of linear amplifiers, which are required to avoid waveform distortion. Thus, linear devices are used in low level amplitude modulated system.

33. Demodulation is done in ________
a) Channel
c) Receiving antenna
d) Transducer
Explanation: Demodulation refers to extracting the original message signal from a transmitted modulated wave. The extraction of the message signal is generally carried out in the receiver. Channel is the medium through which the modulated message signal is transferred and Antenna receives the transmitted signal. Transducer converts the electrical signal to sound waves and vice-versa.

34. What is Fidelity?
a) Equally amplifies all the signal frequencies at receiver
b) Ability of receiver to select wanted signal from various incoming signal
c) Minimum magnitude of input signal required to produced a specified output
d) Ability to amplify weak signals
Explanation: Fidelity is the ability of the receiver to reproduce all modulating signals, equally, without any distortion. The ability of receiver to select wanted signal from various incoming signals is called Selectivity while Sensitivity is the minimum magnitude of input signal required to produce a specified output. It is the ability to amplify weak signals.

35. In a receiver, noise is usually developed at ________
a) Audio stage
b) Receiving antenna
c) RF stage
d) IF stage
Explanation: Ability of receiver to selected only wanted signal, and reject other frequencies, out of the various incoming signals, helps the receiver to operate more efficiently. However, at times, the RF amplifier allows a frequency lying close to the desired frequency, to pass to the next stage. This other frequency is undesired and later on is responsible for production of image frequency. Thus, noise is usually developed at RF stage.

36. Which oscillator is used as a local oscillator in radio receiver?
a) Wien-bridge
b) Hartley
c) Crystal
d) Phase Shift
Explanation: Oscillator which is used as a local oscillator in radio receiver is generally a tuned circuit. This tuned circuit consists of inductors and capacitors to determine the resonant frequency, therefore it is an LC tuned circuit. Out of the four options, only Hartley Oscillator has an LC resonant tank circuit.

37. Process of recovering information signal from received carrier is known as ________
a) Sensitivity
b) Selectivity
c) Demodulation
d) Fidelity
Explanation: Demodulation means extracting information or message signal from the transmitted modulated wave, while minimum magnitude of input signal required to produced a specified output is known as Sensitivity. The ability of receiver to select wanted signal from various incoming signals is called Selectivity. Fidelity means reproducing all modulating frequencies equally, without any distortion.

37. What is the use of a varactor diode in radio receiver?
a) Demodulation
b) Mixing
c) Multiplexing
d) Tuning
Explanation: Varactor diode is a diode working in the reverse-bias because of which no current flows through it. It has variable capacitance which varies with applied voltage. Varactor diodes are mainly used in Voltage Controlled Oscillators (VCOs) and RF Filters for tuning the receiver to the incoming signal or different stations.

a) to detect and amplify information signal from the carrier
b) to modulate a message signal
d) to convert one form of energy into other
Explanation: Receiver is to detect and amplify information signal from the carrier. Transmitter is used to modulate message signal and produce radio waves. Transducer is used to convert one signal form of energy into another form.

39. Figure of merit is ________
a) Ratio of output signal to noise ratio to input signal to noise ratio
b) Ratio of input signal to noise ratio to output signal to noise ratio
c) Ratio of output signal to input signal to a system
d) Ratio of input signal to output signal to a system
Explanation: Figure of merit is a numerical quantity based on the characteristic of system that represents a measure of efficiency or effectiveness. It is defined as the ratio of output signal to noise ratio to input signal to noise ratio.

39. Superheterodyne principle provides selectivity at ________
a) RF stage
b) IF stage
c) Demodulating Stage
d) Audio Stage
Explanation: A superheterodyne receiver uses frequency mixing to convert the received high frequency signal to a fixed lower intermediate frequency (IF), which can be processed more conveniently than original received frequency. Thus, the principle of selectivity is applied at the IF stage as it consists of very efficient filters to only select a wanted signal and pass it to the Demodulating Stage.

40. A heterodyne frequency changer is ________
a) Mixer
b) Demodulator
c) Modulator
d) Local Oscillator
Explanation: A mixer is a nonlinear electrical circuit that multiplies two signal frequencies applied to it, and produces a new frequency. Mixers are widely used to shift signals from one frequency range to other, which is known as heterodyning process. Generally, Local Oscillator generates a frequency to be applied at one of the input terminals of the mixer. Demodulator decodes the message signal from modulated signal, while modulator encodes message signal for transmission.

41. What is the bandwidth required in SSB signal?
a) fm
b) 2fm
c) > 2fm
d) < 2fm
Explanation: In an AM modulated system, total bandwidth required is from fc + fm to fc – fm i.e. bandwidth is equal to 2fm. In SSB-SC transmission, the carrier and one of the sideband gets suppressed, so the bandwidth becomes fm only.

42. One of the advantage of using a high frequency carrier wave is that it dissipates very small power.
a) True
b) False
Explanation: The main advantage of using high frequency signals is that the signal gets transmitted over very long distances and thus dissipates very less power. The antenna height required for transmission also gets reduced at high frequencies. And also it allows less noise interference and enables multiplexing. This is the reason for sending the audio signals at high frequency carrier signals for communication purpose.

43. What is the function of RF mixer?
b) Multiplication of two signals
c) Subtraction of two signals
d) To reduce the amount of noise
Explanation: RF mixer translates the frequencies of the two incoming signals by multiplying them and bringing them to a suitable band which can be processed.

44. The antenna current is 10A. Find the percentage of modulation when the antenna current increases to 10.4A?
a) 50%
b) 30%
c) 28.5%
d) 23%
Explanation:
which gives m = 0.285 or 28.5%.

45. Find the total power, if the carrier of an AM transmitter is 800W and it is modulated to 50%?
a) 100W
b) 800W
c) 500W
d) 900W
Explanation: PT=PC (1+u22), according to the problem PC = 800W and m = 0.5. On substituting values in the equation we get PT=800(1+ 0.522) = 900W.

46. Aliasing refers to ________
a) Sampling of signals less than at Nyquist rate
b) Sampling of signals at Nyquist rate
c) Sampling of signals greater than at Nyquist rate
d) Unsampled the original signal
Explanation: Aliasing refers to the sampling of signals less than at Nyquist rate. Nyquist rate states that the rate of sampling of signals should be greater than or equal to twice the bandwidth of modulating signal. It gets reduced if sampling is done at a higher rate than nyquist rate of sampling. Aliasing can be avoided by using anti-aliasing filters.

47. Baseband compression produces ________
a) a small range of frequencies from low to high
b) a small range of different phases
c) a small range of angles
d) a small range of amplitude
Explanation: A signal compression method in a wireless network provides efficient transfer of compressed signal samples over serial data links in the system. Baseband compression produces a small range of amplitude.

48. Automatic Level Control (ALC) is used to keep the modulation index close to 100%.
a) True
b) False
Explanation: ALC stands for Automatic Level Control. It is a technology which is used for automatic control of output power. It helps in maintaining the output when there are varying changes in the input.

49. A signal that ________ must have linear power amplifier.
a) is complex
b) has variable frequency
c) is linear
d) has variable amplitude
Explanation: If any signal has variable amplitude then its amplifier must be linear. Others may or may not have the same but can possess non-linear amplifiers.

50. Transmitters are designed usually to derive a load impedance of ________
a) 50 ohms resistive
b) 150 ohms resistive
c) 250 ohms resistive
d) 500 ohms resistive
Explanation: Transmitter is an electronic device that produces radio waves with an antenna. The transmitter itself generates a radio frequency alternating current, which is applied to the antenna. Transmitters are usually designed to derive a load impedance of 50 ohms resistive.

51. What we called a resistor if a transmitter is connected to a resistor instead of an antenna?
Explanation: If a transmitter is connected to resistor not antenna than it is called dummy load. Such a load is used for testing purposes to set the parameters of the transmitter, as it would have behaved in presence of an actual antenna.

52. A class D amplifiers is very efficient than other amplifiers.
a) True
b) False
Explanation: Class D amplifier is also known as a switching amplifier. It is operate as electronic switches, and not an electric gain device which is commonly used in most amplifiers. It also has high power conversion efficiency unlike other amplifiers.

53. The carrier is suppressed in ________
a) a mixer
b) a frequency multiplier
c) a transducer
d) a balance modulator
Explanation: A mixer is the one which mixes the audio frequency with the carrier frequency. A transducer converts a signal from one form to another. Balance modulator suppresses the carrier and leaves only the sidebands.

54. What is the full form of AFC?
a) Amplitude to frequency conversion
b) Automatic frequency conversion
c) Automatic frequency control
d) Audio frequency control
Explanation: AFC stands for Automatic frequency control. It is a method to automatically keep a resonant circuit tuned to a frequency of an incoming radio signal. It is used in receivers to tune to the desired frequency.

55. Mixing is used in communication to ________
a) raise the carrier frequency
b) lower the carrier frequency
c) to altered the deviation
d) to change the carrier frequency to any required value
Explanation: Mixing is used to change the frequency of carrier by mixing it with a radio frequency signal or audio signal. The frequency can be changed to any required value in communication.

56. On which factor the bandwidth required for a modulated carrier depends?
a) baseband frequency range
b) signal to noise ratio
c) carrier frequency
d) amplitude of carrier frequency
Explanation: Bandwidth can be seen as a range of frequencies within a band, that is used for transmitting a signal. A signal bandwidth depends on the baseband frequency range.

57. Super heterodyne receivers needs an extra circuitry for frequency conversion.
a) True
b) False
Explanation: Super heterodyne receiver mixes the incoming signal frequency with the locally generated signal frequency, using a mixer, in order to convert the incoming RF signal to a low frequency signal which can be processed easily. It has better sensitivity and selectivity but it needs an extra circuitry for frequency conversion.

58. Neper is ________ decibel.
a) smaller than
b) larger than
c) equal to
d) exactly twice of
Explanation: Neper is a unit used to express ratios, such as gain, loss and relative values. One neper is equal to 8.685 dB. Hence it is greater than dB.

59. The coupling used in stoneman transmission bridge is ________
a) Inductive
b) Resistive
c) Capacitive
d) Combination of inductive and capacitive
Explanation: Stoneman transmission bridge is used in telephone communication to separate voice path from signal path and battery. Capacitive coupling is used in it.

60. Which statement is true about Square Law modulators?
a) it is used for frequency modulation
b) it is used for pulse width modulation
c) it is used for amplitude modulation
d) it is used for phase modulation
Explanation: In amplitude modulation, the amplitude of the carrier wave is varied with respect to the instantaneous amplitude of the wave to be transmitted. Square Law modulators are generally used for generation of amplitude modulation. They have nonlinear current-voltage characteristics. They are highly nonlinear in low voltage region.

61. Which statement is true about ring modulator?
a) it is used for DSB-SC generation
b) it is used for SSB-SC generation
c) it is a summation modulator
d) it is used for AM generation

Explanation: Ring Modulator is used for generating DSB-SC waves. It is a product modulator having four diodes connected in the form of a ring, which suppresses the carrier and allows only the two sidebands to be passed.

62. In ergodic process, ensemble and time averages are identical.
a) False
b) True
Explanation: In ergodic process, ensemble and time averages are identical to each other.

63. What is the amount of information in continuous signal?
a) zero
b) infinite
c) 2 bits
d) 4 bits
Explanation: In any continuous signal, the amount of information that comes out is always infinite.

64. Audio frequency range lies between _________
a) 2 MHz to 20 MHz
b) 20 Hz to 20 KHz
c) 20 KHz to 200 KHz
d) 20 MHz to 200 MHz
Explanation: As per standards audio frequency range lies in between 20 Hz to 20 KHz. This is the human audible range.

65. The image frequency of a superheterodyne receiver ___________
a) is created within the receiver itself
b) is due to insufficient adjacent channel rejection
c) is not rejected by the IF tuned circuits
d) is independent of the frequency to which the receiver is tuned
Explanation: To eliminate the image frequency of a superheterodyne receiver, adequate attenuation on the incoming signal by the RF amplifier filter is needed. It is not rejected by the IF tuned circuits and thus, it produces image frequency.

66. Which one of the following circuits could not demodulate SSB?
a) Phase Shift Method
b) Filter Method
c) Weaver Demodulator
d) Phase discriminator
Explanation: Phase discriminator cannot be used to demodulate SSB. It is used to discriminate between different phases.

67. Which of the following statement is true about a radio detector?
a) the linearity is worse than in phase discriminator
b) there is stabilization against signal strength variations
c) the final output is twice of that obtainable from a similar phase discriminator
d) the circuit is same as in a discriminator, except that the diodes are reversed.
Explanation: Detector, also known as a demodulator, in a radio receiver extracts the original information that is contained in a modulated radio wave. The linearity of a radio detector is worse than in phase discriminator. However, the radio detector does not get affected by amplitude variations and minimizes the limiting required.

a) squelch
b) variable selectivity
c) double conversion
d) variable sensitivity
Explanation: Squelch is a circuit that is preferably used to prevent overloading of the IF amplifier in a receiver.

a) Local oscillator always operates below the frequency of transmitted signal
b) Input of mixer must be tuned to the signal frequency
c) Local oscillator frequency should be double than that of IF
d) RF amplifier normally works at 455 KHz above the carrier frequency
Explanation: For a superheterodyne receiver, one of the inputs of the mixer must be tuned to the signal frequency and another input is of the local oscillator frequency. Tuning means that the circuit works for that one signal frequency only.

70. Which of the following statement is false regarding Armstrong modulation system?
a) The system is basically phase modulated and not frequency modulated
b) Automatic Frequency Control (AFC) is not needed as a crystal oscillator is used
c) Frequency multiplication must be used
d) Equalization is necessary
Explanation: The Armstrong method generates a double sideband carrier signal, phase shifts this signal and then reinserts the carrier to produce a frequency modulated signal. In Armstrong modulation system equalization is not necessary.

71. If the target cross section is changing, the best system for accurate tracking is __________
a) sequential switching
b) lobe switching
c) conical scanning
Explanation: Monopulse radar compares the received signal from a single radar pulse against itself in order to compare the signal in various aspects. If the target cross section is changing then it is the best system used for accurate tracking.

72. After a target has been acquired, the best scanning system for tracking is _________
a) circular
b) spiral
c) conical
d) helical
Explanation: Conical is the best scanning system for tracking. But it is only preferred for tracking after a target has been acquired.

73. Which layer is used for detection of bad frames in wide area network?
a) frame layer
c) physical layer
d) session layer
Explanation: Wide area network covers a long geographical area and thus it also has least connected devices. Link layer is used in WAN for detection of bad frames.

74. TEM stands for Transmitted Electromagnetic wave.
a) True
b) False
Explanation: TEM stands for Transverse Electromagnetic wave. Since both electric and magnetic field are transverse to the direction of wave propagation.

75. What is the reason of attenuation in free space?
a) decrease in energy per square meter due to expansion
b) decrease in energy per square meter due to compression
c) losses in upper atmosphere due to expansion
d) losses in upper atmosphere due to compression
Explanation: In free space, the basic reason of attenuation is the decrease in energy per square meter. It generally happens due to expansion.

76. What is the range of UHF?
a) below the microwave range
b) above the microwave range
c) same as the microwave range
d) it does not include the microwave range in any way
Explanation: Ultra high frequency is the ITU designation for radio frequencies that lies under 300MHz to 3 GHz i.e. its range is below the microwave range.

77. Striplines and microstrips are used to __________
a) couple sections of waveguide
b) couple waveguides to antennas
c) couple components on a circuit board
d) couple electromagnetic waves in an order
Explanation: The main use of Striplines and microstrips in any circuit is to couple components on a circuit board.

78. What is the another term used for a single microwave link?
a) section
b) hop
d) skip
Explanation: Microwave link is a communication system that uses a beam of radio waves in the microwave frequency range. Hop is also used in places of single microwave link.

79. In digital microwave systems, additional repeaters increase the ________
a) selectivity
b) noise level
c) jitter
d) sensitivity
Explanation: In telecommunication, jitter can be seen as a deviation from true periodicity of a presumably periodic signal. Higher level of jitter are more likely to occur on either slow or heavily congested link. In any circuit, adding additional repeaters finally increase the jitter only.

80. What is the full form of SDLC?
b) Synchronous Data Line Control
c) Synchronous Data Line Character
Explanation: SDLC stands for Synchronous Data Link Control. It is a layer 2 protocol for IBM’s Systems Network Architecture. It supports multipoint links as well as error-correction. It helps in the proper flow of data to different levels.

81. Mostly LANs are based on Ethernet.
a) True
b) False
Explanation: LAN stands for Local Area Network and is a computer network that interconnects computers within a limited area. Most LANs are based on Ethernet only. It can also connect devices through wireless link to a server.

82. CSMA stands for _________
a) Carrier Sense Multiple Access
b) Client Sense Multiple Access
c) Carrier Server Multiple Access
d) Client-Server Multiple Access
Explanation: CSMA stands for Carrier Sense Multiple Access. It is a media access control protocol. In it, a node verifies the absence of other traffic before transmission.

83. What we called each computer in a network?
a) node
b) circuit
c) hub
d) switch
Explanation: In networks, each connected computer is seen as a node.

84. Which network never experienced a collision?
a) CSMA
b) Token-passing
c) LAN
d) All networks have collision at some point
Explanation: Token passing is a channel access method. In it, a signal which is seen as a token is passed between nodes to authorize that node to communicate. Token passing never experiences any type of collisions.

85. What is the effect of too many collisions in any network?
a) network slows down
b) network crashes
c) cable overheats
d) data gets lost
Explanation: If a network receives too many collisions, it directly affects its connected nodes. Due to which the network finally slows down.

#### Module 04

1. The frequency shift can be achieved by multiplying the band pass signal as given in equation
x(t) = uc(t)cos2πFctus(t)sin2πFct by the quadrature carriers cos[2πFct] and sin[2πFct] and lowpass filtering the products to eliminate the signal components of 2Fc.
a) True
b) False
Explanation: It is certainly advantageous to perform a frequency shift of the band pass signal by and sampling the equivalent low pass signal. Such a frequency shift can be achieved by multiplying the band pass signal as given in the above equation by the quadrature carriers cos[2πFct] and sin[2πFct] and low pass filtering the products to eliminate the signal components at 2Fc. Clearly, the multiplication and the subsequent filtering are first performed in the analog domain and then the outputs of the filters are sampled.

2. What is the final result obtained by substituting Fc=kB-B/2, T= 1/2B and say n = 2m i.e., for even and n=2m-1 for odd in equation x(nT)= uc(nT)cos2πFcnTus(nT)sin2πFcnT?
a) (1)muc(mT1)us
b) us(mT1T12)(1)m+k+1
c) None
d) (1)muc(mT1)us(mT1T12)(1)m+k+1
Explanation:
x(nT)=uc(nT)cos2πFcnTus(nT)sin2πFcnT → equ1
=uc(nT)cosπn(2k1)2us(nT)sinπn(2k1)2 → equ2
On substituting the above values in equ1, we get say n=2m, x(2mT)xmT(1)=uc(mT1)cosπm(2k1)=(1)muc(mT1)
where T1=2T=1B. For n odd, say n=2m-1 in equ2 then we get the result as follows
us(mT1T12)(1)m+k+1
Hence proved.

3. Which low pass signal component occurs at the rate of B samples per second with even numbered samples of x(t)?
a) uc-lowpass signal component
b) us-lowpass signal component
c) uc & us-lowpass signal component
d) none of the mentioned
Explanation: With the even-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component uc.

4. Which low pass signal component occurs at the rate of B samples per second with odd numbered samples of x(t)?
a) uc – lowpass signal component
b) us – lowpass signal component
c) uc & us – lowpass signal component
d) none of the mentioned
Explanation: With the odd-numbered samples of x(t), which occur at the rate of B samples per second, produce samples of the low pass signal component us.

5. What is the reconstruction formula for the bandpass signal x(t) with samples taken at the rate of 2B samples per second?
a) m=x(mT)sin(π/2T)(tmT)(π/2T)(tmT)cos2πFc(tmT)
b) m=x(mT)sin(π/2T)(t+mT)(π/2T)(t+mT)cos2πFc(tmT)
c) m=x(mT)sin(π/2T)(tmT)(π/2T)(tmT)cos2πFc(t+mT)
d) m=x(mT)sin(π/2T)(t+mT)(π/2T)(t+mT)cos2πFc(t+mT)
Explanation: m=x(mT)sin(π/2T)(tmT)(π/2T)(tmT)cos2πFc(tmT), where T=1/2B

6. What is the new centre frequency for the increased bandwidth signal?
a) Fc‘= Fc+B/2+B’/2
b) Fc‘= Fc+B/2-B’/2
c) Fc‘= Fc-B/2-B’/2
d) None of the mentioned
Explanation: A new centre frequency for the increased bandwidth signal is Fc‘ = Fc+B/2-B’/2

7. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for uc(t) = ?
a) m=uc(mT1)sin(πT1)(tmT1)(π/T1)(tmT1)
b) m=us(mT1T12)sin(πT1)(tmT1+T1/2)(πT1)(tmT1+T12)
c) m=uc(mT1)sin(πT1)(t+mT1)(πT1)(t+mT1)
d) m=us(mT1T12)sin(πT1)(t+mT1+T12)(πT1)(t+mT1+T12)
Explanation: To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B.
uc(t)=m=uc(mT1)sin(πT1)(tmT1)(π/T1)(tmT1).

8. According to the sampling theorem for low pass signals with T1=1/B, then what is the expression for us(t) = ?
a) m=uc(mT1)sin(πT1)(tmT1)(πT1)(tmT1)
b) m=us(mT1T12)sin(πT1)(tmT1+T12)(π/T1)(tmT1+T12)
c) m=us(mT1T12)sin(πT1)(tmT1T12)(πT1)(tmT1T12)
d) m=uc(mT1)sin(πT1)(t+mT1)(πT1)(t+mT1)
Explanation: To reconstruct the equivalent low pass signals. Thus, according to the sampling theorem for low pass signals with T1=1/B .
us(t)=m=us(mT1T1/2)sin(π/T1)(tmT1+T1/2)(π/T1)(tmT1+T1/2)

9. What is the expression for low pass signal component uc(t) that can be expressed in terms of samples of the bandpass signal?
a) n=(1)n+r+1x(2nTT)sin(π/(2T))(t2nT+T)(π/(2T))(t2nT+T)
b) n=(1)nx(2nT)sin(π/(2T))(t2nT)(π/(2T))(t2nT)
c) All of the mentioned
d) None of the mentioned
Explanation: The low pass signal components uc(t) can be expressed in terms of samples of the
band pass signal as follows:
uc(t)=n=(1)nx(2nT)sin(π/(2T))(t2nT)(π/(2T))(t2nT).

10. What is the expression for low pass signal component us(t) that can be expressed in terms of samples of the bandpass signal?
a) n=(1)n+r+1x(2nTT)sin(π/(2T))(t2nT+T)(π/(2T))(t2nT+T)
b) n=(1)nx(2nT)sin(π/(2T))(t2nT)(π/(2T))(t2nT)
c) All of the mentioned
d) None of the mentioned
Explanation: The low pass signal components us(t) can be expressed in terms of samples of the
band pass signal as follows:
us(t)=n=(1)n+r+1x(2nTT)sin(π/(2T))(t2nT+T)(π/(2T))(t2nT+T)

11. What is the Fourier transform of x(t)?
a) X (F) = 12[Xl(FFc)+Xl(FFc)]
b) X (F) = 12[Xl(FFc)+Xl(F+Fc)]
c) X (F) = 12[Xl(F+Fc)+Xl(FFc)]
d) X (F) = 12[Xl(FFc)+Xl(FFc)]
Explanation:
X (F) = x(t)ej2πFtdt
={Re[xl(t)ej2πFct]}ej2πFtdt
Using the identity, Re(ε)=1/2(ε+ε^*)
X (F) = [xl(t)ej2πFct+xl(t)ej2πFct]ej2πFtdt
=12[Xl(FFc)+Xl(FFc)].

12. What is the basic relationship between the spectrum o f the real band pass signal x(t) and the spectrum of the equivalent low pass signal xl(t)?
a) X (F) = 12[Xl(FFc)+Xl(FFc)]
b) X (F) = 12[Xl(FFc)+Xl(F+Fc)]
c) X (F) = 12[Xl(F+Fc)+Xl(FFc)]
d) X (F) = 12[Xl(FFc)+Xl(FFc)]
Explanation: X(F) = 12[Xl(FFc)+Xl(FFc)], where Xl(F) is the Fourier transform of xl(t). This is the basic relationship between the spectrum o f the real band pass signal x(t) and the spectrum of the equivalent low pass signal xl(t).

13. The distortion of information due to low-frequency sampling is known as
a) Sampling
b) Aliasing
c) Inquiry function
d) Anti-aliasing
Explanation: The distortion of information is called aliasing.

14. To avoid losing information from periodic objects we need
a) Sampling frequency twice
b) Nyquist sampling frequency
c) Both a or b
d) Neither a nor b
Explanation: Because nyquist sampling frequency means sampling frequency twice.

15. Nyquist sampling frequency formula is
a) fs=2fmax
b) fs=2fmin
c) fs=fmax
d) fs=fmin
Explanation: None.

16. The sampling of object characteristic at a high resolution and displaying the result at a lower resolution is called?
a) Super-sampling
b) Post-filtering
c) Anti-aliasing
d) a or b
Explanation: Super-sampling is also called Post-filtering by computing intensities and combines results to obtain the pixel intensities.

17. Anti-aliasing by computing overlap areas is referred to as
a) Area-sampling
b) Super-sampling
c) Pixel phasing
d) Only b
Explanation: The intensity of pixel as a whole is determined without calculating sub-pixel intensity.

18. Area-sampling is also known as
a) Pre-filtering
b) Pixel phasing
c) Post-filtering
d) Anti-aliasing
Explanation: None.

19. Raster objects can also be anti-aliased by shifting the display location of pixel areas is known as
a) Super-sampling
b) Pixel shaping
c) Pixel phasing
d) Any of these
Explanation: This technique is applied by micro-positioning the electron beam in relation to object geometry.

20. If we want to use more intensity levels to anti-alias the line, then
a) We increase the number of sampling positions
b) We decrease the number of sampling positions
c) We increase the number of pixels
d) Only c
Explanation: We increase the number of sampling positions across each pixel to use more intensity levels.

21. The procedure that increases the number of intensity levels for each pixel to total number of sub-pixels is
a) Area-sampling
b) Anti-aliasing
c) Super-sampling procedure
d) Only c
Explanation: The super-sampling procedure increases the number of intensity levels for each pixel to total number of sub-pixels.

21. For a 45% line, the line path is________ on the polygon area.
a) Horizontal
b) Centered
c) Vertical
d) Any of these
Explanation: The line path is centered on the polygon area only if a line is 45%.

22. An array of values specifying the relative importance of sub-pixel is referred as________ of sub-pixel weights.
c) Pixel phasing
d) Pixel weighting
Explanation: None.

23. The technique that is more accurate method for anti-aliasing lines is
a) Filtering
b) Area-sampling
c) Super-sampling
d) None
Explanation: In this technique we can imagine a continuous weighting surface covering the pixel.

24. Super-sampling methods can be applied by
a) Sub-dividing the total area
b) Determining the number of sub-pixels inside the area
c) Both a and b
d) Only b
Explanation: Super-sampling methods can be applied by sub-dividing the total area and determining the number of sub-pixels inside the area boundary.

25. Another method for determining the percentage of pixel area within a boundary is
a) Mid-print algorithm
b) Mid-point algorithm
c) Pixel intensity
d) By using inquiry functions
Explanation: This algorithm selects the next pixel along a line by determining which of 2 pixels is closer to the line between 2 pixels.

26. What is the use of Coherence techniques along and between scan lines?
a) To simplify the calculations
b) To determine the area edges
c) To find polygon region
d) To correct interior area
Explanation: Coherence techniques are used along and between scan lines to simplify the calculations.

27. The process of using a pulse signal to represent information is called _______
a) Pulse modulation
b) Frequency modulation
c) Amplitude modulation
d) Phase modulation
Explanation: In pulse modulation, the information to be transmitted is represented by a series of binary pulses. Since the pulse information is binary in nature analog signal shave to be converted to digital before transmitting.

28. Which of the following is false with respect to pulse modulation?
a) Less power consumption
b) Low noise
c) Degraded signal can be regenerated
d) Can transmit analog as well as digital waves
Explanation: Analog values cannot be transmitted as such by pulse modulation since it can only transmit binary data. However, the analog signal can be converted into digital using an ADC and then transmitted via pulse modulation.

29. Which of the following is not a form of pulse modulation?
a) Pulse amplitude modulation
b) Pulse width modulation
c) Pulse position modulation
d) Pulse frequency modulation
Explanation: There are four basic forms of pulse modulation. They are: pulse amplitude modulation, pulse width modulation, pulse position modulation pulse code modulation. In any form of pulse modulation, the frequency of the signal is not changed.

30. How many voltage levels are present in a PWM signal?
a) 0
b) 1
c) 2
d) 3
Explanation: The amplitude of PWM is binary in nature meaning that it has only two levels. The amplitude of the modulating signals varies the width of the pulses generated.

31. Power consumption is low in pulse modulation.
a) True
b) False
Explanation: In pulse modulation, the carrier is not transmitted continuously but in pulses whose width is determined by the amplitude of the modulating signal. The duty cycle is made in such a way that the carrier is off for a longer time than it bursts hence the average power consumption is low.

32. Which pulse modulation technique is least expensive?
a) Pulse amplitude modulation
b) Pulse width modulation
c) Pulse position modulation
d) Pulse code modulation
Explanation: Out of all the pulse modulation techniques, Pulse amplitude modulation is the least expensive and simplest to implement. In pulse amplitude modulation, the amplitude of the pulse varies with the amplitude of the modulating signal.

33. Which of the following is false with respect to pulse position modulation?
a) Can be transmitted in broadband
b) Modulates a high frequency carrier
c) Pulse is narrow
d) Pulse width changes in accordance with the amplitude of modulating signal
Explanation: In PPM, the pulses change position according to the amplitude of the analog signal. The pulses are very narrow. These pulse signals may be transmitted in a baseband form, but in most applications, they modulate a high-frequency radio carrier.

34. Pulse modulation is not used in which of the following?
a) Telemetry systems
b) Remote control models
c) Switch power modes
d) Communication of airplane with ATC
Explanation: Pulse modulation is used in telemetry systems to monitor spacecraft or missile, RC models, for switching power supplies like regulators and also as audio switching power amplifiers. Communication of airplane with ATC is amplitude modulated waves.

35. The process of signal compression and expansion used to reduce distortion and noise is called _____
a) Amplification
b) Companding
c) Compressing
d) Modulating
Explanation: To reduce the effects of noise and distortion in pulse modulation, a process called companding is done. Companding is a process of signal compression and expansion.

36. What type of digital modulation is widely used for digital data transmission?
a) Pulse amplitude modulation
b) Pulse width modulation
c) Pulse position modulation
d) Pulse code modulation
Explanation: The most widely used technique for digitizing information signals for electronic data transmission is pulse code modulation. It has uniform transmission quality and also can be used when the signal traffic is high.

37. What is the output voltage if the input voltage of a compander with a maximum voltage range of 1 V and a μ of 255 is 0.25?
a) 0V
b) 0.25V
c) 0.5V
d) 0.75V

Explanation:

38. What is the output voltage if the input voltage of a compander with a maximum voltage range of 1 V and a μ of 255 is 0.8V0?
a) 0.08V
b) 0.458V
c) 1.02V
d) 1.54V
Explanation:

39. In the single-pulse width modulation method, the output voltage waveform is symmetrical about __________
a) π
b) 2π
c) π/2
d) π/4
Explanation: The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.

40. In the single-pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.
a) 2π
b) 3π/2
c) π/2
d) 3π/4
Explanation: In the negative half the wave is symmetrical about 3π/2.

41. The shape of the output voltage waveform in a single PWM is
a) square wave
b) triangular wave
c) quasi-square wave
d) sine wave
Explanation: Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.

42. In the single-pulse width modulation method, the Fourier coefficient bn is given by
a) (Vs/π) [ sin(nπ/2) sin(nd) ].
b) 0
c) (4Vs/nπ) [sin(nπ/2) sin(nd)].
d) (2Vs/nπ) [sin(nπ/2) sin(nd)].
Explanation: The Fourier analysis is as under:
bn = (2/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (π/2 + d) to (π/2 – d)
2d is the width of the pulse.

43. In the single-pulse width modulation method, the Fourier coefficient an is given by
a) (Vs/π) [ cos(nπ/2) cos(nd) ].
b) 0
c) (4Vs/nπ) [sin(nπ/2) sin(nd)].
d) (2Vs/nπ) [sin(nπ/2) sin(nd)].
Explanation: As the positive and the negative half cycles are identical the coefficient an = 0.

44. In the single-pulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by
a) Vs
b) 4Vs/π
c) 0
d) 2Vs/π
Explanation: The Fourier representation of the output voltage is given by

Put 2d = π & n = 1.

45. In case of a single-pulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression
a) 4Vs/π
b) Vs
c) (4Vs/π) sin 2d
d) (4Vs/π) sin d
Explanation: For the fundamental component put n = 1.

Vo = (4Vs/π) sin (d) sin (ωt)
Hence the peak value is (4Vs/π) sin d.

46. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage
a) d = π
b) 2d = π
c) nd = π
d) nd = 2π
Explanation: To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.
From the below expression,

when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.

47. Find the peak value of the fundamental component of voltage with a pulse width of 2d = 90 and Vs = 240 V for single-pulse modulation in a full wave bridge inverter.
a) 305 V
b) 216 V
c) 0 V
d) 610 V
Explanation: The peak value of the fundamental component of voltage is given by (4Vs/π) sin d.

48. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 3rd harmonic from the output voltage waveform, the value of the pulse width (2d) must be
a) 0°
b) 60°
c) 120°
d) 180°
Explanation: To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/3 = 60°
Hence, 2d = 120°.

49. In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 5th harmonic from the output voltage waveform, the value of the pulse width (2d) must be equal to
a) 72°
b) 86°
c) 91°
d) 5°
Explanation: To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/5 = 36°
Hence, 2d = 72°.

50. Several equidistant pulses per half cycle are used in ___________ type of modulation technique.
a) single-pulse
b) multiple-pulse
c) sine-pulse
d) equidistant-pulse
Explanation: In MPM, several equidistant pulses per half cycle are used.

51. In the multiple-pulse width modulation method, the Fourier coefficient an is
a) (Vs/π) [ cos(nπ/2) cos(nd) ].
b) 0
c) (4Vs/nπ) [sin(nπ/2) sin(nd)].
d) (2Vs/nπ) [sin(nπ/2) sin(nd)].
Explanation: As the positive and the negative half cycles are identical the coefficient an = 0.

52. In the multiple-pulse width modulation method with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, has the Fourier coefficient bn =
a) (8Vs/π) [ cos(ɣπ/2) cos(nd/2) ].
b) 0
c) (4Vs/nπ) [sin(nɣ) sin(nd/2)].
d) (8Vs/nπ) [sin(nɣ) sin(nd/2)].
Explanation: The Fourier analysis is as under:
bn = (4/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (ɣ + d/2) to (ɣ – d/2)
bn = (8Vs/nπ) [sin(nɣ) sin(nd/2)].

53. The amplitude of the nth harmonic of the two-pulse MPM waveform is given by __________
Let d be the width of a single pulse and ɣ be the distance from 0 to the centre of the first pulse.
a) (8Vs/nπ) sin(nɣ) sin(nd/2)
b) (4Vs/nπ) sin(nɣ) sin(nd/2)
c) (8Vs/nπ) sin(nπ) sin(nd/2)
d) (8Vs/nπ) sin(nɣ/2) sin(nd/2)
Explanation: The Fourier series representation of such a wave will be given by

bn = (8Vs/nπ) [sin(nɣ) sin(nd/2)] an = 0
Therefore,
Hence, the amplitude is the term excluding the sin nωt factor.

54. In case of MPM with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, for eliminating the nth harmonic from the output voltage, which of the following condition must be satisfied?
a) d = 2π
b) d = π
c) d = n°
d) d = 2π/n
Explanation:

If d = 2π/n, sin (π) = 0 and the whole term is eliminated.

55. In case of MPM with two pulses per half cycle of width = d each and ɣ as the distance between the first pulse and ωt=0, for eliminating the nth harmonic from the output voltage, the value of gamma (ɣ) must be equal to
a) 0
b) π
c) π/n
d) d/n
Explanation:

If gamma = π/n , the whole nth output voltage comes becomes zero.

56. Find the peak value of the fundamental component of voltage with MPM with two pulses having pulse width = 36° and ɣ = 54°. The Fourier representation of the waveform is as follows.

a) 0.7484 x Vs
b) 1.414 x Vs
c) 0.637 x Vs
d) 2.54 x Vs
Explanation: For the fundamental component put n = 1.
n = 36° and ɣ = 54° . . . (given)
Vo1 = 8Vs/π x sin54 x sin18 = 0.637 Vs.

57. In the MPM method, the comparator is given _______ and _______ types of waveform at its input.
a) square, sine
b) square, quasi-square
c) sine, triangular
d) square, triangular
Explanation: To generate the modulated waves of equal width, square wave which is the reference signal is compared with the triangular wave which is the carrier signal wave.

58. In MPM, the square wave is the ________ signal whereas the triangular wave is the ________ signal.
a) reference, carrier
b) base, reference
c) carrier, reference
d) none of the mentioned
Explanation: To generate the modulated waves of equal width, square wave which is the reference signal is compared with the triangular wave which is the carrier signal wave.

59. In the multiple pulse width modulation method, the firing pulses are generate during the interval when the
a) triangular wave exceeds the square modulating wave
b) square modulating wave exceeds the triangular wave
c) square wave amplitude is same as the triangular wave’s amplitude
d) none of the mentioned
Explanation: The firing pulses to turn on the SCRs (or any other equivalent device) are generated when the triangular carrier signal exceeds the square reference signal.

60. In MPM, ____________ order harmonics can be eliminated by a proper choice of __________ and _________
a) higher, d, ɣ
b) lower, d, ɣ
c) higher and lower, d, ɣ
d) none of the mentioned
Explanation: In multiple pulse width modulation, the lower order harmonics can be eliminated by proper choice of 2d and ɣ.

61. In ___________ type of modulation method, the pulse width is not equal for all the pulses.
a) multiple pulse width modulation
b) single pulse width modulation
c) sinusoidal pulse width modulation
d) none of the mentioned
Explanation: In SPWM, the pulse width is a sinusoidal function of the angular position of the pulse in a cycle.

62. In sinusoidal pulse width modulation, __________ wave is compared with a ___________ type of wave.
a) square, sinusoidal
b) sinusoidal, triangular
c) sinusoidal, quasi-square
d) none of the mentioned
Explanation: In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.

63. In the sinusoidal pulse width modulation, __________ is the carrier wave signal.
a) square wave
b) triangular wave
c) sinusoidal wave
d) quasi-square wave
Explanation: In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.

64. In the sinusoidal pulse width modulation, ____________ is the reference wave signal.
a) square wave
b) triangular wave
c) sinusoidal wave
d) quasi-square wave
Explanation: In SPWM, a high-frequency triangular wave is compared with a sinusoidal reference wave of the desired frequency.

65. In sinusoidal pulse width modulation, the comparator output is high when the
a) triangular wave has magnitude higher than the sinusoidal wave
b) sinusoidal wave has magnitude higher than the triangular wave
c) triangular wave has magnitude equal to the sinusoidal wave
d) none of the mentioned
Explanation: The comparator output is high when the sinusoidal wave (reference signal) has magnitude higher than the triangular wave (carrier signal).

66. In PWM, the comparator output is further given to a ____________
a) integrator
b) scr devices
c) trigger pulse generator
d) snubber circuit
Explanation: The comparator output is processed in a trigger pulse generator in such a manner that the output voltage wave of the inverter has a pulse width in agreement with the comparator output pulse width.

67. The modulation index (MI) is given by
Vr = peak value of the reference wave.
Vc = peak value of the carrier wave.
a) Vr/Vc
b) Vc/Vr
c) (1 + Vc/Vr)
d) 1/(Vc Vr)
Explanation: MI = Vr/Vc.

68. By controlling the modulation index (MI), __________ can be controlled.
a) gain
b) output frequency
c) harmonic content of the output voltage
d) cosine component of the output voltage
Explanation: MI controls the output voltage waveform.

#### Module 05

1. Average energy per bit is given by
a) average energy symbol/log2 M
b) average energy symbol * log2 M
c) log2 M/ Average energy symbol
d) none of the mentioned
Explanation: Average energy per bit is given by average energy symbol divided by log2 M.

2. Which FSK has no phase discontinuity?
a) Continuous FSK
b) Discrete FSK
c) Uniform FSK
d) None of the mentioned
Explanation: Continuous frequency shift keying has no phase discontinuity between symbols.

3. FSK reception is
a) Phase Coherent
b) Phase non coherent
c) Phase Coherent & non coherent
d) None of the mentioned
Explanation: Reception of FSK can be either phase coherent or phase non coherent.

4. FSK reception uses
b) PLL
d) None of the mentioned
Explanation: Frequency shift keying uses correlation receiver and phase locked loop.

5. In non coherent reception _____ is measured.
a) Phase
b) Energy
c) Power
d) None of the mentioned
Explanation: In non coherent reception of FSK, energy in each frequency is measured.

6. Every frequency has ____ orthogonal functions.
a) One
b) Two
c) Four
d) Six
Explanation: Every frequency has two orthogonal functions – sine and cosine.

7. If we correlate the received signal with any one of the two orthogonal function, the obtained inner product will be
a) In phase
c) Zero
d) Cannot be determined
Explanation: If we correlate the received signal with only one of the orthogonal function for example sine, the inner product obtained will be zero.

8. If we correlate the received signal with both orthogonal function the inner product will be
a) In phase
d) Unity
Explanation: If we correlate the received signal with both the orthogonal function, the two inner products obtained will be in phase and quadrature.

9. Simulation is used to determine
a) Bit error rate
b) Symbol error rate
c) Bit error
d) Symbol error
Explanation: A simulation of digital communication system is used to estimate bit error rate.

10. Matched filter is a _____ technique.
a) Modulation
b) Demodulation
c) Modulation & Demodulation
d) None of the mentioned
Explanation: Matched filter is a demodulation technique with LTI.

11. Which is called as on-off keying?
a) Amplitude shift keying
b) Uni-polar PAM
c) Amplitude shift keying & Uni-polar PAM
d) None of the mentioned
Explanation: Amplitude shift keying and uni-polar PAM both schemes are called as on off keying.

12. QAM uses ______ as the dimensions.
a) In phase
d) None of the mentioned
Explanation: QAM uses in phase and quadrature which is cosine and sine respectively as the dimensions.

13. Which has same probability of error?
a) BPSK and QPSK
c) BPSK and PAM
d) BPSK and QAM
Explanation: BPSK is similar to bipolar PAM and both have same probability of error.

14. Which system uses QAM?
a) Digital microwave relay
b) Dial up modem
c) Digital microwave relay & Dial up modem
d) None of the mentioned
Explanation: Digital microwave relay, dial up modem and etc uses QAM modulation technique.

15. Digital communication is _______ to environmental changes?
a) Less sensitive
b) More sensitive
c) Does not depend
d) None of the mentioned
Explanation: Digital communication is less sensitive to environmental changes like temperature etc.

16. Advantages of digital communication are
a) Easy multiplexing
b) Easy processing
c) Reliable
d) All of the mentioned
Explanation: Digital communication is a very reliable communication. It is easy for multiplexing, easy for signalling and processing etc.

17. What is necessary for digital communication?
a) Precision timing
b) Frame synchronization
c) Character synchronization
d) All of the mentioned
Explanation: Bit, character, frame synchronization and precision timing is necessary for digital communication. This is considered as a disadvantage of digital communication.

18. What are the disadvantages of digital communication?
a) Needs more bandwidth
b) Is more complex
c) Needs more bandwidth & Is more complex
d) None of the mentioned
Explanation: Digital communication needs more bandwidth, has higher complexity and little performance degradation occurs during analog to digital conversion and vice versa.

19. Examples of digital communication are
a) ISDN
b) Modems
c) Classical telephony
d) All of the mentioned
Explanation: Some of the examples of digital communication systems are classical telephony, ISDN, Modems, LANs, PCM TDM etc.

20. Which system uses digital transmission?
a) ISDN
b) LANs
c) ISDN & LANs
d) None of the mentioned
Explanation: Though the signal type is analog or digital, the transmission takes place in the digital domain in ISDN and LANs.

21. The interval of frequencies outside which the spectrum is zero is called as ________
a) null to null bandwidth
b) normalized bandwidth
c) absolute bandwidth
d) none of the mentioned
Explanation: The measure of frequencies outside which spectrum is zero is called as absolute bandwidth. It is usually infinite.

22. The attenuation level in bounded power spectral density is
a) 35
b) 50
c) 35 & 50
d) none of the mentioned
Explanation: Bounded power spectral density is the bandwidth outside which the spectrum must have fallen to a stated level below that found at the band center.

23. Synchronization available in digital communication are
a) Symbol synchronization
b) Frame synchronization
c) Carrier synchronization
d) All of the mentioned
Explanation: The synchronization techniques available in digital communication are symbol synchronization, frame synchronization and carrier synchronization.

24. Digital system includes
a) Better encryption algorithm
b) Difficult data multiplexing
c) All of the mentioned
d) None of the mentioned
Explanation: Digital system has an advantage of better encryption algorithm, easier data multiplexing and more reliability.

25. Analog to digital conversion includes
a) Sampling
b) Quantization
c) Sampling & Quantization
d) None of the mentioned
Explanation: Analog to digital conversion is a two step process which includes sampling and quantization.

26. The basic task of the A/D converter is to convert a discrete set of digital code words into a continuous range of input amplitudes.
a) True
b) False
Explanation: The basic task of the A/D converter is to convert a continuous range of input amplitude into a discrete set of digital code words. This conversion involves the processes of Quantization and Coding.

27. What is the type of quantizer, if a Zero is assigned a quantization level?
a) Midrise type
c) Mistreat type
d) None of the mentioned
Explanation: If a zero is assigned a quantization level, the quantizer is of the mid treat type.

28. What is the type of quantizer, if a Zero is assigned a decision level?
a) Midrise type
c) Mistreat type
d) None of the mentioned
Explanation: If a zero is assigned a decision level, the quantizer is of the midrise type.

29. What is the term used to describe the range of an A/D converter for bipolar signals?
a) Full scale
b) FSR
c) Full-scale region
d) FS
Explanation: The term Full-scale range (FSR) is used to describe the range of an A/D converter for bipolar signals (i.e., signals with both positive and negative amplitudes).

30. What is the term used to describe the range of an A/D converter for uni-polar signals?
a) Full scale
b) FSR
c) Full-scale region
d) FSS
Explanation: The term Full scale (FS) is used for uni-polar signals.

31. What is the fixed range of the quantization error eq(n)?
a) –Δ6 < eq(n) ≤ Δ6
b) –Δ4 < eq(n) ≤ Δ4
c) –Δ2 < eq(n) ≤ Δ2
d) –Δ16 < eq(n) ≤ Δ16
Explanation: The quantization error eq(n) is always in the range – Δ2 < eq(n) ≤ Δ2, where Δ is quantizer step size.

32. If the dynamic range of the signal is smaller than the range of quantizer, the samples that exceed the quantizer are clipped, resulting in large quantization error.
a) True
b) False
Explanation: If the dynamic range of the signal, defined as xmax-xmin, is larger than the range of the quantizer, the samples that exceed the quantizer range are clipped, resulting in a large (greater than Δ2) quantization error.

33. What is the relation defined by the operation of quantizer?
a) xq(n) ≡ Q[x(n)] = x^k
b) xq(n) = Q[x(n)] = x^k, if x(n) ∈ Ik
c) xq(k) ≡ Q[x(k)] = x^k
d) none of the mentioned
Explanation: The possible outputs of the quantizer (i.e., the quantization levels) are denoted as x^1,x^2,x^L. The operation of the quantizer is defined by the relation, xq(n) ≡ Q[x(n)]= x^k, if x(n) ∈ Ik.

34. What is the step size or the resolution of an A/D converter?
a) Δ = (R)/2(b+1)
b) Δ = (R)/2(b-1)
c) Δ = (R)/3(b+1)
d) Δ = (R)/2
Explanation: The coding process in an A/D converter assigns a unique binary number to each quantization level. If we have L levels, we need at least L different binary numbers. With a word length of b + 1 bits we can represent 2b+1 distinct binary numbers. Hence we should have 2(b+1) > L or, equivalently, b + 1 > log2 L. Then the step size or the resolution of the A/D converter is given by
Δ = (R)/2(b+1), where R is the range of the quantizer.

35. In the practical A/D converters, if the first transition may not occur at exactly + 1/2 LSB, then such kind of error is known as ____________
a) Scale-factor error
b) Offset error
c) Linearity error
d) All of the mentioned
Explanation: We note that practical A/D converters may have offset error (the first transition may not occur at exactly + 1/2 LSB).

36. In the practical A/D converters, if the difference between the values at which the first transition and the last transition occur is not equal to FS – 2LSB, then such error is known as _________
a) Scale-factor error
b) Offset error
c) Linearity error
d) All of the mentioned
Explanation: We note that practical A/D converters scale-factor (or gain) error (the difference between the values at which the first transition and the last transition occur is not equal to FS — 2LSB).

37. In the practical A/D converters, if the differences between transition values are not all equal or uniformly changing, then such error is known as?
a) Scale-factor error
b) Offset error
c) Linearity error
d) All of the mentioned
Explanation: We note that practical A/D converters, linearity error (the differences between transition values are not all equal or uniformly changing).

38. Which of the following type of multiplexing uses pulse code modulation?
a) Frequency division multiplexing
b) Time division multiplexing
c) Code division multiplexing
d) Amplitude limited multiplexing
Explanation: The most popular form of TDM uses pulse-code modulation (PCM), in which multiple channels of digital data are transmitted in serial form. Each channel is assigned a time slot in which to transmit one binary word of data. The data streams from the various channels are interleaved and transmitted sequentially.

39. Which of the following statements is true with respect to PCM?
a) The parallel binary data is converted into serial before transmission
b) Analog data is transmitted directly
c) Analog signal is amplified before transmission
d) The analog signal is converted into parallel binary data before transmission
Explanation: The analog signal is converted into a digital signal. Since the converted digital signal is parallel bits of data, it has to be converted to serial before it can be multiplexed and transmitted.

40. Which of the following is false with respect to PCM?
a) Reliable
b) Inexpensive
c) Resistant to noise
d) Not easily recoverable
Explanation: When signals have been degraded because of noise, attenuation, or distortion, all the receiver has to do is to determine whether a pulse was transmitted. Amplitude, width, frequency, phase shape, and so on do not affect reception. Thus PCM signals are easily recovered and rejuvenated, no matter what the circumstances.

41. A special PCM system uses 16 channels of data, one whose purpose is identification (ID) and synchronization. Find (a) the number of available data channels.
a) 15
b) 16
c) 14
d) 18
Explanation: 16 (total no. of channels) – 1 (channel used for ID) = 15 (for data).

42. A special PCM system uses 16 channels of data, one whose purpose is an identification (ID) and synchronization. The word length is 6 bits. Find the number of bits per frame.
a) 94
b) 95
c) 96
d) 125
Explanation: Here, Data channels = 16 and word length is 6 bits.
Bits per frame = 6 x 16 = 96.

43. A special PCM system uses 16 channels of data, one whose purpose is an identification (ID) and synchronization. The sampling rate is 3.5 kHz. The word length is 6 bits. Find the serial data rate.
a) 451kHz
b) 326kHz
c) 152kHz
d) 336kHz
Explanation: Serial data rate = sampling rate x no. bits/frame = 3.5 kHz x 96 = 336 kHz.

44. All local and long-distance connections are digital.
a) True
b) False
Explanation: All modern telephone systems use digital transmission via PCM and TDM. The only place where analog signals are still used is in the local loop—the connection between a telephone company’s central office (CO) and the subscriber’s telephone, known as the customer premises equipment (CPE). All local and long-distance connections are digital.

45. In a TDD one channel is sufficient for transmission.
a) True
b) False
Explanation: Time-division duplexing (TDD) means that signals are transmitted simultaneously on a single channel by interleaving them in different time slots. Each time slot may contain one data word, such as 1 byte from an A/D converter or a D/A converter. As long as the serial data rate is high enough, a user will never know the difference. The primary benefit of TDD is that only one channel is needed. It saves spectrum space and cost.

46. Properties used to determine stream’s fidelity
a) Sampling rate
b) Bit depth
c) Sampling rate & Bit depth
d) None of the mentioned
Explanation: Two basic properties to determine stream’s fidelity are bit depth and sampling rate, number of times per second that samples are taken.

47. In bipolar codes, pulses can be
a) Positive
b) Negative
c) Absent
d) All of the mentioned
Explanation: In bipolar codes, the pulses can be positive, negative or absent.

48. Delta modulation is ______ conversion.
a) Analog to digital
b) Digital to analog
c) Analog to digital and digital to analog
d) None of the mentioned
Explanation: Delta modulation is the process of analog to digital and digital to analog conversion technique used for transmission of voice signals.

49. To achieve high signal to noise ratio, delta modulation must use
a) Under sampling
b) Over sampling
c) Aliasing
d) None of the mentioned
Explanation: To achieve high signal to noise ratio, delta modulation must use over sampling techniques.

50. The demodulator in delta modulation technique is
a) Differentiator
b) Integrator
c) Quantizer
d) None of the mentioned
Explanation: The demodulator used in delta modulation is a simple form of integrator.

51. Source of noise in delta modulation is
a) Granularity
d) None of the mentioned
Explanation: Sources of noise in delta modulation are granularity and slope overload.

52. When probability of receiving a symbol is 1 then how much information will be obtained?
a) Little information
b) Much information
c) No information
d) None of the mentioned
Explanation: When the probability of receiving a symbol is 1 then the information obtained is zero.

53. In channel encoding procedure
b) Errors are corrected
c) Redundancy bits are added & Errors are corrected
d) None of the mentioned
Explanation: In channel encoding includes addition of redundancy to the signal such that any bit errors can be corrected.

54. Modulation process includes
a) Analog to digital conversion
b) Digital to analog conversion
c) All of the mentioned
d) None of the mentioned
Explanation: Modulation procedure includes digital to analog conversion which produces a continuous time signal that can be sent through the channel.

55. Switching exists in
a) Point to point communication
c) Point to point & Broadcast communication
d) None of the mentioned
Explanation: Circuit connection in point to point communication is called as switching. Switching does not exist in broadcast communication or network.

56. Space division has dedicated
a) Paths
b) Time slots
c) Paths & Time slots
d) None of the mentioned
Explanation: Space division has dedicated paths and time division has dedicated time slots.

57. In time division system, the actual switch is called as
a) Speech memory
b) Cross point
c) Connecting point
d) None of the mentioned
Explanation: In time division, actual switch is called as speech memory and in space division actual switch is called as cross point.

58. In time division, connection is established using
a) Data exchange
b) Galvanic connection
c) Data exchange & Galvanic connection
d) None of the mentioned
Explanation: In time division, connection is established using data exchange and in space division it is established using galvanic connections.

59. Analogue switches provides
a) Good bandwidth
b) Low distortion
c) Low cost
d) All of the mentioned
Explanation: Some of the features of analogue switches are good bandwidth, low cost, low distortion, lower reliability.

60. Operations performed by switching system
a) Path establishment
b) Information exchange
c) Tariff computation
d) All of the mentioned
Explanation: Operations performed by switching network are path establishment, information exchange, tariff computation, maintenance, billing etc.

61. Time division multiplexing includes ___________________
d) None of the mentioned
Explanation: Time division multiplexing includes both radio or wired link according to application.

62. Which of the following data is correct for TDM?
a) Analog data is transmitted
b) Digital data is transmitted
c) Both analog and digital data transmitted
d) None of the mentioned
Explanation: Using TDM analog data is transmitted.

63. PAM stands for _______________
a) Pulse Amplitude Modulation
b) Power Amplitude Modulation
Explanation: Term PAM stands for process Pulse Amplitude Modulation.

64. Commutators are mechanical switches in operation.
a) True
b) False
Explanation: Commutators perform an operation of on and off and they function as mechanical switches.

65. Which of the following represents a number of samples per second?
a) Product of frame rate and number of samples per frame
b) Frame rate
c) Ratio of samples per frame and frame rate
d) None of the mentioned
Explanation: Number of samples taken per second will be equal to the product of frame rate and number of samples per frame.

66. What will be the general number of samples per frame?
a) 18
b) 30
c) 18 or 30
d) None of the mentioned
Explanation: Generally a number of samples per frame will be 18 or 30.

67. Maximum rate of commutation will be _________________
a) 800 sample per second
b) 900 sample per second
c) 1200 sample per second
d) 1000 sample per minute
Explanation: Maximum rate of commutation in TDM will be 900 samples/sec.

68. Amplitude of each pulse of PAM train conveys the amplitude of particular channel sampled.
a) True
b) False
Explanation: Amplitude data of particular sampled channel can be obtained from the amplitude of PAM train.

69. PTM stand for _________________
a) Pulse train modulation
b) Pulse time modulation
c) Power train modulation
d) None of the mentioned
Explanation: PTM is the short form of pulse train modulation.

70. Which of the following can be generated from the PDM signal?
a) PPM
b) PTM
c) PAM
d) PFM
Explanation: If PDM signals are available, PPM signal can be generated from PDM signals.

71. Multiplexing increases the number of communication channels for transmission.
a) True
b) False
Explanation: Multiplexing is the process of simultaneously transmitting two or more individual signals over a single communication channel, cable or wireless. In effect, it increases the number of communication channels so that more information can be transmitted.

72. In which of the following systems multiplexing is not necessary?
a) Telemetry
c) Satellites
d) Continuous wave transmission
Explanation: Continuous wave transmission such as morse code, multiplexing is not necessary since only two voltage levels are present and each bit is sent one by one. Also, only one information signal is transmitted whereas in telemetry, TV and satellite communications numerous information is transmitted hence multiplexing is required.

73. Time division multiplexing: Digital signal:: Frequency division multiplexing:?
a) Pulse code modulated signal
b) Continuous wave signals
c) Analog signal
d) Pulse position modulated signal
Explanation: The two most common types of multiplexing are frequency-division multiplexing (FDM) and time-division multiplexing (TDM). Two variations of these basic methods are frequency-division multiple access (FDMA) and time-division multiple access (TDMA). In general, FDM systems are used for analog information and TDM systems are used for digital information.

74. What type of multiplexing is widely used in cellphones?
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) Spatial multiplexing
Explanation: Another form of multiple accesses is known as code-division multiple access (CDMA). It is widely used in cell phone systems to allow many cell phone subscribers to use a common bandwidth at the same time. This system uses special codes assigned to each user that can be identified. CDMA uses a technique called spread spectrum to make this type of multiplexing possible.

75. The transmission of multiple signals in a common frequency without interference is called _______
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) Spatial multiplexing
Explanation: Spatial multiplexing is the term used to describe the transmission of multiple wireless signals on a common frequency in such a way that they do not interfere with one another. One way of doing this is to use low-power transmissions so that the signals do not interfere with one another. When very low power is used, the signals do not travel very far. The transmission distance is a function of the power level, frequency, and antenna height.

76. For frequency division multiplexing who defines the channel bandwidth?
a) FCC
b) ARNIC
c) FAA
d) CCA
Explanation: The bandwidths of radio channels vary, and are usually determined by FCC regulations and the type of radio service involved. Regardless of the type of channel, a wide bandwidth can be shared for the purpose of transmitting many signals at the same time.

77. What is the individual carrier frequency of each signal called?
a) Subcarrier
b) Frequency carrier
c) Modulated carrier
d) Coded carrier
Explanation: Each signal to be transmitted feeds a modulator circuit. The carrier for each modulator (fc) is on a different frequency. The carrier frequencies are usually equally spaced from one another over a specific frequency range. These carriers are referred to as subcarriers.

78. Which circuit does the actual multiplexing process in frequency division multiplexing?
a) Linear mixer
b) Oscillator
c) RF amplifier
d) Duplexer
Explanation: The modulator outputs containing the sideband information are added algebraically in a linear mixer; no modulation or generation of sidebands takes place. The resulting output signal is a composite of all the modulated subcarriers. This signal can be used to modulate a radio transmitter or can itself be transmitted over the single communication channel. Alternatively, the composite signal can become one input to another multiplexed system.

79. Which of the following device is used to demultiplex the received signal?
a) Allpass filters
b) Bandpass filters
c) Bandstop filters
d) Differential filters
Explanation: A receiver picks up the signal and demodulates it, recovering the composite signal. This is sent to a group of bandpass filters, each centered on one of the carrier frequencies. Each filter passes only its channel and rejects all others. A channel demodulator then recovers each original input signal.

80. The system which uses FM for the subcarriers is called _____
a) FM II system
b) FM/FM system
c) FM/AM system
d) 2 stage FM system
Explanation: Generally the individual signals which require multiplexing are frequency modulated. These signals are then added up by the mixer and the resulting output signal is again frequency modulated before transmission.

81. A cable TV service uses a single coaxial cable with a bandwidth of 860 MHz to transmit multiple TV signals to subscribers. Each TV signal is 6 MHz wide. How many channels can be carried?
a) 143
b) 123
c) 100
d) 150
Explanation: Total channels = 860/6 = 143.33 or 143.

#### Module 06

1. Modern EDM uses which among the following waves?
a) Visible rays
b) Thermal infra-red
c) Modulated infra-red
Explanation: Modern EDM uses the modulated infra-red waves, which are capable of receiving the reflected waves from a distance of 100km.

2. Which property of an electromagnetic wave, depends on the medium in which it is travelling?
a) Velocity
b) Frequency
c) Time period
d) Wave length
Explanation: The frequency, wavelength and time period can all vary according to the wave producing source. But, the velocity of an electromagnetic wave depends upon the medium through which it is travelling. The velocity of wave in a vacuum is termed as speed of light, which is assumed to be 3 x 108 m/s.

3. The distance in EDM is measured by______________
a) Frequency of the wave
b) Wave length
c) Phase difference
d) Amplitude
Explanation: In general, the various EDM systems available do not measure the transit time directly. The distance is determined by measuring the phase difference between the transmitted and reflected signals, which possible only when the process is done without errors.

4. Tellurometer, a type of EDM uses which of the following waves?
a) Visible rays
b) Infra-red waves
c) Micro waves
Explanation: Tellurometer is generally having a range of 100km i.e., it is capable of measuring distance up to 100km. In order to function properly it requires a high range propagating wave, which can reflect back to the instrument.

5. Find the value of D if the wave length of the wave is 40m, n=2 m and the angles are given as θ1 = 0˚, θ2 = 180˚.
a) 50m
b) 40m
c) 20m
d) 10m
Explanation: The value of D can be found by,
2D = n*λ + Δ*λ. On substitution, we get
2D = 40*(2 + (180-0/360))
D = 50 m.

6. Electromagnetic waves are represented in which of the following format?
a) Longitudinal waves
b) Transverse waves
c) Sinusoidal waves
d) Surface waves
Explanation: A sinusoidal wave describes the oscillation periodically. Since the electromagnetic wave is represented in the form of sine curve it is named after it as sinusoidal wave.

7. For increasing accuracy, high frequency of propagation is used.
a) True
b) False
Explanation: In order to increase accuracy, it is recommended to use an extremely high frequency of propagation. However, the available phase comparison techniques cannot be used at frequencies greater than 5* Hz, which corresponds to a wavelength of 0.6 m.

8. What would be the value of length if the distance is given as 30m, m= 3 and the change in length is 8m.
a) 7.43m
b) 7.34m
c) 6.34m
d) 5.43m
Explanation: The value of length can be found out by using the formula,
D = m*l + Δ l. On substitution, we get
30 = 3*l + 8
L = 7.34 m.

9. Phase difference can be expressed in which of the following format?
a) Meters per second
b) Meters
c) Cycles
d) Seconds
Explanation: In the total station, the distance between two point scan be determined by measuring the phase difference between transmitted and reflected signals. Phase difference can be expressed in terms of the fraction of cycles, which provides the best output for quick calculation and for easy understanding.

10. Which of the following represents the correct sequence for the basis of EDM propagation?
a) Propagation, generation, reflection and reception
b) Generation, reception, reflection and propagation
c) Generation, propagation, reception and reflection
d) Generation, propagation, reflection and reception
Explanation: Electronic Distance Measurement method is based on generation, propagation, reflection and subsequent reception of electromagnetic waves. The type of electromagnetic waves generated depends on many factors but mainly, on the nature of the electrical signal used to generate the waves. It follows the above-mentioned sequence without any disturbance in it.

11. Up to which frequency the ground wave propagation is used?
a) 2MHz
b) 2GHz
c) 30MHz
d) 30GHz
Explanation: Ground wave propagation also known as surface wave propagation is used up to 2MHz. sky wave propagation is used at 2MHz to 30MHz.

12. In a ground wave propagation, which component of electric field is short circuited when it’s in contact by earth?
a) Horizontal
b) Vertical
c) Both horizontal and vertical
d) Neither horizontal nor vertical
Explanation: Any horizontal component of electric field which is in contact with earth is short circuited by earth. Usually ground wave propagation is done by vertical antennas so it is vertically polarized.

13. During ground wave propagation earth behaves like a __________
a) Leaky capacitor
b) Leaky Inductor
c) Series combination of capacitor and inductor
d) Parallel combination of capacitor and inductor
Explanation: Any horizontal component of electric field which is in contact with earth is short circuited by earth. So earth behaves like a leaky capacity. It forms a resistor in shunt with a capacitor.

14. Sky wave propagation reflects the frequencies ___________
a) 2MHz
b) 2 MHz to 30MHz
c) 2 GHz to 30 GHz
d) 30 GHz to 50GHz
Explanation: Ground wave propagation also known as surface wave propagation is used up to 2MHz. sky wave propagation is used at 2MHz to 30MHz.

15. At what distance the sky wave propagation is present from the earth surface?
a) 50 to 400km
b) Below 50 km
c) 600 to 750km
d) 50 to 400 m
Explanation: Sky wave propagation also known as ionosphere propagation reflects the waves with frequency 2 to 30MHz. It is present at 50 to 400km from earth surface.

16. Space wave propagation reflects the waves with frequencies _________
a) Below 2 GHz
b) 2 to 30MHz
c) Above 30GHz
d) Above 30MHz
Explanation: Ground wave propagation also known as surface wave propagation is used up to 2MHz. sky wave propagation is used at 2MHz to 30MHz. Space wave propagation reflects frequencies above 30MHz.

17. Space wave propagates at which frequency band?
a) VHF
b) HF
c) MF
d) EHF
Explanation: Space wave propagation reflects frequencies at 30 to 300MHz range. So it propagates at VHF band.
MF- 300 KHz – 3MHz
EHF- 30GHz-300GHz

18. Communication through LOS can be increased by decreasing the height of antenna.
a) True
b) False
Explanation: Line of sight provides a direct communication link from transmitter to receiver. If the height of antennas is increased then the LOS is also improved.

19. In which of the following mode of propagation the waves are guided along the surface of the earth?
a) Ground wave
b) Sky wave
c) LOS
d) Space wave
Explanation: In ground wave or surface wave propagation the waves are guided along the surface of the earth. In sky wave they are reflected at different layers in the ionosphere. Space wave uses either direct or indirect method of propagation from transmitter to receiver directly.

20. In which of the following modes of propagation the ionosphere acts as the reflecting surface for the waves?
a) Ground wave
b) Sky wave
c) Space wave
d) LOS
Explanation: In Sky wave or Ionospheric wave propagation waves are reflected from the ionosphere layers depending upon different frequencies.

21. Which of the following is the lowest layer of atmosphere?
a) Troposphere
b) Stratosphere
c) Ionosphere
d) Outer atmosphere
Explanation: Troposphere is the lowest layer in the atmosphere ranging up to 15km. stratosphere lies at 50 to 90km and up to 400km ionosphere. Above 400km is the outer atmosphere.

22. What is the range of the region of calm from the earth surface?
a) 20- 70km
b) 2-15km
c) 150-400km
d) Above 400km
Explanation: Region of calm is also known as stratopause. It is at 20 to 70km. Troposphere is up to 15km and above 400km is the outer atmosphere.

23. In which of the following layer of atmosphere free electrons and ions are present?
a) Troposphere
b) Stratosphere
c) Ionosphere
d) Outer atmosphere
Explanation: In ionosphere ionization occurs so there are free electrons, positive and negative ions are also present. It is at 110- 400km range from earth surface.

24. In stratosphere, the excess refractive index is given as ________ in terms of refractive index μ.
a) N=(μ-1)×10-6
b) N=(μ-1)×106
c) N=(1+μ)×10-6
d) N=(μ+1)×106
Explanation: The excess refractive index over unity in millionths is given by N=(μ-1)×10-6
Where refractive index μ=0.0003% greater than unity for troposphere.

25. G region is also known as _______
a) Outer atmosphere
b) Appleton layer
c) Kennelly Heaviside layer
d) Absorption layer
Explanation: Outer atmosphere is also known as G region. It lies above 400km from the earth surface. E layer is called as Kennelly Heaviside layer and F layer is called as Appleton layer. D layer is also known as absorption layer for short wave signal at HF.

26. Which of the following is the top region during night hours in ionosphere?
a) D layer
b) E layer
c) F2 layer
d) F layer
Explanation: During night time D layer disappears and F1, F2 will combine to form a single F layer. So F layer is top in the Ionosphere during night hours and F2 is top at day time.

27. Ionospheric reflections in the shape of M –type occurs between Es and D layers
a) True
b) False
Explanation: Ionospheric reflections in the shape of M –type occurs between Es (layer formed due to the ions of clouds, thunderstorms etc) and higher F layers if the layers are smaller compared to the wavelengths. These M-type reflections take a single path in E layer.

28. Which of the following layer is used mostly for the long distance communication?
a) Troposphere
b) Stratosphere
c) Ionosphere
d) D layer during night hours
Explanation: Ionosphere is most widely used for long distance communication and it lies at 90 to 400km from earth surface. D layer is not present at night hours.

29. Stratosphere is present at _____ from earth surface.
a) 20- 70km
b) 2-15km
c) 150-400km
d) Above 400km
Explanation: Stratosphere is present at 20 to 70km from earth surface. Troposphere is up to 15km and above 400km is the outer atmosphere.

30. During night hours, Ionosphere consists of how many layers?
a) 2
b) 3
c) 4
d) 5
Explanation: During night time D layer is not present. F1 and F2 layers get combined to form an F layer during night time. Therefore only two layers namely E and F are present at night time.

31. What is the frequency used for tropospheric scatter propagation?
a) Above 30GHz
b) Above 30MHz
c) Above 300MHz
d) Above 300GHz
Explanation: Tropospheric scatter propagation is used for UHF and microwaves. So it is used at frequencies above 300MHz. Sky wave propagation is used at frequencies above 30MHz (VHF).

32. Which of the following scattering occurs through the fine layers in the troposphere?
a) Forward Scatter propagation
b) Ionosphere
c) Space wave
d) LOS
Explanation: Forward scatter propagation or Tropospheric propagation occurs at frequencies above 300MHz through the fine layers or blobs in the troposphere. UFH and microwaves propagate much beyond the LOS through forward scattering in tropospheric irregularities.

33. Height of the troposphere ranging from earth surface is up to _____
a) 15m
b) 15km
c) 50m
d) 50km
Explanation: The Troposphere portion of earth extends up to 15km from the earth surface. Ionosphere from 50 to 400km and Outer atmosphere extends above 400km.

34. At which region of the troposphere the temperature remains constant throughout the narrow belt?
a) Tropopause
b) Region of change
c) Region of calm
d) Stratopause
Explanation: At Tropopause starts after the top of troposphere and ends at stratosphere. Above critical height called Tropopause the temperature remains constant throughout belt and increases thereby. Entire belt of troposphere is called Region of change.

35. The lowest layer in the structure of atmosphere extends to a distance of _____km from earth surface
a) 15m
b) 15km
c) 50m
d) 50km
Explanation: The lowest layer of the atmosphere is troposphere. So it ranges up to 15km from the surface of the earth. The gas components remain almost constant in this region.

36. The actual height of troposphere is least at ____
a) poles
b) equator
c) both equator and poles
d) other composition other than poles and equator
Explanation: The actual height of troposphere is least at poles and maximum at equator. At other compositions it is almost remains same. The entire belt of troposphere is called Region of change.

37. Which of the following is a property of troposphere?
a) Temperature decreases with increase in height
b) Temperature increases with increase in height
c) Gas components don’t remain in constant percentage with increase in height
d) Water vapor components remain same with height
Explanation: The property of troposphere is that temperature decreases with increase in height. Gas components remains in constant percentage with increase in height but water vapor components decrease with increase in height.

38. Which of the following is the nearest region from the earth surface in the atmosphere?
a) Troposphere
b) Stratosphere
c) Ionosphere
d) Outer atmosphere
Explanation: Troposphere is the lowest layer and nearest to earth surface in the atmosphere ranging up to 15km. Stratosphere lies at 50 to 90km and up to 400km ionosphere. Above 400km is the outer atmosphere.

39. The region between the top of troposphere and start of stratosphere is called ____
a) tropopause
b) stratopause
c) ionosphere
d) region of calm
Explanation: The region between the top of troposphere and start of stratosphere is called Tropopause. Stratopause is also known as region of calm and ranges from 20 to 70km.

40. Troposphere scatter propagation is used for point to point communication
a) True
b) False
Explanation: Since the troposphere propagation takes at 2- 15km from earth surface there is a possibility for great attenuation. It is used for point to point communication.

41. Ground wave is always __________ polarized.
a) Vertically
b) Horizontally
c) Either vertical or horizontal
d) Neither horizontal nor vertical
Explanation: If the wave is horizontally polarized, then the electric field is short circuited by the earth. So the ground wave is always vertically polarized and vertical antennas are used.

42. Ground wave propagation is used for signals up to frequency __________
a) 2MHz
b) 2GHz
c) 30MHz
d) 30GHz
Explanation: Ground wave propagation is used for signals up to frequency of 2MHz. Ground waves are vertically polarized and transmitting and receiving antennas are placed closely and the wave follows the curvature of the earth.

43. The broadcast signals received at low frequencies during day-time are due to _________
a) Ground wave
b) Space wave
c) Sky wave
d) Tropospheric waves
Explanation: Ground wave propagation is used for signals up to frequency of 2MHz. It is useful for the broadcast and low frequency signals. Space waves are also known as tropospheric waves useful for FM reception. Sky wave propagation is used for long distance communication.

44. Ground wave propagation is also known as _________
a) Surface wave
b) Tropospheric wave
c) Ionospheric wave
d) Stratospheric waves
Explanation: Ground waves are also known as Surface waves as the wave propagates close to the surface of earth. Space waves are known as tropospheric waves and Sky waves as Ionospheric waves.

45. The ground wave propagation uses horizontal polarized antennas
a) True
b) False
Explanation: If the wave is horizontally polarized, then the electric field is short circuited by the earth. So the ground wave is always vertically polarized and vertical antennas are used.

46. Which of the following is particularly used for VLF?
a) Surface wave
b) Tropospheric wave
c) Ionospheric wave
d) Stratospheric waves
Explanation: Ground waves are also known as Surface waves are used for low frequencies and broadcasting. Tropospheric waves are used for Mf and HF signals. Ionospheric waves are used for long distance communications.

47. Ground wave field strength is given by E= _________
a) AEo/d
b) A(Eo)2/d
c) dEo
d) AEo/d2
Explanation: The ground wave field strength at a point E=AEodV/m
Where Eo is field strength of wave at unit distance from transmitting antenna
A is factor of ground losses
D is distance of point from Transmitting antenna.

48. Which of the following propagates by gliding over the surface of earth?
a) Surface wave
b) Tropospheric wave
c) Ionospheric wave
d) Stratospheric waves
Explanation: Ground waves are also known as Surface waves. Ground waves are vertically polarized and transmitting and receiving antennas are placed closely and the wave follows the curvature of the earth.

49. How the ground wave losses vary with high frequencies?
a) Increases
b) Decreases
c) Does not depend on frequency
d) Increase or decrease
Explanation: Ground wave propagation is used at below 2MHz frequency. Ground wave propagation is used for short distance. By using sufficient power and low frequencies it can be used effectively. Ground wave losses increase rapidly with frequencies.

50. The electric field of the component increases in ground wave when it tilts more at the surface.
a) True
b) False
Explanation: As the wave front tilts more and more towards the surface then the electric field gets short circuited. As it gets reduced with tilt, at some point it is completely attenuated.

51. What is the frequency at which tropospheric scatter occurs?
a) Above 30MHz
b) Below 30 MHz
c) < 3MHz
d) > 3 MHz and < 30MHz
Explanation: Tropospheric scatter is also known as forward scatter propagation. It occurs at a frequency above 30MHz. It belongs to the UHF and microwave range.

52. What is the range of frequency at which tropospheric scatter occurs?
a) UHF and Microwave range
b) HF
c) MF and VHF
d) MF and HF
Explanation: Tropospheric scatter occurs at a frequency above 30MHz. It belongs to the UHF and microwave range.
MF- 300 KHz – 3MHz
HF – 3 to 30 MHz
VHF -30 to 300MHz.

53. Tropospheric propagation is also known as forward scatter propagation.
a) True
b) False
Explanation: Tropospheric scatter is also known as forward scatter propagation, occurs at a frequency above 30MHz. It belongs to the UHF and microwave range. Its scattering occurs above radio horizon and travels beyond the LOS.

54. The tropospheric scattering occurs at _________
a) Beyond the LOS
b) In ground wave propagation
c) In sky wave propagation
Explanation: Tropospheric scattering occurs at a frequency above 30MHz. So, this occurs in space wave propagation. Signal scatters in forward direction above the radio horizon and it travels beyond the LOS. Ground wave propagation occurs at less the 2MHz and sky wave at 2 to 30 MHz frequency.

55. Tropospheric scattering, if falls under low frequencies then it leads to Ionosphere scattering.
a) True
b) False
Explanation: The tropospheric scattering occurs at frequency above 30MHz. Under low frequencies it falls below 30MHz and sometimes thus leads to the Ionospheric propagation.

56. The turbulences in the atmosphere lead to ___________
a) tropospheric scattering
b) ground wave propagation
c) sky wave propagation
d) constant velocity of signal
Explanation: The main cause of the tropospheric scattering is the turbulences in the atmosphere. When it meets the turbulences, then there will be an abrupt change in velocity leading to the tropospheric scattering.The tropospheric scattering occurs at frequency above 30MHz.

57. When the signal take-off angle increases, the height of scatter volume ________
a) increases
b) decreases
c) it remains constant
d) may increase or decrease
Explanation: The signal take-off angle determines the height of the scatter volume. As the signal take-off angle increases, the height of scatter volume also increases. A low take-off produces low scatter volume.

58. Which of the following statements regarding tropospheric scattering is false?
a) It occurs in the region near to the mid-point of the transmitter and receiver
b) It occurs above the radio horizon
c) It travels beyond the line of sight distance
d) It occurs below the radio horizon
Explanation: Signal scatters in forward direction above the radio horizon and it travels beyond the LOS. It occurs in the region near to the mid-point of the transmitter and receiver.

59. In tropospheric scattering, rapid fading occurs due to _______ and long-term fading occurs due to ______
a) multipath propagation, turbulences in atmosphere
b) low signal strength, signal travelling below horizon
c) signal travelling below horizon, low signal strength
d) daily and seasonal variations, multi-path propagation
Explanation: Rapid fading occurs due to the multipath propagation. As the turbulent conditions changes constantly, the path length and signal strength levels also change. The turbulence in the atmosphere gives rise to daily and seasonal variations in signal strength so results in long-term fading.

60. Which of the following is used for communication in rugged terrain where normal propagation methods fail?
a) Tropospheric Scattering
b) Ground wave
c) LOS