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

2. Modern phase techniques are capable of __________
a) Resolving modulation
b) Resolving amplitude
c) Resolving frequency
d) Resolving wave length
Answer: d
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.

3. Lower frequency is not suitable in_________
a) Direct transmission
b) Distance calculation
c) Determination of wavelength
d) Determination of frequency
Answer: a
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.

4. Which of the following represents the correct set of modulation classification?
a) Frequency, time period
b) Frequency, amplitude
c) Amplitude, wavelength
d) Wavelength, frequency
Answer: b
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.

5. 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
Answer: d
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.

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

7. Which can’t be done in high frequency zones?
a) Phase comparison
b) Super imposition of waves
c) Distance measurement
d) Wavelength measurement
Answer: a
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.

8. Modulating wave can also be known as ______
a) Total wave
b) Measuring wave
c) Super wave
d) Incubation wave
Answer: b
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.

9. 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
Answer: c
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.

10. Frequency modulation is equipped in all EDM instruments.
a) True
b) False
Answer: a
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.

11. In Amplitude Modulation, the instantaneous values of the carrier amplitude changes in accordance with the amplitude and frequency variations of the modulating signal.
a) True
b) False
Answer: a
Explanation: In Amplitude Modulation, the amplitude of the carrier sine wave is varied by the value of the information signal. The instantaneous value of the carrier amplitude changes in accordance with the amplitude and frequency variations of the modulating signal. The carrier frequency remains constant during the modulation process, But its amplitude varies in accordance with the modulating signal.

12. What is the line connecting the positive and negative peaks of the carrier waveform called?
a) Peak line
b) Maximum amplitude ceiling
c) Modulation index
d) Envelope
Answer: d
Explanation: An imaginary line connecting the positive peaks and negative peaks of the carrier waveform gives the exact shape of the modulating information signal. This line is known as the envelope.

13. What is the reference line for the modulating signal?
a) Zero line
b) Carrier peak line
c) Modulated peak line
d) Un-modulated peak line
Answer: b
Explanation: The modulating signal uses the peak value of the carrier rather than zero as its reference point. The envelope varies above and below the peak carrier amplitude. The zero reference line of the modulating signal coincides with the peak value of the unmodulated carrier.

14. What happens when the amplitude of the modulating signal is greater than the amplitude of the carrier?
a) Decay
b) Distortion
c) Amplification
d) Attenuation
Answer: b
Explanation: The zero reference line of the modulating signal coincides with the peak value of the unmodulated carrier. Because of this, the relative amplitudes of the carrier and modulating signal are important. When the amplitude of the modulating signal is greater than the amplitude of the carrier, distortion will occur.

15. What is the effect of distortion?
a) Total information loss
b) Error information
c) Attenuated information
d) Amplified information
Answer: b
Explanation: Distortion occurs when the modulating signal amplitude is greater than the amplitude of the carrier, causing incorrect information to be transmitted. In amplitude modulation, it is particularly important that the peak value of the modulating signal be less than the peak value of the carrier.

16. What is the circuit used for producing AM called?
a) Modulator
b) Transmitter
c) Receiver
d) Duplexer
Answer: a
Explanation: The circuit used for producing AM is called a modulator. It has two inputs, the carrier and the modulating signal, and the resulting output is the modulated signal. Amplitude modulators compute the product of the carrier and modulating signals.

17. The ratio between the modulating signal voltage and the carrier voltage is called?
a) Amplitude modulation
b) Modulation frequency
c) Modulation index
d) Ratio of modulation
Answer: c
Explanation: For undistorted modulation to occur, the voltage of modulating signal Vm must be less than the carrier voltage Vc. Therefore, the relationship between the amplitude of the modulating signal and the amplitude of the carrier signal is important. This relationship, known as the modulation index m, is the ratio m = ^V_m⁄_Vc.

18. When does over-modulation occur?
a) Modulating signal voltage < Carrier voltage
b) Modulating signal voltage > Carrier voltage
c) Modulating signal voltage = Carrier voltage
d) Modulating signal voltage =0
Answer: b
Explanation: Over-modulation is a condition in which the modulating signal voltage is much greater than the carrier voltage. The received signal will produce an output waveform in the shape of the envelope, whose negative peaks have been clipped off.

19. What is the condition for greatest output power at the transmitter without distortion?
a) Modulating signal voltage > Carrier voltage
b) Modulating signal voltage < Carrier voltage
c) Modulating signal voltage = Carrier voltage
d) Modulating signal voltage = 0
Answer: c
Explanation: When the modulation index is 1 or the percentage of modulation is 100, modulating signal voltage is equal to the carrier voltage. This results in the greatest output power at the transmitter and the greatest output voltage at the receiver, with no distortion.

20. Which of the following modulating signal voltage would cause over-modulation on a carrier voltage of 10v?
a) 9.5
b) 9.99
c) 10
d) 12
Answer: d
Explanation: When the voltage of the modulating signal exceeds the voltage of the carrier signal over-modulating occurs. Here, 12/10 = 1.2 which is greater than 1 and hence would cause over-modulation.

21. In Amplitude Modulation, the instantaneous values of the carrier amplitude changes in accordance with the amplitude and frequency variations of the modulating signal.
a) True
b) False
Answer: a
Explanation: In Amplitude Modulation, the amplitude of the carrier sine wave is varied by the value of the information signal. The instantaneous value of the carrier amplitude changes in accordance with the amplitude and frequency variations of the modulating signal. The carrier frequency remains constant during the modulation process, But its amplitude varies in accordance with the modulating signal.

22. What is the line connecting the positive and negative peaks of the carrier waveform called?
a) Peak line
b) Maximum amplitude ceiling
c) Modulation index
d) Envelope
Answer: d
Explanation: An imaginary line connecting the positive peaks and negative peaks of the carrier waveform gives the exact shape of the modulating information signal. This line is known as the envelope.

23. What is the reference line for the modulating signal?
a) Zero line
b) Carrier peak line
c) Modulated peak line
d) Un-modulated peak line
Answer: b
Explanation: The modulating signal uses the peak value of the carrier rather than zero as its reference point. The envelope varies above and below the peak carrier amplitude. The zero reference line of the modulating signal coincides with the peak value of the unmodulated carrier.

24. What happens when the amplitude of the modulating signal is greater than the amplitude of the carrier?
a) Decay
b) Distortion
c) Amplification
d) Attenuation
Answer: b
Explanation: The zero reference line of the modulating signal coincides with the peak value of the unmodulated carrier. Because of this, the relative amplitudes of the carrier and modulating signal are important. When the amplitude of the modulating signal is greater than the amplitude of the carrier, distortion will occur.

25. What is the effect of distortion?
a) Total information loss
b) Error information
c) Attenuated information
d) Amplified information
Answer: b
Explanation: Distortion occurs when the modulating signal amplitude is greater than the amplitude of the carrier, causing incorrect information to be transmitted. In amplitude modulation, it is particularly important that the peak value of the modulating signal be less than the peak value of the carrier.

26. What is the circuit used for producing AM called?
a) Modulator
b) Transmitter
c) Receiver
d) Duplexer
Answer: a
Explanation: The circuit used for producing AM is called a modulator. It has two inputs, the carrier and the modulating signal, and the resulting output is the modulated signal. Amplitude modulators compute the product of the carrier and modulating signals.

27. The ratio between the modulating signal voltage and the carrier voltage is called?
a) Amplitude modulation
b) Modulation frequency
c) Modulation index
d) Ratio of modulation

28. What is the percentage of modulation if the modulating signal is of 7.5V and carrier is of 9V?
a) 100
b) 91
c) 83.33
d) 0
Answer: c
Explanation: modulation index m = ^V_m⁄_Vc = ^7.5⁄₉* 100 = 83.33.

29. When does over-modulation occur?
a) Modulating signal voltage < Carrier voltage
b) Modulating signal voltage > Carrier voltage
c) Modulating signal voltage = Carrier voltage
d) Modulating signal voltage =0
Answer: b
Explanation: Over-modulation is a condition in which the modulating signal voltage is much greater than the carrier voltage. The received signal will produce an output waveform in the shape of the envelope, whose negative peaks have been clipped off.

30. What is the condition for greatest output power at the transmitter without distortion?
a) Modulating signal voltage > Carrier voltage
b) Modulating signal voltage < Carrier voltage
c) Modulating signal voltage = Carrier voltage
d) Modulating signal voltage = 0
Answer: c
Explanation: When the modulation index is 1 or the percentage of modulation is 100, modulating signal voltage is equal to the carrier voltage. This results in the greatest output power at the transmitter and the greatest output voltage at the receiver, with no distortion.

31. What is the purpose of peak clipper circuits in radios?
a) prevent overmodulation
b) reduce bandwidth
c) increase bandwidth
d) regulate oscillator input voltage
Answer: a
Explanation: Clipper is used to prevent the output of a circuit from exceeding a predetermined voltage. It does not distorted the remaining part of the applied waveform.

32. In FM receiver, role of amplitude limiter is to reduce the amplitude of signals.
a) True
b) False
Answer: b
Explanation: Role of amplitude limiter in frequency modulation is to eliminate any change in amplitude of received FM signals.

33. Balanced modulator is used to suppress carrier signal to create an SSB or DSB.
a) True
b) False
Answer: b
Explanation: Balanced modulator is used to produce 100% modulation.

34. What is the main function of a balanced modulator?
a) to limit the noise picked by a receiver
b) to produce balanced modulation of a carrier wave
c) to suppress carrier signal
d) to produce 100% modulation
Answer: d
Explanation: For achieving 100% modulation, balanced modulator is mainly used in circuits.

35. For classifying a modem as low speed its data rate is ________
a) upto 100bps
b) upto 200bps
c) upto 400bps
d) upto 600bps
Answer: d
Explanation: According to standard data for modem, if the data rate is upto 600bps then modem is classified as having a low speed.

36. Varacter diode modulator is an indirect way of generating FM.
a) True
b) False
Answer: b
Explanation: Varacter diode modulator is not an indirect way of generating FM. It is a Armstrong modulator which is an indirect way of generating FM.

37. Which of the following is an indirect way of generating FM?
a) By reactance modulator
b) By bipolar transistor
c) By varacter diode
d) By armstrong modulator
Answer: d
Explanation: Armstrong modulator which is the only way for indirect generating of frequency modulation.

28. Convex lens has negative focal power and concave lens have positive focal power.
a) True
b) False
Answer: b
Explanation: Convex lens has positive focal power and concave lens have negative focal power.

39. All lens have positive focal power.
a) True
b) False
Answer: b
Explanation: Convex lens has negative focal power and concave lens have positive focal power.

40. Which of the following modulating signal voltage would cause over-modulation on a carrier voltage of 10v?
a) 9.5
b) 9.99
c) 10
d) 12
Answer: d
Explanation: When the voltage of the modulating signal exceeds the voltage of the carrier signal over-modulating occurs. Here, 12/10 = 1.2 which is greater than 1 and hence would cause over-modulation.

41. What is the reason of “envelope” in an amplitude modulated signal?
a) noise signal
b) carrier signal
c) nematic signal
d) baseband signal
Answer: d
Explanation: Envelope is basically a smooth curve that outlines the extremes of any baseband signal. So basically it is message or baseband signal that determines the envelope.

42. AM stands for ________
a) Amplitude Modulation
b) Audio Modulation
c) Antenna Modulation
d) Amplified Modulation
Answer: a
Explanation: AM stands for amplitude modulation. Amplitude modulation is the change in the amplitude of carrier wave in proportion to the instantaneous amplitude of the message signal.

43. What is the equation for full-carrier AM?
a) V(t) = (E_c + E_m) × (sin⁡ ω_c t)
b) V(t) = (E_c + E_m) × (sin⁡ ω_m t) + (sin⁡ ω_c t)
c) V(t) = (E_c × E_m) × (sin ⁡ω_m t) × (sin⁡ ω_c t)
d) V(t) = (E_c + E_m sin⁡ ω_m t) × (sin ⁡ω_c t)
Answer: d
Explanation: Amplitude modulation is the change in the amplitude of carrier wave in proportion to the instantaneous amplitude of the message signal. A carrier can be seen as a waveform with frequency higher than the message signal frequency, that is modulated with respect to input signal for the purpose of transmitting information. The equation for full-carrier AM is V(t) = (E_c + E_m sin⁡ω_m t) × (sin ⁡ω_c t).

44. What is the cause of Overmodulation?
a) distortion
b) splatter
c) both distortion and splatter
d) half reception of signals
Answer: c
Explanation: Overmodulation is the process in which the modulation index is greater than 1 such that the modulating signal voltage exceeds the required voltage to produce 100% modulation. It results out of both distortion and splatter of waveform and causes distortion and overlapping

45. If AM radio station increases its modulation index then the audio gets louder at the receiver.
a) True
b) False
Answer: a
Explanation: Modulation index tells us the amount by which the carrier wave is varied with respect to the message signal. If we increase the modulation index then audio signal gets louder.

46. The modulation index can be derived from ________
a) frequency-domain signal
b) time-domain signal
c) both frequency and time domain signal
d) a highly modulated carrier wave
Answer: c
Explanation: Modulation index tells us the amount by which the carrier wave is varied with respect to the message signal. It can be derived for frequency-domain signals as well as for time-domain signals.

47. A single sideband modulation system is more efficient than a plain amplitude modulated system.
a) True
b) False
Answer: a
Explanation: In Single Sideband transmission, the carrier is suppressed and only either of the two sidebands is transmitted. Thus, it reduces the total power consumption and also reduces the bandwidth required. Whereas, in AM, the carrier being transmitted along with both the sidebands entails more power and larger bandwidth.

48. At peak modulation an SSB transmitter radiate 1000W, what will it radiate with no modulation?
a) 1000 watts
b) 500 watts
c) 250 watts
d) 0 watts
Answer: d
Explanation: Power of a modulated wave is directly proportional to modulation index. Thus, if there is no modulation in any SSB transmitter than, it will not radiate. So it will radiate 0 watts when there is no modulation.

49. Why AM stations has “low-fidelity”?
a) AM is susceptible to noise
b) Commercial AM stations use low power
c) Commercial AM stations have a narrow bandwidth
d) High quantization to noise ratio
Answer: c
Explanation: Fidelity is the ability of receivers to reproduce all modulating signals eually. Low fidelity can be seen as sound recording that contain technical flaws to make sound better compared with the sound that is recorded live. High fidelity refers to the equipment that very accurately produces without any harmonic or resonance. AM stations have low fidelity to have narrow bandwidth.

50Calculate the power in one of the side band in SSBSC modulation when the carrier power is 124W and there is 80% modulation depth in the amplitude modulated signal.
a.) 89. 33 W
b.) 64. 85 W
c.) 79.36 W
d.) 102 W
Answer: c) 79.36 W
Explanation:
Modulation Index= 0.8
Pc = 124W
Power in sidebands may be calculated as = m2 Pc/4
= (0.8)2 * 124/4
= 79.36 W

51Calculate the total modulation Index when a carrier wave is being modulated by two modulating signals with modulation indices 0.8 and 0.3.
a.) 0.8544
b.) 0.6788
c.) 0.9999
d.) 0.5545
Answer: a) 0.8544
Explanation:
Here, m1 = 0.8
m2 = 0.3

total modulation index mt = √( m12 + m22 )
= √( 0.82 + 0.32 )
= √ 0.73
= 0.8544 = 85.44%

52..Calculate the frequencies available in the frequency spectrum when a 2MHz carrier is modulated by two sinusoidal signals of 350Hz and 600Hz.
a.) 2000.35, 1999.65 and 2000.6, 1999.4
b.) 1999.35, 1999.65 and 2000.6, 2000.4
c.) 2000.35, 2000.65 and 2000.6, 2000.4
d.) 1999.35, 1999.65 and 1999.6, 1999.4
Answer: a) 2000.35, 1999.65 and 2000.6, 1999.4
Explanation:
The frequencies obtained in the spectrum after the amplitude modulation are
fc+fm and fc+fm
therefore,
the available frequencies after modulation by 0.350 KHz are
2000KHz+ .350 KHz = 2000.35 and 2000KHz – 0.350 KHz = 1999.65

the available frequencies after modulation by 0.6 KHz are
2000KHz+0 .6 KHz = 2000.6 and 2000KHz – 0.6 KHz = 1999.4
If an AM signal is represented by
v = ( 15 + 3 Sin( 2Π * 5 * 103 t) ) * Sin( 2Π * 0.5 * 106 t) volts

i.) Calculate the values of the frequencies of carrier and modulating signals.
ii.) Calculate the value of modulation index.
iii.) Calculate the value of bandwidth of this signal.

a.) 1.6MHz and 8KHz, 0.6, 16 MHz
b.) 1.9MHz and 18KHz, 0.2, 16 KHz
c.) 2.4 MHz and 18KHz, 0.2, 16 KHz
d.) 1.6MHz and 8KHz, 0.2, 16 KHz
Answer: d) 1.6MHz and 8KHz, 0.2, 16 KHz

Explanation:

The amplitude modulated wave equation is
v = ( 10 + 2 Sin( 2Π * 8 * 103 t) ) * Sin( 2Π * 1.6 * 106 t) volts

Instantaneous value of AM signal is represented by the equation
v = {Vc + Vm Sin ( ωm t )} * Sin ( ωc t )
comparing it with the given equation,

Vc = 10 V
Vm = 2V
ωc (= 2Π fc) = 2Π * 1.6 * 106
ωm (= 2Π fm) = 2Π * 8 * 103

(i) The carrier frequency fc is = 1.6 * 106 = 1.6 MHz
The modulating frequency fm is = 8* 103 = 8 kHz

(ii) The modulation index m = Vm/Vc = 2/10 = 0.2

(iii) The bandwidth BW = 2 fm = 16 kHz

53.An AM signal has a total power of 48 Watts with 45% modulation. Calculate the power in the carrier and the sidebands.

a.) 39.59 W, 4.505W
b.) 40.59 W, 4.205W
c.) 43.59 W, 2.205W
d.) 31.59 W, 8.205W
Answer: c) 43.59 W, 2.205W
Explanation:

Given that Pt = 48 W

Modulation index m= 0.45

The total power in an AM is given by

Pt= Pc ( 1 + m2/2)
= Pc ( 1 +0.452/2)
48 = Pc * 1.10125

Therefore, Pc = 48/ 1.10125
= 43.59 W

The total power in two sidebands is 48 – 43.59 = 4.41 W

So the power in each sideband is 4.41/2 = 2.205 W

54.Calculate the power saved in an Amplitude Modulated wave when it is transmitted with 45% modulation
– Without carrier
– Without carrier and a sideband

a.) 90%, 95%
b.) 82%, 91%
c.) 82%, 18%
d.) 68%, 16%
Answer: a) 90%, 95%
Explanation:
The total power in an AM is given by
Pt= Pc ( 1 + m2/2)

Given: m= 0.45
Therefore Pt= Pc ( 1 +0.452/2 )
Pt= Pc *1.10125
Pc/ Pt = 1/1.10125
= 0.908
= 90%
This shows that the carrier occupies 90% of total power. So 90% of total power may be saved if carrier is suppressed in the AM signal.
(ii) If one of the sidebands is also suppressed, half of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed.

55.Calculate the power saved in an Amplitude Modulated wave when it is transmitted with 45% modulation
– Without carrier
– Without carrier and a sideband
a.) 90%, 95%
b.) 82%, 91%
c.) 82%, 18%
d.) 68%, 16%
Answer: a) 90%, 95%

Explanation:
The total power in an AM is given by
Pt= Pc ( 1 + m2/2)
Given: m= 0.45
Therefore Pt= Pc ( 1 +0.452/2 )
Pt= Pc *1.10125
Pc/ Pt = 1/1.10125
= 0.908
= 90%

This shows that the carrier occupies 90% of total power. So 90% of total power may be saved if carrier is suppressed in the AM signal.

(ii) If one of the sidebands is also suppressed, half of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed.

56.What is the carrier frequency in an AM wave when its highest frequency component is 850Hz and the bandwidth of the signal is 50Hz?

a.) 80 Hz
b.) 695 Hz
c.) 625 Hz
d.) 825 Hz
Answer: d) 825 Hz
Explanation:
Upper frequency = 850Hz
Bandwidth = 50Hz
Therefore lower Frequency = 850-50= 800 Hz
Carrier Frequency = (850-800)/2
= 825 Hz

57.Noise figure of merit in SSB modulated signal is
a.) 1
b.) Less than 1
c.) Greater than 1
d.) None of the above
Answer: a) 1
Explanation:

A figure of merit used to describe the performance of a system. The figure of merit ‘γ’ is the ratio of output signal to noise ratio to input signal to noise

ratio of a receiver system. Figure of merit for SSB modulation is always 1.

58.For low level modulation , amplifier used is
a.) Class A
b.) Class C
c.) Class A and C
d.) None of the above
Answer: a) Class A
Explanation:
When the modulation takes place prior to the output element of the final stage of the amplifier, it is low level modulation. Class A amplifiers are used for this purpose.

59.For high level modulation , amplifier used is
a.) Class A
b.) Class C
c.) Class A and C
d.) None of the above
Answer: b) Class C

Explanation:
When the modulation takes place in the final element of the final stage of the amplifier,
it is high level modulation. Class C amplifiers are used for this purpose.

60.The antenna current of the transmitter is 10A. Find the percentage of modulation when the antenna current increases to 10.4A.
a.) 32%
b.) 28.5%
c.) 64%
d.) 40%
Answer: b) 28.5%
Explanation:
It= Ic √(1+ m2/2)
10.4= 10 √(1+ m2/2)
√ (1+ m2/2) = 1.04
Therefore m= 0.285
= 28.5%

61.Demodulation is:
a.) detection
b.) recovering information from modulated signal
c.) a) and b)
d.) none of the above
Answer: c.) a) and b)
Explanation:
Demodulation is the process of recovering the original information from a modulated carrier wave. Systems are designed to be used as demodulators that detect the information signal from the carrier. The envelope detector and product detector are few of the AM detectors.

61.Calculate the side band power in an SSBSC signal when there is 50% modulation and the carrier power is 50W.
a.) 50W
b.) 25 W
c.) 6.25 W
d.) 12.5 W
Correct Answer : 6.25 W
Explanation:
The side band power is given by
Pc m2/2
= 50* (0.5) 2/2
=6.25W

62.Calculate the modulation index when the un modulated carrier power is 15KW, and after modulation, carrier power is 17KW.
a.) 68%
b.) 51.63%
c.) 82.58%
d.) 34.66%
answer: b.) 51.63%
Explanation:
The total power in an AM is given by
Pt= Pc ( 1 + m2/2)
17= 15(1 + m2/2)
m2/2= 0.134
m=0.5163
= 51.63%

63.Calculation of modulation index using antenna current
An AM transmitter has an antenna current changing from 5 A un modulated to 5.8 A. What is the percentage of modulation?
a.) 38.8%
b.) 83.14 %
c.) 46.8%
d.) 25.2%
Correct Answer: 83.14 %
Explanation:
Modulation index m is given by
m= √ (2{It/Ic}2-1)
= √ (2 (5.8/5)2 -1)
= √ (2 (5.8/5)2 -1)
= 0.8314
= 83.14%

64.The antenna current of the transmitter is 10A. Find the percentage of modulation when the antenna current increases to 10.4A.
a.) 32%
b.) 28.5%
c.) 64%
d.) 40%
Answer: b) 28.5%

Explanation:
It= Ic √(1+ m2/2)
10.4= 10 √(1+ m2/2)
√ (1+ m2/2) = 1.04
Therefore m= 0.285
= 28.5%

65.What is the change in the value of transmitted power when the modulation index changes from 0 to 1?
a.) 100%
b.) Remains unchanged
c.) 50%
d.) 80%
Answer: c) 50%
Explanation:
Pt= Pc ( 1 + m2/2)
Pt= Pc ( 1 + 02/2) = Pc……………..(1)
New total power Pt1= Pc ( 1 + 12/2)
= Pc *3/2……………..(2)
(2) / (1),
We get , Pt1/ Pt= 3/2= 1.5
Pt1= 1.5 Pt
i.e. there is increase in total power by 50%

66.An AM signal has a total power of 48 Watts with 45% modulation. Calculate the power in the carrier and the sidebands.
a.) 39.59 W, 4.505W
b.) 40.59 W, 4.205W
c.) 43.59 W, 2.205W
d.) 31.59 W, 8.205W
Answer: c) 43.59 W, 2.205W
Explanation:
Given that Pt = 48 W
Modulation index m= 0.45
The total power in an AM is given by
Pt= Pc ( 1 + m2/2)
= Pc ( 1 +0.452/2)
48 = Pc * 1.10125
Therefore, Pc = 48/ 1.10125
= 43.59 W
The total power in two sidebands is 48 – 43.59 = 4.41 W
So the power in each sideband is 4.41/2 = 2.205 WNumerical – Power saving in AM when carrier and side band are suppressed

67.Calculate the power saved in an Amplitude Modulated wave when it is transmitted with 45% modulation
– Without carrier
– Without carrier and a sideband
a.) 90%, 95%
b.) 82%, 91%
c.) 82%, 18%
d.) 68%, 16%
Answer: a) 90%, 95%
Explanation:
The total power in an AM is given by
Pt= Pc ( 1 + m2/2)
Given: m= 0.45
Therefore Pt= Pc ( 1 +0.452/2 )
Pt= Pc *1.10125
Pc/ Pt = 1/1.10125
= 0.908
= 90%

This shows that the carrier occupies 90% of total power. So 90% of total power may be saved if carrier is suppressed in the AM signal.
(ii) If one of the sidebands is also suppressed, half of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed.

69.The antenna current of the transmitter is 10A. Find the percentage of modulation when the antenna current increases to 10.4A.
a.) 32%
b.) 28.5%
c.) 64%
d.) 40%
Answer: b) 28.5%
Explanation:
It= Ic √(1+ m2/2)
10.4= 10 √(1+ m2/2)
√ (1+ m2/2) = 1.04
Therefore m= 0.285
= 28.5%

69.What is the change in the value of transmitted power when the modulation index changes from 0 to 1?
a.) 100%
b.) Remains unchanged
c.) 50%
d.) 80%
Answer: c) 50%
Explanation:
Pt= Pc ( 1 + m2/2)
Pt= Pc ( 1 + 02/2) = Pc……………..(1)
New total power Pt1= Pc ( 1 + 12/2)
= Pc *3/2……………..(2)
(2) / (1),
We get , Pt1/ Pt= 3/2= 1.5
Pt1= 1.5 Pt
i.e. there is increase in total power by 50%

70.An AM broadcast station transmits modulating frequencies up to 6 kHz. If the AM station is transmitting on a frequency of 894 kHz, the values for maximum and minimum upper and lower sidebands and the total bandwidth occupied by the AM station are:
a.) 900 KHz, 888KHz, 12 KHz
b.) 894 KHz, 884 KHz, 12 KHz
c.) 894 KHz, 888 KHz, 6 KHz
d.) 900 KHz, 888 KHz, 6 KHz
Correct Answer: a) 900 KHz, 888KHz, 12 KHz
Explanation:
Maximum Frequency fUSB = 894 + 6 = 900 kHz
Minimum Frequency fLSB = 894 – 6 = 888 kHz
Bandwidth BW = fUSB fLSB = 900 888 = 12 kHz OR = 2(6 kHz) = 12 kHz

1. 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
Answer: d
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.

2. Automatic Level Control (ALC) is used to keep the modulation index close to 100%.
a) True
b) False
Answer: a
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.

3. A signal that ________ must have linear power amplifier.
a) is complex
b) has variable frequency
c) is linear
d) has variable amplitude
Answer: d
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.

4. 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
Answer: a
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.

5. What we called a resistor if a transmitter is connected to a resistor instead of an antenna?
a) a test load
b) a temporary load
c) a dummy load
d) a heavy load
Answer: c
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.

6. A class D amplifiers is very efficient than other amplifiers.
a) True
b) False
Answer: a
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.

7. The carrier is suppressed in ________
a) a mixer
b) a frequency multiplier
c) a transducer
d) a balance modulator
Answer: d
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.

8. What is the full form of AFC?
a) Amplitude to frequency conversion
b) Automatic frequency conversion
c) Automatic frequency control
d) Audio frequency control
Answer: c
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.

9. 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
Answer: d
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.

10. 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
Answer: a
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.

21. When two or more signals share a common channel, it is called ________
a) sub-channeling
b) channeling
c) switching
d) multiplexing
Answer: d
Explanation: Multiplexing is a method by which multiple analog or digital signals are combined to form a single complex signal. In multiplexing, many signals are combined into one signal and passed through a single channel, thus sharing the same medium.

22. One of the reason of distortion is shift in phase relationships between baseband frequencies.
a) True
b) False
Answer: a
Explanation: Phase distortion is the change in shape of the waveform. Thus, if we shift the phase relationships between baseband frequencies, it will ultimately result in distortion.

23. Miller effect can cause an amplifier to oscillate.
a) True
b) False
Answer: a
Explanation: Miller effect is responsible for limiting the gain of an amplifier at higher frequencies due to Miller capacitance between the output and input. Also, oscillations in amplifier occur due to input/output feedback.

24. How can we successfully avoid Miller effect?
a) using a common-base amplifier
b) using a common-emitter amplifier
c) by increasing the Q factor
d) by decreasing the Q factor
Answer: a
Explanation: Miller effect is responsible for the increase in equivalent input capacitance of an inverting voltage amplifier. The increase equivalent input capacitance is given by, C_M = C(1 + A_V). Miller effect can be avoided by using a common-base amplifier

1.The noise generated by a resistor depends upon
A. its resistance value
B. its operating temperature
C. both its resistance value and operating temperature
D. none of the above
Answer: Option B
Explanation:No answer description available for this question.

2.In a superheterodyne receiver
A. the IF stage has better selectivity than RF stage
B. the RF stage has better selectivity than IF stage
C. the RF stage has same selectivity than IF stage
D. none of the above
Answer: Option A
Explanation:
No answer description available for this question.

3.The function of an AM detector circuit is to
A. rectify the input signal
B. discard the carrier
C. provide audio signal
D. all of the above
Answer: Option D
Explanation:No answer description available for this question

4.Which of the following should be used in order to prevent overloading or the last IF amplifier in a receiver?
A. Variable selectivity
B. Variable sensitivity
C. Double conversion
D. Squelch
Answer: Option B
Explanation:No answer description available for this question.

5.Most popular IF for receivers tuning to 540 to 1650 kHz is
A. 433 kHz
B. 455 kHz
C. 545 kHz
D. 555 kHz
Answer: Option B
Explanation:No answer description available for this question

6.In a broadcast superheterodyne receiver
A. the local oscillator operates below the signal frequency
B. local oscillator frequency is normally double the IF
C. RF amplifier normally works at kHz above the carrier frequency
D. mixer input must be tuned to the signal frequency
Answer: Option D
Explanation:No answer description available for this question

7.A duplexer is a device used to
A. feed more than one receiver from a single antenna
B. connect two transmitters to the same antenna
C. connect a receiver and a transmitter to the same antenna
D. none of these
Answer: Option C
Explanation:No answer description available for this question.

8.A heterodyne frequency changer is called a
A. Modulator
B. Mixer
C. Demodulator
D. Frequency translator
Answer: Option B
Explanation:No answer description available for this question.

9.RF amplifiers are used in radio receivers for
A. improved image frequency rejection
B. improved rejection of adjacent unwanted signals
C. prevention of re-radiation of the local oscillator through the antenna of the receiver
D. all of the above
Answer: Option D
Explanation:No answer description available for this question.

10.Which are the popular IF frequencies for microwave and radar receivers operating on frequencies in the range 1 to 10 GHz
A. 30, 60 and 70 kHz
B. 3, 6 and 7 kHz
C. 30, 60 and 70 MHz
D. 3, 6 and 7 GHz
Answer: Option C
Explanation:No answer description available for this question

11.In a ratio detector
A. the linearity is worse than in a phase discriminator
B. the output is twice that obtainable from a similar phase discriminator
C. stabilization against signal strength variations is provided
D. the circuit is the same as in a discriminator, except that the diodes are reversed
Answer: Option A
Explanation:No answer description available for this question.

12.Neutralization is used in RF amplifiers to
A. stop oscillation
B. increase bandwidth
C. improve selectivity
D. all of the above
Answer: Option A
Explanation:No answer description available for this question..

13.It is known that noise phase modulates the FM wave. As the noise side band frequency approaches the carrier frequency, the noise amplitude
A. will increase
B. will decrease
C. will remain constant
D. will reduce to negligible value
Answer: Option B
Explanation:No answer description available for this question..

14.In a receiver, which of the following device has IF input but RF output?
A. Demodulator
B. Loudspeaker
C. Audio amplifier
D. Frequency changer
Answer: Option A
Explanation:No answer description available for this question.

15.Transistor are free from which type of noise?
A. Resistance noise
B. Partition noise
C. Flicker noise
D. Shot noise
Answer: Option B
Explanation:No answer description available for this question.

16.Standard AM radio broadcasts are confined to
A. MF
B. HF
C. VHF
D. UHF
Answer: Option A
Explanation:No answer description available for this question.

17.In a radio receiver with simple AGC
A. the highest AGC voltage is produced between stations
B. the faster the AGC time constant, the more accurate the output
C. an increase in signal strength produces more AGC
D. the audio stage gain is normally controlled by AGC
Answer: Option C
Explanation:No answer description available for this question.

18.Which of the following cannot be used to demodulate SSB?
A. Complete phase-shift generator
B. Product detector
C. Diode balanced modulator
D. Bipolar transistor balanced modulator
Answer: Option D
Explanation:No answer description available for this question.

19.A transmitter serial current contains
A. carrier frequencies
B. audio frequencies
C. radio frequencies
D. none of the above
Answer: Option C
Explanation:No answer description available for this question.

20.In a broadcast superheterodynes receiver having no RF amplifier, the loaded Q of the antenna coupling circuit is 100. If the intermediate frequency is 455 kHz. The image frequency at 25 MHz will be
A. 24.09 MHz
B. 24.54 MHz
C. 25.45 MHz
D. 25.91 MHz
Answer: Option D
Explanation:No answer description available for this question.

21. The first radio receivers invented by
A.Marconi
B.Oliver Lodge
C.Alexander Popov
D.All of the above
Answer: D All of the above

22. We should use …………. to prevent overloading of the IF amplifier in a receiver.
A. Squelch
B. Variable selectivity
C. Variable sensitivity
D. Double conversion
Answer: C Variable sensitivity

23. Which of the following circuits can not demodulate SSB?
A. Product modulator
B. Balance modulator
C. Phase discriminator
D. None of the above
Answer: C Phase discriminator

24. ………… is not a useful quantity for comparing the noise performance of receivers.
A. Noise figure
B. Noise temperature
C. Input noise voltage
D. Equivalent noise resistance
Answer: C Input noise voltage

25. Why A notch filter is sometimes used in communication receivers?
A. Spread the bandwidth
B. Made selectivity more precise
C. Reduce receiver gain at some specific frequency
D. Increase receiver gain at some specific frequency
Answer: C Reduce receiver gain at some specific frequency

26. For which purpose EM 84 tube is used in radio receivers?
Magic eye
RF amplifier
Audio amplifier
Full wave rectifier
Answer: A Magic eye

27. What is the selectivity of a radio receiver?
Its ability to suppress noise
Its ability to amplify weak signals
Its ability to reject adjacent unwanted signals
None of the above
Answer: B Its ability to reject adjacent unwanted signals

28. …………. does not happen in transistors?
Shot noise
Flicker noise
Partition noise
Resistance noise
Answer: D Partition noise

29. What happens, if the intermediate frequency is too low in a radio receiver?
A. Selectivity will be too sharp
B. Image-frequency rejection will improve
C. The frequency selectivity of the local oscillator will have to be lowered
D. All of the above
Answer: A Selectivity will be too sharp

30. The local oscillator is tuned to a frequency …………… In a radio receiver.
A. Equal to incoming frequency
B. Lower than the incoming frequency
C. Higher than the incoming frequency
D. None of the above
Answer: C  Higher than the incoming frequency

31. The selectivity of most receivers is determined largely by …….
A. Sensitivity
B. Antenna direction
C. Characteristics of IF section
D. All of the above
Answer: C Characteristics of IF section

32. What does a transmitter serial current contain?
A. Audio frequencies
B. carrier frequencies
C. Radio frequencies
D. All of the above
Answer: Radio frequencies

33. What happens, if the intermediate frequency is too high in a radio receiver?
A. Selectivity will be poor
B. Tracking difficulties will be least
C. Adjacent channel rejection will improve
D. None of the above
Answer: A Selectivity will be poor

34. Which of the following device has IF input but RF output in a receiver?
A. Loudspeaker
B. Demodulator
C. Audio amplifier
D. Frequency changer
Answer: B Demodulator

35. For which purpose, the neutralization is used in RF amplifiers?
A. Stop oscillation
B. Improve selectivity
C. Increase bandwidth
D. None of the above
Answer:  A Stop oscillation

46. A duplexer is a device used to ……………..
Connect two transmitters to the same antenna
Feed more than one receiver from a single antenna
Connect a receiver and a transmitter to the same antenna
None of the above
Answer: C

37. RF amplifiers are used in radio receivers for which purpose?
A. Improved image frequency rejection
B. Improved rejection of adjacent unwanted signals
C. Prevention of re-radiation of the local oscillator through the antenna of the receiver
D. All of the above
Answer: D All of the above

38. ………….. should be used in order to prevent overloading or the last IF amplifier in a receiver.
A. Squelch
B. Double conversion
C. Variable sensitivity
D. Variable selectivity
Answer: D Variable sensitivity

39. Which of the following oscillator is used as a local oscillator in a radio receiver?
A. Crystal
B. Hartley
C. Phase Shift
D. Wien-bridge
Answer: B Hartley

40. Which of the following is the function of radio receiver?
A. Produce radio waves
B. Modulate a message signal
C. Convert one form of energy into other
D. Detect and amplify information signal from the carrier
Answer: D Detect and amplify information signal from the carrier

41. Indicate which of the following statements about the advantages of the phase discriminator over the slope detector is false
a. Much easier alignment
b. Better linearity
c. Greater limiting
d. Fewer tuned circuits
Answer: Option C

42. Show which of the following statements about the amplitude limiter is untrue:
a. The circuit is always biased in class C, by virtue of the leak-type bias.
b. When the input increases past the threshold of the limiting, the gain decreases to keep the output constant.
c. The output must be tuned
d. Leak-type bias must be used
Answer: Option A

43. In a radio receiver with simple AGC
a. an increase in signal strength produces more AGC
b. the audio stage gain is normally controlled by the AGC
c. the faster the AGC time constant the more accurate the output
d. the highest AGC voltage is produced
Answer: Option A

44. In a broadcast superheterodyne receiver, the
a. local oscillator operates below the signal frequency
b. mixer input must be tuned to the signal frequency
c. local oscillator frequency is normally double the IF
d. RF amplifier normally works at 455 kHz above the carrier frequency
Answer: Option B

45. To prevent overloading of the IF amplifier in a receiver, one should use
a. squelch
b. variable sensitivity
c. variable selectivity
d. double conversion
Answer: Option B

46. A superheterodyne receiver with an IF of 450 kHz is tuned to a signal at 1200 kHz. The image frequency is
a. 750 kHz
b. 900 kHz
c. 1650 kHz
d. 2100 kHz
Answer: Option D

47. In a ratio detector
a. the linearity is worse than in phase discriminator
b. stabilization against signal strength variations is provided
c. the output is twice that obtainable from a similar phase discriminator
d. the circuit is the same as in a discriminator, except that the diodes are reversed
Answer: Option A

48. The typical squelch circuit cuts off
a. an audio amplifier when the carrier is absent
b. RF interference when the signal is weak
c. An IF amplifier when the AGC is maximum
d. An IF amplifier when the AGC is minimum
Answer: Option A

49. Indicate the false statement in connection with communications receivers.
a. The noise limiter cuts off the receiver’s output during a noise pulse.
b. A product demodulator could be used for the reception of Morse code.
c. Double conversion is used to improve image rejection
d. Variable sensitivity is used to eliminate selective fading
Answer: Option D

50. The controlled oscillator synthesizer is sometimes preferred over the direct one because
a. it is a simpler piece of equipment
b. its frequency stability is better
c. it does not require crystal oscillator
d. it is relatively free of spurious frequency
Answer: Option D

51. The frequency generated by each decade in a direct frequency synthesizer is much higher than the frequency shown; this is done to
a. reduce the spurious frequency problem
b. increase the frequency stability of the synthesizer
c. reduce the number of decades
d. reduce the number of crystals required
Answer: Option A

52. Indicate which of the following circuits could not demodulate SSB:
a. Balance modulator
b. Product modulator
c. BFO
d. Phase discriminator
Answer: Option D

53. If a FET is used as the first AF amplifier in a transistor receiver, this will have the effect of
a. improving the effectiveness of the AGC
b. reducing the effect of negative-peak clipping
c. reducing the effect of noise at low modulation depths
d. improving the selectivity of the receiver
Answer: Option B

54. Indicate the false statement. The superheterodyne receiver replaced the TRF receiver because the latter suffered from
a. gain variation over the frequency coverage range
b. insufficient gain and sensitivit
c. inadequate selectivity at high frequencies
d. instability
Answer: Option B

55. 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 be the IF tuned circuits
d. is independent of the frequency to which the receiver is tuned
Answer: Option C

56. One of the main functions of the RF amplifier in a superheterodyne receiver is to
a. provide improved tracking
b. permit better adjacent-channel rejection
c. increase the tuning range of the receiver
d. improve the rejection of the image frequency
Answer: Option D

57. A receiver has poor IF selectivity. It will therefore also have poor
a. blocking
b. double-spotting
c. diversion reception
d. sensitivity
Answer: Option A

58. Three-point tracking is achieved with
a. variable selectivity
b. the padder capacitor
c. double spotting
d. double conversion
Answer: Option B

59. The local oscillator of a broadcast receiver is tuned to a frequency higher than the incoming frequency
a. to help the image frequency rejection
b. to permit easier tracking
c. because otherwise an intermediate frequency could not be produced
d. to allow adequate frequency coverage without switching
Answer: Option D

60. If the intermediate frequency is very high (indicate false statement)
a. image frequency rejection is very good
b. the local oscillator need not be extremely stable
c. the selectivity will be poor
d. tracking will be improved
Answer: Option D

61. In a superheterodyne receiver, the IF is 455 kHz. If it is tuned to 1200 kHz, the image frequency will be
A. 1655 kHz
B. 2110 kHz
C. 745 kHz
D. 910 kHz
Answer. B
Explanation:
A signal (image) can interfere with a superheterodyne receiver if fits the following equation.
Image = Signal +/- 2 x I.F.
Which says that a signal has the capacity to interfere with a superhet receiver if its frequency is equal to the signal frequency (1200 kHz in our question) plus or minus twice the IF (455 kHz in our question).
So one possible image is: 1200 + ( 2 x 455 ) = 2110 kHz (possible).
And the other: 1200 – ( 2 x 455 ) = 290 kHz (not likely).

62. The selectivity of most receivers is determined largely by …………
A. sensitivity
B. characteristics of IF section
C. antenna direction
D. all of the above
Answer. B

63. In a superheterodyne receiver with an IF of 450 kHz is tuned to a signal at 1200 kHz. The image frequency is ……….
A. 750 kHz
B. 990 kHz
C. 1650 kHz
D. 2100 kHz
Answer. D

64. In a broadcast superheterodynes receiver having no RF amplifier, the loaded Q of the antenna coupling circuit is 100. If the intermediate frequency is 455 kHz. The rejection ratio at 25 MHz will be ………
A. 1.116
B. 1.386
C. 2.116
D. 2.386
Answer. A

65. As compared to tuned radio frequency receivers which of the following is the advantage of using superheterodyne receivers?
A. High gain and better sensitivity
B. Better selectivity at high frequencies
C. Stability
D. Noise suppression
Answer. A

66. In a radio receiver, if the intermediate frequency is too low
A. Image-frequency rejection will improve
B. Selectivity will be too sharp
C. The frequency selectivity of the local oscillator will have to be lowered
D. All of the above
Answer. B

67. FM amplifier in a superheterodyne receiver
A. increases selectivity
B. suppresses noise
C. provides improved tracking
D. improves the rejection of the image frequency
Answer. D

68. Padders are used in a receiver to
A. discard the carrier
B. facilitate tracking
C. filter the input signal
D. suppress noise
Answer. B

69. In a radio receiver, the local oscillator is tuned to a frequency
A. lower than the incoming frequency
B. higher than the incoming frequency
C. equal to incoming frequency
D. any of the above
Answer. B

70. Which of the following oscillator is generally not used at VHF?
A. Colpitts oscillator
B. Clapp oscillator
C. Armstrong oscillator
D. Ultra Audio oscillator
Answer. C

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

72. One of the advantage of using a high frequency carrier wave is that it dissipates very small power.
a) True
b) False
Answer: a
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.

73. What is the function of RF mixer?
a) Addition of two signals
b) Multiplication of two signals
c) Subtraction of two signals
d) To reduce the amount of noise
Answer: b
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.

74. 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%
Answer: c
Explanation:
 which gives m = 0.285 or 28.5%.

a) 100W
b) 800W
c) 500W
d) 900W
Answer: d
Explanation: P_T=P_C (1+^u2⁄₂), according to the problem P_C = 800W and m = 0.5. On substituting values in the equation we get P_T=800(1+ ^0.52⁄₂) = 900W.

76. 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
Answer: a
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.

77. Which of the following statement about the advantage of phase discriminator over the slope detector is false?
A. Fewer tuned circuits
B. Better linearity
C. Greater limiting
D. Much easier alignment
Answer. C

78. Fidelity of a receiver represents
A. the sensitivity expressed in terms of voltage that must be applied to the receiver input to give a standard output
B. the extent to which the receiver is capable of distinguishing between the desired signal and other frequencies
C. the variation of the output with the modulation frequency when the output impedance is a resistance
D. none of the above
Answer. C

79. Which of the following produces upper and lower side frequencies?A. Microphone
B. Demodulator in a superheterodyne receiver
C. Modulator in a. radio transmitter
D. Oscillator in a receiver
Answer. A

80. The passband of the tuned circuits of a radio receiver should be equal to
A. 20 kHz
B. 455 kHz
C. 1455 kHz
D. more than 455 kHz
Answer. A

1)Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt
a.100 Hz
b.200 Hz
c.400 Hz
d.250 Hz
ANSWER:400 Hz
Explanation:In the given signal, the highest frequency is given by f = 400 π/ 2π = 200 Hz
The minimum sampling rate required to avoid aliasing is given by Nyquist rate. The nyquist rate is = 2 * f
= 2 * 200
= 400 Hz.

2)In Pulse Position Modulation, the drawbacks are
a. Synchronization is required between transmitter and receiver
b. Large bandwidth is required as compared to PAM
c. None of the above
d. Both a and b
ANSWER:Both a and b
Explanation:In Pulse Position Modulation, the position of the pulse of the carrier is varied with reference to the position of a reference pulse. The position is varied in accordance with the sampled modulating signal. In PPM, synchronization is required between the transmitter and the receiver. Large bandwidth is required in Pulse position Modulation as compared to the Pulse amplitude modulation.

3)In PWM signal reception, the Schmitt trigger circuit is used
a. To remove noise
b. To produce ramp signal
c. For synchronization
d. None of the above
ANSWER: To remove noise
Explanation:In pulse width modulation, the width of the carrier varies with the amplitude of the modulating signal at the time of sampling. In PWM signal reception, the received PWM signal is applied to the Schmitt trigger circuit. The Schmitt trigger circuit is used to remove noise in the PWM waveform. This output is the supplied further for detection of the original information.

4)In pulse width modulation
a. Synchronization is not required between transmitter and receiver
b. Amplitude of the carrier pulse is varied
c. Instantaneous power at the transmitter is constant
d. None of the above
ANSWER: Synchronization is not required between transmitter and receiver
Explanation: In pulse width modulation, the width of the carrier varies with the amplitude of the modulating signal at the time of sampling. Pulse width modulation is a type of Pulse Time Modulation. As there is no variation in the amplitude of the carrier, the noise may be easily removed at the receiver. It does not require synchronization between the transmitter and the receiver.

5)In different types of Pulse Width Modulation,
a. Leading edge of the pulse is kept constant
b. Tail edge of the pulse is kept constant
c. Centre of the pulse is kept constant
d. All of the above
ANSWER: All of the above
Explanation: There are types of Pulse Width Modulation. In one of the variations, leading edge of the pulse is kept constant and pulse width is measured with respect to leading edge. In second type, tail edge of the pulse is kept constant and pulse width is measured with respect to it. And the third type has a constant centre of the pulse and the pulse width changes on both the sides of the centre of the pulse.

6)In Pulse time modulation (PTM),
a. Amplitude of the carrier is constant
b. Position or width of the carrier varies with modulating signal
c. Pulse width modulation and pulse position modulation are the types of PTM
d. All of the above
ANSWER: All of the above
Explanation:In Pulse time modulation (PTM), amplitude of the carrier is kept constant and the Position or width of the carrier varies with the amplitude of the modulating signal at the time of sampling. Pulse width modulation and pulse position modulation are the types of Pulse Time Modulation.
As there is no variation in the amplitude of the carrier, the noise may be easily removed at the receiver.

7)Drawback of using PAM method is
a. Bandwidth is very large as compared to modulating signal
b. Varying amplitude of carrier varies the peak power required for transmission
c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver
d. All of the above
8) Pulse time modulation (PTM) includes
ANSWER: All of the above
Explanation:In PAM, Bandwidth is very large as compared to modulating signal frequency. In PAM, the amplitude of the rectangular pulse train is varied according to the instantaneous value of the modulating signal. Due to this, the required power for transmission is also varied. Due to varying amplitude of carrier, the interference of noise is very high in PAM. So it is difficult to remove noise at receiver.

8)Pulse time modulation (PTM) includes
a. Pulse width modulation
b. Pulse position modulation
c. Pulse amplitude modulation
d. Both a and b
ANSWER: Both a and b
Explanation:In pulse modulation systems, the carrier is a train of pulses rather than a continuous signal. The parameters of the pulses are varied according to the instantaneous value of the modulating signal. The carrier is a train of pulses rather than a continuous signal. In PTM, the timing of the pulses of the carrier is varied in accordance with modulating signal. PTM includes:
– Pulse width modulation
– Pulse position modulation

9)In pulse amplitude modulation,
a. Amplitude of the pulse train is varied
b. Width of the pulse train is varied
c. Frequency of the pulse train is varied
d. None of the above
ANSWER: Both a and b
Explanation:
In pulse modulation systems, the carrier is a train of pulses rather than a continuous signal. The parameters of the pulses are varied according to the instantaneous value of the modulating signal. The carrier is a train of pulses rather than a continuous signal. In PTM, the timing of the pulses of the carrier is varied in accordance with modulating signal. PTM includes:
– Pulse width modulation
– Pulse position modulation

10)Types of analog pulse modulation systems are
a. Pulse amplitude modulation
b. Pulse time modulation
c. Frequency modulation
d. Both a and b
ANSWER: Both a and b
Explanation:In pulse modulation systems, the carrier is a train of pulses rather than a continuous signal. The parameters of the pulses are varied according to the instantaneous value of the modulating signal. There are two types of pulse modulation systems:
– Pulse amplitude modulation
– Pulse time modulation

11)The sampling technique having the minimum noise interference is
a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above
ANSWER: Natural sampling
Explanation:The natural sampling is the technique that has the minimum noise interference to the sampled signal. It is obtained by multiplying the input signal with the sampling function. It is a practical method used for sampling of signals. Chopping principle is used to sample the signal in natural sampling and it satisfies the Nyquist criteria for sampling of signals.

12)The instantaneous sampling
a. Has a train of impulses
b. Has the pulse width approaching zero value
c. Has the negligible power content
d. All of the above
ANSWER: All of the above
Explanation:The instantaneous sampling is also called ideal sampling or impulse sampling. The instantaneous sampling has a train of impulses. The pulse width of the samples has almost zero value. Therefore it has negligible power content and thus may not be used for transmission purpose.

13)The techniques used for sampling are
a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above
ANSWER: All of the above
Explanation:The techniques used for sampling are:
a) Instantaneous sampling
b) Natural sampling
c) Flat top sampling
The natural sampling and the flat top sampling techniques are used practically to sample a signal.

14)A low pass filter is
a. Passes the frequencies lower than the specified cut off frequency
b. Rejects higher frequencies
c. Is used to recover signal from sampled signal
d. All of the above
ANSWER: All of the above
Explanation:The low pass filter should have the cut off frequency equal to wm so that it allows only lower frequencies up to the cut off frequency to pass through. The other higher frequencies in the sampled signal are rejected by the low pass filter. The original or desired signal may be recovered from the sampled signal by passing the signal through a low pass filter.

15)Calculate the Nyquist rate for sampling when a continuous time signal is given by
x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt
a. 300Hz
b. 600Hz
c. 150Hz
d. 200Hz
ANSWER: 300Hz
Explanation:For the given signal,
f1 = 100π/2π = 50Hz
f2 = 200π/2π = 100Hz
f3= 300π/2π = 150Hz
The highest frequency is 150Hz. Therefore fmax = 150Hz
Nyquist rate = 2 fmax
= 2 * 150
= 300Hz.

16)A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is
a. fs > 2fm
b. fs < 2fm
c. fs = 2fm
d. fs ≥ 2fm
ANSWER: fs < 2fm
Explanation: If the signal of frequency fm is sampled at the rate fs ≥ 2fm only then the spectrum of the sampled signal is obtained without overlapping. For fs < 2fm the sampled signal spectrum overlap each other, and therefore the signal cannot be recovered easily. For reconstruction of signal to be free from distortion, the important condition is
fs ≥ 2fm

17)The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if
a. The sampled signal is passed through low pass filter
b. Filter has the cut off frequency wm
c. Both a and b
d. None of the above
ANSWER: Both a and b
Explanation:The original or desired signal may be recovered from the sampled signal by passing the signal through a low pass filter. The low pass filter should have the cut off frequency equal to wm so that it allows only low frequencies up to the cut off frequency to pass through. The other higher frequencies in the sampled signal are rejected by the low pass filter.

18)The spectrum of the sampled signal may be obtained without overlapping only if
a. fs ≥ 2fm
b. fs < 2fm
c. fs > fm
d. fs < fm
ANSWER: fs ≥ 2fm
Explanation:
If the signal of frequency f_m is sampled at the rate f_s ≥ 2f_m only then the spectrum of the sampled signal is obtained without overlapping. The spectrum obtained is repetitive in nature and is completely without overlapping.

19)The process of using a pulse signal to represent information is called _______
a)Pulse modulation
b)Frequency modulation
c)Amplitude modulation
d)Phase modulation
Answer:a
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.

20)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
Answer:d
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.

21)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
Answer:d
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.

22)How many voltage levels are present in a PWM signal?
a)0
b)1
c)2
d)3
Answer:c
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.

23)Power consumption is low in pulse modulation.
a)True
b)False
Answer:a
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.
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24)Which pulse modulation technique is least expensive?
a)Pulse amplitude modulation
b)Pulse width modulation
c)Pulse position modulation
d)Pulse code modulation
Answer:a
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.

25)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
Answer:d
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.

26)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
Answer:d
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.

27)The process of signal compression and expansion used to reduce distortion and noise is called _____
a)Amplification
b)Companding
c)Compressing
d)Modulating
Answer: b
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.

28)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
Answer: d
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.

29)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
Answer:d
Explanation:avionics-questions-answers-pulse-modulation-q11

30)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
Answer:c
Explanation:avionics-questions-answers-pulse-modulation-q12

31)Flat top sampling of low pass signals
a)Gives rise to aperture effect
b)Implies over sampling
c)Leads to aliasing
d)Introduces delay distortion
Answer:a
Explanation:Flat top sampling of low pass signals gives rise to aperture effect.

32)In a delta modulation system, granular noise occurs when the
a)Modulating signal increases rapidly
b)Pulse rate decreases
c)Pulse amplitude decreases
d)Modulating signal remains constant
Answer:d
Explanation:In a delta modulation system, granular noise occurs when the modulating signal remains constant.

33)A PAM signal can be detected using
a)Low pass filter
b)High pass filter
c)Band pass filter
d)All pass filter
Answer:a
Explanation:A PAM signal can be detected by using low pass filter.

34)Coherent demodulation of FSK signal can be performed using
a)Matched filter
b)BPF and envelope detectors
c)Discriminator
d)None of the mentioned
Answer:a
Explanation:Coherent demodulation of FSK signal can be performed using matched filter.

35)The use of non uniform quantization leads to
a)Reduction in transmission bandwidth
b)Increase in maximum SNR
c)Increase in SNR for low level signals
d)Simplification of quantization process
Answer:c
Explanation:The use of non uniform quantization leads to increase in SNR for low level signals.
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36)Which of the following requires a synchronizing signal?
a)Single channel PPM system
b)PAM
c)DM
d)All of the mentioned
Answer:b
Explanation:PAM requires a synchronizing signal.

37)A PWM signal can be generated by
a)An astable multi vibrator
b)A monostable multi vibrator
c)Integrating a PPM signal
d)Differentiating a PPM signal
Answer:b
Explanation:A PWM signal can be generated by a mono stable multi vibrator.

38)TDM is less immune to cross-talk in channel than FDM.
a)True
b)False
Answer: b
Explanation:False because different message signals are not applied to the channel simultaneously.

39)In an ideal TDM system, the cross correlation between two users of the system is
a)1
b)0
c)Infinity
d)-1
Answer: b
Explanation: In an ideal TDM system, the cross correlation between two users of the system is 0.

40)TDM requires
a)Constant data transmission
b)Transmission of data samples
c)Transmission of data at random
d)Transmission of data of only one measured
Answer:b
Explanation:TDM requires transmission of data samples.

41)Which principle specifies the relationship between enclosure of poles & zeros by s-plane contour and the encirclement of origin by q(s) plane contour?
a)Argument
b)Agreement
c)Assessment
d)Assortment
Answer:a
Explanation:Argument principle specifies the relationship between enclosure of poles & zeros by s-plane contour and the encirclement of origin by q(s) plane contour.

42)If a Nyquist plot of G (jω) H (jω) for a closed loop system passes through (-2, j0) point in GH plane, what would be the value of gain margin of the system in dB?
a)0 dB
b)2.0201 dB
c)4 dB
d)6.0205 dB
Answer:d
Explanation:Gain Margin is calculated by taking inverse of the gain where the Nyquist plot cuts the real axis.

43)For Nyquist contour, the size of radius is _______
a)25
b)0
c)1
d)∞
Answer:d
Explanation:For Nyquist contour, the size of radius is ∞.

44)Consider a feedback system with gain margin of about 30. At what point does Nyquist plot crosses negative real axis?
a)-3
b)-0.3
c)-30
d)-0.03
Answer:b
Explanation: Gain Margin is always inverse of the point which cuts the Nyquist on the real axis.

45)According to Nyquist stability criterion, where should be the position of all zeros of q(s) corresponding to s-plane?
a)On left half
b)At the center
c)On right half
d)Random
Answer:a
Explanation:According to Nyquist stability criterion zeroes must lie on the left half on the s plane.

46)If the system is represented by G(s) H(s) = k (s+7) / s (s +3) (s + 2), what would be its magnitude at ω = ∞?
a)0
b)∞
c)7/10
d)21
Answer:a
Explanation:On calculating the magnitude of the system and putting the value of frequency one gets the magnitude as 0.

47)Consider the system represented by the equation given below. What would be the total phase value at ω = 0?
200/[s3 (s + 3) (s + 6) (s + 10)].
a)-90°
b)-180°
c)-270°
d)-360°
Answer:c
Explanation:The phase can be calculated by the basic formula for calculating phase angle.

48)Due to an addition of pole at origin, the polar plot gets shifted by ___ at ω = 0 ?
a)-45°
b)-60°
c)-90°
d)-180°
Answer:c
Explanation:Addition of pole causes instability to the system.

49)In polar plots, if a pole is added at the origin, what would be the value of the magnitude at Ω = 0?
a)Zero
b)Infinity
c)Unity
d)Unpredictable
Answer:b
Explanation:Addition of pole causes instability to the system.

50)In polar plots, what does each and every point represent w.r.t magnitude and angle?
a)Scalar
b)Vector
c)Phasor
d)Differentiator
Answer:c
Explanation:Each and every point on the polar plot is the phasor where value of frequency varies.

51)Nyquist criterion helps in
a)Transmitting the signal without ISI
b)Reduction in transmission bandwidth
c)Increase in transmission bandwidth
d)Both a) and b)
ANSWER:d) Both a) and b)

52)The Nyquist theorem
a)Relates the conditions in time domain and frequency domain
b)Helps in quantization
c)Limits the bandwidth requirement
d)Both a) and c)
ANSWER:d) Both a) and c)

53)The difficulty in achieving the Nyquist criterion for system design is
a)There are abrupt transitions obtained at edges of the bands
b)Bandwidth criterion is not easily achieved
c)Filters are not available
d)None of the above
ANSWER:a)There are abrupt transitions obtained at edges of the bands

54)The digital modulation technique in which the step size is varied according to the variation in the slope of the input is called
a)Delta modulation
b)PCM
c)Adaptive delta modulation
d)PAM
ANSWER:c)Adaptive delta modulation

55)The digital modulation scheme in which the step size is not fixed is
a)Delta Modulation
b)Adaptive delta modulation
c)DPCM
d)PCM
ANSWER:b)Adaptive delta modulation

56)In Adaptive Delta Modulation, the slope error reduces and
a)Quantization error decreases
b)Quantization error increases
c)Quantization error remains same
d)None of the above
ANSWER:b)Quantization error increases

1. Multiplexing increases the number of communication channels for transmission.
a) True
b) False
Answer: a
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.

2. In which of the following systems multiplexing is not necessary?
a) Telemetry
b) TV broadcasting
c) Satellites
d) Continuous wave transmission
Answer: d
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.

3. 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
Answer: c
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.

4. What type of multiplexing is widely used in cellphones?
a) Time division multiplexing
b) Frequency division multiplexing
c) Code division multiplexing
d) Spatial multiplexing
Answer: c
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.

5. 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
Answer: d
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.6. For frequency division multiplexing who defines the channel bandwidth?

6. For frequency division multiplexing who defines the channel bandwidth?
a) FCC
b) ARNIC
c) FAA
d) CCA
Answer: a
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.

7. What is the individual carrier frequency of each signal called?
a) Subcarrier
b) Frequency carrier
c) Modulated carrier
d) Coded carrier
Answer: a
Explanation: Each signal to be transmitted feeds a modulator circuit. The carrier for each modulator (f_c) 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.

8. Which circuit does the actual multiplexing process in frequency division multiplexing?
a) Linear mixer
b) Oscillator
c) RF amplifier
d) Duplexer
Answer: a
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.

9. Which of the following device is used to demultiplex the received signal?
a) Allpass filters
b) Bandpass filters
c) Bandstop filters
d) Differential filters
Answer: b
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.

10. 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
Answer: b
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.

11. Frequency division multiplexing:Frequency slots::time division multiplexing:?
a) Time slots
b) Coded information
c) Pulsed information
d) Band slots
Answer: a
Explanation: In FDM, multiple signals are transmitted over a single channel, each signal being allocated a portion of the spectrum within that bandwidth. In time-division multiplexing (TDM), each signal occupies the entire bandwidth of the channel. However, each signal is transmitted for only a brief time. In other words, multiple signals take turns transmitting over the single channel.

12. Serial transmission is not possible without time division multiplexing.
a) True
b) False
Answer: a
Explanation: In serial transmission, the data is sent via a single cable. When a clock pulse is applied to the shift register it transmits the information bit by bit in allocated time slots.

13. What device is used to demodulate a time division multiplexed analog wave?
a) High pass filter
b) Low pass filter
c) Band stop filter
d) Attenuator
Answer: b
Explanation: the analog signal is converted to a series of constant-width pulses whose amplitude follows the shape of the analog signal. The original analog signal is recovered by passing it through a low-pass filter. In TDM using PAM, a circuit called a multiplexer (MUX or MPX) samples multiple analog signal sources; the resulting pulses are interleaved and then transmitted over a single channel.

14. Which of the following device was used in early TDM/PAM telemetry systems?
a) Commuter
b) Linear switch
c) Logic gates
d) DSP
Answer: a
Explanation: Multiplexers in early TDM/PAM telemetry systems used a form of rotary switch known as a commutator. Multiple switch segments were attached to the various incoming signals while a high-speed brush rotated by a dc motor rapidly sampled the signals as it passed over the contacts.

15. What is the time allocated for each channel if the number of samples per frame is 4 and the frame rate is 100frames/sec?
a) 1.2ms
b) 3ms
c) 2.5ms
d) 0.54ms
Answer: c
Explanation: Time period for one frame = 1/100 = 0.01s = 10ms. During that 10-ms frame period, each of the four channels is sampled once. Each channel is thus allotted 10/4 = 2.5 ms.

16. What is the purpose of one shot multivibrator?
a) Trigger all AND gates at clock frequency
b) Trigger all OR gates at clock frequency
c) Trigger all AND gates at signal frequency
d) Trigger all OR gates at signal frequency
Answer: a
Explanation: The one-shot multivibrator is used to trigger all the decoder AND gates at the clock frequency. It produces an output pulse whose duration has been set to the desired sampling interval.

17. The circuit used to regenerate clock pulses from the transmitted PAM signals is called ____
a) Clock demodulator circuits
b) Timer circuits
c) Clock receiving circuits
d) Clock recovery circuits
Answer: d
Explanation: Instead of using a free-running clock oscillator set to the identical frequency of the transmitter system clock, the clock for the demultiplexer is derived from the received PAM signal itself. A circuit called the clock recovery circuits are typical of those used to generate the demultiplexer clock pulses.

18. What are used to reduce or stop synchronization problems while receiving?
a) Clock recovery circuits
b) Demodulators
c) Synchronizer
d) Band pass filter
Answer: a
Explanation: Clock recovery circuits are used to remedy the synchronization problem encountered in demultiplexing. The clock pulse is derived from the transmitted signal so that synchronization errors are reduced.

19. In a four channel system, all four signals transmitted contain information.
a) True
b) False
Answer: b
Explanation: After clock pulses of the proper frequency have been obtained, it is necessary to synchronize the multiplexed channels. This is usually done with a special synchronizing (sync) pulse applied to one of the input channels at the transmitter. In the four-channel system discussed previously, only three actual signals are transmitted. The fourth channel is used to transmit a special pulse whose characteristics are unique in some way so that it can be easily recognized.

20. Which of the following is not an advantage of time division multiplexing?
a) Signal interference is less
b) More flexible
c) Full channel can be used for every signal
d) Fast data transfer
Answer: d
Explanation: Since the time available for transmission is shared by all the signals that are modulated, Time division multiplexing is not fast when compared with other multiplexing techniques. However, for applications such as telemetry, with a high sampling rate, the speed of time division multiplexing is sufficient to meet the requirements.