Updated Analog Communications MCQs – Amplitude Modulation ( Analog Communications ) MCQs
Latest Analog Communications MCQs
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Analog Communications MCQs – Amplitude Modulation ( Analog Communications ) MCQs
The most occurred mcqs of Amplitude Modulation ( ) in past papers. Past papers of Amplitude Modulation ( Analog Communications ) Mcqs. Past papers of Amplitude Modulation ( Analog Communications ) Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Amplitude Modulation ( Analog Communications ) Mcqs. The Important series of Amplitude Modulation ( Analog Communications ) Mcqs are given below:
Frequency Domain Description
1. What is Amplitude Modulation?
a) Change in amplitude of carrier according to modulating signal amplitude
b) Change in frequency of carrier according to modulating signal amplitude
c) Change in amplitude of carrier according to modulating signal frequency
d) Change in amplitude of modulating signal according to carrier signal amplitude
Answer: a
Explanation: Amplitude modulation is a modulation process in which amplitude of carrier wave is varied with respect to amplitude of the message signal to be transmitted. Whereas, Frequency modulation is a modulation process in which frequency of carrier wave is varied with respect to amplitude of the message signal to be transmitted.
2. Frequency components of an AM wave are?
a) Carrier frequency (ωc) with amplitude A
b) Lower side band (ωc + ωm) having amplitude mA⁄2
c) Upper side band (ωc – ωm) having amplitude mA⁄2
d) Carrier frequency (ωc/2) with amplitude A
Answer: a
Explanation: The frequency components of AM waves are: Carrier frequency (ωc) with amplitude A, Lower sideband (ωc – ωm) having amplitude mA/2 and Upper side band (ωc + ωm) having amplitude mA⁄2 (where m represents modulation index, A is the amplitude of the carrier signal, ωm is the amplitude of the message signal).
3. In amplitude modulation frequency and phase of carrier ________
a) varies simultaneously
b) varies alternately
c) initially varies but become same after sometime
d) remains constant
Answer: d
Explanation: In AM, amplitude of carrier signal varies according to instantaneous amplitude of baseband signal. The frequency and phase of carrier signal remains constant.
4. Envelope of AM wave has the same shape as the message of baseband signal.
a) True
b) False
Answer: a
Explanation: Modulation Envelope of an AM wave is said to follow the outline of the message signal when carrier, upper sideband and lower sideband are combined in a single impedance and observed on time versus amplitude plot. Thus, AM wave has same shape as the message signal.
5. When aliasing takes 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
Answer: a
Explanation: Aliasing causes different signals to become indistinguishable during sampling. It happens because sampling rate is less than Nyquist rate. In order to avoid aliasing, signals should be sampled at a rate twice of Nyquist rate.
6. Why the Synchronous detection of AM signals is considered as a disadvantage?
a) Needs additional system for synchronization of carrier
b) Receiver is available at cheap prices
c) Needs less number of system as estimated for generation of carrier
d) Receiver is not complex
Answer: a
Explanation: Disadvantage of synchronous detection of AM signal are that it needs an additional system for generation of carrier. It also needs additional system for synchronization of carrier. Moreover receiver is complex and costly.
7. Which devices did we use for AM Demodulation?
a) Envelope detector and Square law demodulator
b) PLL detector and Foster-Seeley discriminator
c) Ratio detector and Slope detector
d) Only quadrature detector
Answer: a
Explanation: AM signals can be demodulated using Square law demodulator or Envelope detector. Demodulator extracts information from received AM signal by decoding it. Other options are the demodulators for FM signal.
8. Square Law modulators ________
a) used for amplitude modulation
b) have non linear current-voltage characteristics
c) have non linear current-voltage characteristics as well as used for Amplitude Modulation
d) used for frequency modulation
Answer: c
Explanation: Square law modulators are used for amplitude modulation and have non-linear current-voltage characteristics. The output of Square law demodulator is said to vary with respect to square of the input. They are highly linear in low voltage region.
9. What do you understand by low level AM?
a) Output power is low
b) Modulation is done at high power of carrier and modulating signal
c) Collector Modulation Method is low level AM
d) Output power is high
Answer: a
Explanation: In low level AM, modulation is done at low power of carrier and modulating signal therefore output power is low. Therefore, power amplifiers are used to boost the carrier and modulating signal. Collector Modulation Method is high level AM. At high-level AM, output power is high.
10. What do you understand by high level AM?
a) Output power is low
b) Modulation is done at high power of carrier and modulating signal
c) No need to boost the carrier and modulating signal
d) Modulation is done at low power of carrier and modulating signal
Answer: b
Explanation: For high level AM, modulation is done at high power of carrier and modulating signal, so power amplifiers are not used to boost the carrier and modulating signal. Therefore, output power is high. At low-level AM, output power is low.
Single Tone Modulation
1. For Amplitude Modulation, Emitter modulator ________
a) Operates in class C mode
b) Has a low efficiency
c) Output power is high
d) Operates in class B mode
Answer: b
Explanation: Emitter Modulator operates in class A region. It has very low efficiency. The output power is low so for modulation at high level, it is not suitable.
2. Why AM is used for broadcasting?
a) More immune to noise
b) Less transmitting power is required
c) It has high fidelity
d) Avoids Receivers Complexity
Answer: d
Explanation: AM detectors are generally, square law demodulators or envelope detectors at the receiver. As AM detectors at the receiver end are simple circuits and avoid any kind of complex structure, therefore, AM used for broadcasting.
3. Singletone amplitude modulation ________
a) consists of only one frequency component
b) contains a large number of frequency components
c) contains no frequency components
d) contains infinite number of frequency components
Answer: a
Explanation: Single tone modulation consists of only one frequency component in the baseband or message signal. Thus, modulation of carrier wave is done by a single frequency component only.
4. AM spectrum consists of ________
a) Carrier frequency
b) Upper sideband
c) Lower sideband
d) Carrier frequency with both upper and lower sideband
Answer: d
Explanation: Spectrum of Am wave consists of a carrier with upper sideband and lower sideband. If carrier frequency is Wc, then the two sidebands produced by it are (Wc+Wm) and (Wc-Wm), where Wm is the frequency of the message signal. The amplitude of the carrier is A and that of the two sidebands are mA/2, where m is the modulation index.
5. The minimum channel Bandwidth is used by which modulation technique?
a) VSB
b) SSB-SC
c) DSB-SC
d) AM
Answer: b
Explanation: A signal has two sidebands which are exactly the mirror images of each other. So we can remove one side band which further reduces its bandwidth. In SSB-SC modulation technique, the carrier is suppressed and only either of the sidebands is transmitted. Thus, SSB-SC has minimum channel Bandwidth.
6. Neper is ________ decibel.
a) 20/(ln10)
b) 20ln10
c) Same as
d) Exactly twice of
Answer: a
Explanation: Neper is a logarithmic unit used for finding ratios of power quantities. Like decibel it is also a dimensionless unit. 1Np = 20/ln10 = 8.686 db.
7. AM broadcast station transmits modulating frequency upto 6KHz. If transmitting frequency is 810KHz, then maximum and lower sidebands are ________
a) 816KHz and 804KHz
b) 826KHz and 804KHz
c) 916KHz and 904KHz
d) 822KHz and 816KHz
Answer: a
Explanation: Maximum frequency = 810 + 6 = 816KHz and Minimum frequency = 810 – 6 = 804KHz. Moreover it has a bandwidth of (816 – 804) = 12KHz.
( Formula: Fmax = fc + fm,
Fmin = fc-fm,
Bandwidth = 2*fm = Fmax = Fmin,
fc = Carrier frequency, fm = message signal frequency).
8. Find lower frequency component in AM wave, given that highest frequency component is 900KHz and bandwidth is 12KHz?
a) 832KHz
b) 600KHz
c) 868KHz
d) 888KHz
Answer: d
Explanation: Highest frequency component is 900KHz and bandwidth is 12KHz. So lower frequency component is 900 – 12 = 888KHz.
(Formula: Fmin = Fmax-2*fm = Fmax-Bandwidth, where fm = Message Signal Frequency).
9. Amplitude Modulated wave is ________
a) Sum of carrier and modulating wave
b) Product of carrier and modulating wave
c) Difference of carrier and modulating wave
d) Sum of carrier and its product with modulating wave
Answer: a
Explanation: The modulation of a carrier wave by varying its amplitude with respect to amplitude of baseband signal is known as amplitude modulation. It is represented as,
s(t) = [1 + mx(t)] c(t),
where, x(t) = Modulating Wave, m=Modulating Index
c(t) = Carrier Wave = Ac (Cos ωc) t
Thus, Amplitude Modulated wave is the Sum of carrier and its product with modulating wave.
Power Relations in AM Waves
1. Calculate power in each sideband, if power of carrier wave is 176W and there is 60% modulation in amplitude modulated signal?
a) 13.36W
b) 52W
c) 67W
d) 15.84W
Answer: d
Explanation: Modulation index = 0.6 and Pc = 176W. Power in sidebands may be calculated as
2. For 100% modulation, power in each sideband is ________ of that of carrier.
a) 50%
b) 70%
c) 60%
d) 25%
Answer: d
Explanation: Modulation index = 1. Power in sidebands may be calculated as
3. Overmodulation results in ________________
a) Distortion
b) Weakens signal
c) Strengthens the signal
d) provides immunity to noise
Answer: a
Explanation: When instantaneous level of modulating signal exceeds the value necessary to provide 100% modulation, the signal is said to over-modulated. In other words, when modulation index is greater than 1, it results in Overmodulation. Thus, Overmodulation results in distortion of the modulating signal.
4. The maximum power efficiency of an AM modulator is?
a) 25%
b) 33%
c) 66%
d) 100%
Answer: b
Explanation: Efficiency (ή) = m2 / (m2 + 2), m=Modulation Index
For maximum efficiency m = 1 so, ή = 1/(1+2) = 1/3
and ή% = (1/3)x100 = 33%.
5. Noise performance of a square law demodulator of AM signal is?
a) Better than that of synchronous detector
b) Weaker than that of synchronous detector
c) Better than that of envelope detector
d) Weaker than that of envelope detector
Answer: a
Explanation: Process of recovering message signal from received modulated signal is called demodulation. It is exactly opposite to modulation. There are two most used AM demodulators: Square Law Demodulator and Envelope Demodulator. Noise performance of Square Law Demodulator is far better than that of Synchronous Detector.
6. For getting 100% modulation, carrier amplitude should ________
a) exceed signal amplitude
b) be equal to signal amplitude
c) be lesser than signal amplitude
d) be equal to 0
Answer: b
Explanation: Modulation index is the amount of modulation present in a carrier wave. It is also described as the ration of the amplitude of message signal to that of carrier signal.
Modulation Index (m) = Vm/Vc, where Vm is maximum baseband or message signal amplitude and Vc is maximum carrier signal amplitude. So for m = 1, Vm should be equal to Vc.
7. For 100% modulation, total power is?
a) same as the power of unmodulated signal
b) twice as the power of unmodulated signal
c) four times as the power of unmodulated signal
d) one and half times as the power of unmodulated signal
Answer: d
Explanation: Total power, Pt = Pc (1 + m2⁄2), where m is Modulated Signal, Pc is Power of Unmodulated Signal or Carrier Signal.
So, for m=1,
Pt = Pc (1 + 12/2) = 1.5 Pc.
8. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. What is the value of modulation index?
a) 0.7
b) 0.066
c) 0.341
d) 0.916
Answer: b
Explanation: Given equation can be written as 30(1 + 0.066 Sin(700πt)).
Comparing it with general AM equation, s(t) = Ac(1 + mAm cos(wmt)) cos(wct),
Where, Ac = Amplitude of Carrier Signal, Am = Amplitude of Message Signal
m=Modulation Index
So modulation index(m) = 0.066.
9. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. Carrier power of the wave is?
a) 555W
b) 675W
c) 450W
d) 310W
Answer: c
Explanation: 
Hence Pc = 450W.
10. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. Find the total power of amplitude modulated wave?
a) 453W
b) 675W
c) 789W
d) 451W
Answer: d
Explanation: Pt = Pc (1 + bm2⁄2) * pc
So, here, m = 0.066, Pc = 450W
Pt = (1+(0.0662/2))*450 = 451W.
11. An AM signal is represented by x(t) = (30 + 2Sin(700πt)) Cos(2πt x 102t)V. What is its sideband power?
a) 4W
b) 1W
c) 3W
d) 2W
Answer: b
Explanation: Sideband power i = (m2* Pc)/2 = (Pt – Pc) i.e. 451 – 450 = 1W.
AM Wave Representation
1. AM waves is represented by which equation?
a) [1 + m(t)].c(t)
b) [1 – m(t)].c(t)
c) [1 + m(t)].2c(t)
d) [1 + 2m(t)].c(t)
Answer: a
Explanation: Amplitude wave is represented by [1 + um(t)].c(t), where c(t) is carrier signal, m(t) is message signal, u is Modulation Index.
Generally, c(t) = Accos(wct), Ac = Amplitude of Carrier Signal.
2. For attenuation of high frequencies we can use ________
a) inductor
b) shunt capacitance
c) series capacitance
d) combination of inductor and resistor
Answer: b
Explanation: Capacitive reactance, Xc is low for high frequencies as Xc=(1/wC), where w is the frequency and C is the capacitance. Due to this, the high frequencies are shunted to ground and are not transmitted.
Elements Of Communication Systems MCQs
3. If each element of signal occupies 40ms, determine its speed?
a) 20 bauds
b) 25 bauds
c) 40 bauds
d) 0.05 bauds
Answer: b
Explanation: 
4. A modem is classified as low if speed of data rate is ________
a) upto 600bps
b) upto 200bps
c) upto 400bps
d) upto 500bps
Answer: a
Explanation: Modem is device that can modulate carrier wave as well as demodulate transmitted modulated signal to decode the original information. When data rate in bps is upto 600, modem is considered having low speed.
5. Modulation is also called detection.
a) True
b) False
Answer: b
Explanation: Modulation is encoding the message signal for efficient transmission. Whereas, Demodulation is the process to extract or decode the original message signal from the transmitted modulated signal. Demodulation is also called detection.
6. It is suitable to connect woofer from the input through ________
a) band pass filter
b) band stop filter
c) low pass filter
d) high pass filter
Answer: c
Explanation: Woofer is considered as a low frequency loud speaker. It has a range from 10Hz to 500Hz. So it is better to use low pass filter for connection. Los Pass filter allows low frequencies to pass while high pass filter allows high frequencies to pass. Band pass filter allows a range of frequencies to pass whereas. Band stop filter does the opposite.
7. The radiation at right angles is zero means ________
a) l = ʎ
b) l = ʎ⁄4
c) l = 2 ʎ
d) l = ʎ⁄2
Answer: a
Explanation: Antenna is used for converting electromagnetic radiation into electric currents or vice-versa. If the length of antenna is equal to the whole wavelength then the radiation at right angles is zero.
8. Power of carrier wave is 500W and modulation index is 0.25. Find its total power?
a) 500W
b) 415W
c) 375W
d) 516W
Answer: d
Explanation: Total Power (Pt) = (1+m2/2)*Pc
So, here, m = 0.25, Pc = 500W
Pt = (1+(0.252/2))*500 = 516W.
9. Commercial frequency deviation is ________
a) 75 KHz
b) 80 KHz
c) 85 KHz
d) 65 KHz
Answer: a
Explanation: Frequency deviation is the difference between an FM modulated frequency and the carrier frequency. Commercial FM broadcasting uses a maximum frequency deviation of 75 KHz and maximum modulating frequency is approximately 15KHz.
10. Which of the following cannot be the Fourier series of a periodic signal?
a) x(t) = 2Cost + 4Cos7t
b) x(t) = 2Cosπt + 4Cos7t
c) x(t) = Cost + 0.7
d) x(t) = 2Cost + 4Sin7t
Answer: c
Explanation: x(t) = Cost + 0.7 does not follow Dirchlet’s conditions. Fourier series of a periodic signal will only be applicable if it satisifies Dirchlet’s conditions firstly. Dirichlet integral states that a finite number should have finite number of discontinuities and also have finite number of extrema. x(t) doesn’t satisfy any of the criteria, whereas other options satisfy Dirchlet’s conditions. Thus, x(t) = Cost + 0.7, cannot be the Fourier series of a periodic signal.
Generation of AM Waves
1. Which of the following devices is used to generate AM waves?
a) Square-law modulator
b) Reactance modulator
c) Transmitter
d) Transducer
Answer: a
Explanation: AM signals are generated by Square-law modulators. The input for generating AM signal should be of type (A + m(t)) instead of m(t). Square-law modulator sums carrier and information signal, then passes them through a non-linear device.
2. What is the disadvantage of FM over AM?
a) high modulating power is needed
b) requires high output power
c) large bandwidth required
d) high noise is produced
Answer: c
Explanation: Advantage of FM over AM is that the amplitude of an FM wave remains constant. In FM, the power of transmitted wave depends on amplitude of unmodulated carrier wave and hence it is constant. FM is less prone to noise compared to AM. However, wide-band FM has a wider bandwidth than AM as it’s BW is given by Carson’s rule which = 2*(β+1)fm, where β = Frequency Modulation Index and fm is frequency of modulating signal. And BW of AM = 2*fm.
3. For low level amplitude modulation, amplifier must be ________
a) Class C amplifier
b) Class B amplifier
c) Class D amplifier
d) class A amplifier
Answer: a
Explanation: In low level AM, modulation is done at low power of carrier and modulating signal therefore output power is low. Therefore, power amplifiers are used to boost the carrier and modulating signal. Thus, Class C amplifier is used.
4. Approximate range of a cordless telephone is ________
a) 900m
b) 100m
c) 1000m
d) 475m
Answer: b
Explanation: Cordless telephone is one in which the handset is portable and communicates with the body of the phone by radio waves instead of being connected by a cord. The range depends on the quality of instrument and is about 90m to 100m.
5. A wave is modulated by two sin waves having modulation indices of 0.3 and 0.5. Find the total modulation index?
a) 0.1
b) 0.7
c) 0.58
d) 0.35
Answer: c
Explanation: Given that m1 = 0.3 and m2 = 0.5. Total modulation index will be equal to 
. By substituting values we have
which is equal to 0.58.
6. Carrier wave carries information.
a) True
b) False
Answer: b
Explanation: It is a high frequency electro-magnetic wave. A carrier wave does not have any information. One of the properties like amplitude, frequency or phase of the carrier are modulated with respect to an input signal for the purpose of conveying information.
7. What can we do to eliminate distortion in the picture?
a) use a longer transmission line
b) change the antenna orientation
c) use a short transmission line
d) connect a booster
Answer: b
Explanation: According to modulation, length of antenna should be one fourth of wavelength. If distortion occurs whether in television or at any other place, simplest way to eliminate this is to change the orientation of antenna.
8. For 100% amplitude modulation, the power in upper sideband when carrier power is to be 100W?
a) 100W
b) 75W
c) 25W
d) 50W
Answer: c
Explanation: Modulation index, m = 100% = 1. Power in sidebands is (Pcxm2⁄4). By substituting the values, we have (100 x 1⁄4) which is equal to 25W.
Modulation index, m = 100% = 1.
Power in sidebands = (Pcx m2/4), Pc = Power of Carrier
By substituting the values, we have (100 x 1/4) = 25W.
9. Modem is considered as high speed if data rate is ________
a) 10000
b) 20000
c) 5000
d) 30000
Answer: c
Explanation: For data rate in the range 2000 to 10500 bps, modem is designated as high speed. Modem is a combination of modulator and demodulator. Modulator tends to encode message signal and demodulator decodes the message signal from the modulated waveform.
10. Scanning always takes place from ________
a) right to left
b) left to right
c) left to right for odd fields and right to left for even fields
d) left to right for even fields and right to left for odd fields
Answer: b
Explanation: Scanning is mostly used in TV to obtain continuous frames and to provide motion of picture. Scanning always takes place from left to right.
Detection of AM Waves
1. Detection is same as ________
a) Modulation
b) Mixing
c) Filtering
d) Demodulation
Answer: d
Explanation: Process of extracting the original message signal from the received modulated signal is called demodulation. Demodulation is also known as detection.
2. Most commonly used modulation system for telegraphy is ________
a) Multi-tone modulation
b) Single tone modulation
c) PCM
d) FSK
Answer: d
Explanation: FSK (Frequency Shift Keying) is a method to transmit digital signals, which are in form of binary data. In telegraphy, a modem is used to convert binary data to FSK and vice-versa, for communication between computers, over telephone lines. Single tone modulations deals with message signal having single frequency component, whereas, Multi-tone modulation deals with message signal having multiple frequency components.
3. What do you understand by the term “carrier”?
a) waveform with constant frequency, phase and amplitude
b) waveform for which frequency, amplitude or phase is varied
c) waveform with high amplitude, low frequency and constant phase
d) waveform to be transmitted
Answer: b
Explanation: A carrier wave is modulated in terms of amplitude, frequency or phase, with respect to an input signal for conveying information. Carrier wave has generally higher frequency than the input signal.
4. According to Fourier analysis, square wave can be represented as ________
a) fundamental sine wave and even harmonics
b) fundamental sine wave and odd harmonics
c) fundamental sine wave and harmonics
d) fundamental and sub harmonic sine wave
Answer: b
Explanation: A square wave is a non-sinusoidal periodic waveform with steady frequency, in which the amplitude alternates between a fixed maximum and minimum value. According to Fourier analysis, an ideal square wave, with amplitude one, can be represented as an infinite sum of sinusoidal waves. An ideal square wave contains only components of odd integer harmonics.
5. PDM is generated by ________
a) Bistable multivibrator
b) Monostable multivibrator
c) Astable multivibrator
d) Schmitt trigger
Answer: b
Explanation: PDM (Pulse Duration Modulation) is used to produce an analog signal with the help of a digital source. Monostable multivibrators are used to generate a single output pulse of a particular width. Its output is either high or low when a suitable external pulse is applied.
6. Mass of any moving system is represented by ________
a) a resistance
b) a conductance
c) a resistivity
d) an inductance
Answer: d
Explanation: Inductance can also be seen as electromagnetic inertia, a property that opposes changes in electric currents and magnetic fields. Mass has inertia so it is represented by an inductance.
7. Contrast function in TV is done by ________
a) Luminance amplifier
b) Schmitt trigger
c) Band Pass Filter
d) Chroma amplifier
Answer: a
Explanation: Luminance amplifier is mostly used by the Contrast Function, in TV sets. It helps to changes the contrast or picture quality of television with respect to the received visual signals, by making the contrast optimum.
8. Bandwidth of RF amplifier for a color TV receiver is ________
a) equal to channel width
b) more than channel width
c) less than channel width
d) twice of channel width
Answer: b
Explanation: RF amplifiers are tuned amplifiers in which the frequency of operation is controlled by a tuned circuit. Bandwidth of RF amplifier is generally kept a little more than channel so that there is no problem in reception.
9. FSK system involves ________
a) frequency modulation
b) pulse modulation
c) amplitude modulation
d) phase modulation
Answer: a
Explanation: FSK (Frequency Shift Keying) is a method to transmit digital signals, which are in form of binary data. FSK i.e. Frequency Shift Keying involves frequency modulation that assigns bit values to discrete frequency while ASK assigns bit values to discrete amplitude.
10. Diameter of antenna is doubled. The maximum range will ________
a) be doubled
b) be halved
c) become four times
d) decrease to one fourth
Answer: a
Explanation: Range of antenna is directly proportional to antenna diameter. So if diameter of antenna is doubled its range also doubled.
Problems of AM
1. If the modulating frequency of a carrier wave varies between 700Hz and 7KHz, find it’s bandwidth?
a) 10 KHz
b) 23 KHz
c) 17.3 KHz
d) 12.6 KHz
Answer: d
Explanation: Modulating Frequency (fm) = Vmax-Vmin, where,
Vmax = Maximum Amplitude of an amplitude modulated,
Vmin = Minimum amplitude of an amplitude modulated,
fm = 7KHz – 700Hz = 6.3KHz
Bandwidth = 2fm = 2 x 6.3 = 12.6 KHz.
2. A 400W carrier wave is modulated to a depth of 65%. Find the total power of modulated wave?
a) 512.5W
b) 493W
c) 484.5W
d) 609.6W
Answer: c
Explanation: Total power, Pt = Pc (1 + µ2⁄2), where Pc = Carrier Power = 400W
where Modulation Index (µ) = 0.65,
so Pt = 400 (1 +0.652⁄2). On solving it we get Pt = 484.5W.
3. Inductance and capacitance of a line is 0.8 μH⁄m and 32 pF⁄m. Find Z0?
a) 158
b) 166
c) 143
d) 127
Answer: a
Explanation: The characteristic impedance or surge impedance (Z0) of a uniform transmission line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; which is, a wave travelling in one direction in the absence of reflections in the other direction.
.
4. Find VSWR of a line having maximum and minimum value equals to 120mV and 40mV respectively?
a) 3
b) 2
c) 1
d) 4
Answer: a
Explanation: The voltage standing wave ratio (VSWR) is defined as the ratio of the maximum to minimum voltage on a loss-less line. In this way, the VSWR is measured at a particular point and the voltage maxima and minima do not need to be determined along the length of the line.
5. A 1000 KHz carrier is modulated with 300 Hz, 8000 Hz and 2 KHz waves. Determine the frequencies whose chances of occurring in output is least?
a) 1000 KHz
b) 1002 KHz
c) 998 KHz
d) 999.2 KHz
Answer: a
Explanation: Frequencies present in output are of the form fc ± fm, fc ± 2fm, fc ± 3fm. And 1000 KHz is a multiple of none in the range. Whereas, rest options are one of the multiples in the range.
6. If average power of a transmitter is 4kW and maximum power is 20000KW. What is its duty cycle?
a) 2 x 10-3
b) 2 x 10-4
c) 3 x 10-3
d) 0.05
Answer: b
Explanation: Duty cycle is the proportion of time during which a component, device, or system is operated. The duty cycle can be expressed as a ratio or as a percentage.
7. The voltage and electric current of a line are respectively 5kV and 1000μA. Find its power?
a) 12W
b) 13W
c) 5W
d) 10W
Answer: c
Explanation: Electrical power is the rate at which electrical energy is converted to another form, such as motion, heat, or an electromagnetic field.
Power, P = V x I. By substituting the values of voltage(V) and current(I), we have P = (5×103)(1000×10-6) = 5W.
8. ________ microphones works on piezoelectric effect.
a) Crystal
b) Carbon
c) Moving coil
d) Condenser
Answer: a
Explanation: Piezoelectric effect is the ability of materials to generate an electric charge in response to applied mechanical stress. Crystals which demonstrate the piezoelectric effect produce voltages when they are deformed. The crystal microphone uses a thin strip of piezoelectric material attached to a diaphragm. The two sides of the crystal acquire opposite charges when the crystal is deflected by the diaphragm.
9. If modulation index of an AM wave is increased from 1.5 to 2, then the transitted power ________
a) remains same
b) increases by 20%
c) increases by 41%
d) increases by 50%
Answer: c
Explanation: When m=1.5, transmitted power
.
= 0.41 x 100
= 41%
Therefore, there is an increase in total power by 41%.
FDM System
1. Frequency difference between WDM (Wavelength Division Multiplexing) and FDM (Frequency Division Multiplexing) is ________
a) 1
b) 0
c) slightly greater than 1
d) infinity
Answer: d
Explanation: WDM (Wave Division Multiplexing) is partly similar to FDM (Frequency Division Multiplexing). FDM is used for signal transmission through a twisted-pair cable while WDM is used to transmit signals through optical fiber cables. Frequency difference between WDM and FDM is infinity.
2. FDM stands for ________
a) Frequency Density Multiplexing
b) Frequency Difference Multiplexing
c) Frequency Division Multiplexing
d) Frequency Data Manager
Answer: c
Explanation: FDM stands for Frequency Division Multiplexing. The total bandwidth is divided into a series of non-overlapping frequency bands and each band is used to carry a separate signal.
3. TDM stands for ________
a) Time Division Multiplexing
b) Time Difference Multiplexing
c) Time Duration Multiplexing
d) Time Data Manager
Answer: a
Explanation: TDM stands for Time Division Multiplexing. It is used to transmit and receive independent signals over a common signal path with the help of synchronized switches.
4. TDM and FDM are used to multiplex multiple signals into a single carrier.
a) False
b) True
Answer: b
Explanation: Multiplexing is the method of combining two or more than two signals into one, in such a way that each individual signal can be retrieved at the receiver. Both TDM and FDM are used to multiplex different signals into a single carrier. However, TDM is used to transmit and receive independent signals over a common signal path with the help of synchronized switches. Whereas, in case of FDM, the total bandwidth is divided into a series of non-overlapping frequency bands and each band is used to carry a separate signal.
5. What is Synchronous TDM?
a) gives same amount of time to each device
b) gives same amount of frequency to each device
c) gives variable time to each device
d) gives variable frequency to each device
Answer: a
Explanation: TDM is a method of putting multiple data streams in a single signal by separating the signal into many segments, each having a very short duration. Each individual data stream is reassembled at the receiving end based on the timing. Synchronous TDM gives exactly the same amount of time to each connected device. It allocates time to every device even if a device has nothing to transmit.
6. FDM is an analog multiplexing technique used to combines ________
a) analog signals
b) digital signals
c) both analog and digital signals
d) alternatively passes analog and digital signals
Answer: a
Explanation: Frequency Division Multiplexing (FDM) is a networking technique in which multiple data signals are combined for simultaneous transmission via a shared communication medium. FDM uses a carrier signal at a discrete frequency for each data stream and then combines many modulated signals. FDM is used to multiplex multiple analog signals. FDM is applied when the bandwidth of the link is greater than the combined bandwidth of signals to be transmitted.
7. Frequency Division Multiplexing (FDM) is based on orthogonality.
a) True
b) False
Answer: a
Explanation: Frequency Division Multiplexing (FDM) is a networking technique in which multiple data signals are combined for simultaneous transmission via a shared communication medium. FDM uses a carrier signal at a discrete frequency for each data stream and then combines many modulated signals. Orthogonal frequency division multiplexing is used to encode multiple carrier frequencies. It is mainly used for latest wireless telecommunications.
8. Which multiplexing technique transmits digital signals?
a) FDM
b) TDM
c) WDM
d) Both FDM and TDM
Answer: b
Explanation: Time Division Multiplexing is used to transmit digital signals. FDM and WDM techniques are used to transfer analog signals. WDM (Wave Division Multiplexing) is partly similar to FDM (Frequency Division Multiplexing). FDM is used for signal transmission through a twisted-pair cable while WDM is used to transmit signals through optical fiber cables.
9. To get constant time delay, we should use ________
a) FDM technique
b) WDM technique
c) Synchronous TDM
d) Non synchronous TDM
Answer: c
Explanation: TDM is a method of allowing multiple data streams in a single signal by separating the signal into many segments, each having a very short duration. Each individual data stream is reassembled at the receiving end based on the timing. Synchronous TDM gives exactly the same amount of time to each connected device. It allocates time to every device even if a device has nothing to transmit. Thus, Synchronous Time Division Multiplexing is used to provide constant time delay. It can also provide constant bandwidth.
10. Time Division Multiplexing (TDM) is based on orthogonality.
a) True
b) False
Answer: a
Explanation: Both Frequency Division Multiplexing (FDM) and Time Division Multiplexing (TDM) are based on orthogonal multiplexing. Orthogonal multiplexing is a method of encoding digital data on multiple carrier frequencies.
11. As distance increases ________
a) Packet size also increases
b) Packet size decreases
c) Packet size becomes twice of its previous value
d) Packet size becomes half
Answer: a
Explanation: Distance is directly proportional to packet size. So as distance increases packet size also increases.
12. If 32 equally probable events exists and we have to select one of them, number of required bits is?
a) 2
b) 4
c) 8
d) 5
Answer: d
Explanation: 25 = 32, so number of bits required is 5.