Latest Amplitude Modulation ( Analog Communications ) MCQs – Analog Communications MCQs
Latest Analog Communications MCQs
By practicing these MCQs of Amplitude Modulation ( Analog Communications ) MCQs – Latest Competitive MCQs , an individual for exams performs better than before. This post comprising of objective questions and answers related to “ Amplitude Modulation ( Analog Communications ) Mcqs “. As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Analog Communications to approach the success. Tab this page to check ” Amplitude Modulation ( Analog Communications )” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others.
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:
Methods of Reducing Distortion
1. Quantization error occurs in ________
a) Time Division Multiplexing
b) Frequency Division Multiplexing
c) Pulse Code Modulation
d) Pulse Width Modulation
Answer: c
Explanation: The difference between an input value and its quantized value is called quantization error or we can also see it as that when an analog number is not converted into an exact digital number. It takes place in pulse code modulation (PCM). Pulse Code Modulation is a technique in which the amplitude of an analog signal is converted to a binary value represented as a series of pulses.
2. If power transmitted is 90kW, what is the field at a distance of 45km?
a) 2
b) 1
c) 0.036
d) 0.01
Answer: c
Explanation: Electric Field a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
3. Find the number of pulses, if the number of quantization levels is 32 in PCM?
a) 3
b) 6
c) 4
d) 5
Answer: d
Explanation: Quantization is a process in which the continuous range of values of an analog signal is sampled and divided into non-overlapping (but not necessarily equal) subranges, and a discrete, unique value is assigned to each sub-range.
Therefore, number of pulse in quantisation level = 2n = 32, so n = 5, where n = Quantization Level.
4. Which transistor amplifier is most commonly used?
a) Common Base
b) Common Emitter
c) Common Collector
d) None of these
Answer: b
Explanation: Common Emitter is most widely used because it has high power gain and voltage gain. In this method of connection, small changes in base/emitter current cause large changes in collector/emitter current. Therefore, the circuit is that of a current amplifier. Moreover, it also provides maximum transconductance for a given load.
5. A device uses a 8 bit word then the maximum number of words it can transmit is ________
a) 64
b) 256
c) 1024
d) 512
Answer: b
Explanation: Maximum number of words which an n-bit device can transmit = 2n
Therefore, when n = 8,
Number of words it can transmit is 28 = 256.
6. Peak voltage of a carrier is 8kV, each sideband has an amplitude of 800V. Find its modulation index?
a) 0.28
b) 0.09
c) 0.02
d) 0.08
Answer: c
Explanation: (mPc)/2 = Psb, where m = Modulation Index, Pc = Carrier Power and,
Psb = Power of each sideband
Therefore, (mVc 2)/2 = Vsb 22 (Vc = 8000 V, and Vsb = 800)
Therefore, m = (8002 X 2) / 80002 = 0.02.
7. What do you understand by the term Cross talk?
a) generation of closely lying sidebands
b) a method of increasing bandwidth
c) a method of decreasing bandwidth by converting DSB-SC signal to SSB-SC
d) disturbance caused in nearby circuit due to transmitted signal
Answer: d
Explanation: Crosstalk is a disturbance caused by the electric or magnetic fields of one telecommunication signal affecting a signal in an adjacent circuit. It is generally caused by undesired coupling. In a telephone circuit, crosstalk can result in the hearing part of a voice conversation from another circuit.
8. AM waves is represented by Et cos ωc t. What does Et denotes?
a) envelope
b) carrier signal
c) modulating signal
d) amplitude
Answer: a
Explanation: AM wave is represented by (A + xt) cosωc t or Et cosωc t, where xt is modulating signal, A is amplitude and therefore, Et is the envelope of the AM wave.
9. Wireless communication takes place through ________
a) EM waves
b) X-Rays
c) Optical Fibres
d) Infrared Waves
Answer: a
Explanation: Wireless communication takes place through electromagnetic waves. Waves are transmitted through an antenna.
10. Which modulation technique uses minimum bandwidth?
a) DSB-SC
b) SSB-SC
c) FM
d) VSB
Answer: b
Explanation: SSB-SC modulation technique uses minimum bandwidth as it suppresses the carrier and transmits either of the two side-bands while DSB-SC suppresses the carrier and transmits both the side-bands.
Power Calculation
1. If peak voltage of a carrier wave is 10V, what is the peak voltage of modulating signal if modulation index is 50%?
a) 10V
b) 20V
c) 8V
d) 5V
Answer: d
Explanation: From the relation, Modulation Index (µ) = Vm/Vc = 50% = 0.5,
where Vm = Peak voltage of modulating signal,
Vc = Peak voltage of a carrier wave = 10V,
Therefore, Vm = 10 X 0.5 = 5V.
2. Maximum Amplitude of an amplitude modulated 10V and minimum amplitude is 5V. Find its modulation index?
a) 0.65
b) 0.9
c) 0.33
d) 1
Answer: c
Explanation: We know, Modulation Index(µ) = (Vmax-Vmin)/(Vmax+Vmin),
Where, Vmax = Maximum Amplitude of an amplitude modulated = 10V
Vmin = Minimum amplitude of an amplitude modulated = 5V
Therefore, µ = (10-5)/(10+5) = 0.33.
3. 24 channels, each band limited to 3.4 KHz, are to be time division multiplexed. Find the bandwidth required for 128 quantization level? (Given that sampling frequency is 8 KHz)
a) 2436 KHz
b) 1002 KHz
c) 1536 KHz
d) 1337 KHz
Answer: c
Explanation: N = 24, fm = 3.4 kHz
m = 128,
2n = m = 128, n = 7
But fs = 2fm, where, fs = sampling frequency
instead at 2fm 2 x 3.4 kHz 6.8 KHz.
B.W. = N(n+1)X fs = [24(7 + 1)] 8 kHz = 1536 KHz.
4. Sampling frequency of a signal is 6 KHz and is quantized using 7 bit quantizer. Find its bit rate?
a) 48kbPs
b) 64kbPs
c) 16kbPs
d) 8kbPs
Answer: a
Explanation: Bit rate refers to the rate at which data is processed or transferred. It is usually measured in seconds, ranging from bps for smaller values to kbps and mbps.
Bit rate is also known as bitrate or data rate.
Bit rate, Rb = 1⁄Tb where 
where n = number of bits and fs = Sampling Frequency
Tb = 1/42, therefore Bit rate = 42 Kbps.
5. Calculate power in each sideband, if power of carrier wave is 96W and there is 40% modulation in amplitude modulated signal?
a) 11.84W
b) 6.84W
c) 3.84W
d) 15.84W
Answer: c
Explanation: Modulation index = 0.4 and Pc = 96W. Power in sidebands may be calculated as
6. For 50% modulation, power in each sideband is ________ of that of carrier.
a) 10%
b) 4.32%
c) 5%
d) 6.25%
Answer: d
Explanation: Modulation index = 0.5. Power in sidebands may be calculated as 
7. For 100% modulation, total power is ________
a) 1.5Pc
b) 2Pc
c) 3.75Pc
d) 1.25Pc
Answer: a
Explanation: Total power, Pt = Pc (1 + µ2⁄2), where Pc = Carrier Power
where µ = 1 (for 100% modulation),
so Pt = Pc(1 +(12/2)). On solving it we get Pt = 1.5Pc.
8. If each element of signal occupies 70ms, what will its speed?
a) 11.23 bauds
b) 14.28 bauds
c) 17.39 bauds
d) 13.33 bauds
Answer: b
Explanation: The carrier signal is characterized by the number of signal intervals, or pulses, that are transmitted per second. Each pulse is called a baud. Bps stands for bits per second. Bps is a measure of how many bits can be transmitted during one pulse (one baud).
9. Power of carrier wave is 300W and modulation index is 0.75. Find its total power?
a) 465W
b) 384W
c) 323W
d) 502W
Answer: b
Explanation: Total power, Pt = Pc (1 + µ2⁄2), where Pc = Carrier Power = 300W
where Modulation Index (µ) = 0.75,
So Pt, 
10. If a wave is modulated by two waves. One of them has modulation index equal to 0.75 and other has 0.2, the total modulation index will be ________
a) 0.67
b) 0.58
c) 0.77
d) 0.35
Answer: c
Explanation: Given that m1 = 0.75 and m2 = 0.2. Total modulation index will be equal to 
By substituting values we have 
which is equal to 0.77.
11. Find the power saving for DSB-SC wave with 100% modulation?
a) 66%
b) 86%
c) 50%
d) 33%
Answer: a
Explanation: In DSB-SC carrier is suppressed.
So Total Power required in DSBSC Modulation = (u2XPc)/2 = Pc/2
In normal AM, carrier is not suppressed.
So total power required in AM Modulation = (1+(u2/2))XPc = 3Pc/2
Therefore, Power saving = ((Pc/2)/(3Pc/2)) x 100% = 66%.
12. If power transmitted is 45kW, field at a distance of 23km will be ________
a) 0.02
b) 0.75
c) 0.05
d) 0.03
Answer: c
Explanation: Field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
13. Find the number of pulses, if the number of level is 128 in PCM?
a) 3
b) 6
c) 4
d) 7
Answer: d
Explanation: A pulse in signal processing is a rapid, transient change in the amplitude of a signal from a baseline value to a higher or lower value, followed by a rapid return to the baseline value.
Therefore,
2n = 128, so n = 7.
Spectrum of Multitone Signal
1. Video signals are transmitted through ________
a) Frequency Modulation
b) Amplitude Modulation
c) Pulse Modulation
d) Either frequency or amplitude modulation
Answer: b
Explanation: Video signals require a large transmission bandwidth for transmission. So modulation of video signals is possible only by amplitude modulation. It is executed using Vestigial Sideband modulation, where either of the two sidebands is fully transmitted and the other sideband is partly transmitted.
2. What you understand by the term discone antenna?
a) combination of disc and cone
b) combination of disc and cone with a spacing of ʎ/4
c) combination of disc and cone with a spacing of ʎ/2
d) same as a simple antenna
Answer: a
Explanation: Discone antenna is a combination of disc and cone. A discone antenna is a version of a biconical antenna in which one of the cones is replaced by a disc. It is usually mounted vertically, with the disc at the top and the cone beneath. It is usually vertically mounted, with disc at top and the cone beneath.
3. A system has a voltage of 15V and it produces a total current of 1300μA. Find the power of system?
a) 0.025W
b) 0.035W
c) 0.0195W
d) 0.045W
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 = 15 x 1300 x 10-6 = 0.0195W.
4. An input resistance of 20 Kῼ causes a noise voltage of 10μV. Suppose two input resistance each of 20 Kῼ are connected in parallel, then find the total noise voltage?
a) 8μV
b) 10μV
c) 20μV
d) 13μV
Answer: a
Explanation: Thermal noise is distinct from shotnoise, which consists of additional current fluctuations that occur when a voltage is applied and a macroscopic current starts to flow. 
5. Modem is combination of both modulator and demodulator.
a) True
b) False
Answer: a
Explanation: A modem (modulator-demodulator) is used to modulate one or more message signals using one or more carrier waves on the transmitting side. It also demodulates the received encoded signal to get the original information signal.
6. Any signal and its Hilbert transform are mutually orthogonal.
a) True
b) False
Answer: a
Explanation: Hilbert Transform of any signal produces a signal which has same amplitude as that of the original signal but suffers a phase-shift of -90 degree. Consider f(t) be any signal and fh(t) is it’s Hilbert transform.
If we integrate both of them under proper boundary conditions, we get 0 which shows that both are mutually perpendicular i.e.
7. What Squelch circuit does?
a) helps to receive signals that are useful for the system
b) helps to make channel noise free
c) helps to receive two or more than two signals at once
d) helps to suppress the audio output in the absence of sufficiently strong desired signal
Answer: d
Explanation: A Squelch is a circuit which is a mean to impede, but not destroy a signal. It circuit blocks the output in the absence of sufficiently strong desired signal. It is widely used in two-way radios to suppress the annoying sounds.
8. Mostly used radio receivers are?
a) pulsed receivers
b) CW receivers
c) TRF receivers
d) super heterodyne receivers
Answer: d
Explanation: Superheterodyne Receivers are radio receivers that use frequency mixing to convert a received high frequency signal to a fixed lower intermediate frequency which can be processed more conveniently. It is also the most used receivers. It is used in a variety of applications from broadcast receivers to two-way radio communication links as well as many mobile radio communications systems.
9. Find the step size of a signal with peak to peak amplitude of 4V and is quantized into 64 levels?
a) 32.3 x 10-3
b) 30.23 x 10-3
c) 62.5 x 10-3
d) 53.3 x 10-3
Answer: c
Explanation: Step size is the voltage difference between one digital level (i.e. 0001) and the next one (i.e. 0010 or 0000). Quantization step size is the smallest possible difference in amplitude.
Therefore,
Step size, 
10. If the step size of a system is 0.0625, find its quantization power?
a) 3.25 x 10-3
b) 3.25 x 10-4
c) 3.25 x 10-2
d) 3.25 x 10-1
Answer: b
Explanation: Quantization noise is a model of quantization error introduced by quantization in the analog-to-digital conversion (ADC) in telecommunication systems and signal processing. It is a rounding error between the analog input voltage to the ADC and the output digitized value.
Therefore, 
Modulation Using Non-Linear Devices (Balanced Modulator)
1. The approximate aspect ratio for television is ________
a) 2 : 1
b) 3 : 2
c) 4 : 3
d) 3 : 4
Answer: c
Explanation: Aspect ratio is an image projection attribute that describes the proportional relationship between the width of an image and its height. The most common video-graphic aspect ratio is 4:3.
2. Signal to Noise ratio for a hi-fi system is equal to ________
a) 50 dB
b) 100 dB
c) 75 dB
d) 10 dB
Answer: a
Explanation: Signal to Noise ratio compares signal power to noise power. It determines how much signal is present in a transmitted wave compared to presence of noise. It is generally expressed in decibels(dB). For a hi-fi system, it should be equal to 50dB.
3. What is the use of SSB?
a) It has lesser bandwidth
b) It has large bandwidth
c) It has infinite bandwidth
d) It has zero bandwidth
Answer: a
Explanation: In SSB-SC(Single Side Band Suppressed Carrier), the carrier is suppressed and either of the upper side-band and lower-sideband, are transmitted. This reduces it’s bandwidth to the frequency of the message signal.
4. Approximate how many times odd and even fields of television system are scanned?
a) 25 times each
b) 50 times each
c) 50 times each but alternately
d) 25 times each but alternately
Answer: d
Explanation: In television system in India, each of odd and even fields are scanned 25 times but alternately.
5. Any signal and its Hilbert transform have same energy density spectrum.
a) True
b) False
Answer: a
Explanation: Hilbert transform of a signal results in a signal with same amplitude but a phase shift of -90 degrees. Thus, a signal and it’s Hilbert transform have same amplitude spectrum i.e. they have same energy density spectrum and also correlation spectrum.
6. If the spectrum of a narrow band noise is symmetrical and it has a power density spectrum 4×10-6. Power density of quadrature component is ________
a) 2 x 10-6
b) 5 x 10-6
c) 3 x 10-6
d) 4 x 10-6
Answer: a
Explanation: Power density is the amount of power per unit volume. Quadrature component is a amplitude modulated sinusoid which is offset in phase by one quarter cycle.
Power density of quadrature component is 
7. What is the capacitive reactance for DC signals?
a) zero
b) very low
c) equal to one
d) infinite
Answer: d
Explanation: Frequency of DC signals is zero. For zero frequency, capacitive reactance XC is infinite as XC = 1/(f*C), where f is the frequency and C is the capacitance.
8. If a FM signal having modulation index mf is passed through a frequency tripler, then the modulation index of output of frequency tripler is ________
a) mf
b) 3mf
c) 1⁄3 mf
d) 1⁄9 mf
Answer: b
Explanation: A frequency tripler is a frequency multiplier in which an electronic circuit generates an output signal whose output frequency is a harmonic (multiple) of its input frequency. When a FM signal is passed through a frequency tripler, it increases its modulation index 3 times. So the modulation index of output is 3mf.
9. What are the two major drawbacks of delta modulation?
a) Slope Overload and Granular noise
b) Slope Overload and Serration noise
c) Serration noise and Granular noise
d) Slope Overload and Channel Noise
Answer: a
Explanation: Delta modulation is an analog-to-digital and digital-to-analog signal conversion technique used for transmission of voice information where quality is not of primary importance. It is the simplest form of differential pulse-code modulation (DPCM) where the difference between successive samples are encoded into n-bit data streams. The two major drawbacks in delta modulation are slope overload and granular noise. When step size becomes too small to track the original waveform it generates an error known as slope overload and granularity occurs when step size is too large.
10. Beam width of antenna is expressed in ________
a) metres
b) degrees
c) radian
d) volt
Answer: b
Explanation: Angular separation between the two half-power (-3dB) points is known as beam width. It is usually expressed in degrees.
Generation of AM Using Amplifiers
1. Consider a wave, v = 15 sin (3πt + 5sin1300t), what is the carrier and signal frequency?
a) 2.5Hz and 200Hz
b) 1.5Hz and 100Hz
c) 1.5Hz and 207Hz
d) 5.5Hz and 500Hz
Answer: c
Explanation: Comparing with the general FM equation,
FM(t) = Ac sin(2πfct + βsin(2πfmt)),
Therefore,
Signal and Carrier frequency are respectively 
2. SWR is 1 for ________
a) Open or Short Circuit
b) Open circuit only
c) Short circuit only
d) Series or Parallel Circuit
Answer: a
Explanation: Standing Wave Ratio (SWR) is defined as the ratio of the maximum radio-frequency (RF) voltage to the minimum RF voltage along the line. If the line is open or short circuited, SWR is always 1.
3. The resonance frequency of an amplifier is 7MHz and it is having a bandwidth of 10KHz. What is its Q factor?
a) 7000
b) 70
c) 700
d) 7
Answer: c
Explanation: Q factor = Frequency/Bandwidth
= 7MHz/10KHz = 700.
4. Which chart is used for calculations of transmission lines?
a) Andre chart
b) Smith chart
c) Wilson chart
d) Federer chart
Answer: b
Explanation: Smith chart is highly used in radio frequency engineering. Because it is highly useful in solving problems related with transmission lines and matching circuits. It is highly useful for displaying multiple parameters like impedance, admittance etc simultaneously.
5. What is the effective power if the transmitted power of 5W is increased to 50dB?
a) 100KW
b) 300KW
c) 500KW
d) 700KW
Answer: c
Explanation: According to given problem, 10log10 (P/Pt) = 50,
10log10 (P/5) = 50.
So, P = 500KW.
Where, P = Effective Power, Pt = Transmitted Power.
6. Field intensity follows ________
a) Bragg’s law
b) Inverse square law
c) Coulomb’s law
d) Gauss’s law
Answer: b
Explanation: Strength of an electric field due to a charge Q is inversely proportional to the square of the distance from the source. In electrostatics, inverse square law is also known as Coulomb’s Law.
7. If the power density at a given distance ‘R’ is one microwatt per square meter and the effective area of antenna is one square meter then the power captured by antenna is one microwatt.
a) True
b) False
Answer: a
Explanation: The above statement is true because power density is equal to 
where,
Power Density = 1 mW
Pt = Power captured by antenna,
Area of antenna = 4πR2 = 1m2.
Elements Of Communication Systems MCQs
8. The gain of antenna (G) is the ratio of power radiated in a particular direction to the power radiated from an isotropic antenna.
a) True
b) False
Answer: a
Explanation: Gain of antenna,
9. EM waves are attenuated as they travel.
a) True
b) False
Answer: a
Explanation: Electromagnetic radiation can travel through many forms of medium. They travel at the speed of light in air or free space, which are ideal media. However conductive media like metals form a barrier through which they do not travel. There are also some media through which when they travel, get attenuated. EM waves are attenuated as they travel because both field and power decrease as distance increases.
Envelope Detector
1. Envelope Detector is a/an ________
a) Coherent detector
b) Asynchronous Detector
c) Synchronous Detector
d) Product Demodulator
Answer: b
Explanation: An envelope detector is used to demodulate a previously modulated signal by removing all high frequency components of the signal. The capacitor and resistor form a low-pass filter to filter out the carrier frequency. Envelope detectors are asynchronous in nature. The advantage of asynchronous over synchronous is that it is simple, cheap and setup is faster.
2. Effective radiated power (ERP) is equal to ________
a) Pt×Gt×Pd
b) Pt×Pd
c) Gt×Pd
d) Pt×Gt
Answer: d
Explanation: Effective Radiated Power (ERP) measures the combination of the power emitted by the transmitter and the ability of the antenna to direct that power in a given direction. Thus, Effective Radiated Power is equal to the product of antenna gain (Gt) and power transmitted (Pt).
3. Power density from an isotropic antenna is equal to ________
a) Pt4πR2
b) Pc4πR2
c) Pt2πR2
d) Pt4πR
Answer: a
Explanation: Power of a transmitter that is radiated from an isotropic antenna will have a uniform power density in all direction. An isotropic antenna is an ideal antenna that radiates its power uniformly in all directions. There is no actual physical isotropic antenna. However, an isotropic antenna is often used as a reference antenna for the antenna gain. The power density at any distance ‘R’ for an isotropic antenna is equal to Pt4πR2.
4. Compact discs mainly use ________
a) optical recording
b) magnetic recording
c) magnetic retrieval
d) both optical and magnetic recording
Answer: a
Explanation: Compact disk (CD) is mainly used for storage of data. It was developed only for store and play only audio but was later adapted for storage of data also. It mainly used optical recording and retrieval.
5. If the distance between antenna and source doubles, the received signal power decreases by 1⁄4.
a) True
b) False
Answer: a
Explanation: It is due to the R2 term in formula, 
. Thus, the received signal power is inversely proportional to the distance between antenna and source.
6. Power density follows inverse square law.
a) True
b) False
Answer: a
Explanation: From the given formula, With increase in distance, power density decreases. So, we can also say that power density is inversely related to distance. Thus, Power density follows inverse square law.
7. There is no requirement of IF amplifier stages for a video monitor.
a) True
b) False
Answer: a
Explanation: IF amplifiers are used to change the frequency levels in circuits that are too selective, difficult to tune, and unstable. However, Video amplifier is used to amplify video signals before passing them to a Cathode ray tube. And there is no need of IF amplifier stages for a video monitor.
8. Helical antenna is circularly polarized.
a) True
b) False
Answer: a
Explanation: A helical antenna is an antenna consisting of one or more conducting wires (monofilar, bifilar, or quadrifilar with 1, 2, or 4 wires respectively) wound in the form of a helix. In helical antenna, polarization is equally divided between vertical and horizontal components and thus it is circularly polarized.
9. The intermediate frequency of a superheterodyne receiver is 500KHz. What is the image frequency at 1200KHz?
a) 600KHz
b) 500KHz
c) 700KHz
d) 200KHz
Answer: c
Explanation: The image frequency is an undesired input frequency which is demodulated by superheterodyne receivers along with the desired incoming signal. This results in two stations being received at the same time, thus producing interference. In the given problem, Image frequency is equal to (1200 – 500) which is equal to 700KHz.
10. Which of these amplifiers is used for impedance matching?
a) Common Base
b) Common Emitter
c) Common Collector
d) Common Base & Emitter
Answer: c
Explanation: An amplifier is an electronic circuit that is used to amplify the voltage signal or a current signal. Output impedance of common collector amplifier is low and thus it is suitable for impedance matching.
Linear Modulation
1. Discone antenna is mainly used in UHF range.
a) True
b) False
Answer: a
Explanation: A discone antenna is a version of a biconical antenna in which one of the cones is replaced by a disc. It is usually mounted vertically, with the disc at the top and the cone beneath. Discone antenna is a combination of disc and cone. Discone antenna is omnidirectional so it is mainly used in UHF range. It is especially used in airports.
2. Squelch circuit is normally inserted in receiver ________
a) after detector
b) before detector
c) before mixer
d) after power amplifier
Answer: a
Explanation: Squelch is a circuit that acts to suppress the audio (or video) output of a receiver in the absence of a sufficiently strong desired input signal. Squelch is added to suppress noise. It is added after detector when there is no carrier.
3. Each frequency gives rise to four side bands.
a) True
b) False
Answer: b
Explanation: Each frequency in communication gives rise to two side bands, Upper Sideband and Lower Sideband.
4. Field strength (E) is directly proportional to the square of transmission power.
a) True
b) False
Answer: b
Explanation: Field Strength is the intensity of an electric or magnetic field. Actually the Field strength (E) is directly proportional to the square root of transmission power i.e. E ∝ √Pt.
5. If the transmitted power is 100KW then the field at a distance ‘R’ is 60mV/m. Suppose if the transmitted power is reduced to 50KW then the field at same distance ‘R’ will be equal to ________
a) 50mV/m
b) 42mV/m
c) 45mV/m
d) 55mV/m
Answer: b
Explanation: Since, Field strength (E) is directly proportional to the square root of transmission power i.e. E ∝ √Pt. Therefore 
which gives x = 42mV/m
6. If the value of resistor becomes 16 times than its previous value then its noise voltage will become ________
a) 16 times
b) 8 times
c) 4 times
d) 2 times
Answer: c
Explanation: Since the noise voltage is directly proportional to square root of resistance so if the value of resistor increased to 16 times then its noise voltage will become 4 times.
7. What is the modulation index for a single tone modulation, given that positive peak of AM wave is 20V and minimum value is 2V?
a) 0.81
b) 0.91
c) 0.73
d) 1
Answer: a
Explanation: Modulation Index (u) = (Vmax – Vmin)/(Vmax + Vmin)
= (20-2)/(20+2) = 9/11 = 0.81.
8. TV remote control used ultra violet light.
a) True
b) False
Answer: b
Explanation: The infrared (IR) and the ultraviolet(UV) rays represent the two extremities of the visible spectrum (400-700nm). While IR represents electromagnetic radiation with wavelengths longer than those of visible light, UV represents wavelengths shorter than visible light. Each time we press any button of TV remote, it sends a beam of infrared rays. For different commands, it sends out different beams that are detected by a microchip lying inside TV.
9. How capacitance is related to thickness?
a) it is inversely proportional to thickness
b) it is directly proportional to thickness
c) it is inversely proportional to twice of thickness
d) it is directly proportional to twice of thickness
Answer: a
Explanation: The thickness of capacitor affects the value of capacitance because capacitance is dependent on structure and distance between two plates.
Thus, Capacitance, C = A×∈⁄t so it is inversely proportional to thickness.
10. Noise gets mixed with signal at ________
a) receiver
b) transmitter
c) transducer
d) channel
Answer: d
Explanation: Channel acts as a path for taking signals to the receiver. It is a medium of transmission of data from source to destination. There is a high probability of involving noise at the channel.
Average Power For Sinusoidal AM
1. A 1000KHz carrier is modulated by 100Hz and 200Hz waves, then which of the following frequencies cannot be present at the output?
a) 1000.1KHz
b) 999.9KHz
c) 1000.2KHz
d) 999.7KHz
Answer: d
Explanation: Frequencies present at output will be 1000 ± 0.1 KHz = 1000.1KHZ and 999.9KHz and 1000 ± 0.2 KHz = 1000.2KHZ and 999.8KHz. Therefore, Only, 999.7KHz is out of the range.
2. Aspect ratio is the ratio of ________
a) width to height
b) height to width
c) power to width
d) width to power
Answer: a
Explanation: Aspect ratio, being one of the crucial attributes in image projection, describes the varying relationship between the width of the image and it’s height. Thus, Aspect ratio is the ratio of width to the height of image (screen).
3. If the levels in PCM are transmitted in 10 unit code having sampling frequency equals to 20KHz. What is its bandwidth?
a) 150KHz
b) 125KHz
c) 100KHz
d) 110KHz
Answer: c
Explanation: PCM bandwidth is half of Bit rate.
Bit Rate = 10 X 20 = 200
Bandwidth = 200/2 = 100 KHz.
4. Carrier frequency is given by the formula ________
a) 12πLC√
b) 12πL√
c) 12πC√
d) 1πLC√
Answer: a
Explanation: Carrier wave is a high frequency signal whose certain characteristics, like amplitude, phase and frequency, are varied with respect to the instantaneous amplitude of the baseband signal (message signal). This variation is called modulation. Carrier frequency is given by the formula, fc = 12πLC√ where L is inductance and C is Capacitance.
5. Among all amplifiers, Common Collector has highest power gain.
a) True
b) False
Answer: b
Explanation: An amplifier is an electronic device that increases the voltage, current, or power of a signal. Amplifiers are used in wireless communications and broadcasting, and in audio equipment of all kinds. It is common emitter which has highest power gain. Common emitter has a very high voltage gain and also has good amount of current, so its power gain is high.
6. Find the carrier frequency if the inductance and capacitance are 30μH and 10nF respectively?
a) 244KHz
b) 523KHz
c) 290KHZ
d) 300KHz
Answer: c
Explanation: Carrier frequency is a waveform (usually sinusoidal) that is modulated (modified) with an input signal for the purpose of conveying information. 
7. Helical antenna is used to obtain ________ polarization.
a) elliptical
b) circular
c) linear
d) perpendicular
Answer: b
Explanation: A helical antenna is an antenna made of one or more conducting wires wound in the form of a helix. It produces circular polarization, because of which it is highly used in radio telemetry.
8. What is number of possible outputs if there is 6 line digital input?
a) 64
b) 32
c) 16
d) 128
Answer: a
Explanation: Total possible outputs will be 2n = 26 which is equal to 64.
9. Push pull amplifier is a class B amplifier.
a) True
b) False
Answer: a
Explanation: A push pull amplifier is an amplifier which has an output stage that can drive a current in either direction through through the load. Push pull amplifier uses a pair of active devices that alternately supply current to a connected load and also absorb current from a connected load. It is a class B amplifier.
10. For removing unwanted signals we should use ________
a) mixing method
b) filter method
c) modulation method
d) detecting method
Answer: b
Explanation: Filteration method is used to remove unwanted components from the signal. It is mainly used to eliminate noise.