Data-Communication
Question 1 |
The period of a signal is 100ms. then the frequency of this signal in kilohertz is ____?
10 | |
10-1 | |
10-2
| |
10-3 |
Question 1 Explanation:
T=100ms
= (100)10-3 sec
=10-1 sec
F=1/T
F=1/10-1 Hz
F=1/(10-1 *103) KHz
F = 10-2 KHz
= (100)10-3 sec
=10-1 sec
F=1/T
F=1/10-1 Hz
F=1/(10-1 *103) KHz
F = 10-2 KHz
Question 2 |
Consider a noiseless channel with a bandwidth of 5000Hz transmitting a signal with two signal
levels. The maximum bit rate is
2500 bps | |
10000 bps | |
5000 bps | |
20000 bps |
Question 3 |
In which modulation discrete values of carrier frequencies is used to transmit binary data?
Phase Shift Keying | |
Amplitude Shift Keying | |
Frequency Shift Keying | |
Disk Shift Keying |
Question 3 Explanation:
Frequency-shift keying (FSK) is a frequency modulation scheme in which digital information is transmitted through discrete frequency changes of a carrier signal. The technology is used for communication systems such as telemetry, weather balloon radiosondes, caller ID, garage door openers, and low frequency radio transmission in the VLF and ELF bands. The simplest FSK is binary FSK (BFSK). BFSK uses a pair of discrete frequencies to transmit binary (0s and 1s) information. With this scheme, the "1" is called the mark frequency and the "0" is called the space frequency.
Question 4 |
What is the basis of KVL?
Conservation of charge | |
Conservation of energy | |
Conservation of power | |
All of the options |
Question 4 Explanation:
Kirchhoff’s Voltage Law
Kirchhoff’s Voltage Law (KVL) is Kirchhoff’s second law that deals with the conservation of energy around a closed circuit path.
Kirchhoff’s Voltage Law (KVL) is Kirchhoff’s second law that deals with the conservation of energy around a closed circuit path.
Question 5 |
Calculate the modulation percentage if the modulating signal is 8 V and carrier is of 12 V?
50 | |
67 | |
150 | |
33 |
Question 5 Explanation:
Modulation index(m) = Vm/Vc
= (8/12)*100
= 66.66.
= (8/12)*100
= 66.66.
Question 6 |
A special PCM system uses 32 channels of data, one whose purpose is an identification (ID) and synchronization. The sampling rate is 4 kHz. The word length is 5 bits. Find the serial data rate.
1280 kHz
| |
160 kHz | |
320 kHz | |
640 kHz |
Question 7 |
The resistance to be connected across terminal a, b for maximum power transfer to it is:
40Ω | |
5Ω | |
2.5Ω | |
10Ω |
Question 8 |
What is the transmission signal coding method for a T1 carrier called ?
Binary | |
NRZ | |
Bipolar | |
Manchester |
Question 8 Explanation:
The transmission signal coding method of T1 carrier is called Bipolar. Bipolar encoding is a type of return-to-zero (RZ) line code, where two nonzero values are used, so that the three values are +, −, and zero. Such a signal is called a duobinary signal.
Question 9 |
How much bandwidth is required to send 132 voice - grade channels by FDM on an international satellite system ?
500 MHz | |
10 MHz | |
1320 MHz | |
50 MHz |
Question 9 Explanation:
In telecommunications, frequency-division multiplexing (FDM) is a technique by which the total bandwidth available in a communication medium is divided into a series of non-overlapping frequency bands, each of which is used to carry a separate signal.
The most natural example of frequency-division multiplexing is radio and television broadcasting, in which multiple radio signals at different frequencies pass through the air at the same time.
Frequency Division Multiple Access(FDMA):
FDMA is simply another example of the familiar data and voice transmission technique called FDM. This technique is used to allocate small portions of a large bandwidth (500MHz for satellite transponders) to individual users. For instance, a telecommunications common carrier in a particular country like Brazil. It might want 132 voice grade channels for sending voice and analog coded data to various other countries. The bandwidth required on the current international satellite systems for this many channels is 10MHz. Because 1 transponder has a bandwidth of 500Mhz, it could accommodate 50 users, each requiring 132 channels. The Brazilian user might be allocated the frequency band between 5990 and 6000MHz for the uplink to the satellite and the corresponding downlink frequencies would be 3765 to 3775Mhz.
Frequency Division Multiple Access(FDMA):
FDMA is simply another example of the familiar data and voice transmission technique called FDM. This technique is used to allocate small portions of a large bandwidth (500MHz for satellite transponders) to individual users. For instance, a telecommunications common carrier in a particular country like Brazil. It might want 132 voice grade channels for sending voice and analog coded data to various other countries. The bandwidth required on the current international satellite systems for this many channels is 10MHz. Because 1 transponder has a bandwidth of 500Mhz, it could accommodate 50 users, each requiring 132 channels. The Brazilian user might be allocated the frequency band between 5990 and 6000MHz for the uplink to the satellite and the corresponding downlink frequencies would be 3765 to 3775Mhz.
Question 10 |
Which of the following is/are true w.r.t. applications of mobile computing ?
(A) Travelling of salesman
(B) Location awareness services
(A) Travelling of salesman
(B) Location awareness services
(A) true; (B) false. | |
Both (A) and (B) are true. | |
Both (A) and (B) are false. | |
(A) false; (B) true. |
Question 10 Explanation:
Mobile computing main applications:
1.Portability
2. Connectivity
3.Social Interactivity
4. Individuality
5. IMobility
Travelling of salesman applicable to mobility of application.
Location awareness services applicable to Connectivity.
1.Portability
2. Connectivity
3.Social Interactivity
4. Individuality
5. IMobility
Travelling of salesman applicable to mobility of application.
Location awareness services applicable to Connectivity.
Question 11 |
ECL is the fastest of all logic families. High speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into ____.
Race condition | |
Saturation | |
Delay | |
High impedance |
Question 11 Explanation:
→ ECL is the fastest of all logic families. High speed in ECL is possible because transistors are used in difference amplifier configuration, in which they are never driven into Saturation.
→ Emitter-coupled logic (ECL) is a high-speed integrated circuit bipolar transistor logic family. ECL uses an overdriven BJT differential amplifier with single-ended input and limited emitter current to avoid the saturated (fully on) region of operation and its slow turn-off behavior.
→ As the current is steered between two legs of an emitter-coupled pair, ECL is sometimes called current-steering logic (CSL), current-mode logic (CML) or current-switch emitter-follower (CSEF) logic.[
→ Emitter-coupled logic (ECL) is a high-speed integrated circuit bipolar transistor logic family. ECL uses an overdriven BJT differential amplifier with single-ended input and limited emitter current to avoid the saturated (fully on) region of operation and its slow turn-off behavior.
→ As the current is steered between two legs of an emitter-coupled pair, ECL is sometimes called current-steering logic (CSL), current-mode logic (CML) or current-switch emitter-follower (CSEF) logic.[
Question 12 |
A data cube C, has n dimensions, and each dimension has exactly p distinct values in the base cuboid. Assume that there are no concept hierarchies associated with the dimensions. What is the maximum number of cells possible in the data cube, C?
p n | |
p | |
(2 n - 1)p+1 | |
(p + 1) n |
Question 12 Explanation:
Option-A: This is the maximum number of distinct tuples that you can form with p distinct values per dimensions.
Option-B: You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.
Option-C: The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.
Option-D: We have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.
Option-B: You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.
Option-C: The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.
Option-D: We have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.
Question 13 |
Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is __________.
400 bps | |
800 bps | |
1600 bps | |
3200 bps |
Question 13 Explanation:
Given data,
-- 4 channels are multiplexed using TDM
-- Multiplex 1 byte per channel. So, total 4 bytes=4 channels.
-- Each channel sends 100 bytes/sec
-- Bit rate=?
Bit rate= 100 * 4 bytes(1 byte=8 bits)
= 100 * 4 *8 (we are converting into bits because we are finding bit rate)
= 3200 bps
-- 4 channels are multiplexed using TDM
-- Multiplex 1 byte per channel. So, total 4 bytes=4 channels.
-- Each channel sends 100 bytes/sec
-- Bit rate=?
Bit rate= 100 * 4 bytes(1 byte=8 bits)
= 100 * 4 *8 (we are converting into bits because we are finding bit rate)
= 3200 bps
Question 14 |
In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in adjacent cells. If 840 frequencies are available, how many can be used in a given cell?
280 | |
110 | |
140 | |
120 |
Question 14 Explanation:
-- Mobile phone system with hexagonal( equals to 6) cells
-- 840 frequencies are available.
-- How many can be used in a cell?
In this we need maximum 3 unique cells are required.
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-- 840 frequencies are available.
-- How many can be used in a cell?
In this we need maximum 3 unique cells are required.
Association
Disassociation
Error correction
Integration
Question 15 Explanation:
→ Wireless LANs are use high frequency radio waves instead of cables for connecting the
devices in LAN. Users connected by WLANs can move around within the area of network coverage. Most WLANs are based upon the standard IEEE 802.11 or WiFi.
→ A wireless access point (WAP), or more generally just access point (AP), is a networking hardware device that allows other Wi-Fi devices to connect to a wired network.
→ Wireless access point in 802.11 WLAN services are association, association and Integration.
→ A wireless access point (WAP), or more generally just access point (AP), is a networking hardware device that allows other Wi-Fi devices to connect to a wired network.
→ Wireless access point in 802.11 WLAN services are association, association and Integration.
Question 16 |
The frequency band allocated for the downlink in GSM is :
960 - 985 MHz | |
935 - 960 MHz | |
920 - 945 MHz | |
930 - 955 MHz |
Question 16 Explanation:
There are 2 popular GSM methods.
1. GSM-900
2. GSM-1800
GSM-900 uses 890 - 915 MHz to send information from the Mobile Station to the Base
Transceiver Station (uplink) and 935 - 960 MHz for the other direction (downlink), providing 124 RF channels (channel numbers 1 to 124) spaced at 200 kHz. Duplex spacing of 45 MHz is used.
1. GSM-900
2. GSM-1800
GSM-900 uses 890 - 915 MHz to send information from the Mobile Station to the Base
Transceiver Station (uplink) and 935 - 960 MHz for the other direction (downlink), providing 124 RF channels (channel numbers 1 to 124) spaced at 200 kHz. Duplex spacing of 45 MHz is used.
Question 17 |
Theoretically, how many 1.25 MHz Carriers are there in a CDMA cell ?
18 | |
22 | |
9 | |
64 |
Question 17 Explanation:
→ In CDMA, multiple users share a 1.25 MHz channel by using orthogonal spreading codes (Walsh codes)
→ IS-95a standard designed for AMPS cellular band
→ Each cellular provider is allocated 25 MHz spectrum and ten 1.25-MHz CDMA duplex channels if A AMPS Band provider, 9 if B band provider
→ IS-95a standard designed for AMPS cellular band
→ Each cellular provider is allocated 25 MHz spectrum and ten 1.25-MHz CDMA duplex channels if A AMPS Band provider, 9 if B band provider
Question 18 |
The frequency reuse plan is divided into cell grouping using how many cells, where the number of cells equals N ?
3 | |
10 | |
7 | |
21 |
Question 18 Explanation:
The cluster size N = 7 and the frequency reuse factor is 1/7 since each cell contains one-seventh of the total number of available channels.
Total cells=7*3
=21
Frequency reuse concept, cells with the same letter use the same set of frequencies. A cell cluster is outline in blue color and replicated over the converge area.
Total cells=7*3
=21
Frequency reuse concept, cells with the same letter use the same set of frequencies. A cell cluster is outline in blue color and replicated over the converge area.Question 19 |
Which interim standard describes inter-switching networking ?
IS - 54 | |
IS - 95 | |
DS - 45 | |
ANSI - 41 |
Question 19 Explanation:
→ Interim Standard 95 (IS-95) was the first ever CDMA-based digital cellular technology.
→ The IS-95 standards describe an air interface, a set of protocols used between mobile units and the network.
→ IS-95 is widely described as a three-layer stack, where L1 corresponds to the physical (PHY) layer, L2 refers to the Media Access Control (MAC) and Link-Access Control (LAC) sublayers, and L3 to the call-processing state machine.
→ The IS-95 standards describe an air interface, a set of protocols used between mobile units and the network.
→ IS-95 is widely described as a three-layer stack, where L1 corresponds to the physical (PHY) layer, L2 refers to the Media Access Control (MAC) and Link-Access Control (LAC) sublayers, and L3 to the call-processing state machine.
Question 20 |
The mechanism with which several uses can share a medium without interference is :
Frequency modulation | |
Amplitude modulation | |
Multiplexing | |
None of these |
Question 20 Explanation:
Multiplexing is a method by which multiple analog or digital signals are combined into one signal over a shared medium. It can share a medium without interference.
Question 21 |
The _____ measures the relative strengths of two signals or a signal at two different points
frequency | |
decibel | |
Attenuation | |
throughput |
Question 21 Explanation:
Decibel (dB), unit for expressing the ratio between two physical quantities, usually amounts of acoustic or electric power, or for measuring the relative loudness of sounds. One decibel (0.1 bel) equals 10 times the common logarithm of the power ratio
Question 22 |
The process of converting the analog sample into discrete form is called
Modulation | |
Multiplexing | |
Quantization | |
Sampling |
Question 22 Explanation:
Sampling is the process of recording an analog signal at regular discrete moments of time.
sampling is the reduction of a continuous-time signal to a discrete-time signal. A common example is the conversion of a sound wave (a continuous signal) to a sequence of samples (a discrete-time signal).
Quantization in digital signal processing, is the process of mapping input values from a large set (often a continuous set) to output values in a (countable) smaller set, often with a finite number of elements.
The sampling rate fs is the number of samples per second.
The time interval between samples is called the sampling interval Ts=1/fs
sampling is the reduction of a continuous-time signal to a discrete-time signal. A common example is the conversion of a sound wave (a continuous signal) to a sequence of samples (a discrete-time signal).
Quantization in digital signal processing, is the process of mapping input values from a large set (often a continuous set) to output values in a (countable) smaller set, often with a finite number of elements.
The sampling rate fs is the number of samples per second.
The time interval between samples is called the sampling interval Ts=1/fs
Question 23 |
The astable multivibrator has
two quasi stable states | |
two stable states | |
one stable and one quasi-stable state | |
none of these |
Question 23 Explanation:
A multivibrator is an electronic circuit used to implement a variety of simple two-state devices such as relaxation oscillators, timers and flip-flops. It consists of two amplifying devices (transistors, vacuum tubes or other devices) cross-coupled by resistors or capacitors.
Question 24 |
A stable multivibrator are used as
comparator circuit | |
squaring circuit | |
frequency to voltage converter | |
voltage to frequency converter |
Question 24 Explanation:
● A multivibrator is a one type of electronic circuit, that is used to implement a two state system like flip-flops, timers and oscillators.
● Multivibrators are classified into three types based on the circuit operation, namely Astable multivibrators, Bistable multivibrators and Monostable multivibrators.
● The astable multivibrator is not stable and it repeatedly switches from one state to the other. In monostable multivibrator, one state is stable and remaining state is unstable.
● When the power is turned ON consider the flip flop is cleared initially, then the o/p of the inverter will be high. The charging of the capacitor will be done using two resistors R1& R2. When the voltage of the capacitor goes above 2/3 Vcc, then the output of the higher comparator will be High, it changes the control flip flop
● Multivibrators are classified into three types based on the circuit operation, namely Astable multivibrators, Bistable multivibrators and Monostable multivibrators.
● The astable multivibrator is not stable and it repeatedly switches from one state to the other. In monostable multivibrator, one state is stable and remaining state is unstable.
● When the power is turned ON consider the flip flop is cleared initially, then the o/p of the inverter will be high. The charging of the capacitor will be done using two resistors R1& R2. When the voltage of the capacitor goes above 2/3 Vcc, then the output of the higher comparator will be High, it changes the control flip flop
Question 25 |
Which of the following is not a transceiver function?
Transmission and receipt of data | |
Checking of line voltages | |
Addition and subtraction of headers | |
Collision detection |
Question 25 Explanation:
A transceiver is a device comprising both a transmitter and a receiver that are combined and share common circuitry or a single housing. When no circuitry is common between transmit and receive functions, the device is a transmitter receiver.
Question 26 |
Vestigial sideband most commonly used in radio transmission
radio transmission | |
telephony | |
television transmission | |
all of the above |
Question 26 Explanation:
Limitation of single-sideband modulation being used for voice signals and not available for video/TV signals leads to the usage of vestigial sideband. A vestigial sideband (in radio communication) is a sideband that has been only partly cut off or suppressed. Television broadcasts (in analog video formats) use this method if the video is transmitted in AM, due to the large bandwidth used. It may also be used in digital transmission, such as the ATSC standardized 8VSB.
Vestigial sideband (VSB) is a type of amplitude modulation ( AM ) technique (sometimes called VSB-AM ) that encodes data by varying the amplitude of a single carrier frequency . Portions of one of the redundant sidebands are removed to form a vestigial sideband signal - so-called because a vestige of the sideband remains.
Vestigial sideband (VSB) is a type of amplitude modulation ( AM ) technique (sometimes called VSB-AM ) that encodes data by varying the amplitude of a single carrier frequency . Portions of one of the redundant sidebands are removed to form a vestigial sideband signal - so-called because a vestige of the sideband remains.
Question 27 |
The diagram below depicts:
Sound in waveform | |
Wind frequency | |
Line of sight | |
Compressions on a map |
Question 27 Explanation:
●The generic term waveform means a graphical representation of the shape and form of a signal moving in a gaseous, liquid, or solid medium.
●For sound, the term describes a depiction of the pattern of sound pressure variation (or amplitude) in the time domain
● The temporal frequencies of sound waves are generally expressed in terms of cycles (or kilocycles) per second. The simplest waveform is the sine wave, since it has only one frequency associated with it. The sound waves associated with, say, music, are constantly varying.
●For sound, the term describes a depiction of the pattern of sound pressure variation (or amplitude) in the time domain
● The temporal frequencies of sound waves are generally expressed in terms of cycles (or kilocycles) per second. The simplest waveform is the sine wave, since it has only one frequency associated with it. The sound waves associated with, say, music, are constantly varying.
Question 28 |
The diagram below represents
Resolution | |
Vibration | |
Reverberation | |
Frequency |
Question 28 Explanation:
● Reverberation, in psychoacoustics and acoustics, is a persistence of sound after the sound is produced.
● A reverberation, or reverb, is created when a sound or signal is reflected causing a large number of reflections to build up and then decay as the sound is absorbed by the surfaces of objects in the space – which could include furniture, people, and air.
● A reverberation, or reverb, is created when a sound or signal is reflected causing a large number of reflections to build up and then decay as the sound is absorbed by the surfaces of objects in the space – which could include furniture, people, and air.
Question 29 |
The process of taking a snapshot of the waveform at regular intervals and representing it as a binary number is known as
Sampling | |
Standard Assessment | |
Sequential Formatting | |
Sound Structure |
Question 29 Explanation:
Digital sampling is the recording of a sound using a string of numbers to represent the sound. 1s and 0s. These numbers are called samples.
Question 30 |
The percentage resolution of an eight bit D/A converter is
0.39% | |
0.38% | |
0.50% | |
0.51% |
Question 30 Explanation:
→ An eight-bit D/A converter have 2 8 -1 resolvable levels which is nothing but 255 levels.
→ Percentage resolution is (1/255)×100=0.39%
→ Percentage resolution is (1/255)×100=0.39%
Question 31 |
A___ multivibrator circuit is one in which both LOW and HIGH output states are stable
Monostable | |
Bistable | |
Multistable | |
Tristable |
Question 31 Explanation:
● The Bistable Multivibrator is another type of two state device.
● Bistable Multivibrators have TWO stable states (hence the name: “Bi” meaning two) and maintain a given output state indefinitely unless an external trigger is applied forcing it to change state.
● Bistable Multivibrators have TWO stable states (hence the name: “Bi” meaning two) and maintain a given output state indefinitely unless an external trigger is applied forcing it to change state.
Question 32 |
Refer to the monostable multibrator circuit in the figure below.
The trigger terminal(pin 2 of the IC) is driven by a symmetrical pulsed waveform of 10kHz. Determine the duty cycle of the output waveform.
The trigger terminal(pin 2 of the IC) is driven by a symmetrical pulsed waveform of 10kHz. Determine the duty cycle of the output waveform.
0.56 | |
0.55 | |
0.57 | |
0.58 |
Question 32 Explanation:
Question 33 |
The diagram below represents
Moore’s law | |
Newton raphson method | |
Boyle’s law | |
Gregor law |
Question 33 Explanation:
Moore's law is the observation that the number of transistors in a dense integrated circuit doubles about every two years.
Question 34 |
The closeness of the recorded version to the original sound is called
Fidelity | |
Digitization | |
Sampling | |
Nyquist Theorem |
Question 34 Explanation:
→ Digitization is the process of converting information into a digital format . In this format, information is organized into discrete units of data (called bit s) that can be separately
addressed (usually in multiple-bit groups called bytes).
→ sampling is the reduction of a continuous-time signal to a discrete-time signal. A common example is the conversion of a sound wave (a continuous signal) to a sequence of samples (a discrete-time signal).
→ "fidelity" denotes how accurately a copy reproduces its source
→ sampling is the reduction of a continuous-time signal to a discrete-time signal. A common example is the conversion of a sound wave (a continuous signal) to a sequence of samples (a discrete-time signal).
→ "fidelity" denotes how accurately a copy reproduces its source
Question 35 |
When the object making the sound is moving towards you, the frequency goes up due to he waves getting pushed more tightly together. The opposite happens when the object moves away from you and the pich goes down. This phenomenon is called
Band Width | |
Doppler effect | |
Sound refraction | |
Vibrations |
Question 35 Explanation:
→ The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.
→ The Doppler effect is observed whenever the source of waves is moving with respect to an observer.
→ The Doppler effect can be described as the effect produced by a moving source of waves in which there is an apparent upward shift in frequency for observers towards whom the source is approaching and an apparent downward shift in frequency for observers from whom the source is receding.
→ The Doppler effect is observed whenever the source of waves is moving with respect to an observer.
→ The Doppler effect can be described as the effect produced by a moving source of waves in which there is an apparent upward shift in frequency for observers towards whom the source is approaching and an apparent downward shift in frequency for observers from whom the source is receding.
Question 36 |
In MOSFET fabrication, the channel length is delayed during the process of
Isolation oxide growth | |
Channel stop implantation | |
Poly-Silicon gate patterning | |
Lithography step leading to the contact pad |
Question 36 Explanation:
In MOSFET fabrication channel length is defined during Poly-Silicon gate patterning process
Question 37 |
A carrier A c Cos(w C )t is frequency modulated by a signal E m Cos(w m )t. The modulation index is m f . the expression for the resulting FM signal is
A c, Cos[w c t+ m f Sin(w m )t] | |
A c Cos[w c t+ m f Cos(w m )t] | |
A c Cos[w c t+ π m f Sin w m t] | |
A c Cos[w c t+ 2 π m f E m Cos(w m )t/w m ] |
Question 37 Explanation:
Question 38 |
An ideal op-amp is an ideal
Voltage controlled current source | |
voltage controlled voltage source | |
current controlled current source | |
current controlled voltage source |
Question 38 Explanation:
An ideal op-amp lead us to an op-amp with input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”).
Ideal operational amplification will have
➝ High input impedance ( R in = ∞ ) and low output impedance ( R out = 0 )
➝ R in = ∞ , so zero input current
➝ Infinite voltage gain
➝ So, it is a voltage controlled voltage source device.
Ideal operational amplification will have
➝ High input impedance ( R in = ∞ ) and low output impedance ( R out = 0 )
➝ R in = ∞ , so zero input current
➝ Infinite voltage gain
➝ So, it is a voltage controlled voltage source device.
Question 39 |
What frequency range is used for microwave communications, satellite and radar?
Low frequency: 30 kHz to 300 kHz | |
Medium frequency: 300 kHz to 3 MHz | |
Medium frequency: 300 kHz to 3 MHz | |
Extremely high frequency: 30000 kHz |
Question 39 Explanation:
→ Microwave signals are often divided into three categories:
1)Ultra high frequency (UHF) (0.3-3 GHz);
2)Super high frequency (SHF) (3-30 GHz); and
3)Extremely high frequency (EHF) (30-300 GHz).
→ In addition, microwave frequency bands are designated by specific letters.
→ The term “P band” is sometimes used for ultra high frequencies below the L-band. For other definitions, see Letter Designations of Microwave Bands
→ Lower Microwave frequencies are used for longer links, and regions with higher rain fade. Conversely, Higher frequencies are used for shorter links and regions with lower rain fade.
1)Ultra high frequency (UHF) (0.3-3 GHz);
2)Super high frequency (SHF) (3-30 GHz); and
3)Extremely high frequency (EHF) (30-300 GHz).
→ In addition, microwave frequency bands are designated by specific letters.
→ The term “P band” is sometimes used for ultra high frequencies below the L-band. For other definitions, see Letter Designations of Microwave Bands
→ Lower Microwave frequencies are used for longer links, and regions with higher rain fade. Conversely, Higher frequencies are used for shorter links and regions with lower rain fade.
Question 40 |
ICA(in computational intelligence) stands for__
Intermediate Computational Algorithms | |
Immediate Computational and Analysis | |
Independent Component Analysis | |
Independent Computational Algorithms |
Question 40 Explanation:
In signal processing, independent component analysis is a computational method for separating a multivariate signal into additive subcomponents. This is done by assuming that the
subcomponents are non-Gaussian signals and that they are statistically independent from each other.
Question 41 |
The sequence of operation in which PCM is done :
Sampling, quantizing, encoding | |
Quantizing, sampling, encoding | |
Quantizing, encoding, sampling | |
None of the above |
Question 41 Explanation:
● Pulse code modulation (PCM) is a digital scheme for transmitting analog data. The signals in PCM are binary; that is, there are only two possible states, represented by logic 1 (high) and logic 0 (low). This is true no matter how complex the analog waveform happens to be. Using PCM, it is possible to digitize all forms of analog data, including full-motion video, voices, music, telemetry, and virtual reality (VR).
● sampling is the reduction of a continuous-time signal to a discrete-time signal. A common example is the conversion of a sound wave (a continuous signal) to a sequence of samples (a discrete-time signal).
● Quantization in digital signal processing, is the process of mapping input values from a large set (often a continuous set) to output values in a (countable) smaller set, often with a finite number of elements.
● The encoder encodes the quantized samples. Each quantized sample is encoded into an 8-bit code word by using A-law in the encoding process.
● sampling is the reduction of a continuous-time signal to a discrete-time signal. A common example is the conversion of a sound wave (a continuous signal) to a sequence of samples (a discrete-time signal).
● Quantization in digital signal processing, is the process of mapping input values from a large set (often a continuous set) to output values in a (countable) smaller set, often with a finite number of elements.
● The encoder encodes the quantized samples. Each quantized sample is encoded into an 8-bit code word by using A-law in the encoding process.
Question 42 |
A low pass filter is:
Passes the frequencies lower than the specified cutoff frequency | |
Used to recover signal from sampled signal | |
Rejects higher frequencies | |
All of the above |
Question 42 Explanation:
A low-pass filter (LPF) is a filter that passes signals with a frequency lower than a selected cutoff frequency and attenuates signals with frequencies higher than the cutoff frequency. The exact frequency response of the filter depends on the filter design.
Question 43 |
T1 carrier system is used:
For delta modulation | |
Industrial moise | |
For frequency modulated signals | |
None of the above |
Question 43 Explanation:
The T-carrier system, introduced by the Bell System in the U.S. in the 1960s, was the first successful system that supported digitized voice transmission. The original transmission rate (1.544 Mbps) in the T1 line is in common use today in Internet service provider (ISP) connections to the Internet. Another level, the T3 line, providing 44.736 Mbps, is also commonly used by Internet service providers.
The T-carrier system is entirely digital, using pulse code modulation (PCM) and time-division multiplexing (TDM). The system uses four wires and provides duplex capability (two wires for receiving and two for sending at the same time). The T1 digital stream consists of 24 64-Kbps channels that are multiplexed.
The T-carrier system is entirely digital, using pulse code modulation (PCM) and time-division multiplexing (TDM). The system uses four wires and provides duplex capability (two wires for receiving and two for sending at the same time). The T1 digital stream consists of 24 64-Kbps channels that are multiplexed.
Question 44 |
Which of the following steps is/are not required for analog to digital conversion?
(a)Sensing
(b)Conversion
(c)Amplification
(d)Conditioning
(e)Quantization
(a)Sensing
(b)Conversion
(c)Amplification
(d)Conditioning
(e)Quantization
(a) and (b) | |
(c) and (d) | |
(a), (b) and (e) | |
None of the above |
Question 44 Explanation:
Question 45 |
Which of the following is not a lossy compression technique?
JPEG | |
MPEG | |
FFT | |
Arithmetic coding |
Question 45 Explanation:
JPEG( Joint Photographic Experts Group) is a commonly used method of lossy compression for digital images, particularly for those images produced by digital photography. The degree of compression can be adjusted, allowing a selectable tradeoff between storage size and image quality. JPEG typically achieves 10:1 compression with little perceptible loss in image quality
MPEG (Moving Picture Experts Group): It is a method of lossy compression for coding audio-visual information (e.g., movies, video, music).
Arithmetic coding is a form of entropy encoding used in lossless data compression. Normally, a string of characters such as the words "hello there" is represented using a fixed number of bits per character, as in the ASCII code. When a string is converted to arithmetic encoding, frequently used characters will be stored with fewer bits and not-so-frequently occurring characters will be stored with more bits, resulting in fewer bits used in total.
MPEG (Moving Picture Experts Group): It is a method of lossy compression for coding audio-visual information (e.g., movies, video, music).
Arithmetic coding is a form of entropy encoding used in lossless data compression. Normally, a string of characters such as the words "hello there" is represented using a fixed number of bits per character, as in the ASCII code. When a string is converted to arithmetic encoding, frequently used characters will be stored with fewer bits and not-so-frequently occurring characters will be stored with more bits, resulting in fewer bits used in total.
Question 46 |
Consider a discrete memoryless channel and assume that H(x) is the amount of information per symbol at the input of the channel; H(y) is the amount of information per symbol at the output of the channel; H(x|y) is the amount of uncertainty remaining on x knowing y; and I (x; y) is the information transmission. Which of the following does not define the channel capacity of a discrete memoryless channel ?
max I(x; y) p(x) | |
max [H(y) – H(y|x)] p(x) | |
max [H(x) – H(x|y)] p(x) | |
max H(x|y) p(x) |
Question 46 Explanation:
Transmission rate over a noisy channel:
→ H(X) is the amount of information per symbol at the input of the channel.
→ H(Y) is the amount of information per symbol at the output of the channel.
→ H(X|Y) is the amount of uncertainty remaining on X knowing Y.
The information transmission is given by:
I(X;Y) = H(X)−H(X|Y) bits/channel use
For an ideal channel X = Y, there is no uncertainty over X when we observe Y. So all the information is transmitted for each channel use: I(X;Y) = H(X) If the channel is too noisy, X and Y are independent. So the uncertainty over X remains the same knowing or not Y, i.e. no information passes through the channel: I(X;Y) = 0.
Capacity of a noisy channel:
→ H(X) is the amount of information per symbol at the input of the channel.
→ H(Y) is the amount of information per symbol at the output of the channel.
→ H(X|Y) is the amount of uncertainty remaining on X knowing Y.
The information transmission is given by:
I(X;Y) = H(X)−H(X|Y) bits/channel use
For an ideal channel X = Y, there is no uncertainty over X when we observe Y. So all the information is transmitted for each channel use: I(X;Y) = H(X) If the channel is too noisy, X and Y are independent. So the uncertainty over X remains the same knowing or not Y, i.e. no information passes through the channel: I(X;Y) = 0.
Capacity of a noisy channel:
Question 47 |
Match the following:
(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) | |
(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) | |
(a)-(i), (b)-(iii), (c)-(ii), (iv)-(iv) | |
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) |
Question 47 Explanation:
Line coding→ Process of converting digital data to digital signal
Block coding→ Provide redundancy to ensure synchronization and inherits error detection
Scrambling→ Provides synchronization without increasing number of bits
Pulse code modulation→ A technique to change analog signal to digital data
Block coding→ Provide redundancy to ensure synchronization and inherits error detection
Scrambling→ Provides synchronization without increasing number of bits
Pulse code modulation→ A technique to change analog signal to digital data
Question 48 |
Which of the following is not a standard RS-232C signal ?
RTS | |
CTS | |
DSR | |
VDR |
Question 49 |
Which of the following is the fastest analog to digital converter
Flash ADC | |
Dual slope ADC | |
Successive approximation ADC | |
Counter type ADC |
Question 50 |
Open circuit test on transformer is conducted to determine
Core losses | |
Copper Losses | |
Eddy Current losses | |
Hysteresis Losses |
There are 50 questions to complete.