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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

The number of filters in CPM receiver structure is(a) LM(b) ML(c) M/L(d) None of the mentionedI got this question during an interview.This interesting question is from Receiver Synchronization topic in chapter Synchronization of Digital Communications

Answer» RIGHT option is (B) ML

The best EXPLANATION: The number of filters in CPM receiver structure is GIVEN by ML.
Discussion
2.

CPM has(a) Narrower bandwidth(b) Wider bandwidth(c) More bandwidth requirement(d) None of the mentionedI had been asked this question in class test.I would like to ask this question from Receiver Synchronization in section Synchronization of Digital Communications

Answer» CORRECT choice is (a) Narrower bandwidth

For EXPLANATION I would say: The increase in smoothness in the signals concentrate the signal ENERGY in a narrower bandwidth, reducing the AMOUNT of bandwidth required.
Discussion
3.

Continuous phase modulation has(a) More bandwidth efficiency(b) More smoothness(c) More bandwidth efficiency & smoothness(d) None of the mentionedThe question was asked in quiz.I would like to ask this question from Receiver Synchronization in chapter Synchronization of Digital Communications

Answer»

Correct answer is (C) More BANDWIDTH efficiency & smoothness

The explanation: CONTINUOUS phase modulation is an IMPORTANT signalling TECHNIQUE because of their bandwidth efficiency. Bandwidth efficiency is obtained by increasing the smoothness of the signal.
Discussion
4.

To avoid the loop to be driven to lock, the sweep rate should be(a) Smaller(b) Larger(c) It does not depend on it(d) None of the mentionedThis question was posed to me in an interview for job.The above asked question is from Receiver Synchronization topic in chapter Synchronization of Digital Communications

Answer» CORRECT answer is (a) Smaller

Best explanation: The sweep rate MUST not be too large, or the loop will be DRIVEN through the lock POINT so FAST that it will fail to acquire.
Discussion
5.

What is the function of the edge detector?(a) Differentiation(b) Rectification(c) Differentiation & Rectification(d) None of the mentionedThe question was asked during an interview.The above asked question is from Receiver Synchronization in chapter Synchronization of Digital Communications

Answer»

Correct answer is (C) Differentiation & Rectification

Easy explanation: The MAIN OPERATION of the EDGE detector is differentiation and rectification by the USE of square law detector.
Discussion
6.

Acquisition is a ______ operation.(a) Linear(b) Non linear(c) Linear & Non linear(d) None of the mentionedI had been asked this question in an interview for internship.This intriguing question originated from Receiver Synchronization topic in portion Synchronization of Digital Communications

Answer»

Right answer is (b) Non linear

Easy explanation: Acquisition is accomplished by EXTERNAL signal or circuits or sometimes it might be unaided (SELF acquisition). This acquisition PROCESS is an inherently non linear OPERATION.
Discussion
7.

When phase noise and phase jitter is doubled, the phase variance increases by(a) Two times(b) Four times(c) 0.5 times(d) 8 timesThis question was addressed to me in homework.I want to ask this question from Receiver Synchronization in division Synchronization of Digital Communications

Answer»

Correct OPTION is (b) Four times

The best explanation: In PLL, when the phase JITTER and phase NOISE increases by 2, the phase variance increases by 4 than that of the original signal.
Discussion
8.

A cycle slip occurs when the magnitude of the original phase error exceeds(a) Π radians(b) 2π radians(c) π/2 radians(d) π2 radiansThis question was addressed to me at a job interview.Query is from Receiver Synchronization topic in portion Synchronization of Digital Communications

Answer»

Right option is (b) 2π RADIANS

For explanation I WOULD say: A CYCLIC slip OCCURS when the magnitude of the original phase ERROR exceeds 2π radians.
Discussion
9.

PLL designs have(a) First order loop(b) Second order loop(c) Higher order loop(d) None of the mentionedThis question was posed to me during an online exam.Question is from Receiver Synchronization in division Synchronization of Digital Communications

Answer»

Correct OPTION is (b) Second order LOOP

The explanation is: Most of the PLL DESIGNS have second order loop as they can be made unconditionally STABLE.
Discussion
10.

In VCO the output frequency is a linear function of its input(a) Frequency(b) Voltage(c) Time period(d) None of the mentionedThis question was posed to me in examination.I want to ask this question from Receiver Synchronization in portion Synchronization of Digital Communications

Answer»

Right choice is (b) VOLTAGE

To elaborate: A VCO is an OSCILLATOR whose OUTPUT FREQUENCY is a linear function of its input voltage over some range of input and output.
Discussion
11.

Phase locked loops consists of(a) Phase detectors(b) Loop filter(c) Voltage controlled oscillator(d) All of the mentionedThis question was addressed to me in homework.Asked question is from Receiver Synchronization topic in portion Synchronization of Digital Communications

Answer» CORRECT option is (d) All of the mentioned

The best explanation: PHASE locked LOOPS have three BASIC components – phase detector, loop filter and voltage controlled oscillator. It measures the phase difference between the incoming signal and local REPLICA.
Discussion
12.

The system which has ______ signal to noise ratio can tolerate larger timing error.(a) Large(b) Small(c) Large & Small(d) None of the mentionedThe question was asked in exam.I'd like to ask this question from Synchronization and Network Synchronization topic in division Synchronization of Digital Communications

Answer»

Correct answer is (a) Large

Explanation: The SYSTEM with INCREASED SIGNAL to noise ratio will allow it to tolerate larger TIMING error so the improvement is error performance is rapid.
Discussion
13.

When a terminal is able to utilize the measurements made on the return link it is called as(a) Open loop method(b) Closed loop method(c) Quasi close loop method(d) None of the mentionedI had been asked this question in an online interview.This intriguing question comes from Synchronization and Network Synchronization in division Synchronization of Digital Communications

Answer»

The correct CHOICE is (C) Quasi close loop method

Easiest explanation: The case where a terminal is able to utilize the measurements MADE on a return link signal is called as quasi closed loop transmitter SYNCHRONIZATION.
Discussion
14.

Which reduces the effective distance between signals in signal space?(a) Time error(b) Frequency offset(c) Time error & Frequency offset(d) None of the mentionedThe question was posed to me during an interview.My question comes from Synchronization and Network Synchronization in section Synchronization of Digital Communications

Answer»

Correct option is (c) TIME error & FREQUENCY OFFSET

To elaborate: Time error or frequency offset or COMBINATION of both will reduce the effective distance between signals in signal space and degrade error performance.
Discussion
15.

In open loop method, the time reference error __________ with respect to time.(a) Increases exponentially(b) Decreases exponentially(c) Increases quadratically(d) Decreases quadraticallyI had been asked this question in exam.The above asked question is from Synchronization and Network Synchronization in section Synchronization of Digital Communications

Answer» RIGHT CHOICE is (a) INCREASES exponentially

Easiest explanation: In open loop transmitter synchronization system, the TIME REFERENCE error increases quadratically with time.
Discussion
16.

Which method needs precorrect time?(a) Open loop method(b) Closed loop method(c) Open & Closed loop method(d) None of the mentionedI had been asked this question in a national level competition.I would like to ask this question from Synchronization and Network Synchronization topic in chapter Synchronization of Digital Communications

Answer»

Correct choice is (a) Open loop method

The EXPLANATION is: In open loop method the FREQUENCY and time are precorrected.
Discussion
17.

Which method requires external authority and is inflexible?(a) Open loop method(b) Closed loop method(c) Open & Closed loop method(d) None of the mentionedThis question was addressed to me in an online interview.My doubt is from Synchronization and Network Synchronization topic in section Synchronization of Digital Communications

Answer» CORRECT answer is (a) Open loop method

Explanation: Open loop method is inflexible, does not adjust quickly to UNPLANNED changes and NEEDS EXTERNAL authority that provides prior knowledge, where as closed loop method does not need all these.
Discussion
18.

Which method has fast acquisition and can work without return link?(a) Open loop method(b) Closed loop method(c) Open & Closed loop method(d) None of the mentionedThis question was addressed to me in an internship interview.Asked question is from Synchronization and Network Synchronization topic in division Synchronization of Digital Communications

Answer»

The correct answer is (a) OPEN LOOP method

Explanation: The main advantage of open loop method is that acquisition is FAST, the procedure can WORK without return link and the amount of real time computation that is required is SMALL.
Discussion
19.

Which method has two way link?(a) Open loop method(b) Closed loop method(c) Open & Closed loop method(d) None of the mentionedThis question was posed to me by my college professor while I was bunking the class.My doubt stems from Synchronization and Network Synchronization in chapter Synchronization of Digital Communications

Answer»
Discussion
20.

Synchronization is used in system which has(a) Non coherent modulation techniques(b) Many users accessing(c) Non coherent modulation techniques & Many users accessing(d) None of the mentionedI had been asked this question during an internship interview.This question is from Synchronization and Network Synchronization in portion Synchronization of Digital Communications

Answer»

Right option is (c) NON COHERENT modulation TECHNIQUES & Many USERS accessing

The explanation: For communication system using non coherent modulation techniques and many users accessing the central communication node, synchronization is necessary.
Discussion
21.

As the synchronization levels increases, cost(a) Decreases(b) Increases(c) Remains the same(d) None of the mentionedI have been asked this question in semester exam.I want to ask this question from Synchronization and Network Synchronization topic in portion Synchronization of Digital Communications

Answer» RIGHT option is (b) Increases

To explain: For each LEVEL increase in synchronization LEVELS, COST increases.
Discussion
22.

Transmitter synchronization implies(a) One way communication(b) Two way communication(c) One & Two way communication(d) None of the mentionedThe question was posed to me by my school principal while I was bunking the class.Query is from Synchronization and Network Synchronization topic in division Synchronization of Digital Communications

Answer»

Right option is (b) TWO way COMMUNICATION

Best explanation: TRANSMITTER PLAYS a role in synchronization by varying the frequency and timing of its transmissions to correspond to expectation of receiver. This transmitter synchronization IMPLIES two way communication.
Discussion
23.

For block codes _____ is necessary.(a) Phase synchronization(b) Frequency synchronization(c) Frame synchronization(d) None of the mentionedI have been asked this question during an interview for a job.I need to ask this question from Synchronization and Network Synchronization topic in chapter Synchronization of Digital Communications

Answer»

Correct answer is (c) Frame synchronization

Easy explanation: Frame synchronization is necessary when the information is ORGANIZED in BLOCKS or messages of some uniform number of symbols. Thus block CODE needs frame synchronization.
Discussion
24.

The synchronization necessary for BPSK is(a) Frequency tracking(b) Bit timing(c) Phase tracking(d) All of the mentionedThe question was asked in an interview for internship.My question comes from Synchronization and Network Synchronization topic in portion Synchronization of Digital Communications

Answer» RIGHT answer is (d) All of the mentioned

To explain I would say: Non coherently detected BPSK is the simplest DIGITAL receiver and it requires frequency TRACKING, BIT timing synchronizations and also phase synchronization.
Discussion
25.

In phase lock which parameter is synchronized(a) Frequency(b) Phase(c) Frequency & Phase(d) None of the mentionedI had been asked this question in an online quiz.Query is from Synchronization and Network Synchronization topic in portion Synchronization of Digital Communications

Answer»

Correct choice is (c) FREQUENCY & Phase

Explanation: Being in phase lock means that the receiver’s local OSCILLATOR is synchronized in both frequency and phase with the received SIGNAL.
Discussion
26.

Non coherent system requires(a) Phase synchronization(b) Frequency synchronization(c) Phase & Frequency synchronization(d) None of the mentionedI got this question during a job interview.This intriguing question comes from Synchronization and Network Synchronization in portion Synchronization of Digital Communications

Answer» CORRECT choice is (c) PHASE & Frequency synchronization

Explanation: NON coherent SYSTEM requires frequency synchronization.
Discussion
27.

Coherent modulation requires ____ level of synchronization.(a) One(b) Two(c) Three(d) None of the mentionedThis question was posed to me during an interview for a job.Query is from Synchronization and Network Synchronization in portion Synchronization of Digital Communications

Answer»

Right answer is (C) Three

Explanation: Coherent modulation requires three LEVELS of SYNCHRONIZATION – phase symbol and frame.
Discussion