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1.

In soft decision approach 0 represents(a) Certainly one(b) Certainly zero(c) Very likely zero(d) Could be either zero or oneI had been asked this question in an internship interview.Question is from Shannon Hartley Theorem and Turbo Codes topic in section Channel Coding of Digital Communications

Answer»

The correct option is (d) Could be either ZERO or one

For explanation: The decoder front end produces an integer for each BIT in the DATA stream. This integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. It ranges from -127 to 127. Here 0 represents ‘could be either zero or one’.

2.

In soft decision approach 100 means?(a) Certainly one(b) Certainly zero(c) Very likely zero(d) Very likely oneThe question was posed to me in exam.The origin of the question is Shannon Hartley Theorem and Turbo Codes in chapter Channel Coding of Digital Communications

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Correct option is (d) Very likely one

Explanation: The decoder front end produces an integer for each BIT in the data stream. This integer is the MEASURE of how likely it is that the bit 0 or 1 and is called as soft bit. It RANGES from -127 to 127. Here 100 represents very likely one.

3.

The inter leaver connected between the two decoders is used to(a) Remove error bursts(b) Scatter error bursts(c) Add error bursts(d) None of the mentionedI had been asked this question in exam.I want to ask this question from Shannon Hartley Theorem and Turbo Codes in section Channel Coding of Digital Communications

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Correct CHOICE is (B) Scatter error BURSTS

Explanation: An inter leaver installed between the TWO decoders connected in series is USED to scatter error bursts.

4.

In soft decision approach what does -127 mean?(a) Certainly one(b) Certainly zero(c) Very likely zero(d) Very likely oneThis question was addressed to me by my college professor while I was bunking the class.This key question is from Shannon Hartley Theorem and Turbo Codes topic in portion Channel Coding of Digital Communications

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The correct choice is (b) Certainly zero

To explain: The decoder FRONT end produces an INTEGER for each bit in the DATA stream. This integer is the measure of how likely it is that the bit 0 or 1 and is CALLED as soft bit. It ranges from -127 to 127. Here -127 REPRESENTS certainly zero.

5.

Decoders are connected in series.(a) True(b) FalseThe question was posed to me in an international level competition.The doubt is from Shannon Hartley Theorem and Turbo Codes topic in section Channel Coding of Digital Communications

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The CORRECT answer is (a) True

Explanation: Two elementary decoders are CONNECTED in SERIAL connection for decoding the turbo codes.

6.

Components used for generation of turbo codes are(a) Inter leavers(b) Punching pattern(c) Inter leavers & Punching pattern(d) None of the mentionedI have been asked this question in class test.I would like to ask this question from Shannon Hartley Theorem and Turbo Codes topic in portion Channel Coding of Digital Communications

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Correct choice is (C) Inter leavers & PUNCHING pattern

For explanation: There are many instances of turbo CODES, using different COMPONENT ENCODERS, input/output ratios, inter leavers, punching patterns.

7.

Turbo codes are(a) Forward error correction codes(b) Backward error correction codes(c) Error detection codes(d) None of the mentionedThe question was asked in an interview for internship.The above asked question is from Shannon Hartley Theorem and Turbo Codes in division Channel Coding of Digital Communications

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8.

Average effective information is obtained by(a) Subtracting equivocation from entropy(b) Adding equivocation with entropy(c) Ratio of number of error bits by total number of bits(d) None of the mentionedI got this question during a job interview.Asked question is from Shannon Hartley Theorem and Turbo Codes topic in section Channel Coding of Digital Communications

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Correct option is (a) Subtracting equivocation from ENTROPY

The explanation is: According to Shannon the average EFFECTIVE INFORMATION is obtained by subtracting the equivocation from the entropy of the source.

9.

For a error free channel, conditional probability should be(a) Zero(b) One(c) Equal to joint probability(d) Equal to individual probabilityI had been asked this question in an interview.The question is from Shannon Hartley Theorem and Turbo Codes topic in chapter Channel Coding of Digital Communications

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Correct ANSWER is (a) Zero

The explanation: For a error free channel conditional probability should be zero, because having RECEIVED Y there is complete CERTAINTY about the MESSAGE X.

10.

Entropy is the measure of(a) Amount of information at the output(b) Amount of information that can be transmitted(c) Number of error bits from total number of bits(d) None of the mentionedI have been asked this question in my homework.My question is based upon Shannon Hartley Theorem and Turbo Codes topic in portion Channel Coding of Digital Communications

Answer» CORRECT answer is (a) Amount of information at the output

Easiest EXPLANATION: ENTROPY is DEFINED as the AVERAGE amount of information per source output.
11.

Which parameter is called as Shannon limit?(a) PB/N0(b) EB/N0(c) EBN0(d) None of the mentionedI had been asked this question in an interview for internship.I would like to ask this question from Shannon Hartley Theorem and Turbo Codes in portion Channel Coding of Digital Communications

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12.

The capacity relationship is given by(a) C = W log2 ( 1+S/N )(b) C = 2W log2 ( 1+S/N )(c) C = W log2 ( 1-S/N )(d) C = W log10 ( 1+S/N )I had been asked this question in semester exam.My question is based upon Shannon Hartley Theorem and Turbo Codes in section Channel Coding of Digital Communications

Answer» CORRECT CHOICE is (a) C = W log2 ( 1+S/N )

Easiest EXPLANATION: The capacity relationship from Shannon-hartley capacity theorem is GIVEN by C = W log2 ( 1+S/N ).
13.

The minimum nyquist bandwidth needed for baseband transmission of Rs symbols per second is(a) Rs(b) 2Rs(c) Rs/2(d) Rs2This question was addressed to me in an interview for internship.This is a very interesting question from Shannon Hartley Theorem and Turbo Codes topic in division Channel Coding of Digital Communications

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Correct option is (c) Rs/2

The explanation: Theoretical minimum nyquist bandwidth NEEDED for the BASEBAND transmission of Rs symbols per second WITHOUT ISI is Rs/2.

14.

Concatenated code is an example of(a) Error detecting code(b) Error correcting code(c) Error detecting & correctingcode(d) None of the mentionedI got this question by my school teacher while I was bunking the class.My question is based upon Concatenated and Interleaving Codes and its Application in CD topic in section Channel Coding of Digital Communications

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The correct answer is (B) ERROR correcting code

To elaborate: Concatenated CODES belong to the class of error correcting codes.

15.

Turbo code is a(a) Serial concatenation code(b) Parallel concatenation code(c) Block code(d) None of the mentionedI got this question in my homework.I'd like to ask this question from Concatenated and Interleaving Codes and its Application in CD in section Channel Coding of Digital Communications

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Right option is (b) Parallel concatenation code

For explanation I would say: Turbo codes implemented a parallel concatenation of two CONVOLUTION codes with an INTER leaver and an ITERATIVE DECODER.

16.

The performance of non binary concatenated code depends on(a) Bit errors(b) Symbols errors(c) Bit & Symbols errors(d) None of the mentionedI have been asked this question during an interview.Asked question is from Concatenated and Interleaving Codes and its Application in CD topic in chapter Channel Coding of Digital Communications

Answer» RIGHT OPTION is (b) Symbols errors

Explanation: The performance of non binary code like reed-Solomon code which is also an example for CONCATENATED code depends only on the number of SYMBOL errors in the BLOCK.
17.

A concatenated code has(a) Low error rate(b) High complexity(c) Low error rate & High complexity(d) None of the mentionedThe question was posed to me during an internship interview.I would like to ask this question from Concatenated and Interleaving Codes and its Application in CD in portion Channel Coding of Digital Communications

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Right option is (a) Low error rate

Explanation: A concatenated CODE has low error rate with an overall implementation complexity which is LESS than that would be required for SINGLE level CODING.

18.

The minimum end to end delay in block interleaving is(a) 2MN+2M+2(b) 2MN-2M+2(c) 2MN-2M-2(d) 2MN+2M-2This question was addressed to me in an interview for job.The question is from Concatenated and Interleaving Codes and its Application in CD topic in portion Channel Coding of Digital Communications

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The correct option is (b) 2MN-2M+2

For explanation I would say: The inter leaver or De-inter leaver delay is 2MN symbol TIMES. M(N-1)+1memory CELLS needs to be FILLED before TRANSMISSION can begin. Thus the minimum end to end delay is 2MN-2M+2 symbol times, not including any channel propagation delay.

19.

A concatenated code uses(a) One level of coding(b) Two levels of coding(c) Three levels of coding(d) None of the mentionedI have been asked this question in an online quiz.The doubt is from Concatenated and Interleaving Codes and its Application in CD topic in division Channel Coding of Digital Communications

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Right answer is (b) Two LEVELS of coding

The best I can explain: A concatenated code is the one that USES two levels of coding, inner code and OUTER code to ACHIEVE DESIRED error performance.

20.

Separating the symbols makes the channel memory-less.(a) True(b) FalseI have been asked this question in class test.This question is from Concatenated and Interleaving Codes and its Application in CD in chapter Channel Coding of Digital Communications

Answer» CORRECT option is (a) True

Best explanation: Separating the symbols in TIME EFFECTIVELY CHANGES a CHANNEL with memory into a memory-less one.
21.

The channel that exhibits multi-path fading(a) Has memory(b) Exhibits mutually dependent signal transmission impairments(c) Received signal will be distorted(d) All of the mentionedI have been asked this question in an internship interview.This interesting question is from Concatenated and Interleaving Codes and its Application in CD topic in chapter Channel Coding of Digital Communications

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Correct answer is (d) All of the mentioned

To elaborate: A channel that exhibits multi-path FADING where the signal arrives at the receiver over two or more DIFFERENT paths. This channel has MEMORY, exhibits mutually dependent signal TRANSMISSION IMPAIRMENT.

22.

In interpolation(a) New samples are added(b) Unreliable samples are removed(c) New samples are added & Unreliable samples are removed(d) None of the mentionedI have been asked this question in unit test.Enquiry is from Concatenated and Interleaving Codes and its Application in CD in division Channel Coding of Digital Communications

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Right OPTION is (c) New samples are added & UNRELIABLE samples are removed

To explain: The function of the INTERPOLATION function is to ADD new samples estimated from the reliable ones in place of unreliable ones.

23.

The ______ the error correcting capability used, the _____ will be the erasure correcting capability.(a) Larger, smaller(b) Smaller, larger(c) Smaller, smaller(d) Larger, largerThis question was addressed to me during an interview for a job.My doubt stems from Concatenated and Interleaving Codes and its Application in CD topic in section Channel Coding of Digital Communications

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Correct answer is (a) LARGER, smaller

Easy explanation: There is a trade off between ERROR CORRECTION and erasure correction, the larger the error CORRECTING capability used the smaller will be the erasure correcting capability.

24.

Decoding step consists of(a) De-interleaving(b) Decoding(c) De-interleaving & Decoding(d) None of the mentionedThis question was addressed to me during an internship interview.I want to ask this question from Concatenated and Interleaving Codes and its Application in CD in chapter Channel Coding of Digital Communications

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25.

The CIRC error control scheme includes(a) Correction(b) Concealment(c) Correction & Concealment(d) None of the mentionedI got this question in my homework.Origin of the question is Concatenated and Interleaving Codes and its Application in CD topic in division Channel Coding of Digital Communications

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The correct CHOICE is (c) Correction & Concealment

For explanation I would say: The cross interleave REED Solomon error control scheme in the CD system INVOLVES both correction and concealment of errors.

26.

Sources of channel errors are(a) Finger prints(b) Air bubbles(c) Unwanted particles(d) All of the mentionedThe question was asked during a job interview.This is a very interesting question from Concatenated and Interleaving Codes and its Application in CD topic in division Channel Coding of Digital Communications

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Correct option is (d) All of the mentioned

Easy EXPLANATION: Some of the sources of channel errors are some UNWANTED particles, air bubbles, finger prints, SCRATCHES ETC.

27.

Compact disc is used for(a) Digital storage(b) Reproduction of audio signals(c) Digital storage & Reproduction of audio signals(d) None of the mentionedThis question was addressed to me in class test.Asked question is from Concatenated and Interleaving Codes and its Application in CD in section Channel Coding of Digital Communications

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The CORRECT choice is (c) Digital STORAGE & REPRODUCTION of audio SIGNALS

Explanation: Compact DISC digital audio system is used for both digital storage and reproduction of audio signals.

28.

If the inversion was performed correctly, then the multiplication of the original matrix with the inverted matrix should yield identity matrix. This is known as(a) Identity theorem(b) Safety theorem(c) Multiplication theorem(d) Inversion theoremI got this question in final exam.This intriguing question originated from Properties of Convolution Codes and Reed-Solomon Codes in division Channel Coding of Digital Communications

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The CORRECT answer is (B) Safety theorem

For explanation I would say: According to safety theorem, if the inversion is performed correctly, the multiplication of ORIGINAL matrix by INVERTED matrix will yield identity matrix.

29.

Greater the redundancy greater will be the error correcting capability.(a) True(b) FalseThe question was asked in my homework.My doubt stems from Properties of Convolution Codes and Reed-Solomon Codes topic in portion Channel Coding of Digital Communications

Answer» CORRECT choice is (a) True

Explanation: In coding, the improving MECHANISM, the greater the redundancy the greater will be the error correcting CAPABILITY of the CODE.
30.

A primitive element are those which yield ______ when raised to higher order exponents.(a) All zero elements(b) Non zero elements(c) Unity elements(d) None of the mentionedThe question was asked in final exam.This intriguing question comes from Properties of Convolution Codes and Reed-Solomon Codes topic in section Channel Coding of Digital Communications

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The CORRECT option is (B) Non zero ELEMENTS

The best explanation: In a primitive polynomial at least one of its roots will be primitive ELEMENT. A primitive elements are those which YIELD non zero elements when raised to a higher order exponent.

31.

The minimum distance for Reed Solomon code is given by(a) n+k+1(b) n-k+1(c) 1-n-k(d) 1+n-kI had been asked this question in an interview for job.Enquiry is from Properties of Convolution Codes and Reed-Solomon Codes in section Channel Coding of Digital Communications

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Right answer is (b) n-k+1

Easiest explanation: REED Solomon code achieves the largest POSSIBLE code MINIMUM DISTANCE for any linear code and it is given by n-k+1.

32.

R-S codes have(a) High code rate(b) Low redundancy(c) High code rate & Low redundancy(d) None of the mentionedThe question was asked by my school principal while I was bunking the class.I'm obligated to ask this question of Properties of Convolution Codes and Reed-Solomon Codes in portion Channel Coding of Digital Communications

Answer» RIGHT choice is (C) High CODE rate & Low redundancy

The explanation is: Reed Solomon codes have high code rate and low redundancy. The complexity of a high SPEED implementation INCREASES with redundancy.
33.

The look ahead length L is given by(a) m+1(b) m-1(c) (m+1)/2(d) 1-mI had been asked this question in homework.Origin of the question is Properties of Convolution Codes and Reed-Solomon Codes in portion Channel Coding of Digital Communications

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Right choice is (a) m+1

The EXPLANATION is: Look ahead LENGTH L is given by m+1 where m is the preselected POSITIVE integer.

34.

On increasing L(a) Coding gain increases(b) Complexity increases(c) Coding gain & Complexity increases(d) None of the mentionedThe question was asked during a job interview.My enquiry is from Properties of Convolution Codes and Reed-Solomon Codes in portion Channel Coding of Digital Communications

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Right option is (c) Coding GAIN & Complexity INCREASES

For explanation: On increasing the look AHEAD length the coding gain increases and also increases the decoder implementation complexity.

35.

Reed Solomon codes are(a) Non binary(b) Cyclic(c) Non binary & Cyclic(d) None of the mentionedThe question was posed to me in exam.My question is from Properties of Convolution Codes and Reed-Solomon Codes topic in section Channel Coding of Digital Communications

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The correct answer is (c) NON binary & Cyclic

For explanation I would say: Reed Solomon codes are non binary cyclic codes with symbols made up of m BIT SEQUENCES where m is any positive integer with value GREATER than 2.

36.

A feedback decoder makes hard decision.(a) True(b) FalseThe question was asked in homework.My question is from Properties of Convolution Codes and Reed-Solomon Codes topic in section Channel Coding of Digital Communications

Answer» CORRECT option is (a) True

The explanation is: The FEEDBACK decoder MAKES hard DECISION on data BITS.
37.

Soft decision results in(a) Increase in complexity(b) Decrease in storage(c) Increase in complexity & Decrease in storage(d) None of the mentionedI had been asked this question during an internship interview.This is a very interesting question from Properties of Convolution Codes and Reed-Solomon Codes in chapter Channel Coding of Digital Communications

Answer» CORRECT answer is (a) Increase in COMPLEXITY

The best EXPLANATION: Soft decisions require more STORAGE and it also increases the complexity. For this reason soft decisions are PREFERRED less when compared to hard decisions.
38.

Which factor is desirable?(a) Minimum correlation(b) Maximum correlation(c) Maximum distance(d) None of the mentionedThe question was asked in an online interview.Question is from Properties of Convolution Codes and Reed-Solomon Codes topic in chapter Channel Coding of Digital Communications

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Right choice is (b) Maximum CORRELATION

To EXPLAIN I WOULD say: While using Euclidean DISTANCE squared METRIC, maximum correlation is desirable rather than minimum distance.

39.

Which parameter is used in soft decision algorithm?(a) Euclidean distance(b) Euclidean distance squared(c) Euclidean distance & distance squared(d) None of the mentionedI got this question in semester exam.This interesting question is from Properties of Convolution Codes and Reed-Solomon Codes in portion Channel Coding of Digital Communications

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The correct CHOICE is (c) Euclidean distance & distance squared

For explanation: As soft decision DECODING algorithm has limited resolution it cannot USE hamming distance metric. Instead it USES Euclidean and monotonic metric such as Euclidean distance squared.

40.

Which has a limited resolution?(a) Hard decision decoder(b) Soft decision decoder(c) Hard & Soft decision decoder(d) None of the mentionedThis question was addressed to me in an interview for job.The doubt is from Properties of Convolution Codes and Reed-Solomon Codes in chapter Channel Coding of Digital Communications

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The correct option is (B) Soft decision decoder

To explain: The soft decision algorithm has LIMITED resolution than hard decision algorithm and thus it cannot use hamming DISTANCE METRIC.

41.

The error correcting capability of a code scheme increases as the(a) Number of channel symbols per information bit increases(b) Bandwidth increases(c) Information per bit increases(d) All of the mentionedThis question was posed to me in a job interview.Origin of the question is Properties of Convolution Codes and Reed-Solomon Codes in chapter Channel Coding of Digital Communications

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

The best explanation: The error CORRECTING capability of a code scheme INCREASES as the number of CHANNEL symbols n per information bit k increases or the rate k/n decreases. As n increases bandwidth and complexity also increases.
42.

A code should be selected such that it should(a) Not have catastrophic error propagation(b) Have minimum free distance(c) All of the mentioned(d) None of the mentionedThe question was posed to me in an online quiz.My enquiry is from Properties of Convolution Codes and Reed-Solomon Codes topic in portion Channel Coding of Digital Communications

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Correct choice is (a) Not have catastrophic ERROR PROPAGATION

The explanation: Generally code’s are selected BASED on the free distance property. The CRITERIA for selecting code is that it should not have catastrophic error propagation and it should have maximum free distance.

43.

A catastrophic error is an event whereby ______ number of code symbol error causes _______ number of decoded data error bits.(a) Finite, finite(b) Finite, infinite(c) Infinite, finite(d) Infinite, infiniteThe question was posed to me during an internship interview.My doubt stems from Properties of Convolution Codes and Reed-Solomon Codes in chapter Channel Coding of Digital Communications

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The correct option is (b) Finite, infinite

To EXPLAIN I would say: A catastrophic error is DEFINED as an event WHEREBY a finite NUMBER of code symbols error causes an infinite number of decoded data bit errors.

44.

Which distance is related to the error correcting capability of the code?(a) Maximum distance(b) Minimum distance(c) Maximum & Minimum distance(d) None of the mentionedThe question was posed to me in an interview for internship.Query is from Properties of Convolution Codes and Reed-Solomon Codes topic in section Channel Coding of Digital Communications

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Correct OPTION is (b) MINIMUM DISTANCE

The best I can explain: Minimum distance between all pair of CODE words is related to the error CORRECTING capability of the code.

45.

Branch word synchronization is the process of determining the(a) Beginning of received sequence(b) Beginning of transmitted sequence(c) End of received sequence(d) End of transmitted sequenceThe question was posed to me by my school teacher while I was bunking the class.Enquiry is from Convolution Encoding and Decoding in division Channel Coding of Digital Communications

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The CORRECT CHOICE is (a) Beginning of received sequence

The best explanation: Branch word synchronization is the PROCESS of determining the beginning of a branch word in the received sequence. Such synchronization can take place WITHOUT adding new information to the transmitted SYMBOLS.

46.

Gaussian channel is a hard decision channel.(a) True(b) FalseThis question was addressed to me during an interview.I'm obligated to ask this question of Convolution Encoding and Decoding in chapter Channel Coding of Digital Communications

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The correct CHOICE is (B) False

Best explanation: The quantized GAUSSIAN channel is referred to as the soft DECISION channel.

47.

Binary symmetric channel is hard decision channel.(a) True(b) FalseI have been asked this question during an internship interview.My question comes from Convolution Encoding and Decoding topic in division Channel Coding of Digital Communications

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Correct option is (a) True

The best I can explain: A binary symmetric CHANNEL is an EXAMPLE for hard decision channel which means that even though CONTINUOUS values may be received by the DEMODULATOR the BSC only allows firm decisions.

48.

A binary symmetric channel is a(a) Discrete memoryless channel(b) Continuous memoryless channel(c) Discrete memory channel(d) Continuous memory channelThis question was addressed to me in an online interview.I want to ask this question from Convolution Encoding and Decoding in section Channel Coding of Digital Communications

Answer» RIGHT ANSWER is (a) Discrete memoryless channel

Easiest explanation: A binary symmetric channel is a discrete memoryless channel that has binary input and output ALPHABETS and symmetric transition PROBABILITIES.
49.

Soft decision decoding requires(a) Less memory(b) More memory(c) More speed(d) None of the mentionedI got this question in an online interview.Query is from Convolution Encoding and Decoding topic in chapter Channel Coding of Digital Communications

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The correct answer is (B) More memory

The best explanation: The PRICE paid for SOFT decision DECODING is an increase in required memory size at the DECODER.

50.

If the quantization level of a demodulator output is ________ then it is called as soft decision decoding.(a) Equal to 2(b) More than 2(c) Less than 2(d) None of the mentionedThe question was posed to me in an online interview.Question is taken from Convolution Encoding and Decoding topic in division Channel Coding of Digital Communications

Answer»

The CORRECT choice is (b) More than 2

Explanation: If the quantization LEVEL of a demodulator output is 2 it is called as HARD decision decoding and if it is greater than 2 it is called as soft decision decoding.