The PYQ practice room
GATE EC 2017 Set 1
All 65 solved GATE EC 2017 Set 1 questions in exam order. Open a question, commit to an answer, and learn from the step-by-step solution. One question at a time.
Go beyond PYQs with Success TrackerAI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply.Questions
65
Paper marks
100
Question formats
2
MCQ · NAT
Revision mode
Self-paced
No timer. Focus on understanding.
Difficulty mixEasy 18Medium 43Hard 4
Explore the questions
General Aptitude (GA)
1056
Q56MCQ2 marksEasyShe has a sharp tongue and it can occasionally turn ________Think it through. Then check your answer.Question
She has a sharp tongue and it can occasionally turn ________Correct answer
(A) hurtful
Step-by-step solution
The idiom "sharp tongue" refers to a person who speaks in a harsh, critical, or sarcastic manner. Therefore, the most appropriate word to complete the sentence is "hurtful", as it describes the effect of such speech. The other options ("left", "methodical", "vital") do not logically fit the context of the idiom.57
Q57MCQ2 marksEasyI ________ made arrangements had I ________ informed earlier.Think it through. Then check your answer.Question
I ________ made arrangements had I ________ informed earlier.Correct answer
(A) could have, been
Step-by-step solution
The sentence follows the structure of a third conditional with an inverted 'if' clause.1.The standard third conditional structure is: If + past perfect, would/could/might + have + past participle.2.When the 'if' clause is inverted, it becomes: Had + subject + past participle, would/could/might + have + past participle.Applying this to the sentence:- Main clause: "I could have made arrangements..."
- Inverted 'if' clause: "...had I been informed earlier."
58
Q58MCQ2 marksMediumIn the summer, water consumption is known to decrease overall by . A Water Board official states that in the summer household consumption decreases by , while other…Think it through. Then check your answer.Question
In the summer, water consumption is known to decrease overall by . A Water Board official states that in the summer household consumption decreases by , while other consumption increases by .Which of the following statements is correct?Correct answer
(D) There are errors in the official's statement.
Step-by-step solution
Let be the initial household consumption and be the initial other consumption. The total initial consumption is .
According to the problem, the actual summer consumption is .
The official states that summer household consumption is and other consumption is .
For the official's statement to be consistent with the overall decrease:
Since and must be positive values (representing consumption), will always be greater than zero. Thus, no positive values of and can satisfy this equation, making the official's statement mathematically impossible. Therefore, there are errors in the official's statement.59
Q59MCQ2 marksEasyof deaths on city roads may be attributed to drunken driving. The number of degrees needed to represent this as a slice of a pie chart isThink it through. Then check your answer.Question
of deaths on city roads may be attributed to drunken driving. The number of degrees needed to represent this as a slice of a pie chart isCorrect answer
(B) 144
Step-by-step solution
A full pie chart represents of the data and corresponds to a total angle of .
To find the angle for :
Angle .60
Q60MCQ2 marksEasySome tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences? i. At least one bench is…Think it through. Then check your answer.Question
Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences?i. At least one bench is a table
ii. At least one shelf is a bench
iii. At least one chair is a table
iv. All benches are chairsCorrect answer
(B) Only ii
Step-by-step solution
Let represent the sets of tables, shelves, chairs, and benches respectively.
Given:1. (Some tables are shelves)2. (Some shelves are chairs)3. (All chairs are benches)Evaluating conclusions:
i. : Not necessarily true. There is no direct information linking tables and benches.
ii. : True. Since some shelves are chairs () and all chairs are benches (), the intersection must contain , which is non-empty.
iii. : Not necessarily true. There is no direct information linking chairs and tables.
iv. : Not necessarily true. While all chairs are benches, the reverse is not necessarily true.Thus, only conclusion ii is correct.61
Q61MCQ2 marksEasy"If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually…Think it through. Then check your answer.Question
"If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages: for though I have spent a lifetime in the country, I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters".Here, the word 'antagonistic' is closest in meaning toCorrect answer
(D) hostile
Step-by-step solution
The word 'antagonistic' refers to showing or feeling active opposition or hostility towards someone or something.
(A) impartial means neutral or unbiased.
(B) argumentative means given to expressing divergent or opposite views in a heated way.
(C) separated means moved or kept apart.
(D) hostile means showing or feeling opposition or dislike; unfriendly.
Thus, 'hostile' is the closest synonym in the context of the subcontinent being divided into opposing parts.62
Q62MCQ2 marksMediumand are seated around a circular table. 's neighbours are and . is seated third to the left of and second to the right of . 's…Think it through. Then check your answer.Question
and are seated around a circular table. 's neighbours are and . is seated third to the left of and second to the right of . 's neighbours are and ; and and are not seated opposite each other. Who is third to the left of ?Correct answer
(A) X
Step-by-step solution
Let the 8 positions be 1 to 8 in a clockwise direction. Assume people face the center.1.Place at position 1.2. is 3rd to the left of . If left is anti-clockwise: . So is at 4.3. is 2nd to the right of . If right is clockwise: . For at 4, must be at 6 ().4.'s neighbours are and . Since is at 6, and must be at 7 and 8 ( sequence).5.'s neighbours are and . Since is at 1 and is at 8, must be at 2.6.The remaining positions are 3 and 5 for and . (pos 1) is opposite pos 5. Since and are not opposite, must be at 3 and must be at 5.7.The final arrangement is: .8.Third to the left of is , which is .63
Q63MCQ2 marksMediumTrucks ( long) and cars ( long) go on a single lane bridge. There must be a gap of at least after each truck and a gap of at least…Think it through. Then check your answer.Question
Trucks ( long) and cars ( long) go on a single lane bridge. There must be a gap of at least after each truck and a gap of at least after each car. Trucks and cars travel at a speed of . If cars and trucks go alternately, what is the maximum number of vehicles that can use the bridge in one hour?Correct answer
(A) 1440
Step-by-step solution
1.Speed .2.Space occupied by one truck unit (length + gap) .3.Space occupied by one car unit (length + gap) .4.Since they go alternately, one cycle (Truck + Car) occupies .5.Total distance covered by the stream in 1 hour () .6.Number of cycles cycles.7.Each cycle contains 2 vehicles (1 truck and 1 car). Total vehicles .64
Q64MCQ2 marksMediumThere are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian?Think it through. Then check your answer.Question
There are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian?Correct answer
(A) 56
Step-by-step solution
The total number of people in the group is 6. The total number of possible subsets (subgroups) that can be formed from a set of 6 people is .To find the number of subgroups that have at least one Indian, we can subtract the number of subgroups that contain no Indians from the total number of subgroups.A subgroup contains no Indians if it is formed entirely from the 3 Chinese people. The number of such subgroups is .Therefore, the number of subgroups containing at least one Indian is:Note: Even if we consider only non-empty subgroups, the result remains the same because the empty set is excluded from both counts ().65
Q65MCQ2 marksMediumA contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at …Think it through. Then check your answer.Question
A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at intervals in this plot.The path from P to Q is best described by
Correct answer
(C) Down-Up-Down
Step-by-step solution
To determine the nature of the path from P to Q, we observe the values of the contour lines crossed:1.Starting at P: Point P is located inside the contour line, indicating a peak or high elevation area ().2.First Segment: Moving towards Q, the path crosses the line, then the line, and enters the region enclosed by the line. This indicates a descent (Down).3.Second Segment: The path then leaves the region, crosses the line, and enters another contour region. This indicates an ascent (Up).4.Third Segment: Finally, the path leaves the region and moves towards point Q, which is located near the contour line. This indicates a final descent (Down).Thus, the sequence of elevation change is Down-Up-Down.
Electronics and Communication Engineering
551
Q1MCQ1 markMediumConsider the matrix [figure]…Think it through. Then check your answer.Question
Consider the matrixIt is given that has only one real eigenvalue. Then the real eigenvalue of is
Correct answer
(C) 15
Step-by-step solution
For any matrix where the sum of elements in each row is a constant , is an eigenvalue of the matrix.
In the given matrix , the sum of elements in each row is:
Row 1:
Row 2:
Row 3:
Row 4:
Row 5:
Since all row sums are equal to 15, is an eigenvalue of .
It is given that has only one real eigenvalue, so the real eigenvalue must be 15.2
Q2MCQ1 markEasyThe rank of the matrix [figure] isThink it through. Then check your answer.Question
The rank of the matrixis
Correct answer
(C) 2
Step-by-step solution
To find the rank of matrix , we check for linear dependence among the rows.
Notice that (or ). This means and are linearly dependent.
Now check and : they are not proportional, so they are linearly independent.
Since there are 2 linearly independent rows, the rank of the matrix is 2.3
Q3MCQ1 markEasyConsider the following statements about the linear dependence of the real valued functions , and , over the field of real numbers. I. and…Think it through. Then check your answer.Question
Consider the following statements about the linear dependence of the real valued functions , and , over the field of real numbers.I. and are linearly independent on
II. and are linearly dependent on
III. and are linearly independent on
IV. and are linearly dependent on Which one among the following is correct?Correct answer
(B) Both I and III are true
Step-by-step solution
The functions are linearly independent on any interval where .
This can be shown using the Wronskian:Since for all , the functions are linearly independent on any interval.
Therefore, they are linearly independent on both and .
Statements I and III are true, while II and IV are false.4
Q4NAT1 markEasyThree fair cubical dice are thrown simultaneously. The probability that all three dice have the same number of dots on the faces showing up is (up to third decimal place)…Think it through. Then check your answer.Question
Three fair cubical dice are thrown simultaneously. The probability that all three dice have the same number of dots on the faces showing up is (up to third decimal place) __________Correct answer
0.027 to 0.028
Step-by-step solution
When three fair cubical dice are thrown simultaneously, the total number of possible outcomes is .The favorable outcomes, where all three dice have the same number of dots, are:
(1, 1, 1)
(2, 2, 2)
(3, 3, 3)
(4, 4, 4)
(5, 5, 5)
(6, 6, 6)
There are 6 such favorable outcomes.The probability is the ratio of favorable outcomes to the total number of outcomes:
To express this probability up to the third decimal place:
Rounding to the third decimal place, we get .Thus, the probability is .5
Q5MCQ1 markMediumConsider the following statements for continuous-time linear time invariant (LTI) systems. I. There is no bounded input bounded output (BIBO) stable system with a pole in the…Think it through. Then check your answer.Question
Consider the following statements for continuous-time linear time invariant (LTI) systems.
I. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane.
II. There is no causal and BIBO stable system with a pole in the right half of the complex plane.
Which one among the following is correct?Correct answer
(D) Only II is true
Step-by-step solution
Let's analyze each statement:Statement I: There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane.
For a continuous-time LTI system to be BIBO stable, its region of convergence (ROC) must include the -axis. If there is a pole in the right half of the complex plane (RHP), the ROC cannot include the -axis if the system is causal. However, a non-causal system can be BIBO stable even with RHP poles, provided its ROC is to the left of the RHP pole and includes the -axis. For example, a system with a pole at can be stable if its ROC is . Such a system would be non-causal. Therefore, statement I is false.Statement II: There is no causal and BIBO stable system with a pole in the right half of the complex plane.
For a causal LTI system, the ROC is always to the right of the rightmost pole. For the system to be BIBO stable, the ROC must include the -axis. If a causal system has a pole in the RHP, its ROC will be to the right of this RHP pole, and thus the -axis will not be included in the ROC. Therefore, a causal system with a pole in the RHP cannot be BIBO stable. This statement is true.Based on the analysis:
Statement I is false.
Statement II is true.Therefore, only statement II is true.The final answer is6
Q6MCQ2 marksMediumConsider a single input single output discrete-time system with as input and as output, where the two are related as…Think it through. Then check your answer.Question
Consider a single input single output discrete-time system with as input and as output, where the two are related asWhich one of the following statements is true about the system?Correct answer
(A) It is causal and stable
Step-by-step solution
Let's analyze the causality and stability of the given discrete-time system.Causality:
A system is causal if its output at any time depends only on the present input and past inputs (where ).The given system is defined as:
for
otherwiseFor , the output depends on (present input) and (past input). For or , the output is , which does not depend on future inputs. Since the output never depends on future inputs, the system is causal.Stability (BIBO Stability):
A discrete-time LTI system is BIBO stable if every bounded input produces a bounded output. This condition is met if and only if the impulse response is absolutely summable, i.e., .To find the impulse response , we set (the unit impulse).
For :
Let's evaluate for different values of within the range :
...
And for or .So, the impulse response is:
Now, let's check for absolute summability:
Since , the system is BIBO stable.Therefore, the system is both causal and stable.The final answer is7
Q7NAT2 marksMediumIn the circuit shown, the positive angular frequency (in radians per second) at which the magnitude of the phase difference between the voltages and equals…Think it through. Then check your answer.Question
In the circuit shown, the positive angular frequency (in radians per second) at which the magnitude of the phase difference between the voltages and equals radians, is ________.Correct answer
0.9 to 1.1
Step-by-step solution
In the given circuit, let the current flowing through the series combination be .The voltage is across the series combination of the resistor and the inductor:The voltage is across the resistor:The phase of is given by:The phase of is given by:The magnitude of the phase difference between and is:Given that the phase difference is radians:Thus, the positive angular frequency is rad/s.8
Q8MCQ2 marksMediumA periodic signal has a trigonometric Fourier series expansion If…Think it through. Then check your answer.Question
A periodic signal has a trigonometric Fourier series expansionIf , we can conclude thatCorrect answer
(A) aₙ are zero for all n and bₙ are zero for n even
Step-by-step solution
The signal has two symmetry properties:1.Odd Symmetry: . For an odd signal, the Fourier series contains only sine terms. Therefore, for all .2.Half-Wave Symmetry (HWS): , where is the period. Here, , so the condition represents HWS. For a signal with HWS, only odd harmonics are present in the Fourier series. This means and for all even .Combining these:- From odd symmetry: for all .
- From HWS: for even .
9
Q9MCQ2 marksMediumA bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statements is…Think it through. Then check your answer.Question
A bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statements is true?Correct answer
(A) Silicon atoms act as p -type dopants in Arsenic sites and n -type dopants in Gallium sites
Step-by-step solution
Silicon (Si) is a group IV element. GaAs is a III-V compound semiconductor, where Gallium (Ga) is from group III and Arsenic (As) is from group V.- When a Si atom (4 valence electrons) replaces a Ga atom (3 valence electrons), it has one extra electron to donate to the conduction band, thus acting as an -type dopant.
- When a Si atom (4 valence electrons) replaces an As atom (5 valence electrons), it has one fewer electron than required for the bonds, thus creating a hole and acting as a -type dopant.
10
Q10MCQ2 marksMediumAn Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of and…Think it through. Then check your answer.Question
An Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of and corresponding to the and regions respectively. At the operational temperature , assume complete impurity ionization, , and intrinsic carrier concentration to be . What is the magnitude of the built-in potential of this device?Correct answer
(D) 0.173 V
Step-by-step solution
The built-in potential () for an junction (which is essentially a homojunction with different doping levels) can be calculated using the formula:Given:
Substitute the values into the formula:Using :Rounding to three decimal places, .Therefore, the magnitude of the built-in potential is approximately 0.173 V.The final answer is11
Q11MCQ1 markMediumFor a narrow base PNP BJT, the excess minority carrier concentrations ( for emitter, for base, for collector) normalized to equilibrium…Think it through. Then check your answer.Question
For a narrow base PNP BJT, the excess minority carrier concentrations ( for emitter, for base, for collector) normalized to equilibrium minority carrier concentrations ( for emitter, for base, for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in?
Correct answer
(C) Inverse active
Step-by-step solution
Let's analyze the given graph for a PNP BJT:1.Emitter Region (P-type): The minority carriers are electrons (). The graph shows is negative at the emitter-base junction. A negative excess minority carrier concentration implies reverse bias at the emitter-base (EB) junction. In a PNP transistor, for the EB junction to be reverse biased, the emitter is positive with respect to the base.2.Collector Region (P-type): The minority carriers are electrons (). The graph shows is positive at the collector-base junction. A positive excess minority carrier concentration implies forward bias at the collector-base (CB) junction. In a PNP transistor, for the CB junction to be forward biased, the collector is negative with respect to the base.Let's summarize the junction biases:- Emitter-Base (EB) junction: Reverse biased
- Collector-Base (CB) junction: Forward biased
- Forward Active: EB junction forward biased, CB junction reverse biased.
- Saturation: EB junction forward biased, CB junction forward biased.
- Inverse Active (Reverse Active): EB junction reverse biased, CB junction forward biased.
- Cutoff: EB junction reverse biased, CB junction reverse biased.
12
Q12MCQ1 markMediumFor the operational amplifier circuit shown, the output saturation voltages are . The upper and lower threshold voltages for the circuit are, respectively,…Think it through. Then check your answer.Question
For the operational amplifier circuit shown, the output saturation voltages are . The upper and lower threshold voltages for the circuit are, respectively,
Correct answer
(B) +7 V and -3 V
Step-by-step solution
The given circuit is an inverting Schmitt trigger with a non-zero reference voltage.
Output saturation voltages are .
Reference voltage .
Feedback resistors are and .1. Upper Threshold Voltage ():
When the output is at (i.e., ), the voltage at the non-inverting input () is determined by the voltage divider formed by , , , and .Substitute :2. Lower Threshold Voltage ():
When the output is at (i.e., ), the voltage at the non-inverting input () is:Substitute :Thus, the upper threshold voltage is and the lower threshold voltage is .Therefore, the correct option is (B).13
Q13MCQ1 markEasyA good transconductance amplifier should haveThink it through. Then check your answer.Question
A good transconductance amplifier should haveCorrect answer
(C) high input and output resistances
Step-by-step solution
A transconductance amplifier is a voltage-controlled current source (VCCS). For an ideal VCCS, the input resistance should be infinite () to avoid loading the input voltage source, and the output resistance should be infinite () so that the output current is independent of the load resistance. Therefore, a good transconductance amplifier should have high input and output resistances.14
Q14MCQ1 markEasyThe Miller effect in the context of a Common Emitter amplifier explainsThink it through. Then check your answer.Question
The Miller effect in the context of a Common Emitter amplifier explainsCorrect answer
(D) a decrease in the high-frequency cutoff frequency
Step-by-step solution
The Miller effect describes the increase in the equivalent input capacitance of an inverting voltage amplifier due to the amplification of the capacitance between the input and output terminals. In a Common Emitter amplifier, the Miller capacitance appears in parallel with the input. This significantly increases the total input capacitance, which creates a dominant pole at a lower frequency, thereby reducing the high-frequency bandwidth (decreasing the high-frequency cutoff frequency ).15
Q15MCQ1 markMediumIn the latch circuit shown, the NAND gates have non-zero, but unequal propagation delays. The present input condition is: . If the input condition is changed…Think it through. Then check your answer.Question
In the latch circuit shown, the NAND gates have non-zero, but unequal propagation delays. The present input condition is: . If the input condition is changed simultaneously to , the outputs and are
Correct answer
(B) either X = '1', Y = '0' or X = '0', Y = '1'
Step-by-step solution
The circuit is a NAND-based SR latch.1.Initial state: . For a NAND gate, if any input is 0, the output is 1. Thus, and .2.Transition: and both change to 1.3.Now, the top NAND gate has inputs and , so its output tends to . The bottom NAND gate has inputs and , so its output tends to .4.Since the propagation delays are unequal, one gate will switch to 0 faster than the other.5.If the top gate is faster, becomes 0 first. This 0 is fed back to the bottom gate, making its inputs and , so stays at .6.If the bottom gate is faster, becomes 0 first. This 0 is fed back to the top gate, making its inputs and , so stays at .7.Therefore, the final state will be either or .16
Q16MCQ1 markEasyThe clock frequency of an 8085 microprocessor is . If the time required to execute an instruction is , then the number of T-states needed for…Think it through. Then check your answer.Question
The clock frequency of an 8085 microprocessor is . If the time required to execute an instruction is , then the number of T-states needed for executing the instruction isCorrect answer
(C) 7
Step-by-step solution
Given:
Clock frequency .
Time period of one T-state .
Total execution time for the instruction .
Number of T-states .17
Q17NAT1 markHardConsider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% duty…Think it through. Then check your answer.Question
Consider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% duty cycle and CLK2 is a one-fifth period delayed version of CLK1. The duty cycle at the output of the latch in percentage is _________.
Correct answer
29.9 to 30.1
Step-by-step solution
1.Analyze the signals: Let the period of the clock signals be .- CLK1 has a 50% duty cycle, so it is high for and low for .
- CLK2 is a one-fifth period delayed version of CLK1, so it is high for and low for .
3.Determine Output :- In the interval , CLK2 is high (transparent mode):
- From to , , so .
- From to , , so .
- At , CLK2 goes low (hold mode). The latch holds the value .
- remains until the next transparency interval starts at .
- The high time of in one period is .
- Duty cycle = .
18
Q18NAT1 markMediumThe open loop transfer function where is an integer, is connected in unity feedback configuration as shown in the figure. [figure]…Think it through. Then check your answer.Question
The open loop transfer function where is an integer, is connected in unity feedback configuration as shown in the figure.Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter is _________.
Correct answer
0.99 to 1.01
Step-by-step solution
1.Steady State Error for Step Input: For a unity feedback system, the steady state error for a unit step input is , where . For , must be . This implies that the system must be at least Type 1, so .2.Steady State Error for Ramp Input: The steady state error for a unit ramp input is , where . Given , we have .3.Solve for :For to be a finite non-zero value (), the power of in the denominator must be zero. Thus, , which gives .4.Verification: If :This matches the given error . Therefore, .19
Q19MCQ1 markEasyConsider a stable system with transfer function where and …Think it through. Then check your answer.Question
Consider a stable system with transfer functionwhere and are real valued constants. The slope of the Bode log magnitude curve of converges to dB/decade as . A possible pair of values for and isCorrect answer
(A) p = 0 and q = 3
Step-by-step solution
The transfer function is given as:At high frequencies (), the magnitude of the transfer function is dominated by the highest power of in the numerator and denominator:The magnitude in dB is:The slope of the Bode log magnitude curve as is given by the coefficient of :Given that the slope converges to dB/decade:Checking the options:- (A) (Matches)
- (B)
- (C)
- (D)
20
Q20MCQ1 markEasyWhich of the following can be the pole-zero configuration of a phase-lag controller (lag compensator)?Think it through. Then check your answer.Question
Which of the following can be the pole-zero configuration of a phase-lag controller (lag compensator)?Correct answer
(A) [figure]
Step-by-step solution
A phase-lag compensator (or lag compensator) is used to improve the steady-state error of a system. It introduces a pole and a zero, with the pole being closer to the origin than the zero. This configuration provides a negative phase shift, hence 'lag'.The transfer function of a lag compensator is typically given by , where . This means the zero is further away from the origin than the pole.Let's analyze the given pole-zero plots:Option (A) shows a pole at (which is not part of the compensator itself but often represents the system type), a zero at a negative real value, and a pole at a negative real value further to the left than the zero. This configuration, specifically the pole-zero pair on the negative real axis, has the pole closer to the origin than the zero. This is characteristic of a phase-lag compensator. For example, if the zero is at and the pole is at , then .Option (B) shows a pole at , a pole at a negative real value, and a zero at a negative real value further to the left than the pole. This configuration, where the zero is further from the origin than the pole, is characteristic of a phase-lead compensator.Option (C) is similar to (A) but the pole-zero pair is located differently. However, the relative positions of the pole and zero for the compensator part are consistent with a lag compensator (pole closer to origin than zero).Option (D) is similar to (B) and represents a lead compensator.Comparing (A) and (C), both show a pole closer to the origin than the zero for the compensator part. However, the standard representation of a lag compensator places the pole and zero on the negative real axis, with the pole being to the right of the zero (closer to the origin). Option (A) correctly depicts this relationship for the compensator pole-zero pair (excluding the pole at origin which is likely part of the plant).Therefore, option (A) is the correct representation of a phase-lag controller's pole-zero configuration.The final answer is21
Q21NAT1 markEasyLet be independent random variables. has mean and variance , while has mean and variance . The mutual information between…Think it through. Then check your answer.Question
Let be independent random variables. has mean and variance , while has mean and variance . The mutual information between and in bits is ________.Correct answer
0 to 0
Step-by-step solution
Mutual information quantifies the amount of information obtained about one random variable through observing the other random variable. It is defined as:
whereH(X)is the entropy of and is the conditional entropy of given .For independent random variables and , the knowledge of one variable does not provide any information about the other. Therefore, the conditional entropy is equal to the entropy , and is equal to .So, .
Similarly, .Thus, for independent random variables, the mutual information is always zero.The given means and variances are irrelevant for calculating mutual information when the variables are independent.The final answer is22
Q22MCQ1 markEasyWhich one of the following statements about differential pulse code modulation (DPCM) is true?Think it through. Then check your answer.Question
Which one of the following statements about differential pulse code modulation (DPCM) is true?Correct answer
(D) The difference of message signal sample with its prediction is quantized
Step-by-step solution
Differential Pulse Code Modulation (DPCM) is a signal digitization technique that quantizes the difference between the actual message signal sample and its predicted value (based on previous samples). This difference is known as the prediction error. By quantizing the error signal, which typically has a smaller dynamic range than the original signal, DPCM achieves better efficiency or lower bit rates for the same quality.23
Q23MCQ1 markMediumIn a digital communication system, the overall pulse shape at the receiver before the sampler has the Fourier transform . If the symbols are transmitted at the rate…Think it through. Then check your answer.Question
In a digital communication system, the overall pulse shape at the receiver before the sampler has the Fourier transform . If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is the inter symbol interference zero?Correct answer
(B) [figure]
Step-by-step solution
The Nyquist criterion for zero Inter-Symbol Interference (ISI) states that the sum of the shifted replicas of the Fourier transform of the pulse shape must be constant:where is the symbol rate. Given symbols/sec, we have kHz. We need to check if is constant for kHz.For Option (B), is 1 for kHz and for kHz.
In the interval kHz:1.For kHz, .2.For kHz, the relevant terms are and . Since is symmetric, .Sum .Since the sum is constant () for all , the ISI is zero. Other options do not satisfy this condition (e.g., in Option C, the sum at is , while at it is 1).24
Q24NAT1 markMediumThe voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is Volts, where is the distance…Think it through. Then check your answer.Question
The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is Volts, where is the distance along the length of the cable in metres, is the complex propagation constant, and is the angular frequency. The absolute value of the attenuation in the cable in dB/metre is ______.Correct answer
0.85 to 0.88
Step-by-step solution
Given the voltage expression , the propagation constant is .
From the given value , we identify the attenuation constant .
To convert the attenuation from Nepers per meter (Np/m) to decibels per meter (dB/m), we use the relation:Rounding to two decimal places, we get approximately 0.87, which is within the range [0.85, 0.88].25
Q25MCQ1 markMediumConsider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the…Think it through. Then check your answer.Question
Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless linkCorrect answer
(C) remains unchanged
Step-by-step solution
The received power in a free-space link is given by the Friis transmission equation:where is the transmitted power, and are the effective areas of the transmitter and receiver antennas, is the wavelength, and is the distance between them.If , , and are all doubled:
The new received power is:Since the received power remains unchanged and all other factors (bandwidth, noise power) are unchanged, the Signal-to-Noise Ratio (SNR) remains the same. Consequently, the maximum capacity remains unchanged.26
Q26MCQ1 markEasyLet for real . From among the following, choose the Taylor series approximation of around , which includes all powers of less than or equal…Think it through. Then check your answer.Question
Let for real . From among the following, choose the Taylor series approximation of around , which includes all powers of less than or equal to 3.Correct answer
(C) 1 + x + (3)/(2)x² + (7)/(6)x³
Step-by-step solution
The Taylor series expansion for around is:Substitute :Expand the terms and keep only powers of up to 3:1.Constant term:2. term:3. term:4. term:Thus, the approximation is .27
Q27NAT1 markMediumA three dimensional region of finite volume is described by where are real. The volume of (up to two decimal places) is…Think it through. Then check your answer.Question
A three dimensional region of finite volume is described bywhere are real. The volume of (up to two decimal places) is ________.Correct answer
0.7 to 0.85
Step-by-step solution
To find the volume of the region , we can use triple integration in cylindrical coordinates .The region is defined by the inequalities:1.2.In cylindrical coordinates, . The first inequality becomes , which implies .
The limits for the variables are:28
Q28NAT1 markMediumLet where are real, and let be the straight line segment from point to point . The value…Think it through. Then check your answer.Question
Let where are real, and let be the straight line segment from point to point . The value of is ________.Correct answer
-11.1
Step-by-step solution
The given line integral is where .
First, we check if the vector field is conservative by calculating its curl:Since the curl is zero, the field is conservative. There exists a potential function such that .
Integrating the components:1.2.3.Combining these, we get the potential function .
The value of the line integral for a conservative field is independent of the path and is given by the difference in the potential function at the endpoints:
Given and :
.29
Q29MCQ1 markMediumWhich one of the following is the general solution of the first order differential equation , where are real?Think it through. Then check your answer.Question
Which one of the following is the general solution of the first order differential equation , where are real?Correct answer
(D) y = 1 - x + tan(x + c), where c is a constant.
Step-by-step solution
Given the differential equation:Let . Differentiating both sides with respect to :Substituting from the original equation:This is a separable differential equation:Integrating both sides:Substituting back :This matches option (D).30
Q30NAT1 markMediumStarting with , the solution of the equation , after two iterations of Newton-Raphson's method (up to two decimal places) is ___________.Think it through. Then check your answer.Question
Starting with , the solution of the equation , after two iterations of Newton-Raphson's method (up to two decimal places) is ___________.Correct answer
0.65 to 0.72
Step-by-step solution
To solve the equation using the Newton-Raphson method, we define the function:
The derivative of the function is:
The Newton-Raphson iteration formula is:
Given the initial guess :Iteration 1:
Iteration 2:
Rounding to two decimal places, the solution is approximately . The official range for the correct answer is to .31
Q31MCQ1 markMediumLet x(t) be a continuous time periodic signal with fundamental period T = 1 seconds. Let {aₖ} be the complex Fourier series coefficients of x(t), where k is integer valued.…Think it through. Then check your answer.Question
Let x(t) be a continuous time periodic signal with fundamental period T = 1 seconds. Let {aₖ} be the complex Fourier series coefficients of x(t), where k is integer valued. Consider the following statements about x(3t):I. The complex Fourier series coefficients of x(3t) are {aₖ} where k is integer valued
II. The complex Fourier series coefficients of x(3t) are {3aₖ} where k is integer valued
III. The fundamental angular frequency of x(3t) is 6π rad/sFor the three statements above, which one of the following is correct?Correct answer
(B) only I and III are true
Step-by-step solution
Let the original signal be with fundamental period s.
The fundamental frequency is Hz.
The fundamental angular frequency is rad/s.
The Fourier series representation of is given by:where are the complex Fourier series coefficients.Now, consider the time-scaled signal . This is a time compression by a factor of 3.Analysis of Period and Frequency (Statement III):
The new fundamental period is given by s.
The new fundamental frequency is Hz.
The new fundamental angular frequency is rad/s.
Therefore, Statement III is true.Analysis of Fourier Coefficients (Statements I and II):
Let's find the Fourier series for . We substitute for in the Fourier series expansion of :The Fourier series of is also defined with respect to its own fundamental frequency :where are the Fourier coefficients of .
By comparing the two expressions for , we can see that the coefficients must be equal for each harmonic index .This is a standard property of the Fourier series: time scaling a signal to does not change the values of the Fourier series coefficients, it only changes the fundamental frequency to .Therefore, the complex Fourier series coefficients of are .
Statement I is true.
Statement II is false.Conclusion:
Statements I and III are true.
Thus, the correct option is (B).32
Q32NAT1 markMediumTwo discrete-time signals and are both non-zero only for , and are zero otherwise. It is given that…Think it through. Then check your answer.Question
Two discrete-time signals and are both non-zero only for , and are zero otherwise. It is given thatLet be the linear convolution of and . Given that and , the value of the expression is ________.Correct answer
31 to 31
Step-by-step solution
Given that and are non-zero only for , we have:
for
for , with .The linear convolution is given by:
Using the given values:
Now, we calculate and :
The value of the expression is:
.33
Q33NAT1 markMediumLet be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by…Think it through. Then check your answer.Question
Let be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by
and for and .
Let be the discrete-time Fourier transform (DTFT) of , where is the normalized angular frequency in radians. Given that and , the value of (in radians) is equal to ________.Correct answer
2.05 to 2.15
Step-by-step solution
The impulse response is given as for .
The Discrete-Time Fourier Transform (DTFT) is defined as:Substituting the given values:Using Euler's formula :We are given for :Since is never zero, we solve for:In the interval , the solution is:The value is approximately .34
Q34NAT1 markMediumThe figure shows an RLC circuit excited by the sinusoidal voltage Volts, where is in seconds. The ratio…Think it through. Then check your answer.Question
The figure shows an RLC circuit excited by the sinusoidal voltage Volts, where is in seconds. The ratio is ________.Correct answer
2.55 to 2.65
Step-by-step solution
Given the excitation voltage , the angular frequency is .The impedances of the components are:- Inductor:
- Resistor:
- Capacitor:
Let be the current flowing through the rightmost branch. Then:The ratio of the amplitudes is the ratio of the magnitudes of these phasors:35
Q35NAT1 markMediumIn the circuit shown, the voltage is described by: where…Think it through. Then check your answer.Question
In the circuit shown, the voltage is described by:where is in seconds. The time (in seconds) at which the current in the circuit will reach the value 2 Amperes is ________.Correct answer
0.3 to 0.4
Step-by-step solution
Note: Although the symbol in the diagram resembles an inductor labeled , the official answer range corresponds to the component being a capacitor . We will proceed with the capacitor model.For , . The circuit consists of a resistor in series with a parallel combination of a capacitor and a resistor. The current is the current through the resistor.1.Find the Thevenin equivalent across the capacitor terminals:
3.Current through the resistor:4.Solve for when :36
Q36NAT2 marksMediumThe dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron concentration of…Think it through. Then check your answer.Question
The dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron concentration of and electronic charge . If a bias of is applied across a region of this semiconductor, the resulting current density in this region, in , is ________.Correct answer
1.5 to 1.7
Step-by-step solution
1.Calculate the applied electric field ():2.Determine drift velocity () from the graph:The graph shows drift velocity is linear up to a saturation point. The saturation velocity is reached at an electric field .
Since our applied field is less than , the electron is in the linear region.
The mobility is:
The drift velocity at is:
3.Calculate current density ():37
Q37NAT2 marksHardAs shown, a uniformly doped Silicon (Si) bar of length with a donor concentration is illuminated at such that electron and hole…Think it through. Then check your answer.Question
As shown, a uniformly doped Silicon (Si) bar of length with a donor concentration is illuminated at such that electron and hole pairs are generated at the rate of , where . Hole lifetime is , electronic charge , hole diffusion coefficient and low level injection condition prevails. Assuming a linearly decaying steady state excess hole concentration that goes to 0 at , the magnitude of the diffusion current density at , in , is ________.
Correct answer
15.9 to 16.1
Step-by-step solution
1.The generation rate is given as .2.The problem states that the steady-state excess hole concentration is linearly decaying and goes to 0 at . Thus, .3.In steady state, for a constant diffusion current density (implied by the linear profile), the generation rate must balance the recombination rate: .4.Substituting the expressions: .5.Calculate : .6.The hole diffusion current density is .7.The gradient is .8.Thus, .9.Convert to cm: .10.Calculate magnitude: .38
Q38NAT2 marksMediumAs shown, two Silicon (Si) abrupt - junction diodes are fabricated with uniform donor doping concentrations of and…Think it through. Then check your answer.Question
As shown, two Silicon (Si) abrupt - junction diodes are fabricated with uniform donor doping concentrations of and in the -regions, and uniform acceptor doping concentrations of and in the -regions of the diodes, respectively. Assuming that the reverse bias voltage is built-in potentials of the diodes, the ratio of their reverse bias capacitances for the same applied reverse bias, is ________.
Correct answer
10 to 10
Step-by-step solution
The junction capacitance for an abrupt - junction is given by:Given that , we can approximate:For Diode 1:
For Diode 2:
The ratio is:39
Q39NAT2 marksMediumIn the figure shown, the transistor acts as a switch. [figure] For the input as shown in the figure, the transistor switches between the cut-off and saturation…Think it through. Then check your answer.Question
In the figure shown, the transistor acts as a switch.For the input as shown in the figure, the transistor switches between the cut-off and saturation regions of operation, when T is large. Assume collector-to-emitter voltage at saturation and base-to-emitter voltage . The minimum value of the common-base current gain () of the transistor for the switching should be ________.
Correct answer
0.89 to 0.91
Step-by-step solution
1.For the transistor to act as a switch, it must enter saturation when .2.Calculate the base current :.3.Calculate the collector current at the edge of saturation :.4.For saturation to occur, we require , so the minimum current gain is:.5.The common-base current gain is related to by .6.Minimum .7.The value falls within the acceptable range of to .40
Q40NAT2 marksMediumFor the circuit shown, assume that the NMOS transistor is in saturation. Its threshold voltage and its transconductance parameter…Think it through. Then check your answer.Question
For the circuit shown, assume that the NMOS transistor is in saturation. Its threshold voltage and its transconductance parameter . Neglect channel length modulation and body bias effects. Under these conditions, the drain current in mA is ________.Correct answer
1.9 to 2.1
Step-by-step solution
The circuit diagram shows an NMOS transistor with a voltage divider biasing the gate and a source resistor. We need to find the drain current .1. Calculate Gate Voltage ():
The gate is biased by a voltage divider formed by and from .
, , .
.2. Apply KVL to the source loop:
For an NMOS transistor, . Also, .
So, .3. Use the saturation region current equation:
The transistor is stated to be in saturation. The drain current in saturation is given by:
Given and .
Substitute into the equation. Note that is in mA and is in k, so will be in V.
4. Solve the quadratic equation for :
Using the quadratic formula :
Two possible values for : or .5. Check for saturation condition:
For saturation, . Also, .
Let's check for both values:
If : . This is less than , so the transistor is in cutoff, not saturation. This solution is invalid.
If : . This is greater than , so the transistor can be in saturation.Now, let's verify the saturation condition for .
.
.
.
.
Since , the transistor is indeed in saturation for .Thus, the drain current .The final answer is .41
Q41NAT2 marksMediumFor the DC analysis of the Common-Emitter amplifier shown, neglect the base current and assume that the emitter and collector currents are equal. Given that ,…Think it through. Then check your answer.Question
For the DC analysis of the Common-Emitter amplifier shown, neglect the base current and assume that the emitter and collector currents are equal. Given that , , and the BJT output resistance is practically infinite. Under these conditions, the midband voltage gain magnitude, V/V, is ________.Correct answer
127 to 129
Step-by-step solution
1.DC Analysis:- Calculate base voltage :
- Calculate emitter voltage :
- Calculate emitter current :
- Since base current is neglected, .
- Calculate transconductance :
3.AC Analysis:- The midband voltage gain magnitude for a bypassed Common-Emitter amplifier is:
- Calculate gain magnitude:
42
Q42NAT2 marksMediumThe amplifier circuit shown in the figure is implemented using a compensated operational amplifier (op-amp), and has an open-loop voltage gain, and an…Think it through. Then check your answer.Question
The amplifier circuit shown in the figure is implemented using a compensated operational amplifier (op-amp), and has an open-loop voltage gain, and an open-loop cut-off frequency, . The voltage gain of the amplifier at , in V/V, is ________.Correct answer
43.3 to 45.3
Step-by-step solution
The circuit shown is a non-inverting amplifier configuration.1. Calculate the ideal closed-loop gain ():
For a non-inverting op-amp, the ideal closed-loop gain is given by:
From the circuit, and .
.2. Understand the op-amp frequency response:
The op-amp is compensated, meaning it behaves like a single-pole system. Its open-loop gain can be described by:
Given and .3. Calculate the gain-bandwidth product (GBW):
For a single-pole compensated op-amp, the gain-bandwidth product is constant:
.4. Calculate the closed-loop bandwidth ():
The closed-loop bandwidth of the non-inverting amplifier is approximately:
.5. Calculate the actual closed-loop gain at the given frequency:
The voltage gain of the amplifier at a frequency is given by:
We need to find the magnitude of the gain at .
Substitute the values: , , .
.The final answer is .43
Q43MCQ2 marksEasyWhich one of the following gives the simplified sum of products expression for the Boolean function , where and are minterms…Think it through. Then check your answer.Question
Which one of the following gives the simplified sum of products expression for the Boolean function , where and are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?Correct answer
(B) A C + AB + ABC
Step-by-step solution
The given Boolean function is .
The inputs are A, B, and C, with A as the Most Significant Bit (MSB) and C as the Least Significant Bit (LSB).The minterms are:- (for input 000)
- (for input 010)
- (for input 011)
- (for input 101)
We can form groups of 1s:BC=00 BC=01 BC=11 BC=10 A=0 1 0 1 1 A=1 0 1 0 0 1.A group of two 1s at and . This covers and . The common term is .2.A group of two 1s at and . This covers and . The common term is .3.A single 1 at , which is . This cannot be grouped with any other adjacent 1.The simplified Sum of Products (SOP) expression is the sum of these terms:
.This matches option (B).44
Q44NAT2 marksHardA 4-bit shift register circuit configured for right-shift operation, i.e. , is shown. If the present state…Think it through. Then check your answer.Question
A 4-bit shift register circuit configured for right-shift operation, i.e. , is shown. If the present state of the shift register is ABCD = 1101, the number of clock cycles required to reach the state ABCD = 1111 is __________.Correct answer
10 to 10
Step-by-step solution
The problem describes a 4-bit shift register with state (A, B, C, D) performing a right-shift operation. The next state is given by .The diagram shows that the input is generated by feedback from the outputs of flip-flops C and D, using an XOR gate. So, the feedback logic shown in the diagram is .Let's trace the state transitions using the logic from the diagram:
Initial state (Cycle 0): (A, B, C, D) = (1, 1, 0, 1).- .
- Now, .
- Cycle 0: (1, 1, 0, 1). .
- Cycle 1: (0, 1, 1, 0). .
- Cycle 2: (0, 0, 1, 1). .
- Cycle 3: (1, 0, 0, 1). .
- Cycle 4: (0, 1, 0, 0). .
- Cycle 5: (0, 0, 1, 0). .
- Cycle 6: (0, 0, 0, 1). .
- Cycle 7: (1, 0, 0, 0). .
- Cycle 8: (1, 1, 0, 0). .
- Cycle 9: (1, 1, 1, 0). .
- Cycle 10: (1, 1, 1, 1). The target state is reached.
45
Q45MCQ2 marksEasyThe following FIVE instructions were executed on an 8085 microprocessor. [code] The Accumulator value immediately after the execution of the fifth instruction isThink it through. Then check your answer.Question
The following FIVE instructions were executed on an 8085 microprocessor.The Accumulator value immediately after the execution of the fifth instruction isMVI A, 33H MVI B, 78H ADD B CMA ANI 32HCorrect answer
(B) 10H
Step-by-step solution
Let's trace the execution of the 8085 microprocessor instructions:1.The Accumulator (A) is loaded with the immediate dataMVI A, 33H(Move Immediate to Accumulator):33H.
A =33H(binary:0011 0011)2.Register B is loaded with the immediate dataMVI B, 78H(Move Immediate to Register B):78H.
B =78H(binary:0111 1000)3.The content of Register B is added to the content of the Accumulator. The result is stored in the Accumulator.ADD B(Add Register B to Accumulator):
A = A + B
A =33H+78H
In decimal:
In hexadecimal:
A =ABH(binary:1010 1011)4.The content of the Accumulator is complemented (1's complement). Each bit is inverted.CMA(Complement Accumulator):
A =
Current A =1010 1011
Complemented A =0101 0100
A =54H(binary:0101 0100)5.The content of the Accumulator is logically ANDed with the immediate dataANI 32H(AND Immediate with Accumulator):32H. The result is stored in the Accumulator.
A = A AND32H
Current A =54H(binary:0101 0100)
Immediate data =32H(binary:0011 0010) Performing bitwise AND:
0101 0100(A)
&0011 0010(32H)
-------------
0001 0000The result is0001 0000in binary, which is10Hin hexadecimal.
A =10HTherefore, the Accumulator value immediately after the execution of the fifth instruction is10H.The final answer is46
Q46MCQ2 marksMediumA finite state machine (FSM) is implemented using the D flip-flops A and B, and logic gates, as shown in the figure below. The four possible states of the FSM are…Think it through. Then check your answer.Question
A finite state machine (FSM) is implemented using the D flip-flops A and B, and logic gates, as shown in the figure below. The four possible states of the FSM are and . Assume that is held at a constant logic level throughout the operation of the FSM. When the FSM is initialized to the state and clocked, after a few clock cycles, it starts cycling throughCorrect answer
(D) (D) only two of the four possible states if X_(IN) = 0
Step-by-step solution
From the circuit diagram, the next-state equations for the flip-flops are:
(NAND gate)Let's analyze the state transitions starting from for both cases of :Case 1:
(always)- Initial state:
- Next state: ,
- Next state: ,
- Next state: ,
- Initial state:
- Next state: ,
- Next state: ,
- Next state: ,
47
Q47MCQ2 marksMediumA linear time invariant (LTI) system with the transfer function is connected in unity feedback configuration as shown in the…Think it through. Then check your answer.Question
A linear time invariant (LTI) system with the transfer functionis connected in unity feedback configuration as shown in the figure.For the closed loop system shown, the root locus for intersects the imaginary axis for . The closed loop system is stable for
Correct answer
(A) K 1.5
Step-by-step solution
To determine the stability of the closed-loop system, we first find the characteristic equation:Substituting the given transfer function :For a second-order system to be stable, all coefficients must have the same sign. Assuming :1.2.3.Combining these conditions for , the system is stable when . At , the coefficient becomes zero, and the roots lie on the imaginary axis, which is consistent with the problem statement.48
Q48MCQ2 marksMediumWhich one of the following options correctly describes the locations of the roots of the equation on the complex plane?Think it through. Then check your answer.Question
Which one of the following options correctly describes the locations of the roots of the equation on the complex plane?Correct answer
(C) Two RHP roots and two LHP roots
Step-by-step solution
The given characteristic equation is .Method 1: Direct Solution
Let . The equation becomes a quadratic in :
The roots for are:
Now, solve for :1.For :
Roots: (Right Half Plane) and (Left Half Plane).2.For :
Roots: (Right Half Plane) and (Left Half Plane).In total, there are two roots in the Right Half Plane (RHP) and two roots in the Left Half Plane (LHP).Method 2: Routh-Hurwitz Criterion
Construct the Routh array for :
: 1 1 1
: 0 (Row of zeros)
Form the auxiliary equation from the row: .
Take the derivative: .
Replace the row with coefficients [4, 2]:
: 1 1 1
: 4 2
: 1
:
: 1There are two sign changes in the first column (0.5 to -6 and -6 to 1), indicating 2 roots in the RHP. Since the polynomial is even, the roots are symmetric about the origin, implying 2 roots in the LHP. No roots are on the imaginary axis as the coefficient is non-zero and the Routh array completed with sign changes.Thus, option (C) is correct.49
Q49MCQ2 marksMediumThe Nyquist plot of the transfer function does not encircle the point for but does encircle the point …Think it through. Then check your answer.Question
The Nyquist plot of the transfer function does not encircle the point for but does encircle the point for . Then the closed loop system (having unity gain feedback) isCorrect answer
(B) stable for K = 10 and unstable for K = 100
Step-by-step solution
The open-loop transfer function is given by .1.Determine Open-Loop Stability:The open-loop poles are the roots of .2.Apply Nyquist Stability Criterion:The number of closed-loop poles in the RHP is given by , where is the number of clockwise encirclements of the critical point by the Nyquist plot.
For the system to be stable, we must have . Since , stability requires (no encirclements).3.Analyze for given K values:- For : The problem states the plot does not encircle . Thus, . . The system is stable.
- For : The problem states the plot encircles . Thus, . . The system is unstable.
50
Q50MCQ2 marksMediumIn binary frequency shift keying (FSK), the given signal waveforms are , and , where…Think it through. Then check your answer.Question
In binary frequency shift keying (FSK), the given signal waveforms are
, and
,
where is the bit-duration interval and is in seconds. Both and are zero outside the interval . With a matched filter (correlator) based receiver, the smallest positive value of (in milliseconds) required to have and uncorrelated isCorrect answer
(B) 0.5 ms
Step-by-step solution
Two signals and are uncorrelated if their correlation is zero. For zero-mean signals, this is equivalent to orthogonality:Substituting the given waveforms:Using the identity :Neglecting the high-frequency term (or noting that for , both terms become zero):For the smallest positive value, :Thus, the smallest positive value of is 0.5 ms.51
Q51MCQ2 marksMediumLet be a wide sense stationary random process with the power spectral density as shown in Figure (a), where is in Hertz (Hz). The random process is…Think it through. Then check your answer.Question
Let be a wide sense stationary random process with the power spectral density as shown in Figure (a), where is in Hertz (Hz). The random process is input to an ideal lowpass filter with the frequency responseas shown in Figure (b). The output of the lowpass filter is .
Let be the expectation operator and consider the following statements:I.
II.
III. Select the correct option:Correct answer
(A) only I is true
Step-by-step solution
1.Mean of the processes: For a Wide Sense Stationary (WSS) process with mean , the power spectral density contains an impulse at . The given is continuous at and has no impulse, implying .2.Mean of the output: For an LTI system with frequency response , the mean of the output is . Since from the definition of the filter, . Thus, is true. Statement I is TRUE.3.Power of : The average power of is given by the area under its PSD:4.Power of : The PSD of the output is . For the ideal lowpass filter with cutoff Hz:5.Evaluation of Statements II and III:- Statement II: and . They are not equal. Statement II is FALSE.
- Statement III: . Statement III is FALSE.
52
Q52NAT2 marksMediumA continuous time signal , where is in seconds, is the input to a linear time invariant (LTI) filter with the impulse response…Think it through. Then check your answer.Question
A continuous time signal , where is in seconds, is the input to a linear time invariant (LTI) filter with the impulse responseLet be the output of this filter. The maximum value of is ______.Correct answer
7.9 to 8.1
Step-by-step solution
The impulse response of the filter is given by .
Taking the Fourier transform to find the frequency response :
Using the property , we can rewrite as:
This is an ideal low-pass filter with a gain of 2 and a cutoff frequency Hz.The input signal is .
The frequencies present in the input are:
Hz
HzEvaluating the filter response at these frequencies:
For Hz: , so .
For Hz: , so .The output signal is:
The maximum value of is 8.53
Q53NAT2 marksHardAn optical fiber is kept along the direction. The refractive indices for the electric fields along and directions in the fiber are and…Think it through. Then check your answer.Question
An optical fiber is kept along the direction. The refractive indices for the electric fields along and directions in the fiber are and , respectively ( due to the imperfection in the fiber cross-section). If the free space wavelength of a light wave propagating in the fiber is . If the lightwave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimetres, is ______.Correct answer
0.36 to 0.38
Step-by-step solution
The fiber is birefringent with refractive indices and . The difference in refractive indices is .
The free space wavelength is cm.A circularly polarized wave has a phase difference of (or ) between its orthogonal components ( and ). For the wave to become linearly polarized, the total phase difference must become an integer multiple of (e.g., ).The additional phase difference introduced by the fiber over a distance is given by:
To change from circular polarization (initial phase difference ) to linear polarization (target phase difference ), the minimum additional phase shift required is :
cm.54
Q54MCQ2 marksMediumThe expression for an electric field in free space is , where represent the spatial…Think it through. Then check your answer.Question
The expression for an electric field in free space is , where represent the spatial coordinates, represents time, and are constants. This electric fieldCorrect answer
(C) represents an elliptically polarized plane wave propagating along the x - y plane.
Step-by-step solution
1.The phase term is . Comparing with , we find , so the propagation vector is . This is in the - plane.2.For a plane wave in free space, . Here, . Thus, it is a plane wave.3.The field has components , , and .4.Let's define a coordinate system relative to . The propagation direction is . One transverse component is . The other transverse component is .5.The field can be written as .6.Since the two orthogonal components have a phase difference () but unequal magnitudes (), the wave is elliptically polarized.55
Q55MCQ2 marksMediumA half wavelength dipole is kept in the - plane and oriented along from the -axis. Determine the direction of null in the radiation pattern for…Think it through. Then check your answer.Question
A half wavelength dipole is kept in the - plane and oriented along from the -axis. Determine the direction of null in the radiation pattern for . Here the angle () is measured from the -axis, and the angle () is measured from the -axis in the - plane.Correct answer
(A) θ = 90^, φ = 45^
Step-by-step solution
A half-wave dipole has its radiation nulls along its axis. The dipole is placed in the - plane, which corresponds to . It is oriented at an angle of with respect to the -axis in the - plane, meaning its axis lies along the direction (and ). Since we are looking for the null in the range , the direction of the null is and .