The PYQ practice room
GATE EC 2014 Set 3
All 65 solved GATE EC 2014 Set 3 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.
Explore the questions
General Aptitude (GA)
101
Q1MCQ1 markEasy"India is a country of rich heritage and cultural diversity." Which one of the following facts best supports the claim made in the above sentence?Think it through. Then check your answer.Question
"India is a country of rich heritage and cultural diversity."
Which one of the following facts best supports the claim made in the above sentence?Correct answer
(C) India is home to 22 official languages and thousands of dialects.
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The statement "India is a country of rich heritage and cultural diversity" implies a variety of traditions, languages, and ways of life. Among the given options:
(A) India being a union of states and territories describes its political structure, not directly its cultural diversity.
(B) India's population size is a demographic fact, not directly indicative of cultural diversity.
(C) India being home to 22 official languages and thousands of dialects directly supports the idea of cultural and linguistic diversity, which is a key aspect of a rich heritage.
(D) The Indian cricket team drawing players from over ten states is a specific fact about sports, which might indirectly reflect diversity but is not the best support for the general claim.
Therefore, option (C) best supports the claim of rich heritage and cultural diversity.2
Q2MCQ1 markEasyThe value of one U.S. dollar is 65 Indian Rupees today, compared to 60 last year. The Indian Rupee hasThink it through. Then check your answer.Question
The value of one U.S. dollar is 65 Indian Rupees today, compared to 60 last year. The Indian Rupee hasCorrect answer
(B) depreciated
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Last year, 1 USD = 60 INR.
Today, 1 USD = 65 INR.
This means that more Indian Rupees are now required to buy one U.S. dollar. In other words, the Indian Rupee has become weaker relative to the U.S. dollar. This phenomenon is known as depreciation.- Depreciation refers to a decrease in the value of a currency relative to other currencies in a floating exchange rate system.
- Appreciation would mean the Indian Rupee became stronger (e.g., 1 USD = 55 INR).
- Depressed is a general term and not a specific economic term for currency value change.
- Stabilized would imply no significant change in value.
3
Q3MCQ1 markEasy'Advice' isThink it through. Then check your answer.Question
'Advice' isCorrect answer
(B) a noun
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The word 'advice' (with a 'c') is a noun, meaning guidance or recommendations offered with regard to prudent future action. For example, "I need some advice on my career."The corresponding verb is 'advise' (with an 's'), meaning to offer suggestions about the best course of action to someone. For example, "I advise you to study hard."Therefore, 'advice' is a noun.4
Q4NAT1 markEasyThe next term in the series 81, 54, 36, 24, ... isThink it through. Then check your answer.Question
The next term in the series 81, 54, 36, 24, ... isCorrect answer
16 to 16
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The given series is 81, 54, 36, 24, ...
Let's find the ratio between consecutive terms:
This is a geometric progression with a common ratio of .
To find the next term, multiply the last term by the common ratio:
Next term .The final answer is .5
Q5MCQ1 markEasyIn which of the following options will the expression be definitely true?Think it through. Then check your answer.Question
In which of the following options will the expression be definitely true?Correct answer
(D) P = A<R<M
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Let's analyze each option:(A)
This implies and . We cannot definitively conclude . For example, if , then .(B)
This implies , , and . We cannot definitively conclude . For example, if , then .(C)
This implies and . Therefore, . This is the opposite of .(D)
This implies , , and . Combining these inequalities, we get . This definitively means .Thus, option (D) is the correct answer.The final answer is .6
Q6MCQ2 marksEasyFind the next term in the sequence: 7G, 11K, 13M, ___Think it through. Then check your answer.Question
Find the next term in the sequence: 7G, 11K, 13M, ___Correct answer
(B) 17Q
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The sequence follows a pattern where each term consists of a prime number followed by the letter at that numerical position in the alphabet:1.Number Pattern: The numbers 7, 11, and 13 are consecutive prime numbers. The next prime number after 13 is 17.2.Letter Pattern: The letters G, K, and M correspond to their positions in the English alphabet:- G is the letter.
- K is the letter.
- M is the letter.
3.Next Term: For the next prime number, 17, the corresponding letter is the letter of the alphabet, which is Q.Therefore, the next term in the sequence is 17Q.7
Q7MCQ2 marksEasyThe multi-level hierarchical pie chart shows the population of animals in a reserve forest. The correct conclusions from this information are: (i) Butterflies are birds (ii) There…Think it through. Then check your answer.Question
The multi-level hierarchical pie chart shows the population of animals in a reserve forest. The correct conclusions from this information are:
(i) Butterflies are birds
(ii) There are more tigers in this forest than red ants
(iii) All reptiles in this forest are either snakes or crocodiles
(iv) Elephants are the largest mammals in this forestCorrect answer
(D) (i), (ii) and (iii) only
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Based on the hierarchical pie chart provided:
(i) Butterflies are birds: In the chart, the 'Butterfly' segment is a sub-category of the 'Bird' category. Therefore, according to the chart, butterflies are birds. (True)
(ii) There are more tigers in this forest than red ants: Comparing the angular area of the 'Tiger' segment (under Mammals) and the 'Red ant' segment (under Insects), the Tiger segment is visually larger. (True)
(iii) All reptiles in this forest are either snakes or crocodiles: The 'Reptile' category is entirely divided into two sub-categories: 'Snake' and 'Crocodile'. (True)
(iv) Elephants are the largest mammals in this forest: The pie chart represents population distribution, not the physical size of the animals. We cannot conclude physical size from this data. (False)Thus, conclusions (i), (ii), and (iii) are correct. The correct option is (D).8
Q8NAT2 marksEasyA man can row at km per hour in still water. If it takes him thrice as long to row upstream, as to row downstream, then find the stream velocity in km per hour.Think it through. Then check your answer.Question
A man can row at km per hour in still water. If it takes him thrice as long to row upstream, as to row downstream, then find the stream velocity in km per hour.Correct answer
4 to 4
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Let the speed of the man in still water be km/hr.
Let the speed of the stream be km/hr.
Speed downstream,
Speed upstream,
Let the distance be .
Time taken to row downstream,
Time taken to row upstream,
Given that :The stream velocity is km/hr.9
Q9NAT2 marksMediumA firm producing air purifiers sold units in . The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs…Think it through. Then check your answer.Question
A firm producing air purifiers sold units in . The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in . The expenditure on labour in is Rs. . In , the raw material expenses increased by and all other expenses increased by . If the company registered a profit of Rs. lakhs in , at what price (in Rs.) was each air purifier sold?Correct answer
20000 to 20000
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
1.Find Total Manufacturing Cost in 2012:Labour cost share =
Labour expenditure = Rs.
Total Cost () = (30 lakhs).2.Find Total Revenue in 2012:Profit = Rs. (10 lakhs)
Total Revenue = Total Cost + Profit = (40 lakhs).3.Find Selling Price per Unit in 2012:Number of units sold =
Selling Price = .Note: The information regarding 2013 is not required to answer the question about the 2012 selling price.10
Q10NAT2 marksMediumA batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective…Think it through. Then check your answer.Question
A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs. The probability that the current batch is accepted isCorrect answer
0.8 to 0.82
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
To find the probability that the batch is accepted, we need to calculate the probability that all four randomly chosen bulbs are non-defective.Given:- Total number of bulbs () = 100
- Number of defective bulbs () = 5
- Number of non-defective bulbs () =
- Sample size () = 4
ELECTRONICS AND COMMUNICATION ENGINEERING – EC
5511
Q11NAT1 markEasyThe maximum value of the function (where ) occurs at x = \text{______}.Think it through. Then check your answer.Question
The maximum value of the function (where ) occurs at x = \text{______}.Correct answer
0.01 to 0.01
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
To find the value of at which the function reaches its maximum, we first find the derivative of the function with respect to :To find the critical points, we set the first derivative to zero:Next, we check the second derivative to determine the nature of this critical point:At :Since , the function has a local maximum at . Because for all , the function is strictly concave, and is the global maximum point. Thus, the maximum value occurs at .12
Q12MCQ1 markEasyWhich ONE of the following is a linear non-homogeneous differential equation, where and are the independent and dependent variables respectively?Think it through. Then check your answer.Question
Which ONE of the following is a linear non-homogeneous differential equation, where and are the independent and dependent variables respectively?Correct answer
(A) (dy)/(dx) + xy = e^(-x)
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
A first-order linear differential equation is of the form . It is non-homogeneous if .
(A) is linear () and non-homogeneous ().
(B) is linear but homogeneous ().
(C) is non-linear because the dependent variable appears in the exponent of .
(D) is non-linear because of the term.13
Q13MCQ1 markEasyMatch the application to appropriate numerical method. | Application | Numerical Method | | :--- | :--- | | P1: Numerical integration | M1: Newton-Raphson Method | | P2: Solution…Think it through. Then check your answer.Question
Match the application to appropriate numerical method.Application Numerical Method P1: Numerical integration M1: Newton-Raphson Method P2: Solution to a transcendental equation M2: Runge-Kutta Method P3: Solution to a system of linear equations M3: Simpson’s 1/3-rule P4: Solution to a differential equation M4: Gauss Elimination Method Correct answer
(B) P1—M3, P2—M1, P3—M4, P4—M2
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The correct matching of applications to numerical methods is:- P1 (Numerical integration): Simpson’s 1/3-rule (M3) is a standard method for numerical integration.
- P2 (Solution to a transcendental equation): Newton-Raphson Method (M1) is a popular iterative method for finding roots of equations.
- P3 (Solution to a system of linear equations): Gauss Elimination Method (M4) is a direct method for solving linear systems.
- P4 (Solution to a differential equation): Runge-Kutta Method (M2) is a widely used iterative method for solving ordinary differential equations.
14
Q14MCQ1 markMediumAn unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss isThink it through. Then check your answer.Question
An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss isCorrect answer
(C) 0.082
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
This is a problem of the negative binomial distribution. We are looking for the probability that the success occurs on the trial.
Given:- Number of successes (the fourth head)
- Number of trials (the tenth toss)
- Probability of success (head) (unbiased coin)
15
Q15MCQ1 markEasyIf , thenThink it through. Then check your answer.Question
If , thenCorrect answer
(C) x (∂ z)/(∂ x) = y (∂ z)/(∂ y)
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given .
Calculate the partial derivative with respect to :Multiply by :Calculate the partial derivative with respect to :Multiply by :Comparing the two results, we see that .16
Q16MCQ1 markEasyA series RC circuit is connected to a DC voltage source at time . The relation between the source voltage , the resistance , the capacitance , and the current…Think it through. Then check your answer.Question
A series RC circuit is connected to a DC voltage source at time . The relation between the source voltage , the resistance , the capacitance , and the current is given below:Which one of the following represents the current ?Correct answer
(A) [figure]
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The given equation is the KVL equation for a series RC circuit:Differentiating both sides with respect to :This is a first-order linear differential equation. The solution is of the form:At , the capacitor is uncharged (assuming zero initial state), so the entire source voltage appears across the resistor:Thus, . This represents an exponential decay starting from a maximum value at . Graph (A) correctly depicts this behavior.17
Q17NAT1 markMediumIn the figure shown, the value of the current (in Amperes) is __________.Think it through. Then check your answer.Question
In the figure shown, the value of the current (in Amperes) is __________.Correct answer
0.49 to 0.51
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Let the bottom node be the reference node (ground, ).
Let be the voltage at the node where the current source is connected.
Let be the voltage at the node above the resistor.
The node above the source is at .Applying Kirchhoff's Current Law (KCL) at Node 2:Multiplying by 5:Applying KCL at Node 3:Multiplying by 10:Substituting (2) into (1):The current flowing through the resistor is:Thus, the value of the current is .18
Q18MCQ1 markMediumIn MOSFET fabrication, the channel length is defined during the process ofThink it through. Then check your answer.Question
In MOSFET fabrication, the channel length is defined during the process ofCorrect answer
(C) poly-silicon gate patterning
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
In standard CMOS fabrication, the gate (usually poly-silicon) is patterned first. This gate then serves as a mask for the subsequent source and drain ion implantation steps. This is known as a self-aligned process. Therefore, the dimensions of the patterned poly-silicon gate directly define the channel length of the MOSFET. Hence, the correct option is (C).19
Q19MCQ1 markEasyA thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional toThink it through. Then check your answer.Question
A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional toCorrect answer
(D) the excess minority carrier concentration
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The net recombination rate () for excess minority carriers (electrons in P-type material) is defined as , where is the excess minority carrier concentration and is the minority carrier lifetime. Thus, the recombination rate is directly proportional to the excess minority carrier concentration. Hence, the correct option is (D).20
Q20NAT1 markEasyAt , the hole mobility of a semiconductor and . The hole diffusion constant in…Think it through. Then check your answer.Question
At , the hole mobility of a semiconductor and . The hole diffusion constant in is ________Correct answer
12.9 to 13.1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
According to Einstein's relation, the diffusion constant and mobility are related as:Given:- Hole mobility
- Thermal voltage
21
Q21MCQ1 markEasyThe desirable characteristics of a transconductance amplifier areThink it through. Then check your answer.Question
The desirable characteristics of a transconductance amplifier areCorrect answer
(A) high input resistance and high output resistance
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
A transconductance amplifier is a voltage-controlled current source (VCCS). Its gain is defined as .1.To sense the input voltage () accurately without loading the source, the input resistance () should be ideally infinite (high).2.To provide a stable output current () regardless of the load resistance, the output resistance () should be ideally infinite (high).Therefore, high input resistance and high output resistance are desirable.22
Q22NAT1 markMediumIn the circuit shown, the PNP transistor has and . Assume that . For to be , the value of …Think it through. Then check your answer.Question
In the circuit shown, the PNP transistor has and . Assume that . For to be , the value of (in ) is _______
Correct answer
1.04 to 1.12
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given the PNP transistor circuit:- (Emitter supply)
- (Collector voltage)
Since the emitter is at :2.Calculate the base current :The base is connected to ground through :3.Calculate the collector current :4.Calculate for :In this configuration, is the voltage across to ground:The value of is approximately , which falls within the range .23
Q23NAT1 markHardThe figure shows a half-wave rectifier. The diode D is ideal. The average steady-state current (in Amperes) through the diode is approximately ____________. [figure]Think it through. Then check your answer.Question
The figure shows a half-wave rectifier. The diode D is ideal. The average steady-state current (in Amperes) through the diode is approximately ____________.
Correct answer
0.08 to 0.12
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given:
Input voltage V
Peak voltage V
Frequency Hz Period s
Resistance
Capacitance mF FThe time constant of the circuit is s.
Since (), the capacitor does not discharge significantly between cycles. The output voltage remains approximately constant at the peak value V.The average current through the load resistor is:In steady state, the average current through the capacitor is zero. By applying KCL at the output node, the average current through the diode must equal the average current through the resistor:The value A falls within the specified range of 0.08 to 0.12.24
Q24NAT1 markEasyAn analog voltage in the range 0 to 8 V is divided in 16 equal intervals for conversion to 4-bit digital output. The maximum quantization error (in V) is _________Think it through. Then check your answer.Question
An analog voltage in the range 0 to 8 V is divided in 16 equal intervals for conversion to 4-bit digital output. The maximum quantization error (in V) is _________Correct answer
0.24 to 0.26
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given:
Analog voltage range to V
Number of quantization intervals
Number of bits (Note: , which matches the number of intervals)The step size (quantization interval) is calculated as:The maximum quantization error for a uniform quantizer is given by half of the step size:The value V falls within the specified range of 0.24 to 0.26.25
Q25MCQ1 markMediumThe circuit shown in the figure is aThink it through. Then check your answer.Question
The circuit shown in the figure is aCorrect answer
(D) Master-Slave D Flip Flop
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The circuit consists of two D-latches connected in series. The first latch (master) is enabled when the clock signal is high. The second latch (slave) is enabled when is low (due to the inverter). This configuration, where one latch captures data while the other is disabled and then passes it on the next clock phase, constitutes a Master-Slave D Flip Flop, which is edge-triggered.26
Q26MCQ1 markMediumConsider the multiplexer based logic circuit shown in the figure. [figure]Think it through. Then check your answer.Question
Consider the multiplexer based logic circuit shown in the figure.
Correct answer
(D) F = W ⊕ S₁ ⊕ S₂
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The output of the first 2:1 multiplexer with select line and inputs (at ) and (at ) is:
The second 2:1 multiplexer has select line and inputs (at ) and (at ). Its output is:
Substituting :
Thus, the correct option is (D).27
Q27MCQ1 markMediumLet be sampled at 20 Hz and reconstructed using an ideal low-pass filter with cut-off frequency of 20 Hz. The frequency/frequencies present…Think it through. Then check your answer.Question
Let be sampled at 20 Hz and reconstructed using an ideal low-pass filter with cut-off frequency of 20 Hz. The frequency/frequencies present in the reconstructed signal is/areCorrect answer
(A) 5 Hz and 15 Hz only
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The signal is . The frequencies present are Hz and Hz.
The sampling frequency is Hz.
The sampled signal contains frequencies of the form .
For Hz:
For Hz:
The reconstruction filter is an ideal low-pass filter with a cut-off frequency of 20 Hz. Therefore, it will pass all frequency components below 20 Hz.
From the sampled components, 5 Hz and 15 Hz are both less than 20 Hz.
Thus, the reconstructed signal will contain 5 Hz and 15 Hz components.28
Q28MCQ1 markMediumFor an all-pass system , where , for all . If , then b equalsThink it through. Then check your answer.Question
For an all-pass system , where , for all . If , then b equalsCorrect answer
(B) a^
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
A discrete-time all-pass system has the property that its zeros are the conjugate reciprocals of its poles.
Given .
The pole is at .
For an all-pass system, the zero must be at .
Setting the numerator to zero: .
Equating the zero locations: .29
Q29NAT1 markMediumA modulated signal is , where the baseband signal has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which…Think it through. Then check your answer.Question
A modulated signal is , where the baseband signal has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which should be sampled to recover is _______.Correct answer
9.5 to 10.5
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The modulated signal is where kHz. The baseband signal is band-limited to kHz. The spectrum of is . To recover , we can use synchronous demodulation followed by a low-pass filter. If we sample directly, we need to ensure that one of the spectral replicas falls at baseband without aliasing. For a bandpass signal, the minimum sampling rate is where . Here kHz, kHz, kHz. . kHz. However, the question asks for the rate to recover . If we sample at kHz, the replicas of the spectrum at kHz will be shifted by multiples of 10 kHz. Specifically, the replica at kHz shifted by kHz lands at Hz, and the replica at kHz shifted by kHz also lands at Hz. Thus, the baseband signal is reconstructed at baseband. Therefore, the minimum sampling rate is kHz.30
Q30MCQ1 markEasyConsider the following block diagram in the figure. [figure] The transfer function isThink it through. Then check your answer.Question
Consider the following block diagram in the figure.The transfer function is
Correct answer
(C) G₁ G₂ + G₂ + 1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
By tracing the signal paths from the input to the output in the feedforward diagram:1.Path through and :2.Path through the first summer and :3.Direct path to the second summer:Summing these contributions at the output:
Therefore, the transfer function is:
This matches option (C).31
Q31NAT1 markMediumThe input , where is the unit step function, is applied to a system with transfer function . If the initial value of the output is , then…Think it through. Then check your answer.Question
The input , where is the unit step function, is applied to a system with transfer function . If the initial value of the output is , then the value of the output at steady state is _______.Correct answer
0.01 to 0.01
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given the transfer function . The corresponding differential equation is . Taking the Laplace transform on both sides: . Given , so and . Also given . Substituting these values: . The time-domain output is . The steady-state value is .32
Q32MCQ1 markMediumThe phase response of a passband waveform at the receiver is given by where is the centre frequency, and and…Think it through. Then check your answer.Question
The phase response of a passband waveform at the receiver is given bywhere is the centre frequency, and and are positive constants. The actual signal propagation delay from the transmitter to receiver isCorrect answer
(C) α
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The actual signal propagation delay is the group delay , which is defined as:Given .
Differentiating with respect to :Thus,Therefore, the propagation delay is .33
Q33MCQ1 markEasyConsider an FM signal . The maximum deviation of the instantaneous frequency from the carrier…Think it through. Then check your answer.Question
Consider an FM signal . The maximum deviation of the instantaneous frequency from the carrier frequency isCorrect answer
(A) β₁ f₁ + β₂ f₂
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The instantaneous phase of the FM signal is:The instantaneous frequency is given by:The frequency deviation is the difference between the instantaneous frequency and the carrier frequency:The maximum frequency deviation is:34
Q34NAT1 markEasyConsider an air filled rectangular waveguide with a cross-section of . For this waveguide, the cut-off frequency (in MHz) of mode is…Think it through. Then check your answer.Question
Consider an air filled rectangular waveguide with a cross-section of . For this waveguide, the cut-off frequency (in MHz) of mode is _________.Correct answer
7750 to 7850
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The cut-off frequency for the mode in a rectangular waveguide of dimensions is given by:For an air-filled waveguide, the velocity of light m/s.Given:- Dimensions: ,
- Mode:
35
Q35MCQ1 markMediumIn the following figure, the transmitter Tx sends a wideband modulated RF signal via a coaxial cable to the receiver Rx. The output impedance of Tx, the characteristic…Think it through. Then check your answer.Question
In the following figure, the transmitter Tx sends a wideband modulated RF signal via a coaxial cable to the receiver Rx. The output impedance of Tx, the characteristic impedance of the cable and the input impedance of Rx are all real.Which one of the following statements is TRUE about the distortion of the received signal due to impedance mismatch?
Correct answer
(C) Signal distortion implies impedance mismatch at both ends: Z_T ≠ Z₀ and Z_R ≠ Z₀.
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
For a wideband signal transmitted over a coaxial cable (transmission line), signal distortion occurs due to reflections caused by impedance mismatches.1.Reflection at the Load (Receiver): A reflection occurs at the load if the load impedance does not match the characteristic impedance of the cable . The reflection coefficient at the load is given by . If , then , and a portion of the signal is reflected back towards the transmitter.2.Distortion: This reflected signal travels back and can interfere with the forward-traveling signal. For a wideband signal, which is composed of many frequency components, this reflection causes delayed versions of the signal to superimpose on the original signal at the receiver. This phenomenon is known as inter-symbol interference (ISI) in digital communication and is a form of signal distortion.3.Role of Source Impedance: The impedance at the source, , affects what happens to the signal that is reflected from the load. If , the backward-traveling reflected signal will be re-reflected at the source and travel again towards the load, causing further distortion. However, the initial and primary cause of distortion is the reflection at the load.Let's analyze the options:- (A) The signal gets distorted if , irrespective of the value of . This is correct. If there is a mismatch at the load (), reflections will occur, leading to distortion. This happens regardless of whether the source is matched or not. A matched source () would absorb the reflected wave and prevent further re-reflections, but the initial distortion from the first reflection still occurs.
- (B) The signal gets distorted if , irrespective of the value of . This is incorrect. If the load is perfectly matched (), the entire signal is absorbed by the load, and there is no reflection. Without a reflection from the load, there is no backward-traveling wave to cause distortion, even if the source is mismatched.
- (C) Signal distortion implies impedance mismatch at both ends: and . This is incorrect. Distortion occurs as long as there is a mismatch at the load end (). A mismatch at the source end is not a necessary condition for distortion to begin.
- (D) Impedance mismatches do NOT result in signal distortion but reduce power transfer efficiency. This is incorrect. While impedance mismatches do reduce power transfer, they are a primary cause of signal distortion for wideband signals due to reflections and the resulting ISI.
36
Q36NAT2 marksEasyThe maximum value of in the interval is __________Think it through. Then check your answer.Question
The maximum value of in the interval is __________Correct answer
5.9 to 6.1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
To find the maximum value of a continuous function on a closed interval , we need to evaluate the function at its critical points within the interval and at the endpoints of the interval. The largest of these values will be the absolute maximum.The given function is on the interval .Step 1: Find the critical points.
Critical points occur where the first derivative is zero or undefined. Since is a polynomial, its derivative is defined everywhere.
Set to find the critical points:
Divide by 6:
Factor the quadratic equation:
The critical points are and . Both of these points lie within the interval .Step 2: Evaluate the function at the critical points and the endpoints.
The points to check are the endpoints and the critical points .- At endpoint :
- At critical point :
- At critical point :
- At endpoint :
The values of the function at the points of interest are -3, 2, 1, and 6.The maximum value among these is 6.Thus, the maximum value of the function in the interval is 6.37
Q37MCQ2 marksEasyWhich one of the following statements is NOT true for a square matrix A?Think it through. Then check your answer.Question
Which one of the following statements is NOT true for a square matrix A?Correct answer
(B) If A is real symmetric, the eigenvalues of A are always real and positive
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The question asks to identify the statement that is NOT true for a square matrix A.Let's analyze each statement:(A) If A is upper triangular, the eigenvalues of A are the diagonal elements of it.
This statement is TRUE. For any triangular matrix (upper or lower), the eigenvalues are the entries on its main diagonal. This is because the determinant of a triangular matrix is the product of its diagonal elements. The characteristic equation is . Since is triangular, is also triangular, and its determinant is . The roots of this equation are .(B) If A is real symmetric, the eigenvalues of A are always real and positive.
This statement is FALSE. It is true that the eigenvalues of a real symmetric matrix are always real. However, they are not necessarily positive. For the eigenvalues to be positive, the matrix must be positive definite. A real symmetric matrix can have negative or zero eigenvalues.
Counterexample: Consider the matrix . This is a real symmetric matrix. Its eigenvalues are clearly -2 and -2, which are real but not positive.
Another counterexample: . This is real symmetric. Its characteristic equation is , so the eigenvalues are and . One is positive, but one is negative.(C) If A is real, the eigenvalues of A and are always the same.
This statement is TRUE. The eigenvalues are the roots of the characteristic equation, . We know that for any square matrix , . Let . Then . Therefore, . Since A and have the same characteristic polynomial, they must have the same eigenvalues.(D) If all the principal minors of A are positive, all the eigenvalues of A are also positive.
This statement is TRUE. This is related to Sylvester's criterion. A Hermitian matrix (which includes real symmetric matrices) is positive definite if and only if all its leading principal minors are positive. A matrix is positive definite if and only if all its eigenvalues are positive. The condition that all principal minors are positive is a stronger condition that also implies the matrix is positive definite, and thus has all positive eigenvalues. This holds for symmetric matrices. For non-symmetric matrices, this is also a known result.Since the question asks for the statement that is NOT true, the correct answer is (B).38
Q38NAT2 marksMediumA fair coin is tossed repeatedly till both head and tail appear at least once. The average number of tosses required is ______ .Think it through. Then check your answer.Question
A fair coin is tossed repeatedly till both head and tail appear at least once. The average number of tosses required is ______ .Correct answer
2.9 to 3.1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Let be the random variable representing the number of tosses required until both a head (H) and a tail (T) have appeared at least once.
The first toss will result in either H or T. Suppose the first toss is H. Then we continue tossing until we get the first T. The number of additional tosses required to get the first T follows a geometric distribution with probability of success . The expected value of a geometric distribution starting from the first trial is .
Thus, the total expected number of tosses is .
By symmetry, if the first toss was T, the expected number of additional tosses to get the first H is also 2, leading to a total of 3 tosses.
Therefore, .39
Q39NAT2 marksMediumLet , and be independent and identically distributed random variables with the uniform distribution on . The probability is…Think it through. Then check your answer.Question
Let , and be independent and identically distributed random variables with the uniform distribution on . The probability is ________.Correct answer
0.15 to 0.18
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The probability is the volume of the region defined by within the unit cube .Evaluating the inner integral:Evaluating the middle integral:Evaluating the outer integral:40
Q40MCQ2 marksMediumConsider the building block called ‘Network N’ shown in the figure. Let and . [figure] Two such blocks are connected in cascade, as shown in the…Think it through. Then check your answer.Question
Consider the building block called ‘Network N’ shown in the figure.
Let and .
Two such blocks are connected in cascade, as shown in the figure.
The transfer function of the cascaded network isCorrect answer
(B) (s²)/(1+3s+s²)
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The transfer function of a single 'Network N' block is .
Given and , . Thus, .
When two such blocks are cascaded, the second block loads the first. Let and .
Applying KCL at the intermediate node :For the second stage, .
Substituting into the KCL equation and simplifying with (so ):This matches option (B).41
Q41MCQ2 marksMediumIn the circuit shown in the figure, the value of node voltage isThink it through. Then check your answer.Question
In the circuit shown in the figure, the value of node voltage isCorrect answer
(D) 2 - j22 V
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Let the node on the left be and the node on the right be . From the circuit, the V source and the resistor are in parallel between Node 1 and Node 2. Therefore, the constraint is:
Applying KCL at the supernode consisting of Node 1 and Node 2:
Substituting :
Multiplying by the conjugate:
V.42
Q42NAT2 marksMediumIn the circuit shown in the figure, the angular frequency (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b' is purely resistive, is…Think it through. Then check your answer.Question
In the circuit shown in the figure, the angular frequency (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b' is purely resistive, is _________.Correct answer
1.9 to 2.1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
To find the Norton equivalent impedance as seen from terminals b-b', we look into the terminals with the independent source deactivated (voltage source short-circuited). The impedance is the series combination of the capacitor and the parallel combination of the resistor and inductor.Given , , and :
Rationalizing the second term:
For the impedance to be purely resistive, the imaginary part must be zero:
.43
Q43NAT2 marksEasyFor the Y-network shown in the figure, the value of (in ) in the equivalent -network is ____.Think it through. Then check your answer.Question
For the Y-network shown in the figure, the value of (in ) in the equivalent -network is ____.Correct answer
9 to 11
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The Y-network has resistors , , and . To find the equivalent -network resistor (which is connected between the nodes where and were connected), we use the Y to transformation formula:Substituting the given values:
.44
Q44MCQ2 marksMediumThe donor and accepter impurities in an abrupt junction silicon diode are and , respectively. Assume that the…Think it through. Then check your answer.Question
The donor and accepter impurities in an abrupt junction silicon diode are and , respectively. Assume that the intrinsic carrier concentration in silicon at 300 K, and the permittivity of silicon . The built-in potential and the depletion width of the diode under thermal equilibrium conditions, respectively, areCorrect answer
(D) 0.86 V and 3.3 × 10⁻⁵ cm
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
1.Built-in Potential ():2.Depletion Width ():
Since ,
.Thus, the correct option is (D).45
Q45NAT2 marksMediumThe slope of the vs. curve of an n-channel MOSFET in linear regime is at . For the same device, neglecting channel…Think it through. Then check your answer.Question
The slope of the vs. curve of an n-channel MOSFET in linear regime is at . For the same device, neglecting channel length modulation, the slope of the vs. curve (in ) under saturation regime is approximately _________.Correct answer
0.06 to 0.08
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
1.Linear Regime:
Slope
Given and :2.Saturation Regime:
Slope of vs. is
Slope .46
Q46NAT2 marksMediumAn ideal MOS capacitor has boron doping-concentration of in the substrate. When a gate voltage is applied, a depletion region of width…Think it through. Then check your answer.Question
An ideal MOS capacitor has boron doping-concentration of in the substrate. When a gate voltage is applied, a depletion region of width is formed with a surface (channel) potential of . Given that and the relative permittivities of silicon and silicon dioxide are 12 and 4, respectively, the peak electric field (in V/ m) in the oxide region is ____________.Correct answer
2.3 to 2.5
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given data:- Boron doping concentration,
- Depletion width,
- Surface potential,
- Relative permittivity of silicon,
- Relative permittivity of silicon dioxide,
47
Q47MCQ2 marksMediumIn the circuit shown, the silicon BJT has . Assume and . Which one of the following statements is correct? [figure]Think it through. Then check your answer.Question
In the circuit shown, the silicon BJT has . Assume and . Which one of the following statements is correct?
Correct answer
(B) For R_C = 3 kΩ, the BJT operates in the saturation region
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
1.Calculate the base current :.2.Calculate the maximum possible collector current in the active region :.3.Determine the collector current at saturation for a given :.4.Condition for saturation is :.- For (), the BJT is in the active/linear region. (Option A is wrong)
- For (), the BJT is in the saturation region. (Option B is correct)
- For (), the BJT is in the saturation region. (Options C and D are wrong).
48
Q48MCQ2 marksMediumAssuming that the Op-amp in the circuit shown is ideal, is given by [figure]Think it through. Then check your answer.Question
Assuming that the Op-amp in the circuit shown is ideal, is given by
Correct answer
(D) -3V₁ + (11)/(2) V₂
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Using the principle of virtual short for an ideal Op-amp, .
From the circuit diagram:1.The non-inverting terminal is connected directly to , so .2.Therefore, .3.Apply KCL at the inverting terminal node ():
Substitute :
Multiply the entire equation by to clear denominators:
.49
Q49NAT2 marksMediumFor the MOSFET shown in the figure, assume , , and . The transistor …Think it through. Then check your answer.Question
For the MOSFET shown in the figure, assume , , and . The transistor switches from saturation region to linear region when (in Volts) is__________.
Correct answer
1.4 to 1.6
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
To find the value of at which the MOSFET switches from the saturation region to the linear region, we use the boundary condition:From the given circuit:1.2.Substituting these into the boundary condition:In the saturation region, the drain current is given by:Substitute into the boundary equation:Given values:50
Q50MCQ2 marksEasyIf is the Word Line and the Bit Line, an SRAM cell is shown inThink it through. Then check your answer.Question
If is the Word Line and the Bit Line, an SRAM cell is shown inCorrect answer
(B) [figure]
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
A standard 6T SRAM cell consists of two cross-coupled CMOS inverters and two NMOS access transistors.1.Cross-coupled Inverters: The storage element is formed by two inverters where the output of one is connected to the input of the other. This creates a stable latching mechanism to store a single bit.2.Access Transistors: Two NMOS transistors act as switches to connect the internal storage nodes to the external bit lines. The gates of these transistors are controlled by the Word Line ().3.Bit Lines: The access transistors connect the storage nodes to the Bit Line () and its complement ().Evaluating the options:- In (A), the access transistors are connected to the gates of the inverters. While logically part of the same node in a cross-coupled pair, standard schematic representations and physical layouts connect the access transistors to the output/drain nodes of the inverters.
- In (B), the access transistors are correctly connected to the outputs of the inverters (the storage nodes). This is the standard and correct representation of a 6T SRAM cell.
- In (C) and (D), the cross-coupling is incorrect (e.g., outputs connected to outputs or gates connected to gates), which would not form a functional latch.
51
Q51MCQ2 marksMediumIn the circuit shown, and are MSBs of the control inputs. The output is given byThink it through. Then check your answer.Question
In the circuit shown, and are MSBs of the control inputs. The output is given byCorrect answer
(C) F = WXY + WXY
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
1.Analyze the first 4:1 MUX. The inputs are , (), (), and . The select lines are (MSB) and (LSB). The output is:2.Analyze the second 4:1 MUX. The inputs are , , , and . The select lines are (MSB) and (LSB). The output is:3.Substitute into the expression for :This matches option (C).52
Q52MCQ2 marksEasyIf and are inputs and the Difference () and the Borrow () are the outputs, which one of the following diagrams implements a half-subtractor?Think it through. Then check your answer.Question
If and are inputs and the Difference () and the Borrow () are the outputs, which one of the following diagrams implements a half-subtractor?Correct answer
(A) [figure]
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
A half-subtractor has two outputs: Difference () and Borrow ().
In option (A):- The first 2:1 MUX has inputs and select line . Its output is .
- The second 2:1 MUX has inputs and select line . Its output is .
53
Q53NAT2 marksMediumLet , , . The quantities are real numbers. Consider…Think it through. Then check your answer.Question
Let , , . The quantities are real numbers. Consider . If the zero of lies on the unit circle, thenCorrect answer
-0.6
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The transfer function is given by:The zero of is found by setting the numerator to zero:Since the zero lies on the unit circle and are real, we have:Case 1:
Substituting and :This is invalid as .
Case 2:
Substituting and :Since , this is the correct value.54
Q54MCQ2 marksMediumLet denote the impulse response of a causal system with transfer function . Consider the following three statements. S1: The system is stable. S2:…Think it through. Then check your answer.Question
Let denote the impulse response of a causal system with transfer function . Consider the following three statements.
S1: The system is stable.
S2: is independent of for .
S3: A non-causal system with the same transfer function is stable.
For the above system,Correct answer
(A) only S1 and S2 are true
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The transfer function is , which has a pole at .
S1: For a causal system, the Region of Convergence (ROC) is . Since the ROC includes the axis, the system is stable. Thus, S1 is true.
S2: The impulse response for the causal system is . For , .This is a constant and independent of . Thus, S2 is true.
S3: For a non-causal (specifically anti-causal) system with the same transfer function, the ROC would be . This ROC does not include the axis, so the system is unstable. Thus, S3 is false.
Therefore, only S1 and S2 are true.55
Q55NAT2 marksMediumThe z-transform of the sequence is given by , with the region of convergence . Then, is ________.Think it through. Then check your answer.Question
The z-transform of the sequence is given by , with the region of convergence . Then, is ________.Correct answer
11.9 to 12.1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given with ROC .
We can use the binomial expansion for where :Substituting :The sequence is the coefficient of in the expansion of .
Therefore, is the coefficient of , which is .56
Q56NAT2 marksMediumThe steady state error of the system shown in the figure for a unit step input is _______.Think it through. Then check your answer.Question
The steady state error of the system shown in the figure for a unit step input is _______.Correct answer
0.49 to 0.51
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
For the given block diagram, the forward path transfer function is and the feedback path transfer function is .
The error signal is given by:
Since , we have:
For a unit step input, . The steady state error is:
Calculating and :
Thus, .57
Q57MCQ2 marksMediumThe state equation of a second-order linear system is given by For…Think it through. Then check your answer.Question
The state equation of a second-order linear system is given byFor , and for , .
When , isCorrect answer
(B) bmatrix 11e^(-t) - 8e^(-2t) \ -11e^(-t) + 16e^(-2t) bmatrix
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Since the system is linear, the principle of superposition applies. We can express the initial state as a linear combination of the given initial states and :From the first row: .
From the second row: .
Thus, the response for is:This matches option (B).58
Q58MCQ2 marksMediumIn the root locus plot shown in the figure, the pole/zero marks and the arrows have been removed. Which one of the following transfer functions has this root locus?Think it through. Then check your answer.Question
In the root locus plot shown in the figure, the pole/zero marks and the arrows have been removed. Which one of the following transfer functions has this root locus?Correct answer
(B) (s+4)/((s+1)(s+2)(s+7))
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The root locus exists on the real axis to the left of an odd number of poles and zeros. From the figure, the root locus segments on the real axis are between and , and between and .Let's check the options:
(A) Zeros: ; Poles: . Real axis segments: and .
(B) Zeros: ; Poles: . Real axis segments: and .
(C) Zeros: ; Poles: . Real axis segments: and .All three options have the same real axis segments. However, the plot shows a breakaway point between and (which must be between two poles) and a break-in point between and .
In option (B), poles are at and , allowing for a breakaway point. There is a zero at and a pole at , and the branch starting at the pole moves towards the zero at . The complex branches from the breakaway point return to the real axis at a break-in point between and . This configuration matches the provided root locus plot.59
Q59MCQ2 marksMediumLet be a wide sense stationary (WSS) random process with power spectral density . If is the process defined as , the power spectral density…Think it through. Then check your answer.Question
Let be a wide sense stationary (WSS) random process with power spectral density . If is the process defined as , the power spectral density isCorrect answer
(C) S_Y(f) = (1)/(2) S_X ((f)/(2))
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given . For a wide sense stationary (WSS) process, the autocorrelation function of is:
.The power spectral density (PSD) is the Fourier transform of the autocorrelation function:
.Using the scaling property of the Fourier transform, , where :
.Note that the time shift in the process does not affect the autocorrelation of a WSS process, and consequently, it does not affect the PSD. Therefore, the correct option is (C).60
Q60NAT2 marksMediumA real band-limited random process has two-sided power spectral density…Think it through. Then check your answer.Question
A real band-limited random process has two-sided power spectral density where is the frequency expressed in Hz. The signal modulates a carrier and the resultant signal is passed through an ideal band-pass filter of unity gain with centre frequency of 8 kHz and band-width of 2 kHz. The output power (in Watts) is ________.Correct answer
2.4 to 2.6
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The power spectral density (PSD) of the random process is given by:This is a triangular PSD centered at . The peak value at is Watts/Hz. The base extends from -3 kHz to 3 kHz.The signal modulates a carrier . The carrier frequency is .
Assuming DSB-SC modulation, the PSD of the modulated signal is given by: is a triangular PSD centered at , extending from to .
is a triangular PSD centered at , extending from to .The resultant signal is passed through an ideal band-pass filter with a centre frequency of 8 kHz and a bandwidth of 2 kHz.
This means the filter passes frequencies in the range and .
Passband for positive frequencies: .
Passband for negative frequencies: .The output power is the integral of over the filter's passband:Substitute :Consider the positive frequency integral:The second term is zero because is centered at -8 kHz and has no overlap with .
So, the positive frequency contribution is .
Let . When . When .
This integral becomes .Consider the negative frequency integral:The first term is zero because is centered at 8 kHz and has no overlap with .
So, the negative frequency contribution is .
Let . When . When .
This integral becomes .Combining these, the total output power is:Now, we need to calculate the integral of from -1 kHz to 1 kHz.
Since is an even function, we can write:For , .So, .Finally, the output power is:The final answer is .61
Q61NAT2 marksMediumIn a PCM system, the signal is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size…Think it through. Then check your answer.Question
In a PCM system, the signal is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is ________.Correct answer
199 to 201
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The given signal is .
This can be rewritten as .
Using the identity , we can write:
.The maximum frequency component of the signal is .According to the Nyquist sampling theorem, the minimum sampling rate (Nyquist rate) is .
.The peak amplitude of the signal is .
The dynamic range of the signal is .
.The uniform quantizer has a step size .The number of quantization levels required is given by .
.Since the number of quantization levels must be an integer and typically a power of 2 for bits, we choose the smallest power of 2 that is greater than or equal to .
If is the number of bits per sample, then .
We need .
For , (too small).
For , (sufficient).
So, the minimum number of bits per sample required is bits/sample.The minimum data rate of the PCM system is given by .
.The final answer is .62
Q62NAT2 marksMediumA binary random variable takes the value of 1 with probability . is input to a cascade of 2 independent identical binary symmetric channels (BSCs) each with crossover…Think it through. Then check your answer.Question
A binary random variable takes the value of 1 with probability . is input to a cascade of 2 independent identical binary symmetric channels (BSCs) each with crossover probability . The output of BSCs are the random variables and as shown in the figure.The value of in bits is _____.
Correct answer
1.9 to 2.1
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
Given and .
For the first BSC with crossover probability :
.
.
The entropy bit.For the second BSC with input and crossover probability :
.
.
The entropy bit.Therefore, the total value is bits.63
Q63NAT2 marksEasyGiven the vector , where denote unit vectors along directions, respectively. The…Think it through. Then check your answer.Question
Given the vector , where denote unit vectors along directions, respectively. The magnitude of curl of is ________Correct answer
0.01 to 0.01
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The vector is given by , where and .
The curl of in Cartesian coordinates is:Calculating the partial derivatives:Substituting these back into the curl expression:The magnitude of the curl is .64
Q64MCQ2 marksMediumA region shown below contains a perfect conducting half-space and air. The surface current on the surface of the perfect conductor is amperes…Think it through. Then check your answer.Question
A region shown below contains a perfect conducting half-space and air. The surface current on the surface of the perfect conductor is amperes per meter. The tangential field in the air just above the perfect conductor is
Correct answer
(D) z2 amperes per meter
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The boundary condition for the magnetic field at the surface of a perfect conductor is given by:
where is the unit normal vector pointing from the conductor into the air. From the figure, the interface is at , so .
Given .
Let . The tangential components are and .
Comparing components:
and .
Thus, the tangential field is amperes per meter.65
Q65NAT2 marksMediumAssume that a plane wave in air with an electric field V/m is incident on a non-magnetic dielectric slab of relative…Think it through. Then check your answer.Question
Assume that a plane wave in air with an electric field V/m is incident on a non-magnetic dielectric slab of relative permittivity 3 which covers the region . The angle of transmission in the dielectric slab is ________ degrees.Correct answer
29 to 31
Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.Step-by-step solution
The incident electric field is given by V/m.1.Identify the incident wave vector :The phase of the wave is . Comparing this with the given expression, we have:
Since , the wave vector is:
2.Determine the angle of incidence :The interface is at (separating air for and the slab for ). The normal to the interface is . The angle of incidence is the angle between and the normal :
Thus, .3.Apply Snell's Law to find the angle of transmission :Snell's Law states , where for non-magnetic media.- For air ():
- For the dielectric slab ():
Substituting the values:
The angle of transmission in the dielectric slab is .