ECE GATE 2014 Past Paper PDF
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2014
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This is a GATE 2014 Electronics Engineering exam paper, focusing on topics like engineering mathematics, communications, and electronic devices. The paper contains various questions with detailed solutions.
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ECE GATE 2014 February 15 2:00 PM S. No.: 01 Marks: 1 Topic: Engineering Mathematics Type: NAT Concept: Linear Algebra Level: easy Sub-Concept: Basics of Matrix Time: 60 sec Concep...
ECE GATE 2014 February 15 2:00 PM S. No.: 01 Marks: 1 Topic: Engineering Mathematics Type: NAT Concept: Linear Algebra Level: easy Sub-Concept: Basics of Matrix Time: 60 sec Concept Field: Determinant 01. The determinant of matrix 𝐴 is 5 and the determinant of matrix 𝐵 is 40. The determinant of matrix 𝐴𝐵 is ________. Sol. The Correct Answer is (200) Given | A | = 5, | B | = 40 According to property of determinant | AB | = | A |. | B | | AB | = 5 x 40 | AB | = 200 S. No.: 2 Marks: 1 Topic: : Communications Type: NAT Concept: Random Variables & Noise Level: Easy Sub-Concept: Random Variables & Probability Time: 60 sec Concept Field: Expectation , Mean and Variance 2. Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E [ X ] , is __________. Sol. Correct Answer is (50) Given X = 1, 3, 5, 7, ………99, n = 50 (Number of observation) 1 n 1 E (x) = n i=1 xi = 50 1+ 3 + 5 + 7 +........+ 99 1 E (x) = ( 50 ) = 50 Ans 2 50 Hence, The value of E(x) = 50. S. No.: 3 Marks: 1 PAG E Topic: Engineering Mathematics Type: MCQ Concept: Calculus Level: easy Sub-Concept: Maxima and Minima Time: 75 sec Concept Field: 1 and 2 Variable Maxima and Minima 3. For 0 ≤ 𝑡 < ∞, the maximum value of the function f(t) = e–t–2e–2t occurs at (A) t = loge 4 (B) t = loge 2 (C) t = 0 (D) t = loge 8 Sol. Correct option is (A) Given, f(t) = e–t–2e–2t ……………….(1) Differentiating both side of function f(t) with respect to t. F’(t) = –e–t + 4e–2t the stationary points are given by, F'(t) = 0 − e −t + 4e−2t =0 4e−2t+t =e−t 4e −t = 1 1 e −t = 4 taking loge on both sides, 1 logee −t = loge 4 − tloge (e ) = 0 − loge 4 t = loge 4 Ans Hence, the correct option is A. S. No.: 4 Marks: 1 Topic: Engineering Mathematics Type: MCQ Concept: Calculus Level:Easy Sub-Concept: Limits Time: 60 sec Concept Field: Continuity 4. The value of x 1 lim 1+ Is x (A) In2 (B) 1.0 (C) e (D) ∞ Sol. Correct option is (C) PAG E x 1 1 lim 1+ = lim 1+ x → x x → = 1 The given limit is in the form of 1 lim f ( x ) = lim g ( x ) = 0 then x →a x →a f( x ) 1 lim lim 1+ f ( x ) (x) x →a g g (x) =e If x →a Hence, Right option is (C) S. No.: 5 Marks: 1 Topic: Engineering Mathematics Type: MCQ Concept: Numerical Methods Level: Moderate Sub-Concept: Differential Equations Time:90 sec Concept Field: Runge Kutta Method 5. If the characteristic equation of the differential equation d 2y dy 2 + 2 +y =0 dx dx has two equal roots, then the values of α are (A) ±1 (B) 0,0 (C) ±j (D)±1/2 Sol. Correct option is (A) Given, D.E. d 2y dy 2 + 2 +y =0 dx dx (D2 +2 α D + 1) y = 0 It is form of homogeneous linear differential equation. [ f(D)] y = 0 According to auxiliary equation by, f(m) = 0 m2 + 2αm+ 1 = 0 −2 + 4 2 − 4 −2 − 4 2 − 4 m1= ,m 2 = 2 2 According to question, both the roots are equal. So m1 = m2 −2 + 4 2 − 4 −2 − 4 2 − 4 = 2 2 4 2 − 4 = − 4 2 − 4 2 4 2 − 4 = 0 4 2 − 4 = 0 2 = 1 = 1 Ans PAG E Hence, Right option is (A) S. No.: 06 Marks: 1 Topic: Network Theory Type: MCQ Concept: Network Theorems Level:Easy Sub-Concept: Norton's Theorem Time:45 sec Concept Field: 6. Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance (A) in series with a current source (B) in parallel with a voltage source (C) in series with a voltage source (D) in parallel with a current source Sol. The correct answer is option (D) According to Norton, a complex network connected to a load can be replaced by an equivalent impedance in parallel with a convert source as shown in figure: Hence, the correct option is (4) S. No.: 7 Marks: 1 Topic: Network Theory Type: NAT Concept: Circuit Basics Level:High Sub-Concept: Ohm's Law, KVL , KCL Time:150 sec Concept Field: Circuit Reduction Problems 7. In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in mA) through the 4 K resistor at t = 0+ is _______ PAG E Sol. (The correct answer is 1.25) According to the question, switch t is open for a long time. So, the capacitor is charged for very long time. So, At t = 0 (before t = 0) switch is open. – Capacitor will behave as open circuit and inductor will behave as short circuit. i(0– ) − For 10 − 5i(0– ) − 4i(0− ) − 1i(0− ) = 0 10 − 10i(0– ) = 0 i(0–) = 1m A And for VC (0−) → VC (0−) = 4i(0−) + 1i(0−) VC (0− ) = 5i(0− ) VC (0−) = 5 1 PAG E VC ( 0− ) = 5V For VC (0−) = 5V i( 0 − ) = 1m A For t = 0+ Capacitor does not allow sudden change in the voltage (0 )− (0 ) − VC = VC − 5V And inductor does not allow sudden change in the current i(0 ) = i(0 ) = 1m A − − + So, for t = 0 , the switch is closed. For current through 4k resistor Apply KVL in the loop. We get – VC − 4i1 = 0 5 − 4i1 = 0 4i1 = 5 5 i1 = 4 i1 = 1.25m A (through 4k resister) So, the correct answer is 1.25 mA S. No.: 8 Marks: 1 Topic: Electronic Devices Type: MCQ Concept: Semiconductor Physics Level:Moderate Sub-Concept: Properties of Si / Ge Time:120 sec Concept Field: Doping and Mass- Action Law PAG E A silicon bar is doped with donor impurities N D = 2.25 10 atoms/cm3. Given 15 8. −3 the intrinsic carrier concentration of silicon at T = 300 K is ni = 1.5 10 cm 10. Assuming complete impurity ionization, the equilibrium electron and hole concentrations are −3 −3 (A) n0 = 1.5 10 cm ,p0 = 1.5 105cm 16 −3 −3 (B) n0 = 1.5 10 cm ,p0 = 1.5 1015cm 10 −3 −3 (C) n0 = 2.25 10 cm ,p0 = 1.5 1010cm 15 −3 −3 (D) n0 = 2.25 10 cm ,p0 = 1 10 cm 15 5 Sol. The correct option is (D) According to the question, Given, N D = 2025 1015atom s /cm 3 ni = 1 1010 cm −3 T = 300 k Electron concentration, n = N D+ N D+ =D N D And Where, N D = Dopant concentration D = Impurity ionization ratio N D+ = How much percentage of N D is ionized So, for complete impurity ionization, D = 1 n0 = N D+ = 1 N D So, n0 = N D+ = N D n = 2.25 1015 /cm 3 Holes concentration, ni2 P0 = ND {for ND >> ni} P0 = ( 1.5 10 ) 10 2 2.25 1020 2.25 1015 2.25 1015 P0 = 1 105 /cm 3 Hence, the correct answer is (D) S. No.: 9 Marks: 1 Topic: Electronic Devices Type: MCQ Concept: Transistors Level:Easy PAG E Sub-Concept: BJT Time:90 sec Concept Field: Alpha, Beta and Gamma Gains 9. An increase in the base recombination of a BJT will increase (1) the common emitter dc current gain (2) the breakdown voltage BVCEO (3) the unity-gain cut-off frequency fT (4) the transconductance gn Sol. (The correct option is (B) ) We know the relation between the voltage between the collector and emitter with base open (BVCEO) and the voltage from collector to base with emitter open (BVCEO) is given by– BVCBO BVCEO = ( ) 1/N ---- eqn (1) As base recombination increases, the base current (IB ) will also increase. As, IC = IE − IB If IB increases, IC will decrease Thus, will also decreased from IC = IB 1 So, will increase 1 If increase thus from equation --- (1) – BVCBO BVCEO = ( decreases) ( ) 1/N So, BVCEO will increase. Hence, the correct answer is (B) S. No.: 10 Marks: 1 Topic: Advanced Electronics Type: MCQ Concept: VLSI Technology Level: easy Sub-Concept: Fabrication Techniques Time: 45 sec Concept Field: 10. In CMOS technology, shallow P-well or N-well regions can be formed using (1) low pressure chemical vapour deposition (2) low energy sputtering (3) low temperature dry oxidation (4) low energy ion-implantation Sol. (The correct option is (D) PAG E An ion implantation is a useful alternative to diffusion process that is both ion-implantation and diffusion method are used to diffuse dopants into an specific area i.e. to introduce departs in a particular area so that n-region as well as p-region both can be formed where n-region can be p-well region. Since diffusion process requires high temperature while ion-implantation can be done at relatively low temperature. So, ion-implantation method is preferred as precise control of doping concentration can be done. Hence, the correct answer is (D). S. No.: 11 Marks: 1 Topic: Analog Electronics & Circuits Type: MCQ Concept: Feedback Amplifiers Level: Moderate Sub-Concept: Types of Feedback Time:120 sec Concept Field: Voltage -current 11. The feedback topology in the amplifier circuit (the base bias circuit is not shown for simplicity) in the figure is (1) Voltage shunt feedback (2) Current series feedback (3) Current shunt feedback (4) Voltage series feedback Sol. (The correct option is (B)) PAG E RE is the feedback element in the circuit. The voltage across RE is Vf (feedback voltage) V0 = VCE + Vf [output node is different from the feedback node] Which implies it as current sampling. In the above circuit, the applied signal is voltage VS. So, Appling KVL at the input loop, we get – VS − IR B S − VBE − Vf = 0 Here, input voltage (VS ) and feedback voltage (Vf ) are in series in the loop. So, it is called as series mixing or voltage mixing. Hence, the feedback topology involved is current series i.e. current sampling and series mixing. The correct answer is (B). S. No.: 12 Marks: 1 Topic: Analog Electronics & Circuits Type: MCQ Concept: Operational Amplifiers Level: Moderate Sub-Concept: OP-AMP Basic Time: 120 sec Concept Field: CMRR 12. In the differential amplifier shown in the figure, the magnitudes of the common-mode and differential-mode gains are Acm and Ad , respectively. If the resistance R1 is increased, then. PAG E (1) Acm increases (2) common-mode rejection ratio increases (3) Ad increases (4) common-mode rejection ratio decreases Sol. (The correct option is (B) common-made rejection ratio increases.) We know, Ad CM RR = ACM Where, Ad = Differential mode gain Acm = common mode gain −RC Ad = Ad does not depend on RE because re PAG E A d for balanced output = −gm RC −R Acm = C 2RE CM M R RE As R E increases CMRR increases. So, the correct answer is (B) S. No.: 13 Marks: 1 Topic: Analog Electronics & Circuits Type: NAT Concept: Multi-stage Amplifiers Level: High Sub-Concept: Need of Cascading Time: 150 sec Concept Field: 13. A cascade connection of two voltage amplifiers A1 and A2 is shown in the figure. The open-loop gain Av0, input resistance Rin, and output resistance Ro for A1 and A2 are as follows: A1: Av 0 = 10,Rin = 10k ,Ro = 1k A1: Av 0 = 5,Rin = 5k ,Ro = 200 The approximate overall voltage gain vout /vm is _________. Sol. (The correct answer is 34.722) Since amplifier model is given by: Therefore, the equivalent circuit is given as– PAG E By voltage divider, (stage 1) 5 Vin 2 = 10Vin 5+1 50 V = Vin in 2 6 By voltage divider (stage – 2) 1 Vout = 5Vin2 1+ 0.2 1 50 Vout = 5 Vin 1.2 6 Vout 250 = Vin 7.2 Vout = 34.722 Vin Hence, the correct answer is 34.722 S. No.: 14 Marks: 1 Topic: Digital Electronics Type: MCQ Concept: Boolean Algebra Level: Easy Sub-Concept: K - Maps Time:75 sec Concept Field: Prime Implicants and Don't Care Conditions 14. For an n-variable Boolean function, the maximum number of prime implicants is (1) 2 (n − 1) (2) n /2 n (3) 2 (n−1) (4) 2 Sol. The correct option is (D) A prime implicant of a function is an implicant that cannot be covered by more general (more reduced meaning with fewer literals) implicant. For example: considering a k-map of variables A, B PAG E In the above k-map, pair or quadrants are not opened for the two cases– Case 1– M1 = m4 = 1, m2 = M3 = 0; prime implication AB ,AB Case 2– M1 = m4 = 0, m2 = M3 = 0; prime implication AB ,AB For all other cases of two or more implicants, k-map has pair or quadrant i.e., function is reduced. Therefore, the maximum number of prime implicant is 2. For more or n-variable cases, the maximum number of prime implicant is 2n−1 Hence, the correct answer is (D) S. No.: 15 Marks: 1 Topic: Digital Electronics Type: NAT Concept: Number Systems Level:Easy Sub-Concept: Digital Binary Codes Time: 90 sec Concept Field: BCD , Excess 3, Gray Codes 15. The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is ________. Sol. (The correct answer is 4) Decimal number is represented into 4 bits of a BCD number ( X )10 → b3 b2 b1 b0 (4 bits) Since decimal number 1856357 has 7 digits. So, total number of required to represent in BCD from =7×4 = 28 bits {1 byte = 8 bits} 28 = 3.5 The required number of bytes 4 4 Ans. Hence, the correct answer is 4. S. No.: 16 Marks: 1 Topic: Digital Electronics Type: MCQ PAG E Concept: Combinational Circuits Level: Easy Sub-Concept: Adders and Subtractors Time: 90 sec Concept Field: Half and Full 16. In a half-subtractor circuit with X and Y as inputs, the Borrow (M) and Difference (N = X – Y) are given by (1) M = X Y , N = XY (2) M = XY , N = X Y (3) M = XY , N = X Y (4) M = XY , N = X Y Sol. (the correct answer is C) Truth Table X Y N M 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 Where N = Difference (N = X – Y) and M = Borrow Therefore, N = xy + xy = x y M = xy Hence, the answer is option (C) S. No.: 17 Marks: 1 Topic: Signals & Systems Type: MCQ Concept: Digital Filter Design Level: Moderate Sub-Concept: Finite Impulse Response(F I R) Time: 150 sec Concept Field: Frequency Sampling 17. An FIR system is described by the system function 7 3 H (z ) = 1+ z−1 + z−2 2 2 The system is (1) maximum phase (2) minimum phase (3) mixed phase (4) zero phase Sol. (The correct option is (C) ) Given, 7 3 H (z ) = 1+ z−1 + z−2 2 2 PAG E Note: [For minimum phase system, all zero are inside the unit circle, for maximum phase system, all zero are outside the unit circle] So, to find the system is whether maximum phase, minimum phase, mixed phase or zero phase, we have to obtain the zero of the function— 7 3 1+ z−1 + z−2 = 0 2 2 Therefore, multiplying numerator by 2z2 2z2 + 7z + 3 = 0 2z2 + 6z + z + 3 = 0 (2z2 + 6z ) + (z + 3) = 0 2z (z + 3) + (z + 3) = 0 (2z + 1) + (z + 3) = 0 −1 The zeros of the –3, 2 −1 Since one zero is 2 which is inside the unit circle and other zero is (–3) which is outside the circle. Hence, the system is mixed phase. So, the correct answer is (C) S. No.: 18 Marks: 1 Topic: Signals & Systems Type: MCQ Concept: Z Transforms Level: Moderate Sub-Concept: Discrete Signals Transforms Time:90 sec Concept Field: Properties and Expression 18. Let x[n]= x[−n], Let X [z] be the z-transform of x (n ). If 0.5 + j 0.25 is a zero of X(z), which one of the following must also be a zero of X(z). (A) 0.5 – j 0.25 (B) 1/ (0.5 + j 0.25) (C) 1/ (0.5 – j 0.25) (D) 2 + j 4) Sol. (The correct answer is (B)) Let x[n]= x[−n] Let x(z) be the z-transform of x[x] Time reversal property of z-transform is given. If x(n) = x (–x) x(z) = x (z–1) [z-transfer time reversal property] If one of the zero given is 0.5 + j0.25 of x(z) x (0.5 + j0.25) = 0 PAG E X ( Z −1 ) , So, for the equation can be written as 1 X =0 0.5 + j0.25 1 So, 0.5 + j0.25 is the other zero of x(z). Hence, the correct option is (B) S. No.: 19 Marks: 1 Topic: Signals & Systems Type: NAt Concept: Fourier Series Level: Moderate Sub-Concept: Exponential Time: 75 sec Concept Field: Coefficients 19. Consider the periodic square wave in the figure shown. The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is closest in value to ________ Sol. (The correct answer is 0.5) 1 n harmonic component for a periodic square wave th n 1 2 The power of nth harmonic component is n Ratio of the harmonic power and 5th harmonic power = 1 P7 72 = P5 1 52 2 5 72 25 49 0.510 0.5 Hence, the answer is 0.5. S. No.: 20 Marks: 1 Topic: Control Systems Type: NAT Concept: Time Domain Analysis Level: Moderate Sub-Concept: Transient Analysis Time:90 sec Concept Field: Zero to Nth Order Responses 20. The natural frequency of an undamped second-order system is 40 rad/s. If the system is damped with a damping ratio 0.3, the damped natural frequency in rad/s is _________ Sol. (The correct answer is 38.157) Given: n = 40r /sec PAG E Damping ratio = 0.3 We know, for damped natural frequency is given by---- d = n 1− 2 = 40 1− (0.3) 2 = 40 1− 0.09 = 40 0.91 = 40 0.9539 = 40 × 0.9539 d = 38.157r /sec The correct answer is d = 38.157r /sec S. No.: 21 Marks: 1 Topic: Control Systems Type: MCQ Concept: Introduction Level: easy Sub-Concept: Examples Time:120 sec Concept Field: Transfer Function or Impulse Response 21. For the following system y ( s) When X 1 ( s) = 0 , the transfer function x2 ( s) is s+1 2 (1) s 1 (2) s + 1 s+2 (3) s ( s + 1) s+1 (4) ( s + 1) s Sol. The correct answer is (D) Given, X 1 ( s) = 0 PAG E X 1 ( s) = 0 The block diagram will be stated as – Therefore, the transfer functions of the given system will be – y ( s) G ( s) T.F = x2 ( s) 1+ G (s) H (s) 1 s 1 s 1+. s s + 1 1 s s 1+ s ( + 1) s 1 s s + 1+ 1 s+1 1 s s+2 s+1 s+1 s ( s + 2) y ( s) s+1 T.F = = x2 ( s) s ( s + 2) PAG E Hence, the correct answer is option (D) S. No.: 22 Marks: 1 Topic: Communications Type: MCQ Concept: Random Variables & Noise Level:High Sub-Concept: Practical Probability Functions Time: 180 sec Concept Field: Gaussian PDF 22. The capacity of a band-limited additive white Gaussian noise (AWGN) channel p C = W log2 1+ 2 is given by w bits per second (bps), where W is the channel 2 bandwidth, P is the average power received and a is the one-sided power spectral density of the AWGN. P 2 = 1000 For a fixed a , the channel capacity (in kbps) with infinite bandwidth (W → ) is approximately. (A) 1.44 (B) 1.08 (C) 0.72 (D) 0.36 Sol. The correct option is A We know, The channel capacity, P W log2 1+ 2 C= w … (1) */N ote loga b = loga c logc b Since, … (2) Thus equation (1) can be expressed as equation 2 p W (log2 e ) loge 1+ 2 C = w 1 P 1 1 P 2 1 P 3 W (log2 e ) 2 − 2 + 2 .... = 1 W 2 W 3 W W P P2 P3 log2 e 2 − + −.... 2 4W 3 4W 2 = W P P2 P3 (log2 e ) − + −.... 2 4W 3 4W 2 2 C = … (3) as given, P = 1000, 2 & w → So putting these value in equation 32 → P2 P3 lim (log2 e ) 1000 − + −.... C = x → 2 4W 3 4W 2 PAG E = log2 e ( 1000) bit /sec C = 1.144 kbps Hence, the correct answer is A S. No.: 23 Marks: 1 Topic: Communications Type: NAT Concept: Amplitude Modulation Level: easy Sub-Concept: Amplitude Modulation Equation Time: 90 s Concept Field: Analysis of AM 23. Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index ( ) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V, is __________ Sol. (The correct answer is 0.5) According to the question- Maximum value envelope, Am ax = 3v Minimum value of envelop, Am in = 1v modulation index m axim um value of envelope+m inim um value of envelope () = m axim um value of envelope + m inim um value of envelope 3−1 = −3 + 1 2 1 = = 4 2 = 0.5 Therefore, the answer is 0.5. S. No.: 24 Marks: 1 Topic: Electromagnetics Type: NAT Concept: Transmission Lines Level: easy Sub-Concept: Types of Lines / Loads Time:150 s Concept Field: Lossless / Distortion Less 24. To maximize power transfer, a lossless transmission line is to be matched to a resistive load impedance via a /4 transformer as shown. PAG E The characteristic impedance (in ) of the /4 transformer is _________. Sol. (The correct answer is 70.7) According to question, Load impedance = Z L = 100 Input impedance = Z in = 50 In this question, impedance is matched by using QWT [Quarter wave transfer] ( /4) Input impedance for quarter wave transfer = Z2 Z in = 0 ZL Z 0 = Zin Zl 2 Z0 = Z in Z l Z 0 = 50 100 Z 0 = 5000 Z 0 = 70.71 Hence, the correct answer is 70.7 S. No.: 25 Marks: 1 Topic: Electromagnetics Type: MCQ Concept: Electromagnetic Waves Level: Moderate Sub-Concept: Material Propagations Time: 150 s Concept Field: Free Space 25. Which one of the following field patterns represents a TEM wave travelling in the positive x direction? (1) E = +8 y ˆ,H = −4z ˆ (2) E = −2y ˆ,H = −3z ˆ PAG E (3) E = +2zˆ,H = +2y ˆ (4) E = −3y ˆ,H = +4z ˆ Sol. (The correct option is (B) ) We know for TEM wave, E H = P Where, E = Electric field H = magnetic field And p = direction of propagation So, for (a) option, +8y , H = −4z E= E H = 8yˆ ( −4zˆ) = −32 ( yˆ zˆ) = −32 ( x ) (negative x − direction ) ˆ For (b) option, E = −2̂y , H = −3z ˆ ˆ ( −3z E H = 2y ˆ) = 6 ( y ˆ) ˆz = 6xˆpositive x − direction For (C) option. E = 2z ˆ,H = 2y ˆ E H = 2 ( zˆ y ˆ ) = 2 ( −x ˆ) (negative x − direction ) For (d) option, E = −3y ˆ, H = +4z ˆ E H = ( −3y ˆ) = −12 ( y ˆ 4z ˆ) = −12x ˆz ˆ (negative x − direction ) Hence, the correction answer is (B) S. No.: 26 Marks: 2 Topic: Engineering Mathematics Type: MCQ Concept: Linear Algebra Level:Moderate Sub-Concept: Solution of System Time: 150 s Concept Field: Homogeneous and Non-Homogeneous System 26.. System of Linear equation 2 1 3 a 5 3 0 1 b = −4 have 1 2 5 b 14 (A) a unique solution (B) infinitely many solutions (C) no solution (D) exactly two solutions Sol. Correct Answer is (B) Given – system of linear equation PAG E 2 1 3 a 5 3 0 1 b = −4 1 2 5 b 14 It is form of non-homogeneous equation Such that augmented matrix is given by 2 1 3 : 5 R 2 → 2R 2 − 3R 1 A :B = 3 0 1 : −4 and 1 2 5 : 14 R 3 → 2R 3 − R 1 2 1 3 : 5 A :B = 0 −3 −7 : −23 R 3 → R 3 + R 1 0 3 7 : 23 2 1 3 : 5 A :B = 0 −3 −7 : −23 0 3 7 : 23 So ( A ) = ( A :B ) = 2 but ( A ) = ( A :B ) n Hence, Right option is (B). S. No.: 27 Marks: 2 Topic: Engineering Mathematics Type: MCQ Concept: Complex Variables Level:Moderate Sub-Concept: Analytic Functions Time:180 s Concept Field: C-R equations 27. The real part of an analytic function f (z) where z = x + jy is given by e–y cos(x). The imaginary part of f(z) is (A) ey cos(x) (B) e–y sin(x). (C) –ey sin(x). (D) –e–y sin(x). Sol. Correct Answer is (B) Given, Real Part of f(z) is, Cl = e–y cos x By Cauchy-Riemann equation u v = x y u = −e− y sin x Thus, x v = −e − y sin x So y PAG E Integrating on both side v y = −e − y.sin x V = e–y sinx + f(x) ……(1) By Cauchy – Riemann equation u v =− x y Putting the value of u and v u −y v −y e sin x + f ( x ) = − e cos x x y ( e − y cos x + f '( x ) = − −e − y cos x ) f '( x ) = 0 f (x) = k Put the value of f(x) in equation (1) V = e–y sin(x) + +k For k = 0 v = e− y sin x Ans Hence, Right option is (B). S. No.: 28 Marks: 2 Topic: Engineering Mathematics Type: NAT Concept: Linear Algebra Level:Moderate Sub-Concept: Basics of Matrix Time: 150 sec Concept Field: Determinant 28. The maximum value of the determinant among all 2 × 2 real symmetric matrices with trace 14 is ________. Sol. Correct Answer is (49) Given – A 2 x 2 real symmetric matrix Let x z A= z y Trace ( A ) = x + y = 14 y = 14 − x D et ( A ) = A = xy − z2 A = x ( 14 − x ) − z2 14x − x2 − z2.......(1) For determinant to be maximum PAG E (n ) P= = 14 − 2x x (n ) W hen = 0 14 − 2x = 0 x x=7 t A q= = −2z z A W hen =0z=0 z 2 A r= = −2 x2 2 A s= =0 x. z 2 A t= = −2 z2 For maxima, r < 0 and rt –s2 > 0 X = 7, 2 = 0 y= 14 –x y=7 Det (A) = |A| = xy – z2 = 7 x 7 –0 = 49 Ans Hence the maximum value of determinant is 49. S. No.: 29 Marks: 2 Topic: Engineering Mathematics Type: NAT Concept: Vector Calculus Level: easy Sub-Concept: Differentiation Time: 150 s Concept Field: Divergence 29. If r = xa ˆx + ya ˆz + za ( ) ˆz and r = r, then div r2 ( 1nr) = _______. Sol. Correct Answer is (3) Given, r = xa ˆx + ya ˆz + za ˆz and r = r, According to formula of gradient 1 r r (1nr)= , = 2......(1) r r r PAG E r then div r2 ( 1hr) = div r2 2 From eq. (1) r = div r = ( Ax ) + ( Ay ) + ( Az ) x y z = 1+ 1+ 1 div r2 ( 1hr) = 3 Ans r2 ( 1hr) = 3 Hence, the value of div . S. No.: 30 Marks: 2 Topic: Network Theory Type: NAT Concept: Transient Analysis Level:High Sub-Concept: R-L-C Transients Time: 200 s Concept Field: Time Constant Calculations 30. A series LCR circuit is operated at a frequency different from is resonant frequency. The operating frequency is such that the current leads the supply voltage. The roagnitude of current is half the value at resonance. If the value of L.C and R are 1 H, 1 F and 1 , respectively the operating angular frequency (in rad/s) is ________ Sol. (The correct answer is 0.45 ) R = 1 C = 1P L=1H At resonance, XL = XC modified circuit will be as shown in figure – 2 V V I = = =V Resonance current = R R 1 According to the question– PAG E Current leads the supply voltage- V = jw L − j/w c + R Since, I V = jw − j/w + 1 I (w − 1/w ) − tan+ 0 1 − tan+ (w − 1/w )0 w − 1/w 0 w2 1 w 1 --------------* According to question- I = iR /2 I = V /2 2 V 1 = w − + 1 From eqn 1 → I w v 1 = w 2 + −2+1 v w2 2 1 2= w 2 + w −1 2 Thus, r2 (ln r) = r Divergence of (r (l r) ) → 2 n Div (r (l r)) = div (r) 2 n x+ y+ z = x y z =1+1+1 div (r2 (ln r) ) = 3 S. No.: 31 Marks: 2 Topic: Network Theory Type: NAT Concept: Two Port Networks Level: Moderate Sub-Concept: H Parameters Time: 180 s PAG E Concept Field: Problems and Analysis 31. In the h-parameter model of the 2-port network given in the figure shown, the value of h22 (in S) is________ Sol. (The correct answer is 1.25) Since hybrid parameters are given by– V1 h11 h12 i1 I = h 2 21 h22 V2 Therefore for h22 , I2 = h21I1 + h22V2 (1) V2 = h11I1 + h12V2 (2) Putting I1 = 0 in eqn (1) [assume I1 = 0 ] I2 = 0 + h22VZ I2 h22 = V2 PAG E from figure 3 we get– V2 6 6 6 = + || I2 5 5 5 12 6 || = 5 5 4 12 6 72 V2 25 5 = 5 5 = = 0.8 I2 12 6 18 + 5 5 5 I2 1 = = 1.25 = h22 V2 0.8 Hence the correct answer is 1.25 S. No.: 32 Marks: 2 Topic: Network Theory Type: MCQ PAG E Concept: Circuit Basics Level: High Sub-Concept: Description of Elements Time: 240 s Concept Field: Circuit Reduction Problems 32. In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t > 0? 5 2 (1) I(t) = 3 ( 1− e−t/ ). = m sec 3 5 2 I(t) = ( 1− e−t/ ) , = m sec (2) 2 3 5 I(t) = ( 1− e−t/ ) , = 3m sec (3) 3 5 I(t) = ( 1− e−t/ ) , = 3m sec (4) 2 Sol. The correct option is (A) According to the equation, Initially, the capacitor is uncharged. So, C ( ) V 0− = 0V Since capacitor does not allow sudden change in voltage. VC (0 ) = VC (0 ) = 0V + − At t = 0+, time constant, = Req C For Req- 2 1 2 Req = 2111 = K 2+1 3 PAG E 2 = 103 1 10−6 3 2 = m sec 3 At t = {capacitor will be fully charged} Therefore, the capacitor will act as open circuit 2 10 VC ( ) = 5 = V 2+1 3 ( (0 ) )e + VC (t) = VC ( ) − VC −t/ + VC ( ) 10 −t/ 10 0 − e + = 3 3 10 −t/ 10 10 VC (t) = − 3 e + 3 3 ( 1− e−t/ ) VC (t) 10 5 I(r) = R2 3 ( 1− e −t/c ) ( 1− e −t/ ) = I(t) 3 Hence, Option (A) is correct option. S. No.: 33 Marks: 2 Topic: Network Theory Type: NAT Concept: Miscellaneous Topics Level:Moderate Sub-Concept: Magnetic Coupled Circuits Time: 180 s Concept Field: 33. In the magnetically coupled circuit shown in the figure, 56% of the total flux emanating from one coil links the other coil. The value of the mutual inductance (in H) is __________ PAG E Sol. (The correct answer is 2.50 ) According to the question- 56% of the total flux emanating from one coil links to other. So, the coupling factor will be- K= 56% i.e. k = 0.56 m utual flux or links flux 56% k= = because total flux 100% so, k = 0.56 Given, L1 = 4H L2 = 5H We know the formula for mutual inductance- M = k L1L2 = 0.56 4 5 = 0.56 20 M = 2.50H Hence, the correct answer is 2.50 H S. No.: 34 Marks: 2 Topic: Electronic Devices Type: NAT Concept: Semiconductor Physics Level:moderate Sub-Concept: Properties of Si / Ge Time:180 s Concept Field: Conductivity and Drift Expressions 34. Assume electronic charge q = 1.6 × 10–19 C, kT/q = 25 mV and electron mobility n = 1000 cm 2 /V − s. If the concentration gradient of electrons injected into PAG E a P-type silicon sample is 1 × 1021/cm4, the magnitude of electron diffusion current density (in A/cm2) is _________. Sol. (The correct answer is 4000 ) According to the question, q = 1.6 10−19C KT = 25m V = VT Thermal Voltage, q Electron mobility n = 1000cm /v − s 2 dn = 1 1021 /cm 4 Electron concentration density, dx Therefore, from Einstein equation, dn VT dx D n = nVT = 1000 0.025 D n = 25cm /5 2 Hence, diffusion current density- dn Jn = qD n dx = 1.6 × 10–19 × 25 × 1021 Jn = 4000A /cm 2 Hence, the correct answer is 4000 A/cm2 S. No.: 35 Marks: 2 Topic: Electronic Devices Type: NAT Concept: P- N Junction Diodes Level: Moderate Sub-Concept: Diode Basics Time: 150 s Concept Field: Cut-in Voltage, Energy Band Diagram 35. Consider an abrupt PN junction (at T = 300 K) shown in the figure. The depletion region width Xn on the N-side of the function is 0.2 m and the permittivity of silicon ( i) is 1.044 10 −12 F/cm. At the junction, the approximate value of the peak electric field (in kV/cm) is _________ Sol. (The correct answer is 30.7 ) According to question– Xn = 0.2m = 0.2 10−4cm si= 1.044 + 10−12F /cm PAG E N D = 1016 /cm 2 Since, N A N D xnN D = xpN A xn N A = cp N D Since, N A N D N A + ND N A xn =1 W xn w Therefore, q xnN D E = = E s E s 1.6 10−19 0.2 10−4 1016 = 1.044 10−12 E = 30651.34 v /cm Hence, the correct answer is 30.7 KV/m S. No.: 36 Marks: 2 Topic: Electronic Devices Type: MCQ Concept: P- N Junction Diodes Level: Moderate Sub-Concept: Diode Basics Time:180 s Concept Field: Construction , Depletion Region −3 When a silicon diode having a doping concentration of N A = 9 10 cm 16 36. on p- −3 side and N D = 1 10 cm 16 on n-side is reverse biased, the total depletion width is found to be 3 m. Given that the permittivity of silicon is 1.04 × 1012 F/cm, the depletion width on the p-side and the maximum electric field in the depletion region, respectively, are (A) 2.7m and 2.3 × 105 V/cm (B) 0.3 m and 4.15 × 105 V/cm (C) 0.3 m and 0.42 × 105 V/cm (D) 2.1m and 0.42 × 105 V/cm Sol. The correct answer is (B) Given, −3 N A = 9 1016cm −3 N D = 1 1D 16cm W = depletion width = 3m = xn + xp PAG E E si = permittivity of silicon = 1.04 × 10–12 F/cm. w = 3m = 3 10 cm −4 Since, we know— xnN D = xpN A xn N A = xp N D xn NA = xn + xp N A + N D Or xp ND Xp = = xn + xp N A + N D w xp 1 1016 = 3 10−4 9 1016 + 1016 1 1016 xp = 16 ( 3 10−4 ) 9 10 + 1 0 16 −5 xp = 3 10 cm xp = 0.3m We know, depletion width, 2E si 1 1 w = + (V0 + VR ) q N A ND 2 1.04 10−12 1 1 3 10−4 = 1.6 10−19 + 16 (V 0 + VR ) 9 10 16 10 (V0 + VR ) = 62.3076 v S. No.: 37 Marks: 2 Topic: Analog Electronics & Circuits Type: MCQ Concept: Diode Applications Level:High Sub-Concept: Rectifiers Time:240 s Concept Field: Half Wave, Full Wave, Bridge 37. The diode in the circuit shown has Von = 0.7 Volts but is ideal otherwise. If Vi = 5sin (w t) Volts, the minimum and maximum values of Vo (in Volts) are, respectively, PAG E (A) –5 and 2.7 (B) 2.7 and 5 (C) –5 and 3.85 (D) 1.3 and 5 Sol. (The correct answer is option (C) According to the question, V0N = 0.7v vi = 5sin (w t) v At Vi(max) = 5v, Diode will be ON as shown in figure below— PAG E Therefore, by voltage divider 5 1+ ( 2 + 0.7) 1 V0 (m ax) = 1+ 1 5 + 2.7 7.7 V0 (m ax) = = 2 2 V0(m ax) = 3.85V V = −5v, At i(m in) diode will be off i.e. (open circuit as shown in figure below: V0(m in) = −5V Therefore, the correct answer is option (C) S. No.: 38 Marks: 2 Topic: Electronic Devices Type: MCQ Concept: Transistors Level: Moderate Sub-Concept: MOSFET Time: 180 s Concept Field: Threshold and Pinch-off 38. For the n-channel MOS transistor shown in the figure, the threshold voltage VTh is 0.8 V. Neglect channel length modulation effects. When the drain voltage VD = 1.6V. the drain current ID was found to be 0.5 mA. I VD is adjusted to be 2V by changing the values of R and VDD, the new value of ID (in mA) is PAG E (A) 0.625 (B) 0.75 (C) 1.125 (D) 1.5 Sol. (The correct option is C) Given: VTh = 0.8 V When drain voltage V0 = 1.6 V then I0 = 0.5 m A Since, drain and gate terminals are common, VG S = VD S VG = VS { VS = O V } VD S VG S − VTh Therefore, mesfet will always be in saturation, kn (VGS − VTh ) 2 ID = Thus, 2 kn ( 1.6 − 0.8) 2 0.5 = = 2 … (1) 1 = kn (0.8) 2 1 kn 0.64 When VD = 2V = VG S new PAG E kn ( ) 2 ID = VGS − VTh 2 new 1 (2 − 0.8) 2 ID = k 0.64 2 ID = 1.25 m A Hence, the correct option is (C) S. No.: 39 Marks: 2 Topic: Analog Electronics & Circuits Type: NAT Concept: FET /MOSFET Biasing Level: High Sub-Concept: MOSFET Biasing Time: 240 s Concept Field: Choice of Bias Voltages 39. For the MOSFETs shown in the figure, the threshold voltage Vt = 2V and 1 W K = C ax 0.1 m A /V 2 2 L. The value of ID (in mA) is ________. Sol. (The correct answer is 0.9 ) Given, Vt = 2V b 1 W k = C ox = 0.1m A /V 2 2 L PAG E Thus, from the circuit, VG 2 = 0 R2 VG 1 = VD D R1 + R2 10 12 = 10 + 10 VG 1 = 6 V Both the N-MOS and P-MOS are in saturation. Applying KVL at M2 – 0 − VG S 2 − ( −5) = 0 VG S 2 = 5V We know 1 W nc0 (VGs2 − Vth ) 2 I02 = I0 = 2 L = 0.1 9 ID = 0.9 m A Hence, the correct answer is 0.9 mA S. No.: 40 Marks: 2 Topic: Digital Electronics Type: MCQ Concept: Sequential Circuits Level: HIgh Sub-Concept: Flip- Flops Time: 240 s Concept Field: D and T Flip Flops 40. In the circuit shown, choose the correct timing diagram of the output (y) from the given waveforms W1, W2, W3 and W4. PAG E (A) W1 (B) W2 (C) W3 (D) W4 Sol. (The correct option is C) PAG E In the above circuit, the two flip=fops FF1 and FF2 are negative edge triggered D-flip-flop. Thus, the output of the flip-flop changes when– Clk (clock) pluse changes from 1 to 0; Q 1 = X 1; w hen clk sw itches from 1 to 0. ie. from high to low Q 2 = X 2 ;w hen clk sw itches from 1 to 0 Thus, Y = Q1 Q2 S. No.: 41 Marks: 2 Topic: Digital Electronics Type: MCQ Concept: Sequential Circuits Level: High Sub-Concept: Flip- Flops Time: 200 s Concept Field: Inter- Conversions, State Diagrams, Truth Tables 41. The outputs of the two flip-flop Q1, Q2 in the figure shown are initialized to 0, 0. The sequence generated at Q1 upon application of clock signal is (A) 01110…. (B) 01010…. (C) 00110…. (D) 01100…. Sol. (The correct answer is D) PAG E From the circuit, J1 = Q 2 K1 = Q2 J2 = Q 1 K2 = Q1 Initially, the output of the flip-flops are– Q1 = 0 Q2 = 0 Thus, the truth table for sequential circuit is– Clock J1 (Q 2 ) K 1 (Q 2 ) J2 (Q 1 ) K 2 (Q 1 ) Q 1 Q 2 Initial → − − − − 0 0 1 CP → st 1 0 0 1 1 0 2 CP → nd 1 0 1 0 1 1 3 CP → rd 0 1 1 0 0 1 4 CP → th 0 1 0 1 0 0 Hence the correct answer is Q1 = 01100 …. S. No.: 42 Marks: 2 Topic: Microprocessors Type: MCQ Concept: Interfacing Level: High Sub-Concept: Memory Interfacing Time: 180 s Concept Field: 42. For the 8085 microprocessors, the interfacing circuit to input 8-bit digital data (DI0 – DI7) from an external device is shown in the figure. The instruction for correct data transfer is. PAG E (A) MVI A, F8H (B) IN F8H (C) OUT F8H (D) LDA F8F8H Sol. (The correct answer is D) The above circuit diagram shows that it is a memory mapped Input/output. Since to enable 3 to 8 Decoder G 2A is required active low signal through Io /M and G 2B is required active low through RD. Now to enable the decoder output of AND gate, must be 1-and DS1 signal is required to be 1 which is the output of multi-inputs AND gate to enable I/O device. Thus, A 15 A 14 A 13 A 12 A 11 A 10 A 9 A 8 A 7 A6 A5 A 4 A 3 A 2 A 1 A 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 F 8 F 8 Therefore, device address = F8F8H PAG E So, the instruction will be – LDA F8F8H Hence, the correct answer is (D) S. No.: 43 Marks: 2 Topic: Signals & Systems Type: NAT Concept: Classifications of Systems Level:Moderate Sub-Concept: Convolution Time: 180 s Concept Field: Properties of Convolution 43. Consider a discrete-time signal n for 0 n 10 x n = 0 otherw ise If y[n ] is the convolution of y[n ] with itself, the value of y is _______ Sol. (The correct answer is 10) Given, n for 0 n 10 x n = 0 otherw ise for convolution → y (n ) = x (n ) * x (n ) y ( 4) = 0 + 3 + 4 + 3 + 0 y ( 4) = 10 [Other method] → or n for 0 n 10 x n = 0 otherw ise y (n ) is the convolution of x (n ) with itself then, y n = x n * x n n x k x n − k = k =0 Therefore, PAG E 4 y 4 = x k x 4 − k k =0 = x 0 x 4 + x 1 x 3 + x 2 + x 3 x ( 1) + x ( 4 ) x (0 ) =0+3+4+3+0 y 4 = 10 Hence, the correct answer is 10 S. No.: 44 Marks: 2 Topic: Signals & Systems Type:MCQ Concept: Classifications of Systems Level:Moderate Sub-Concept: Linear Time Invariant Time: 150 s Concept Field: Discrete LTI 44. The input-output relationship of a causal stable LTI system is given as 𝑦|𝑛| = 𝛼𝑦|𝑛 − 1| + 𝛽𝑥|𝑛| If the impulse response h n of this system satisfies the condition n =0 hn =2 the relationship between and is (A) = 1− /2 (B) = 1+ /2 (C) = 2 (D) = −2 Sol. (The correct answer is (A) Given, y n = y n − 1 + x n …(1) h n = 2 And n =0 By z-transform– x (n − n.) Z.Tx (z ) z −n 0 Thus applying the property of Z-transform in equation 1- y (z ) = z−1y (z ) + (z ) y ( z ) ( 1− z−1) = (z ) y (z ) = H (z ) = x (z ) ( 1− z ) −1 … (2) 1 for x 1 anU (n ) ⎯⎯→ Z.T −1 1− az 1+ x + x +....... 2 1 = Thus applying this property of Z-Transform in eq → n 1− x h (n ) = ( ) (n ) n ( ) = 1 ( ) n h n = =2 n =0 n =0 1− PAG E = 1− 2 = 1− 2 Hence the correct answer (A) S. No.: 45 Marks: 2 Topic: Signals & Systems Type: NAT Concept: Discrete Fourier Analysis Level: Easy Sub-Concept: DT Fourier Transform Time: 120 s Concept Field: Properties of DTFT sin c2 (5t) dt 45. The value of the integral − is __________. Sol. (The correct answer is 0.2) Let x (t) = sin c (5t) sin c (5T ) ⎯⎯ F.t → Thus, by Parsevall’s Theorem, x2 (t) dt = x2 (f) df − − 2 1 2.5 = −2.5 5 df 1 (2.5 − ( −2.5) ) = 25 1 5 = 25 x2 (t) dt = 0.2 − sin2 C (5t) dt = 0.2 Therefore, − Ans. Hence, the correct answer is 0.2 S. No.: 46 Marks: 2 Topic: Signals & Systems Type:MCQ Concept: Laplace Transforms Level:Moderate Sub-Concept: Unilateral LT Time:180 s Concept Field: Properties & Comparison PAG E 46. 𝐴𝑛 𝑢𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑙𝑖𝑛𝑒𝑎𝑟 𝑡𝑖𝑚𝑒 𝑖𝑛𝑣𝑎𝑟𝑖𝑎𝑛𝑡 (𝐿𝑇𝐼) 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑒𝑑 𝑏𝑦 [𝑥̇ 1 𝑥̇ 2 ] = [−1 0 0 − 2 ][𝑥1 𝑥2 ] If the initial conditions are x1 ( 0 ) = 1 and x2 (0) = −1, the solution of the state equation is (A) x1 (t) = −1,x2 (t) = 2 (B) x1 (t) = −e ,x2 (t) = 2e −t −t (C) x1 (t) = e ,x2 (t) = −e −t −2t (D) x1 (t) = −e ,x2 (t) = −2e −t −t Sol. (The correct answer is C) Given, x −1 0 x1 x = 0 −2 x 2 Initial condition: x1 ( 0 ) = 1 x2 (0) = −1 dx1 x1.= = −x1 dt By Laplace Transform – sx1 (s) − x1 ( 0) = −x1 (s) (s + 1) x1 (s) = 1 Property of Laplace Transform 1 −at 1 x1 ( s) = e u (t) ⎯⎯⎯ L.T → s+1 s + a using the Laplace Transform property, we get – x1 (t) = e u (t) −t dx2 x.2 = = −2x2 dt Therefore, By Laplace Transform – s 2 (s) − x2 (0) = x2 (s) ( −2) (s + 2) x2 (s) = −1 −1 x2 (s) = s+2 By using the above property of Laplace Transform, we get x2 (t) = −e−2tu (t) Hence, the correct answer is (C). S. No.: 47 Marks: 2 Topic: Control Systems Type: NAT Concept: Frequency Domain Analysis Level: Moderate PAG E Sub-Concept: Bode Plots Time: 180 s Concept Field: Gain and Phase Response 47. The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure. If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is ________. Sol. (The correct answer is 0.5) For unit ramp input of Type-1 system— 1 ess = KV 26.02 − 6.02 Initial slope of Bode plot = log ( 0.1) − log 1 = –20 dB/dec Initial slope cutting ‘low ’ axis = K V = KV = 2 1 1 ess = KV = 2 ess = 0.5 Hence, the correct answer is 0.5 S. No.: 48 Marks: 2 Topic: Control Systems Type: NAT Concept: Introduction Level: Moderate Sub-Concept: Examples Time: 200 s Concept Field: Block Diagram and Signal Flow PAG E 48. Consider the state space system expressed by the signal flow diagram shown in the figure. The corresponding system is (A) always controllable (B) always observable (C) always stable (D) always unstable Sol. (The correct answer is A) Given, signal flow diagram— Thus, from signal flow graph, x1 = x2 x2 = x3 x3 = x1x1 + a2x2 + a3x3 + u y = c1x1 + c2x2 + c3x3 Thus, in matrix form, it can be written as— x1 0 1 0 x1 0 x = 0 0 1 x + 0 u 2 2 x3 a1 a2 a3 x3 1 y1 x1 y = c c c x 2 1 2 3 2 x3 And y3 Thus, 0 1 0 0 0 1 a1 a2 a3 A= PAG E 0 0 B = 1 C = C 1 C 2 C 3 For controllability Q C = B AB A 2B 0 0 1 Q C 0 1 a1 1 a3 a2+a32 Q c = 1 0 Since, the rank of the matrix is 3; which is equal to the number of variable a1, a2 and a3. Thus, the system is always controllable matrix. For observability C c1 c2 c3 Q 0 = CA c3a1 c1 + c2a2 c2 + c3a3 CA 2 (c2 + c3a3 ) c1 c3a1 + (c2 + c3a3 ) a2 c1c2a2 + (c2 + c3a3 ) a3 Q o depends on the value of the unknown) i.e. a1,a2 ,a3 ,c1,c2 ,c3 Thus, the system is not observable always Hence, the correct answer is (A) S. No.: 49 Marks: 2 Topic: Communications Type: NAT Concept: Pulse Modulation Schemes Level: High Sub-Concept: Pulse Coded Modulation Time: 180 s Concept Field: Stages of Sampling, Quantization, Encoding The input to a 1-bit quantizer is a random variable X with pdf fx ( x ) = 2e −2x 49. for x 0 and fx ( x ) = 0 for x 0. For outputs to be of equal probability, the quantizer threshold should be ________ Sol. (The correct answer is 0.35) Given, fx ( x ) = 2e −2x for x 0 fx ( x ) = 0 for x 0 The input 1-bit quantizer, can be defined as 1, x xT xq = 0, x xT Where, xT → threshold value According to the question, Output probability is equal. Therefore, PAG E p ( xq = 1) = p ( xq = 0) p ( x xT ) = p ( x xT ) For threshold value xT fx ( x ) dx = fx ( x ) dx xT 0 xT xT 2e−2x dx = 0 2e−2x dx xT 2e −2x 2e −2x = −2 x −2 0 T −2xT 0 + e −2xT = −e +1 2e−2x+ = 1 1 xT = ln2 2 xT = 0.35 S. No.: 50 Marks: 2 Topic: Communications Type: MCQ Concept: Random Variables & Noise Level: Moderate Sub-Concept: Noise and Linear Systems Time:240 s Concept Field: Bit Error Rate as Noise 50. Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s1 (t) = cos2 ft 1 and s2 (t) = cos2 ft 2 , where = 4m V. Assume an AWGN channel with two-sided noise power N0 = 0.5 10−12W /H z spectral density 2. Using an optimal receiver and the 1 −u 2 /2 Q (v ) = 2 v e du relation , the bit error probability for a data rate of 500 kbps is (A) Q ( 2) (B) ( Q 2 2 ) (C) Q ( 4) (D) Q 4 2 ( ) Sol. (The correct answer is C) Given, s1 (t) = cos2f1 t s2 (t) = cos2 f1 t Where = 4m V PAG E N0 = 0.5 10−12w /H z and Two-sided noise power spectral density 2 for orthogonal binary fSK , E Q b (Pe ) N Bit error probability = o Where, E b = energy per bit So, E b = 5000 kbps (data rate) Eb = ( 4 10 ) −3 2 2 500 103 −12 E b = 16 10 Now, noise power spectral density, No = 0.5 10−12 w /H z 2 N o = 2 0.5 10−12 w /H z −12 N o = 10 w /H z Thus, (Pe) Bit error probability E Pe = Q b No 16 10−12 Q = Q 16 = 0 − 12 Pe = Q ( 4) Hence, the correct answer is (C) S. No.: 51 Marks: 2 Topic: Communications Type: NAT Concept: Random Variables & Noise Level:High Sub-Concept: Random Process Time: 240 s Concept Field: WSS , SSS ,Ergodicity 51. The power spectral density of a real stationary random process X(t) is given by 1 , fW S x (f) = w 0, f W 1 E X (t) X t − The value of the expectation 4W is ___________. Sol. (The correct answer is 4) Given, PAG E 1 , f w S X (f) = w 0 , f w Since, the autocorrection function— R x ( ) = S x (f) e j2ftdt w 1 j2ft = −w w e ft dfw 1 e j2f = w j2 −w 1 j w (e j2w t − e − j2w ) 2 = sin ( 2 w ) R x ( ) = = w 1 1 E (t) t − = E x (t) t − 4w Now, 4N Thus, auto correlation function can be defined -- R ( )= E x (t) (t − ) 1 1 R = E x (t) t − 4w 4w 1 1 E X (t) t − = R x Thus, 4w 4w sin 2 1 w 4w 1 w = 4w sin ce,sin 2 = sin90 = 1 = 4 sin ( /2) =4×1 1 E x (t) t − =4 4w Hence, the correct answer is 4. S. No.: 52 Marks: 2 Topic: Communications Type: NAT Concept: Amplitude Modulation Level: High Sub-Concept: AM Receivers Time: 240 s Concept Field: Tuned RF Receiver PAG E 52. In the figure, M (f) is the Fourier transform of the message signal m (t) where c ) and w (t) = cos ( 2 fc + A ) t , A = 100 Hz and B = 40 Hz. Given v (t) = cos (2ft where fc A = cos (2fc + A ) t , where fc A. The cutoff frequencies of both the filters are fc. The bandwidth of the signal at the output of the modulator (in Hz) is _________. Sol. (The correct answer is 60) Given, A = 100 Hz B = 40 Hz V (t) = cos (2 ft c ) w (t) = cos ( 2 fc + A ) t, where fc A m (t) ⎯⎯→ F.T M (f) Signal v (t) = cos (2ft c ) PAG E After multiplication of v (t) with m (t) , we get— Let w (t) = m (t) ,v (t) frequency spectrum for w (t) After using high pass filter → Again the signal will be multiplied with ( ( c 2 f + A ) t) and Low pass filter with cut off frequency fe. Therefore, the bandwidth of output signal is – Bandwidth = A – B Bandwidth = 100 – 40 Bandwidth = 60 Hz Hence, the answer is 60 Hz. S. No.: 53 Marks: 2 Topic: Electromagnetics Type: MCQ Concept: Electromagnetic Waves Level:Moderate Sub-Concept: Reflections & Transmissions Time:120 sec Concept Field: S & P Polarizations 53. If the electric field of a plane wave is E ( x,t) = x ˆ3cos (t − kz + 30 ) − y ˆ4 sin (t − kz + 45) (m V /m ) PAG E the polarization state of the plane wave is (1) left elliptical (2) left circular (3) right elliptical (4) right circular Sol. The correct answer is (A). Given, E (z,t) = x ˆ3cos (w t − kz + 30 ) − y ˆ4 sin (w t − kz + 45 ) (m V /m ) E (z,t) = x ˆ3cos (w t − kz + 30 ) − y ˆ4 sin (w t − kz + 135 ) Propagation direction is in ‘+z’ direction. ŷ is leading And x̂ is lagging E x = 3, E y = 4 Ex E y Therefore, it is elliptical polarization and since, ŷ is leading and x̂ is lagging and Q = 30 − 135 = −105 Therefore, it is left elliptical polarization. Hence, the correct answer is (A) S. No.: 54 Marks: 2 Topic: Electromagnetics Type: NAT Concept: Transmission Lines Level:High Sub-Concept: Basics Time:240 sec Concept Field: Characteristic Impedance 54. In the transmission line shown, the impedance Zin (in ohms) bet
