ACE Engineering Academy. Hyderabad Delhi Bhopal Pune Bhubaneswar Bengaluru Lucknow Patna Chennai Vijayawada Visakhapatnam Tirupati

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2 : : Gate 06 ECE Set Q. Q.5 Carry one mark each 0. Based on the given statements, select the appropriate option with respect to grammar a statements. (i) The height of Mr. X is 6 feet. (ii) The height of Mr. Y is 5 feet (A) Mr. X is longer than Mr. Y (B) Mr. X is more elongated than Mr. Y (C) Mr. X is taller than Mr. Y (D) Mr. X is lengthier than Mr. Y 0. Ans: (C) Sol: In degrees of comparison Mr. X is taller than Mr. Y is apt. Positive degree tall Comparative degree taller Superlative degree tallest 0. The students the teachers on teachers day for twenty years of dedicated teaching. (A) facilitated (B) felicitated (C) fantasized (D) facilitated 0. Ans: (B) Sol: Felicitate means honour. 03. After India s cricket world cup victory in 985, Shrotria who was playing both tennis and cricket till then, decided to concentrate only on cricket. And the rest is history. What does the underlined phrase mean in this context? (A) history will rest in peace (B) rest is recorded in history books (C) rest is well known (D) rest in archaic 03. Ans: (C) Sol: rest is history is an idiomatic expression which means rest is well known 04. Given (9 inches) / = (0.5 yards) /, which one of the following statements is TRUE? (A) 3 inches = 0.5 yards (B) 9 inches =.5 yards (C) 9 inches = 0.5 yards (D) 8 inches = yards 04. Ans: (C) Sol: Given (9 inches) / = (0.5 yards) / 9 inches = 0.5 yards 05. S, M, E and F are working in shifts in a term to finish a project. M works with twice the efficiency of other but for half as many days as E worked. S and M have 6 hour shifts in a day, whereas E and F have hours shifts. What is the ratio of contribution of M to contribution of E in the project? (A) : (B) : (C) :4 (D) : 05. Ans: (B) Sol: M efficiency = [ efficiency of S,E, and F] Contribution of M in the project = x days 6 hrs Contribution of E in the project = x days hrs Contribution of M : Contribution of E x 6 : x :

3 : 3 : Questions & Solutions Q.6 Q.0 Carry two marks each 06. The Venn diagram shows the preference of the student population for leisure activities. Read books Watch Tv Play sports From the data given, the number of students who like to read books or play sports is. (A) 44 (B) 5 (C) 79 (D) Ans: (D) Sol: Read books = n(r) = = 76 Play sports = n(s) = = 83 n (R S) = 44 7 = 5 n (R S) = n (R) n (S) n (R S) = = Social science disciplines were in existence in an amorphous form until the colonial period when they were institutionalized. In varying degrees, they were intended to further the colonial interest. In the time of globalization and the economic rise of postcolonical countries like India, conventional ways of knowledge production have become obsolete. Which of the following can be logically inferred from the above statements? (i) Social science disciplines have become obsolete. (ii) Social science disciplines had a pre-colonial origin (iii) Social science disciplines always promote colonialism (iv) Social science must maintain disciplinary boundaries (A) (ii) only (B) (i) and (iii) only (C) (ii) and (iv) only (D) (iii) and (iv) only 07. Ans: (A) Sol: Until the colonial period means pre-colonial origin. Other options can t be inferred.

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5 : 5 : Questions & Solutions 08. Two and a quarter hours back, when seen in a mirror, the reflection of a wall clock without number markings seemed to show :30. What is the actual current time shown by the clock? (A) 8:5 (B) :5 (C) :5 (D) : Ans: (D) Sol: Time back = = hrs 5 min 4 Clock time (C.T) Mirror Time (M.T) = 60 C.T = The actual time shown by the clock = = M and N start from the same location. M travels 0 km East and then 0 km North-East. N travels 5 km South and then 4 km South-East. What is the shortest distance (in km) between M and N at the end of their travel? (A) 8.60 (B).50 (C) 0.6 (D) Ans: (C) Sol: From the given data, the following diagram is possible M A 0 km C B 0 km 5 km 5 km D 45 4 km E N DE cos 45 o 4 DE = cos45 o 4 =.88 km EN sin 45 o 4 EN = sin 45 o 4 =.88 km

6 sin 45 o EN = 4 EN = sin 45 o 4 =.88 km CN = NE CE =.88 5 = 7.88 km CB = AB AC = 0.88 = 7.7 km (NB) = (NC) (BC) = (7.88) (7.7) NB = = 0.66 km NM = NB BN = = 0.6 km : 6 : Gate 06 ECE Set 0. A wire of length 340 mm is to be cut into two parts. One of the parts is to be made into a square and the other into a rectangle where sides are in the ratio of :. What is the length of the side of the square (in mm) such that the combined area of the square and the rectangle is a MINIMUM? (A) 30 (B) 40 (C) 0 (D) Ans: (B) Sol: Length of the wire = 340 m x x 3 Square Rectangle x x Perimeter of rectangle = 3 3 = x Perimeter of square = 340 x 340 x Side of square 4 Total area = Area of square Area of rectangle 340 x x x 340 x Combined area of square rectangle = minimum f(x) = 0 f 340 x x x 9 4 x x f x 0 x x 3 x 9

7 : 7 : Questions & Solutions 4 x 340 x 9 4 = x = x Side of square = 4 40mm Another method: Elimination procedure from alternatives option [C] and [D] are not possible because area may be maximum. Option (A) Side of the square = x = 30 mm Perimeter of the square = = 0 mm Perimeter of the rectangle = = 0 mm x x = 0 x = 37 x = 37 = 74 Area of square = x = (30) = 900 Area of rectangle = x x = = 738 Total area = = 3638 mm Option (B) Side of the square = x = 40 mm Perimeter of the square = = 80 mm x x = 80 mm 6x = 80 mm x = 30 mm Area of the square = = 600 mm Area of the rectangle = = 800 mm Total area = = 3400 mm 3400 mm < 3638 mm Option B is correct. Q. Q.5 Carry one mark each. 0. The value of x for which the matrix A = has zero as an eigen value is x 0. Ans: x = Sol: For eigen value of A is to be zero, det (A) = 0 3 {(63 7x ) 5} { (8 9x)78} 4 { 36 4} = 0 x =

8 : 8 : Gate 06 ECE Set 0. Consider the complex valued function f(z) = z 3 3 b z where z is a complex variable. The value of b for which the function f(z) is analytic is 0. Ans: 0 Sol: f (Z) = z 3 b z 3 for b = 0, f(z) becomes polynomial so it is analytic every where only when b = As x varies from to 3, which of the following describes the behaviour of the function f(x) = x 3 3x? (A) f(x) increases monotonically (B) f(x) increases, then decreases and increases again (C) f(x) decreases, then increases and decreases again (D) f(x) increases and then decreases 03. Ans: (B) Sol: Since, f() = 3, f(0) =, f() =, f() = 3, f(3) = 04. How many distinct values of x satisfy the equation sin(x) = x/, where x is in radians? (A) (B) (C) 3 (D) 4 or more 04. Ans: (C) Sol: Sin x = x touches at 3 points. 05. Consider the time-varying vector I = in Cartesian coordinates, where > 0 is a constant. When the vector magnitude I is at its minimum value, the angle that I makes with the x axis (in degree, such that 0 80) is 05. Ans: 90 o Sol: If = 0 I = 5 If = I = Sin 5 Cos I R ef â 5sin t â I= 5 cost x y

9 : 9 : Questions & Solutions

10 : 0 : Gate 06 ECE Set 06. In the circuit shown below, V s is a constant voltage source and I L is a constant current load V s R I L The value of I L that maximizes the power absorbed by the constant current load is V (A) S V (B) S V (C) S (D) 4R R R 06. Ans: (B) Sol: Maximum power delivered by the source to any load vs Pmax w 4R Here power absorbed by the load P L v L.I L W V S I L.R. I L V.I I.R W If = I L S L VS R L vs vs vs vs vs PL vs..r w R R R 4R 4R PL P max 07. The switch has been in position for a long time and abruptly changes to position at t = 0 If time t is in seconds, the capacitor voltage V C (in volts) for t > 0 given by (A) 4 ( exp (t/0.5)) (B) 0 6exp (t/0.5)) (C) 4 (exp(t/0.6)) (D) 0 6 exp (t /0.6) 07. Ans: (D) 0 Sol: v 0. 4v v 0 c v c c 5 5 0v Req.C 0V sec / T 4 0e t c 0.F 6 V (t) 0 0 6e v for 0 t t 0. t = 0 V c 4 5A

11 08. The figure shows an RLC circuit with a sinusoidal current source. : : Questions & Solutions I m sint I R I L I C R L C 0 0mH 0F At resonance, the ratio I L / I R, i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is 08. Ans: Sol: At resonance, I R =I 0 IL QI 90 A I C = QI90 o A Where Q = W 0 CR R C L IL So, Q I R The z-parameter matrix for the two-port network shown is j j j 3 j Z j R j Where the entries are is. Suppose b b Z a Z b Z c Then the value of R b (in ) equals 09. Ans: 3 s s z A z C z C Sol: z (s) s 3 s zc z B z C Here, z c = s And z B z C = 3 s z B s = 3 s z B = 3s R B jx B = 3 j R B = 3

12 0. The energy of the signal x(t) = 0. Ans: 0.5 Sol: sin at rect a X rect 4 8 /4 sin 4t 4t X() 4 4 : : Gate 06 ECE Set is E x ( t ) X 8 6 d The Ebers Moll model of a BJT is valid (A) only in active mode (B) only in active and saturation modes (C) only in active and cut-off modes (D) in active, saturation and cut-off modes. Ans: (A). A long-channel NMOS transistor is biased in the linear region V DS = 50 mv and is used as a resistance. Which one of the following statements is NOT correct? (A) If the device width W is increased, the resistance decreases (B) If the threshold voltage is reduced, the resistance decrease (C) If the device length L is increased, the resistance increases (D) If V GS is increased, the resistance increases. Ans: (D) Sol: A. TRUE B. TRUE C. TRUE D. FALSE rds on W ncox V gs VT L

13 : 3 : Questions & Solutions 3. Assume that the diode in the figure has V on = 0.7 V, but is otherwise ideal. R k V R 6 k The magnitude of the current i (in ma) is equal to 3. Ans: 0.5 Sol: V k 6k I Thevenin eq k 6k () 4 0.5V 6 8 Diode needs at least 0.7V, with 0.5V at the terminals, the diode is OFF. Therefore the circuit reduces to V k I k 6k 8k = 0.5mA 6k I

14 : 4 : Gate 06 ECE Set 4. Resistor R in the circuit below has been adjusted so that I = ma. The bipolar transistor Q and Q are perfectly matched and have very high current gain, so their base currents are negligible. The supply voltage V cc is 6 V. The thermal voltage kt/q is 6 mv. V CC R I Q Q I R The value of R (in ) for which I = 00 µa is 4. Ans: Vbe Vt Sol: IC ISe I c I Vbel Vt n Vt n Is Is I Vbe Vt n Is From the circuit kt V be =V be I R and V t 6mV q I ma Vt n 6mV n Vbe V I be R 00 A = = I I 00 A 5. Which one of the following statements is correct about an ac-coupled common-emitter amplifier operating in the mid-band region? (A) The device parasitic capacitances behave like open circuits, whereas coupling and bypass capacitances behave like short circuits. (B) The device parasitic capacitances, coupling capacitances and bypass capacitances behave like open circuits. (C) The device parasitic capacitances, coupling capacitances and bypass capacitances behave like short circuits. (D) The device parasitic capacitances behave like short circuits, whereas coupling and bypass capacitances behave like open circuits. 5. Ans: (A) Sol: The parasitic capacitances are in PF and the coupling and bypass capacitors are in F. Therefore for the mid frequency band, parasitic capacitance act like open circuits and coupling and bypass capacitances act like short circuits.

15 : 5 : Questions & Solutions 6. Transistor geometries in a CMOS inverter have been adjusted to meet the requirement for worst case charge and discharge times for driving a load capacitor C. This design is to be converted to that of a NOR circuit in the same technology, so that its worst case charge and discharge times while driving the same capacitor are similar. The channel length of all transistors are to be kept unchanged. Which one of the following statements is correct? V DD V DD In out In In C C (A) Widths of PMOS transistors should be doubled, while widths of NMOS transistors should be halved. (B) Widths of PMOS transistors should be doubled, while widths of NMOS transistors should not be changed. (C) Widths of PMOS transistors should be halved, while widths of NMOS transistors should not be changed. (D) Widths of PMOS transistors should be unchanged, while widths of NMOS transistors should be halved. 6. Ans: (C) Sol: Width of PMOS transistors should be halved. while width of NMOS transistors should not be changed, because NMOS transistors are in parallel. If any one transistor ON, output goes to LOW. 7. Assume that all the digital gates in the circuit shown in the figure are ideal, the resistor R = 0 k and the supply voltage is 5V. The D flip-flops D, D D 3, D 4 and D 5 are initialized with logic values, 0,, 0, and 0, respectively. The clock has a 30% duty cycle. D Clock Q D Q D Q D Q D Q D D D 3 D 4 D 5 R =0k The average power dissipated (in mw) in the resistor R is

16 : 6 : Gate 06 ECE Set 7. Ans:.5 Sol: Clk Q Q Q 3 Q 4 Q 5 Y= Q 3 Q The waveform at OR gate output, Y is [A = 5V] Average power V Ao Lt P T R R T T A dt RT T A 0 T T 3T 4T 5T T = 5T y (t) dt A dt A RT T o 5T P (T T) (5T 3T).5mw 3T A.3T R(5T) A 4: multiplexer is to be used for generating the output carry of a full adder. A and B are the bits to be added while C in is the input carry and C out is the output carry. A and B are to be used as the select bits with A being the more significant select bit. I 0 I I I 3 4: MUX S S 0 A B Which one of the following statements correctly describes the choice of signals to be connected to the inputs I 0, I, I and I 3 so that the output is C out? (A) I 0 = 0, I = C in, I = C in and I 3 = (B) I 0 =, I = C in, I = C in and I 3 = (C) I 0 = C in, I = 0, I = and I 3 = C in (D) I 0 = 0, I = C in, I = and I 3 = C in 8. Ans: (A) Sol: C i (A, B, C i ) = m (3, 5, 6, 7) using 4: max C out C in C in I 0 I I I C in C in

17 : 7 : Questions & Solutions 9. The response of the system s G(s) to the unit step input u(t) is y(t). (s ) (s 3) dy The value of at t=0 is dt 9. Ans: Sol: Method : Given Y(s) = Y(S) = s dy L sy(s) dt sy(s) = s s s 3 (s ) (s )(s 3) s s s 3 u(s) [Given u(s) = s ] dy dt t 0 Lt s (s ) (s )(s 3) = Method: (s ) Y(s) = = s(s )(s 3) y(t) = /3 3/ e t 5/6e -3t dy (t dt 3 5 3s (s ) 6(s 3) 5 6 t 3t ( 3)e = 3/ 5/ = 0 ) 3/ ( )e 0. The number and direction of encirclements around the point j0 in the complex plane by the s Nyquist plot of G(S) is 4 s (A) zero (B) one, anti-clockwise (C) one, clockwise (D) two, clockwise 0. Ans: (A) Sol: Im GH Re GH N = 0 Number of Encirclements about (, j0) is Zero

18 : 8 : Gate 06 ECE Set. A discrete memoryless source has an alphabet (a, a, a 3, a 4 ) with corresponding probabilities,,,. The minimum required average codeword length in bits to represent this source for error-free reconstruction is. Ans:.75 Sol: H log log 4 log 8 log =.75. A speech signal is sampled at 8 khz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in khz) required for transmission is. Ans: 6 Sol: R b = 64 kbps BW = 3 khz R b 6 4 khz

19 : 9 : Questions & Solutions 3. A uniform and constant magnetic field B = ẑb exists in the ẑ direction in vacuum. A particle of mass m with a small charge q is introduced into this region with an initial velocity v = xˆ vx ẑv. Given that B, m, q, v x and v z are all non-zero, which one of the following describes the eventual trajectory of the particle? (A) Helical motion in the ẑ direction (B) Circular motion in the xy plane (C) Linear motion in the ẑ direction (D) Linear motion in the xˆ direction 3. Ans: (A) Sol: Given B BẐ V VX Xˆ VZẐ x component of V is perpendicular to magnetic field B. A change moving perpendicular to the magnetic field experience a radial force causing circular motion shown in figure. z- component of V is parallel to magnetic field B. A change moving parallel to the field generates no force shown in figure (b). Motion with components perpendicular and parallel to the field causes the change to move in a helical path along z direction. Show in figure (c). B q V o F F F B v v q B Fig (a) Fig (b) q v Fig (c) 4. Let the electric field vector of a plane electromagnetic wave propagating in a homogenous medium j(wtz) be expressed as E = xˆe e, where the propagation constant is a function of the angular x frequency. Assume that () and E x are known and are real. From the information available, which one of the following CANNOT be determined? (A) The type of polarization of the wave (B) The group velocity of the wave (C) The phase velocity of the wave (D) The power flux through the z = 0 plane 4. Ans: (D) Sol: Given E xˆe x e j tz As medium properties and area of z = 0 plane is not given in the data, hence Average power flow (or) power flux cannot be determined.

20 : 0 : Gate 06 ECE Set 5. Light from the free space is incident at an angle I to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are n =.5 and n =.4, respectively. Free space i Light n (cladding) n (core) The maximum value of i (in degrees) for which the incident light will be guided in the core of the fibre is 5. Ans: 3.58 Sol: Given n =.5, n =.4 The maximum angle over which the incident light rays entering the fiber is called acceptance angle, A. sin n n (or) A A sin n n sin.5.4 A o Q. 6 Q. 55 carry two marks each dx 6. The ordinary differential equation 3x, with x(0) = is to be solved using the forward dt Euler method. The largest time step that can be used to solve the equation without making the numerical solution unstable is 6. dy Sol: = 3y, y(0) = dx If 3h < then solution of differential equation is stable < < 3h < < 3h < 0 0 < 3h < 0 < h < 3 If 0 < h < 0.66 then we get stable 7. Suppose C is the closed curve defined as the circle x y = with C oriented anti-clockwise. The value of (xydx xydy) over the curve C equals 7. Ans: 0 (Zero) Sol: Using Green s Theorem xy dx x ydy xy xydxdy C R 0

21 8. Two random variables X and Y are distributed according to f X,Y (x, y) (xy), 0 x, 0 y 0, otherwise. The probability P(X Y ) is : : Questions & Solutions 8. Ans: 0.33 R Sol: P(X Y) = f x, y x dxdy x y x0 y0 y = xy 0 = x x 0 dxdy x 0 dx x dx Y (0, ) x = 0 y = xy = R y = 0 (, 0) x = X = The matrix a A has det (A) = 00 and trace (A) = b The value of a b is. 9. Ans: 3 Sol: trace (A) = 4 a b 7 = 4 a b = 7 det (A) = 00 a b 0 ab = 00 ab = 0 a = 5, b = (or) a =, b = 5 a b = 3

22 30. In the given circuit, each resistor has a value equal to. a : : Gate 06 ECE Set R eq b What is the equivalent resistance across the terminals a and b? (A) /6 (B) /3 (C) 9/0 (D) 8/5 30. Ans: (D) Sol: a 4/3 4/3 4/3 a 4/ 60/33 b 4/ a 4/ b

23 : 3 : Questions & Solutions so, = R ab 8 = 0.53 = In the circuit shown in the figure, the magnitude of the current (in amperes) through R is R R 60V 5 3 R v x 5 V x 3. Ans: 5 v 60 v Sol: Nodal 0.04v x v Where v x 8 v = 40v V 40 So, IR I3 A 5A s 6 A continuous-time filter with transfer function H(S) is converted to a discrete time s 6s 8 filter with transfer function z z G(Z) so that the impulse response of the z z k continuous-time filter, sampled at Hz, is identical at the sampling instants to the impulse response of the discrete time filter, The value of k is. 3. Ans: Sol: s 6 s 6 H s s 6s 8 s s 4 s s 4 h t t e u t 4t e.u t T h s F s nts 4nTs nts e unts e,unts e n un e n.un z z z z H z e z e z z z 0.35

24 : 4 : Gate 06 ECE Set z 0.35z z 0.367z H z z 0.503z z 0.503z z 0.503z k = The Discrete Fourier Transform (DFT) of the 4-point sequence X[n] = {x[0], x[], x[], x[3]} = {3,, 3, 4} is X[k] = {X[0], X[], X[], X[3]} = {, j, 0, j} If X [k] is the DFT of the -point sequence x [n] = {3, 0, 0,, 0, 0, 3, 0, 0, 4, 0, 0}, x[8] The value of is X [] 33. Ans: 6 Sol: Interpolation in time domain equal to replication in frequency domain. n x n x 3 X (k) = [, j, 0, j,, j, 0, -j,, j, 0, j] X (8) =, X () = j X 8 6 X j

25 : 5 : Questions & Solutions 34. The switch S in the circuit shown has been closed for a long time. It is opened at time t = 0 and remains open after that. Assume that the diode has zero reverse current and zero forward voltage drop. t = 0 0V S mh 0F V C The steady state magnitude of the capacitor voltage V C (in volts) is 34. Ans: 00 0 Sol: il 0 0A il0 v c (0 ) = 0v = v c (0 ) For diode, R r = and R f = 0(given) V(t) 0A mh 0mH V C 0V For t 0 Transform the above network in Laplace domain V(s) 0 s sl sc V C (s) 0V Nodal 0 5 s v sl 0L (s) s LC v v(s) 0L. v(s) 0 sc. LC s LC LC

26 : 6 : Gate 06 ECE Set 0 (t) 0 L. C s n v n n L sin t C rad / sec Where n LC 4 v(t) 00sin0 t where n for 0 t for 0 t LC By KVL v(t) v c (t) = 0 v c (t) = v(t) = 00sin(0000t) V for 0 t V C (t) 00V t 00V So, v c = 0090v v c 00v 35. A voltage V G is applied across a MOS capacitor with metal gate and p-type silicon substrate at T = 300 K. The inversion carrier density (in number of carriers per unit area) for V G = 0.8 V is 0 cm. For V G =.3 V, the inversion carrier density is 4 0 cm. What is the value of the inversion carrier density for V G =.8 V? (A) cm (B) cm (C) 7. 0 cm (D) cm 35. Ans: (B) 36. Consider avalanche breakdown in a silicon p n junction. The n-region is uniformly doped with a donor density N D. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical filed E crit. Assume E crit to be independent of N D. If the built-in voltage of the p n junction is much smaller than the breakdown voltage, V BR, the relationship between V BR and N D is given by (A) VBR N D = constant (B) N D VBR = constant (C) N D V BR = constant (D) N D / V BR = constant 36. Ans: (C) Sol: V V 0 BR E q CRIT N D N A

27 : 7 : Questions & Solutions P N N A ND V BR.N D = CONSTNAT is CONSTANT E CRIT N A N D 37. Consider a region of silicon devoid of electrons and holes, with an ionized donor density of 7 3 N d 0 cm. The electric filed at x = 0 is 0 V/cm and the electric filed at x = L is 50 kv/cm in the positive x direction. Assume that the electric filed is zero in the y and z directions at all points. N d = 0 7 cm 3 X = 0 X = L Given q = coulomb, 0 = F/cm, r =.7 for silicon, the value of L in nm is. 37. Ans: [3.358 nm] end de end Sol: E x dx kv / cm L L cm = m = m = 3.358nm 38. Consider a long-channel NMOS transistor with source and body connected together. Assume that the electron mobility is independent of V GS and V DS. Given, g m = 0.5A/V for V DS = 50 m V and V GS = V, g d = 8A/V for V GS = V and V DS = 0 V, I D I D Where g m and g d VGS VDS The threshold voltage (in volts) of the transistor is 38. Ans:. Volts W Sol: I o ncox Vgs VT V DS VDS L g m di dv D gs n C ox W. V L DS

28 : 8 : Gate 06 ECE Set 6 W g m 0.50 ncox 00 3 L VDS 500 did W g d ncox V gs VT dvds L 80 6 = [ V gs V T ] 6 80 Vgs VT 6 00 V T = V gs 0.8 V T = V 0.8V =. V V T =.V The figure shows a half-wave rectifier with a 475 F filter capacitor. The load draws a constant current I O = A from the rectifier. The figure also shows the input voltage V t, the output voltage V C and the peak-to-peak voltage ripple u on V C. The input voltage V t is a triangle-wave with an amplitude of 0 V and a period of ms. 0V V i 47F V C I O = A 0 0V u 0 t The value of the ripple u (in volts) is. 39. Ans:.V Sol: (x) u (y) V i VC R T T u (V ipeak VD ) (Vipeak VD ) RC Peak to peak Amplitude of ripple Vipeak VD u.t RC Vipeak VD If the load is represented by a constant current source I0 R I0.T A.(m) u C 475F V i 475F I 0 = A =.V V i V C t

29 : 9 : Questions & Solutions 40. In the opamp circuit shown, the Zener diodes Z and Z clamp the output voltage V O to 5 V or 5 V. The switch S is initially closed and is opened at time t = 0 0V S t = 0 00F 0V k 0V Z V O 0V k 4 k Z 0V 0V The time t = t (in seconds) at which V O changes state is. 40. Ans: sec Sol: At t = 0 V N = 0V 470K V 0 = 5V V P = V K 4K The output V 0 changes state when V N = V for t 0 V C 0V 0K V N 0V / RC V (0) V ( ) e t V ( ) 0 0e 4 4 t /0, 0 0 VC (t) C C C V N = 0 V c = 0 [00e -t ] = 0 0e -t For op-amp to change state V N = V P 0 0e -t = 0e -t = 9 9 t = ln 0 t = 0.798sec = 0 0e -t

30 : 30 : Gate 06 ECE Set 4. An opamp has a finite open loop voltage gain of 00. Its input offset voltage V ios ( = 5mV) is modeled as shown in the circuit below. The amplifier is ideal in all other respects. V input is 5 mv. k 5 k A 0 =00 V Ios = 5mV V Input The output voltage (in millivolts) is. 4. Ans: A Sol: V0 V ios Vinput A V mV 5mV 30mV = 43.79mV k k k 4. An 8 Kbyte ROM with an active low Chip Select input ( CS) is to be used in an 8085 microprocessor based system. The ROM should occupy the address range 000H to FFFH. The address lines are designated as A 5 to A 0, where A 5 is the most significant address bit. Which one of the following logic expression will generate the correct CS signal for this ROM? (A) A5 A4 (A3.A A3.A) (B) A 5.A 4.(A 3 A ) (C) A5.A4.(A3.A A3. A (D) A5 A4 A3. A 4. Ans: (A) Sol: Address Range given is cs 3 A A 0 8 kb ROM To provide cs as low, The condition is A 5 = A 4 = 0 and A 3 A = 0 (or) (0) i.e A 5 = A 4 = 0 and A 3 A shouldn t be 00,. Thus it is A 5 A 4 [A 3 A A, A ] 3 8 A 5 A 4 A 3 A A A 0 A 9 A 8 A 7 A 6 A 5 A 4 A 3 A A A 0 000H FFFH 0 0 0

31 : 3 : Questions & Solutions 43. In an N bit flash ADC, the analog voltage is fed simultaneously to N comparators. The output of the comparators is then encoded to a binary format using digital circuit. Assume that the analog voltage source V in (whose output is being converted to digital format) has a source resistance of 75 as shown in the circuit diagram below and the input capacitance of each comparator is 8 pf. The input must settle to an accuracy of ½ LSB even for a full scale input change for proper conversion. Assume that the time taken by the thermometer to binary encoder is negligible. Vref55 Vin Vref Vref Thermometer Digital code Output to binary conversion If the flash ADC has 8 bit resolution, which one of the following alternatives is closest to the maximum sampling rate? (A) megasamples per second (B) 6 megasamples per second (C) 64 megasamples per second (D) 56 megasamples per second 43. Ans: (B) Sol: R V in V RC in eq T V in C eq V in Vin has to settle down within LSB of full scale value. i.e 509 Vin. T 509 V T ( in ) (5580 ) 50 T 0.5 sec Thus sample period T s T T s 0.5 m sec fs, max Hz 6 Megasamples 6 Ts 0.50,min

32 : 3 : Gate 06 ECE Set 44. The state transition diagram for a finite state machine with states A, B and C, and binary input X, Y and Z, is shown in the figure. Y= X=0, Y=0, Z=0 Y= A B Y=0, Z=0 X=0, Z= X=, Y=0 X= Z= Which one of the following statements is correct? (A) Transitions from State A are ambiguously defined (B) Transition from State B are ambiguously defined (C) Transitions from State C are ambiguously defined (D) All of the state transitions are defined unambiguously. 44. Ans: (C) Sol: In state (C), when XYZ =, then Ambiguity occurs Because, from state (C) When X =, Z = N.S is (A) When Y =, Z = N.S is (B) C Y=, Z= Z= 0 Y=0, Z=

33 45. In the feedback system shown below GS : 33 : Questions & Solutions. s s The step response of the closed-loop system should have minimum setting time and have no overshoot. r K G(s) y The required value of gain k to achieve this is 45. Ans: Sol: Given G(s) = (s s) From Diagram CE KG(s) = 0 s s K = 0 Minimum Settling Time is obtain. For Critical Damped System For Critical Damped System ( ) the % m p = 0% n n = n = rad/sec K = 46. In the feedback system shown below GS r K s s s 3 G(s) y The positive value of k for which the gain margin of the loop is exactly 0 db and the phase margin of the loop is exactly zero degree is 46. Ans: 60 Sol: Given Forward path TF = (s )(s )(s 3) Given GM = 0dB, PM = 0 0 That Means Given System is Marginal Stable KG(s) = 0 CE = s 3 s 6s 6 K = 0 S 3 S S S K 0 (6K) 6 6K K = 60 For Marginal Stable

34 47. The asymptotic Bode phase plot of shown below. : 34 : Gate 06 ECE Set k G(S), with k and p both positive, is (s 0.) (s 0) (s p ) 0 o rad/s 45 o 35 o 5 o 70 o The value of p is 47. Ans: Sol: From the Bode Diagram at =, the phase Angle is = = tan tan 0. 0 tan p 35 = tan tan tan 0. 0 P 35 = tan - P 45 = tan - p P P = 48. An information source generates a binary sequence { n }. n can take one of the two possible values and with equal probability and are statistically independent and identically distributed. This sequence is pre-coded to obtain another sequence { n }, as n = n k n 3. The sequence { n } is used to modulate a pulse g(t) to generate the baseband signal X(t) =, 0 t T n g(t nt), where g(t) n 0 otherwise If there is a null at f in the power spectral density of X(t), then k is 3T 48. Ans: Sol: The Auto correlation function is k 0 R b ( ) k 3 0 otherwise

35 Power spectral density S b (f) = k k cos(f3t) Null will occur at f 3T So at f Sb (f ) k k cos 3T 0 3T 3T k k = 0 (k ) = 0 k = : 35 : Questions & Solutions 49. An ideal band-pass channel 500 Hz-000 Hz is deployed for communication. A modem is designed to transmit bits at the rate of 4800 bits/s using 6-QAM. The roll-off factor of a pulse with a raised cosine spectrum that utilizes the entire frequency band is 49. Ans: 0.5 Sol: Bw = 500 H z R b[ ] 500Hz 6 log R b [ ] = = ( ) Consider random process X(t) = 3V(t) 8, where V (t) is a zero mean stationary random process with autocorrelation R v () = 4e 5. The power is X(t) is 50. Ans: 00 W Sol: R v () = 4.e 5 R v (0) = 4 Mean square value = 00 W 5. A binary communication system makes use of the symbols zero and one. There are channel errors. Consider the following events: x 0 : a zero is transmitted x : a one is transmitted y o : a zero is received y : a one is received 3 The following probabilities are given: P(x o), P(yo x o ),and P(yo x). The 4 information in bits that you obtain when you learn which symbol has been received (while you know that a zero has been transmitted) is

36 : 36 : Gate 06 ECE Set 5. Ans: Sol: P(y/x) y o y x o 3/4 /4 x / / x 0 T BSC / /4 3/4 y 0 R P(xy) y o y x o 3/8 /8 x /4 /4 x T / y T H(y / x o ) P x y log P(y / x ) P(x y ) log Py / x o 3 log 8 o 3 log o 5. The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that the plate separation become d. d o At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects? (A) E (B) E (C) E (D) E/ 5. Ans: (A) Sol: Energy stored when spacing is d is given by Energy stored = Energy density volume E = Ed V V = d A = da When spacing between the plated is doubled, d = d Then, V = d A = da E = Ed da = Ed (da) E = E There with the modified capacitor energy stored is doubled. d = d d 53. A lossless microstrip transmission line consists of a trace of width w. It is drawn over a practically infinite ground plane and is separated by a dielectric slab of thickness t and relative permittivity r >. The inductance per unit length and the characteristic impedance of this line are L and Z o, respectively. A

37 : 37 : Questions & Solutions w = 0 = 0 r ; r > t Which one of the following inequalities is always satisfied?` (A) Z 53. Ans: (A) o Lt w (B) Z o r o Lt w (C) Z o r o Lw t (D) o r Z o Lw t o r 54. A microwave circuit consisting of lossless transmission lines T and T is shown in the figure. The plot shows the magnitude of the input reflection coefficient as a function of frequency f. The phase velocity of the signal is transmission lines is 0 8 m/s. Input length = m T Length = L Z 0 = 50 Z 0 = 50 Open 50 (reflection coefficient) f (in GHZ) 54. Ans: 0. Sol: l= oc Z o 50 O.C Zo Z Z L = 50 // j 50 cot β Z Z 0 = 0 only when Z L = o Z Z0 oc 50 // j 50 cot β l oc = 50 Z

38 This satisfied only when j 50 cot β l oc = i.e., β l oc = m oc m m oc mv oc f 8 m 0 [ f in GHZ,Here f = 0,,,3.. GHZ] 9 f 0 m oc 0f f = GHz (Here f= GHz m = for minimum length l oc ) m oc 0 oc [for m =] 0 L oc = 0. : 38 : Gate 06 ECE Set 55. A positive charge q is placed at x = 0 between two infinite metal plates placed at x = d and at x = d respectively. The metal plates lie in the yz plane. at x = d q x = 0 at x = d The charge is at rest at t = 0, when a voltage V is applied to the plate at d and voltage V is applied to the plate at x = d. Assume that the quantity of the charge q is small enough that it does not perturb the field set up by the metal plates. The time that the charge q takes to reach the right plate is proportional to (A) d/v (B) d /V (C) d / V (D) d / V

39 : 39 : Questions & Solutions 55. Ans: (C) Sol: yz=plane x= d,v=v d q x=0 x=d, v = v When there is no external field, Change at rest having potential energy only P.E = qv Fig By an application of an external field, change carries acquire some kinetic energy, with velocity V. qv = ½ mv ev V m Time taken to reach x = d plate is known as g tr Gap tramit time d d t g v ev m d t d V